Elementary Linear Algebra 2nd Edition

Elementary Linear Algebra: A Matrix Approach, 2nd Edition No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

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PREFACE In response to concern that the first course in linear algebra was not meeting the needs of the students who take it, the Lineru· Algebra Curriculum Study Group was formed in January 1990. With the support of the National Science Foundation. it sponsored a workshop on the undergraduate linear algebra cu rricu lum at the College of William and Mary in August of that year. Participants at the workshop included representatives from mathematics departments and other disciplines whose students study linear algebra. Its recommendations, 1 published in January 1993, have inspired a number of textbooks, including this one. ln its core syllabus. abstract vector spaces are regarded as a supplementary topic. What dist inguishes this book from many others is its complete development of the concepts of linear algebra in R" before the introduction of abstract vector spaces. We agree wi th the following statement by the Linear Algebra Curriculum Study Group:

Furthermore. an overemphasis on abstraction may overwhelm beginning students the poim where they leave the course with liu/e understanding or mastery of the basic concepts they may need in later courses and their careers.

10

Although we bel ieve that the first linear algebra course is a natural one in which to introduce mathematica l theory and proofs, this ca n be accomplished without discussion of abstract vector spaces. In addition, if abstract vector spaces are included in the first linear algebra course, we bel ieve that they can be taught more effectively after a thorough presentation of the essential topics in the more familiar context of R". Never1heless. we have written this book so that it is possible to teach concepts in the context of abstract vector spaces immediately after the correspond ing topics are discussed in R". (Suggestions for doing so are included in the sample course descriptions on page xiv.) Although there is no use of calcu lus until the introduction of vector spaces in Chapter 7, the material is aimed at students who have the mathematical maturity obta ined by having taken one year of calculus. The core topics can be comfortably covered within one semester, but there is adequate material for a two-quarter course.

PEDAGOGICAL APPROACH This text is written for a matrix-oriented course. as reconunended by the Linear Algebra CutTiculum Study Group. In our experience, such a course results in greater understanding of the concepts of linear algebra and serves the needs of students in many disciplines. It begins with the study of matrices, vectors, and systems of linear equations and gradually leads to more complicated concepts and general principles. such as linear independence, subspaces, and bases. As mentioned, this text develops all the core content of linear algebra in R" before introducing abstract vector spaces. This provides students add itional opportuni ties to visualize concepts in the fami liar context of the Euclidean plane and 3-space before encountering the abstraction of vector spaces. Our approach is based on an early introduction of the rank of a matrix. This concept is then encountered in other contexts throughout the book. For example, the 1

David Carlson, Charles R. Johnson, David C. Lay, and A. Duane Porter. "The Linear Algebra Curriculum StudyGroup Recommendations for the First Course in Linear Algebta." 1J,. Coll•g• Mllthemlltic.< JOLmw/, 24 (1993), pp. 41 - 46.

xi


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xi i Preface rank of a matrix is used initially to check if solutions of a system of linear equations exist and are unique. Later, it is used to test if sets are linearly independent or are spanning sets for R". Then, in Chapter 2, it is used to determine whether linear transformations are one-to-one or onto. Even though a course taught from th is book may devote less time to the study of abstract mathematics than a more traditional course. we have found that it s till serves as an excellent prerequisite to abstract algebra and an abstract second comse in linear algebra. such as one using our text Linear Algebra.

TECHNOLOGY The Linear Algebra Study Group recommends that technology be used in the first course in linear algebra. In our experience, the use of technology, whether through computer software or supcrcalculators. greatly enhances a course taught from this book by freeing students from tedious computations and enabl ing them to concentrate on conceptual understanding. Most sections include exercises designed to be worked by means of MATLAB or similar technology. Additional technology exercises are found at the end of each chapter. For MATLAB users, our website contains data files and M-files that can be downloaded. For the convenience of those wish ing to use technology, we have added an appendix (Appendix D) with an introduction to MATLAB. It provides sufficient back· ground to prepare students to perform the calcu lations required for this book and to work the technology exercises.

EXAMPLES AND PRACTICE PROBLEMS Our examples motivate and illustrate definitions and theoretical results. These are written to be understood by students so that an instmctor need not discuss each example in class. In fact, in our own teaching, we almost never d iscuss examples from the text, but rather present simi lar examples, leaving the text examples to be read by students. Many examples are accompanied by similar practice problems that enable students to test their understanding of the material in the text. Complete solutions of these practice problems are included within the text in order to help prepare students to work the exercises in each section.

EXERCISES We felt that the first edition of this book had an ample number of computational exercises. Yet some of our conscientious students asked us for more. Therefore we have significantly increased the number of computational exercises in this edition. As in the first edition, except for the specially marked exercises that use technology, all our computational exercises are designed so that the calculations involve "nice" numbers. This permits students to concentrate on the linear algebra concepts rather than the computations. Most sections include approximately 20 true/fa lse exercises that are designed to test a student's understanding of the conceptual ideas in each section. l n response to reviewer suggestions, we have moved these to follow the straightforward computational exercises so that students can gain confidence by working computational exercises before encountering the true/false exercises. Answers to a ll true/false questions arc given in the book. so students will know if they have misunderstood some concept.


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Elementary Linear Algebra: A Matrix Approach, 2nd Edition, page: xiii No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

Preface xiii

For a proof-ori ented course, we have included a signi ficant number of accessible exercises requiring proofs. These are ordered according to difficulty. Finally, each chapter ends with a set of review exerc ises that provide practice wi th all the main topics in each chapter. The exerc ises are designed for students to use as preparation for a chapter exam ination. As noted, there are a lso technology exercises at the end of each c hapter and in the exercise sets of most sections. For a list of suggested exercises for each section, see our website.

APPLICATIONS This book includes a wide variety of applications, and they are given when the necessary prerequisites have been introduced. There arc applications to economics (the Leontief inpu t-output model), electrical networks (cu1Tent flow through an electrical network). population change (the Leslie matrix). traffic flow, scheduling ((0, I )matrices), anthropology, Google searches, counting problems (d ifference equations), predator-prey models and harmonic motion (systems of differential equations). leastsquares approx imation, computer graph ics, principal component analysis, and music (applying tri gonometric polynomials). These applications are entirely independent and so may be taught according to the interests of the instructor and students. Although we do not assume that any pmticular applications will be covered, we believe thai the core material is greatly enhanced when its use is demonstrated through applications.

ORGANIZATION OF THE TEXT In Chapter I, students are introduced to matrices. vectors, and systems of linem· equations. The study of linear combinations of vectors leads naturally to the span of a set of vectors and the concepts of linear dependence and independence. The rank and nullity of a matrix are introduced early m1d used to detennine when a system of equations has solutions, and how many solutions there are. In Chapter 2, we introduce matrix multiplication, along with matrix inverses and linear transformations. The relationship between linear transfom1ations and matrices is emphasized so that matrix techniques can be used to answer questions such as whether a linear transformation is one-to-one or o nto. Because we use detenninants mainly in the context of eigenvalues, we provide a short but complete treatment in Chapter 3. The important topics of subspaces, bases, and dimension are covered in Chapter 4 . We follow these concepts with a discussion of coord inate systems and matrix representations of linear transformations. Although Chapters 3 and 4 can be interchanged, we prefer to cover determ inants before subspaccs so that there is no delay between the discussions of change of coordinates and d iagonalization in Chapter 5. Chapter 5 contains the important results about eigenvalues, eigenvectors, and diagonalization of matrices and linear operators. With the introduction of orthogonality in Chapter 6, we are able to emphasize the geometry of vectors, matrices, and linear transformations. The important applications of orthogonal projections and least-squares lines illustrate the usefulness o f these ideas. We continue with a discussion of orthogonal matrices and operators and the study of symmetric matrices. The chapter concludes with the singular value decomposition of a matrix and applications to principal component analysis and computer graphics. Chapter 7 introduces abstract vector spaces. Because of the careful foundation that has been developed in Eucl idean spaces. most of the concepts in earlier

chapters- such as span, linear independence, and subspace- are easily genemli zed.


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Elementary Linear Algebra: A Matrix Approach, 2nd Edition, page: xiv No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

xiv Preface We focus mainly on function and matrix spaces. For example, a nice application in this context arises when Lagrange interpolating polynomials are used to find a basis for a polynomial space. Differential operators and integrals are now explored as special cases of linear transformations. Finally. inner product spaces are introduced. and the Gram-Schmidt process is applied to produce the Legendre polynomials. Leastsquares theory and trigonometric polynomials arc used to explore periodic motion in the selling of musical notes. Although we prefer to introduce abstract vector spaces only after the complete development of all the core topics in R", the book is wrillen to allow abstract vector spaces to be introduced earlier. alongside the cotTesponding topics in R". For details, see the sample syllabi that follow. We have included a number of optional topics, for examp le. the LU. QR. and singular value decomposition, of a matrix, the spectr.JI decomposition of a symmetric matrix, Lagrange interpolating polynomials. block multiplication, the Moore-Penrose generalized inverse, and quadratic forms. In addition, there are many applications of the subject matter throughout the book.

SAMPLE SYLLABI The following list shows the sections of this book that cover the Linear Algebra Curriculum Study Group's core syllabus:

Linear Algebra Cu"iculum Study Group's Core Syllabus Matrix Addition and Multiplication (4 days: Sections 1.1, 1.2, 2.1, and 2.5) Systems of Linear Equations (5 days: Sections 1.3, 1.4. 2.3. 2.4. 2.6) Determinants (2 days: Sections 3.1 and 3.2) Properties of R" (10 days: Sections 1.2. 1.6. 1.7, 4.1. 4.2. 4.3, 2.7. 2.8. 6.1, and 6.2) Eigenvalues and Eigenvectors (4 days: Sections 5.1, 5.2, 5.3, and 6.5) More on Orthogonality (3 days: Sections 6.2. 6.3, and 6.4) Suggested Syllabus for a One-Semester 3-Credit Course Chapter I: Sections I. I, 1.2, 1.3, 1.4, 1.6, 1.7 Chapter 2: Sections 2.1. 2.3, 2.4. 2.7, 2.8 Chapter 3: Sections 3. I, 3.2 (omitt ing optional material) Chapter 4: Sections 4.1. 4.2, 4.3, 4.4 Chapter 5: Sections 5. 1, 5.2, 5.3 Chapter 6: Sections 6.1. 6.2, 6.3, 6.4, 6.5 Supplementary materia l selected from Sections 1.5, 2.2, 2.6, 5.5, 6.6, 6.7, 7.1, 7.2, 7.3. 7.5 Suggested Syllabus for a One-Semester 3-Credit Course that Integrates Abstract Vector Spaces with R" Chapter I: Sections 1.1, 1.2, 1.3. 1.4. 1.6, 1.7 Chapter 2: Sections 2. 1, 2.3, 2.4, 2.7, 2.8 Chapter 3: Sections 3. 1. 3.2 (omitting optional material) Chapter 4: Sections 4. 1, 7.1, 7.2, 4.2, 4.3, 4.4, 7.3 Chapter 5: Sections 5.1. 5.2, 7.4. 5.3 Chapter 6: Sections 6. 1, 6.2, 6.3, 6.4, 7.5, 6.5 Suggested Syllabus for a Two-Quarter Sequence of 3-Credit Courses Chapter I: Sections 1.1. 1.2, 1.3, 1.4, 1.5, 1.6. 1.7 Chapter 2: Sections 2. 1, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8


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Elementary Linear Algebra: A Matrix Approach, 2nd Edition, page: xv No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

Preface xv

Chapter 3: Sections 3. 1, 3.2 Chapter 4: Sections 4. 1, 4.2, 4.3, 4.4, 4.5 Chapter 5: Sections 5. 1. 5.2, 5.3, 5.5 Chapter 6: Sections 6. 1, 6.2, 6.3, 6.4. 6.5, 6.6 Supplementary mate rial se lected from Sections 2.2, 6.6, 6.7, 7. 1, 7.2, 7.3, 7.4, 7.5

SUPPLEMENTS A Swdem Solulions Manual (ISBN 0-1 3-239734-X) is ava ilable for purchase by students. It includes solutions of all the true/false exercises and over hal f of the odd-numbered exercises. An lnslruc/or's Solutions Manual that includes solutions of all the exercises is available from the publ isher.

ACKNOWLEDGMENTS This second edition has been greatly improved by the comments of the following reviewers: Adam Avilez (Mesa Community College). Roc Goodman (Rutgers University), Chungwu Ho (Evergreen Valley College), Steve Kaliszewski (Arizona State University), Noah Rhec (U niversity of Missouri-Kansas City). and Edward Soares (College of the Holy Cross). We especia ll y appreciate the many detailed comments provided by Jane Day (San Jose State University). In addition, we are indebted to Development Ed itor Paul Trow, who provided significant editorial and mathematical suggestions about several chapters of mtr manuscript and to W.R. Winfrey (Concord College) for wri ting the introductory application for each chapter. Finall y. we would like to express our appreciation to Senior Editor Holl y Stark and Senior Managing Editor Scott Disan no of Pearson Pre ntice Hall for the ir he lp during the wri ting and production of thi s book. To find the latest information about this book, consult our home page on the World Wide Web. We encourage comments, which can be sent to us by e-mail or ordinary post. Our home page and e-mail addresses are as follows : home page: e-mail:

http://www.cas.ilstu.edu/math!matrix matri [email protected] LAWRENCE

E.

ARNOLD STEPHEN


H.

SPENCE

J.

l NSEL

FRIEDBERG

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TO THE STUDENT Linear algebra is concerned with vectors and matrices, and with special functions called linear tra11.1j'omwrions that are defined on vectors. Since vectors arise in a wide variety of settings. linear algebra can be applied to many disciplines. ln fact, it is one of the most imp011ant tools of applied mathematics. In th is book, we present applications of linear algebra to economics. physics. biology, statistics, computer graphics, and other fields. Like most areas of mathematics, linear algebra has its own terminology and notation. To be successful in you r study of linear algebra, you must be able to solve problems and communicate ideas with its language and symbolism. Developing these ab ilities requ ires more than just altend ing lectures or C
xvii


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xviii To the Student

• Prepare regularly for each class. You cannot expect to learn to play the piano merely by attending lessons once a week-long and carefu l practice between lessons is necessary. So it is wi th linear algebra. At the least. you should study the material presented in class. Often. however, the current lesson bu ilds on previous material that must be reviewed . Usually, there arc exercises to work that deal with the material presented in class. In add ition, you should prepare for the next class by reading ahead. Each new lesson usually introduces several important concepts or definitions that must be learned in order for subsequent sections to be understood. As a result, falling behind in your study by even a single day prevents you from understanding the new material that follows. Many of the concepts of linear algebra are deep; to fully understand these ideas requires time. It is simply not possible to absorb an entire chapter· s worth of material in a single day. To be successful, you must !cam the new material as it arises and not wait to study unti l you arc less busy or until an exam is imm inent. • Ask questions of yourself and others . Mathematics is not a spectator sport. It can be learned only by the interaction of study and questioning. Certain natural questions arise when a new topic is introduced: What purpose does it serve? How is the new topic related to previous ones? What examples illustrate the topic? For a new theorem. one might also ask: Why is it true? How does it relate to previous theorems? Why is it usefu l? Be sure you don't accept something as true unless you believe it. If you are not convinced that a statement is correct, you should ask for further details. • Review often . As you attempt to understand new material, you may become aware that you have forgotten some previous materia l or that you haven 't understood it well enough. By relearning such material, you not only gain a deeper understanding of prev ious topics, but you also enable you rself to leam new ideas more qu ickly and more deeply. When a new concept is introduced , search for related concepts or results and write them on paper. This enables you to see more easily the connections between new ideas and previous ones. Moreover, expressing ideas in writing helps you to learn, because you must think carefully about the materia l as you write. A good test of your understanding of a section of the textbook is to ask yourself if, with the book closed. you can expla in in writing what the section is about. If not, you will benefit from reading the section again. We hope that your study of linear algebra is successful and that you take from the subject concepts and techniques that arc useful in future courses and in your career.


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1

INTRODUCTION

Ideal Edge

Real Edge

For computers to process digital images, whether satellite photos or x-rays, there is a need to recognize the edges of objects. Image edges, w hich are rapid changes or d iscontinuities in image intensity, reflect a boundary bet ween dissimilar regions in an image and thus are important basic characteristics of an image. They often indicate the physica l extent of objects in the image or a bou ndary between light and shadow on a sing le su rface or other reg ions of interest. The lowermost two figures at the left indicate the changes in image intensity of t he ideal and real edges above, when moving from right to left. We see that rea l intensities can change rapidly, but

___j __/

not instantaneously. In principle, the edge may be found by looking for very large changes over small distances. However, a digital image is discrete rather t han continuous: it is a matrix of nonnegative entries

that provide numerical descriptions of the shades of gray for t he pixels in the image, where the entries vary from 0 for a white pixel to 1 for a black pixel. An analysis must be done using the discrete analog of the derivative to measure the rate of change of image intensity in two directions.


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2

1 Introduction

The Sobel matrices, 51

=[

=~ ~ ~ - 1 0 I

]

and S2

=

~ ~ ~

] provide a method for measuring [ - 1 - 2 - 1 these intensity changes. Apply the Sobel matrices 51 and 52 in turn to the 3x3 subimage centered on each pixel in the original image. The resu lts are the changes of intensity near the pixel in the horizontal and the vertical directions, respectively. The ordered pair of numbers that are obtained is a vector in the plane that provides

Original Image

Notice how the edges are emphasized in the thresholded image. In regions where image intensity is constant, these vectors have length zero, and hence the corresponding regions appear white in the thresholded


the direction and magnitude of the intensity change at the pixel. This vector may be thought of as the discrete analog of the gradient vector of a function of two variables studied in calculus. Replace each of the original pixel values by the lengths of these vectors, and choose an appropriate threshold value. The final image, called the thresho/ded image, is obtained by changing to black every pixel for which the length of the vector is greater than the threshold value, and changing to white all the other pixels. (See the images below.)

Thre. holded Image

image. Likewise, a rapid change in image intensity, which occurs at an edge of an object, results in a relatively dark colored boundary in the thresholded image.

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CHAPTER

1

MATRICES, VECTORS, AND SYSTEMS OF LINEAR EQUATIONS he most common usc of linear algebra is to solve systems of linear equations, which arise in appl ications to such diverse disciplines as physics, biology, econom ics. engineering. and sociology. ln this chapter, we describe the most efficient algorithm for solv ing systems of linear equations, Gaussian eliminmion. This algorithm, or some variation of it, is used by most mathematics software (such as MATLAB). We can write systems of linear equations com pactly, using arrays called matrices and vectors. M ore importantly, the arithmetic properties of these arrays enable us to compute solutions of such systems or to determine if no solutions exist. This chapter begins by developing the basic properties of matrices and vectors. In Sections 1.3 and 1.4. we begin our st\1dy of systems of linear equations. In Sections 1.6 and 1.7. we introduce two other important concepts of vectors, namely, generating sets and linear independence. which provide information about the existence and uniqueness of solutions of a system of linear equations.

T

[!}]

MATRICES AND VECTORS

Many types of numerical data are best displayed in two-dimensional arrays, such as tables. For example, suppose that a company owns two bookstores, each of w hich sells newspapers , magazines, and books. Assume that the sales (i n hundreds of dollars) of the two bookstores for the months of July and August are represented by the following tables: Ju ly S tore Ne\\'spapcrs Magazines Books

I

6 15 45

2 8 20 64

and

Store New, pape rs Magazi nes Books

Augu>t I 2 7 9

18

52

31 68

The first column of the July table shows that store I sold $ 1500 worth of magazines and $4500 wonh of books during July. We can represent the infom1ation on July sales more s imply as

[ I~ 45


2~].

64

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4

CHAPTER 1 Matrices, Vectors, and Systems of Linear Equations

Such a rectangular array of real numbers is called a matrix.• It is customary to refer to real numbers as sca la rs (originally from the word scale) when working with a matrix. We denote the set of real numbers by 1?.. Defin itions A matrix (plural, matrices) is a rectangular array of scalars. If the matrix has m rows and 11 columns, we say that the size of the mat rix is m by 11 , written m x 11. The matrix is square if m = n. The scalar in the ith row and j th column is called the (i, j )·entry of the matrix.

If A is a matrix, we denote its (i ,j)-entry by aij . We say that two matrices A and

B are eq ual if they have the same size and have equal corresponding entries: that is, a;1 = bij for all i and j. Symbolically, we write A = 8.

In our bookstore example, the July and August sales are contained in the matrices

8 =

[~~45 642~]

c

and

=

[~~52 683~] ·

Note that b 12 = 8 and c 12 = 9, so 8 :/; C. Both 8 and C are 3 x 2 matrices. Because of the context in which these matrices arise, they are called in vel/lory matrices. Other examples of matrices are

[!

-4

and

[!l

[-2

0

I] .

The first matrix has size 2 x 3, the second has size 3 x I, and the third has size I x 4. Let A

Practice Problem 1 .,..

= [~ ~l

(a) What is the ( I, 2)-ent ry of A? (b) What is azz? Sometjmes we are interested in only a part of the information contained in a matrix. For example, suppose that we are interested in only magazine and book sales in July. Then the relevant information is contruned in the last two rows of 8: that is, in the matrix E defined by

E-

['5 45

?Q] 64

E is called a submarrix of B. In general. a s ubmatrix of a matrix M is obtained by deleting from M entire rows, entire columns, or both. It is permissible, when fonn.ing a submatrix of M. to delete none of the rows or none of the columns of M. As another example, if we delete the first row and the second column of 8 , we obtain the submatrix

1 Jame~ Jo~eph Sylve~ter (1814-


1897) coined the term matrix in the 1850~.

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1.1 Matrices and Vectors

5

MATRIX SUMS AND SCALAR MULTIPLICATION Matrices are more than conven ient devices for storing information. Their usefulness lies in their arithmetic. As an example, suppose that we want to know the total numbers of newspapers, magazines, and books sold by both stores during July and August. It is natural to form one matrix whose entries are the sum of the corresponding entries of the matrices 8 and C , namely, Store I Newspapers 13 33 Magazines [ 97 Books

2

17]

51 132

.

If A and 8 are 111 x 11 matrices. the sum of A and 8. denoted by A + 8. is the m x 11 matrix obtained by adding the corresponding entries of A and 8; that is, A + B is the m x 11 matrix whose (i .j)-entry is a;i + bij . Notice that the matrices A and 8 must have the same size for their sum to be defined. Suppose that in our bookstore example, July sales were to double in all categories. Then the new matrix of Jul y sales would be 12 30

16] 40 128

[ 90

We denote this matrix by 28. Let A be an m x 11 matrix and c be a scalar. The sca lar multiple cA is the m x 11 matrix whose entries are c times the corresponding entries of A; that is, cA is the 111 x 11 matrix whose (i ,j)-entry is ca;i . Note that lA =A. We denote the matrix ( - I )A by - A and the matrix OA by 0. We call the m x 11 matrix 0 in which each entry is 0 the m x 11 zero matrix.

Example 1

Compute the matrices A + 8, 3A, - A, and 3A + 48, where

A=[~ Solution

4 -3

~]

8= [ -4 5

and

I

-6

~].

We have

7 - 9 ~].

A + 8 = [- 1

5

3A =

[~

12 -9

~].

-A=

-4 [-3 -2 3

- 2] 0 •

and

3A + 48 =

[~

12 6] -9 0

+

16 OJ_ [-7 26 -33 !].

4 [ - 16 20 -24 4 -

Just as we have defined addition of matrices, we can also define s ubtraction. For any matrices A and 8 of the same size, we define A - 8 to be the matrix obtained by subtracti ng each entry of 8 from the corresponding e ntry of A. Thus the (i ,j)-entry

of A - 8 is


aij -

b;i . Notice that A - A

= 0 for all matrices A.

Page 8 of 10


6


If, as in Example I, we have

4

A=[~

-3

~l

B=

[-45

~l

I

-6

and

0

=[~

~l

0 0

then - B

Practice Problem 2 .,.

=[-~

Let A = [ ;

- I

6

-

-~l

A- B

= [ -~

~ -~] and B = [ ~

_

3 3

-~l

and A - 0

= [~

4 - 3

~l

~ ~l Compute the following matrices:

(a) A - B

(b) 2A (c) A+ 38

We have now defined the operations of matrix addition and scalar multiplication. The power of linear algebra lies in the natural re lations between these operations, which are described in our first theorem.

THEOREM 1.1 (Properties of Matrix Addition and Scalar Multiplication)

mx

11

Let A, B, and C be

matrices, and let .\' and r be any scalars. Then

(a) A+ 8 = B +A. (b) (A+ B) + C =A+ (B +C).

(commutative law of matrix addition) (associative law of matrix addition)

(c) A+ 0 =A.

(d) A+ (- A)=O . (e) (s1)A

= S(IA).

(f) s(A+B)=sA + sB. (g) (s

+ r)A =sA + rA .

We prove pans (b) and (f). The rest arc left as exercises. (b) The matrices on each side of the equati on are m x 11 matrices. We must show that each entry of (A + B) + C is the same as the co1Tcsponding entry of A + (B + C). Consider the (i ,))-entries. Because of the defin ition of matrix addition, the (i ,j )-entry of (A + B) + C is the sum of the (i .})-entry of A + B, which is aij + b;i, and the (i ,j)-entry of C , which is cij . Therefore this sum equals (aij + bij) + Cij . Similarly. the (i .})-entry of A + (B + C) is Oij + (bij + Cij ). Because the assoc iative law holds for add ition of scalars, (aij + b;i ) + cij = aij + (bij + Cij ). Therefore the (i.j)-entry of (A+ B)+ C equals the (i .j)-entry of A + (B + C), proving (b). (f) The matrices on each side of !he equation are m x 11 matrices. As in the proof of (b), we consider the (i ,})-entries of each matrix. The (i ,j)-entry of s(A + B) is defined to be the product of s and the (i .})-entry of A + B. which is a 1i + b;i . This product equals s(aij + b;j ). The (i ,j)-entry of sA + sB is the sum of the (i ,j)-entry of sA. which is saij , and the (i ,j)-entry of sB, which is sbij . Tllis sum is saij + sb;j . Since s(aij + b;j ) = sa;j + sb;j . (f) is proved. • PROOF

Because of the associative law of matrix addition, sums of three or more matrices can be written unambiguously without parentheses. Thus we may write A 8 C instead of either (A+ B) + C or A + (B + C).

+ +


Page 9 of 10



7

MATRIX TRANSPOSES In the bookstore example, we could have recorded the in formation about July sales in the following form: Store I 2

New!>papcr' 6

Maga7incs 15

8

20

Books

45 64

This representation produces the matrix

6 15 45] [ 8 20 64 . Compare this with

B

=

[~~45 642~] ·

The rows of the first matrix are the columns of 8, and the columns of the first mat rix are the rows of B. This new matrix is called the transpose of B. ln general, the t ranspose of an m x 11 matrix A is the 11 x m matrix de noted by AT whose (i ,j)-entry is the (j, i )·entry of A. The matrix C in our bookstore example and its transpose are

c

Practice Problem 3 •

Let A = [;

-

=

[~~52 683~]

and

~ -~] and B = [ ~

_

cr =

[79 3I81 52] 68 .

~ ~l Compute the following matrices:

(a) AT

(b) (3 Bf ~

(c) (A + B f

The following theorem shows that the transpose preserves the operations of matrix addition and sca lar multipl ication:

THEOREM 1.2 (Properties ofthe Transpose)

Let A and 8 be m x n matrices, and let s be any

scalar. Then (a) (A + B)T =AT+ B T (b) (sAl = sAT (c) (A r)r =A.

We prove part (a). The rest arc left as exercises. (a) The matrices on each side of the equation are n x m matrices. So we show that the (i.j)-entry of (A + B)' equals the (i,j)-entry of AT+ Br. By the definition of transpose, the (i ,j)-entry of (A + B )T equals the (j, i)-entry of A + 8 , which is aji + bji. On the other hand , the (i,j)-entry of AT + Br equals the sum of the (i ,j)-entry of AT and the (i ,j)-entry of s r, that is, aji bji· Because the • (i ,j )-entries of (A + B)' and A r + B r arc equal, (a) is proved. PROOF

+


Page 10 of 10


8


VECTORS A matrix that has exactly one row is called a row vector , and a matrix that has exactly one column is called a column vector . The tenn vector is used to refer to either a row vector or a column vector. The entries of a vector are called components. In this book. we normally work with column vectors, and we denote the set of all column vectors with n components by 1<.". We write vectors as boldface lower case letters such as u and v. and denote the ith component of the vector u by u; . For example, if u =

[-~l then

112 =

- 4.

Occasionally, we ident ify a vector u in 1(." with an 11-tuple, (11 1 , 112, .. . , u,). Because vectors are special types of matrices, we can add them and multiply them by scalars. In th is context, we call the two arithmetic operations on vectors vector addition and scala r multiplication. These operations satisfy the propenies listed in Theorem 1. 1. In particular, the vector in 1<." with all zero components is denoted by 0 and is called the zero vector . It satisfies u + 0 = u and Ou = 0 for every u in 1<.".

IJifi ..!.!tfW Let u - [

-n

and v -

u

[~l

n.

+ v = [_

Then

u_ v

= [ =~l

and

For a given matrix. it is often advantageous to consider its rows and columns as vectors. For example. for the matrix

[o

- 2), and the columns are

_;J.

[~ ~

[n [n

and [

the rows arc [2

4

3) and

-n

Because the columns of a matrix play a more imponant role than the rows, we introduce a special notation. When a capital letter denotes a matrix, we use the corresponding lower case letter in boldface with a subscript j to represent the jth column of that matrix. So if A is an m x 11 matrix, its jth column is

Otjl

azj

[

Omj

.r

.

GEOMETRY OF VECTORS (ll, b)

For many applications,2 it is usefu l to represent vectors geometrically as directed line segments, or arrows. For example, if v =

X

[~]

is a vector in 1<.2 , we can represent v

as an arrow from the origin to the point (a. b) in the xy-plane. as shown in Figure 1.1. 2 The importance of vectors in physics was recognized late in the nineteenth century. The algebra of

Figure 1.1 A vector in

n2


vectors, developed by Oliver Heaviside (1850- 1925) and Josiah Willard Gibbs (1 839- 1903). won out over the algebra of quaternion.s to become the language of physicists.

Page 1 of 10



Example 3

9

Velocity Vectors A boat cmises in still water toward the northeast at 20 miles per hour. The velocity u of the boat is a vector that points in the direction of the boat's motion, and whose length is 20, the boat's speed. If the positive y-axis represents north and the positive x-axis represents east, the boat's direction makes an angle of 45° with the x-axis. (See Figure 1.2.) We can compute the components of u = [

11 •] 112

by using trigonometry: lit

RIVER

= 20cos45° =

Therefore. u =

Figure 1.2

[

wh

and

20 sin 45° =

112 =

wh.

10J2l JOJ2j' where the units are in miles per hour.

VECTOR ADDITION AND THE PARALLELOGRAM LAW We can represent vector addi tion graph ically, using arrows, by a result called the parallelogram law.3 To add nonzero vectors u and v, first form a parallelogram with adjacent sides u and v. Then the sum u + v is the arrow along the diagonal of the parallelogram as shown in Figure 1.3. (a + c, b + d)

Figure 1.3 The parallelogram law of vector addition

Velocities can be combined by adding vectors that represent them.

Example4

Imagine that the boat from the previous example is now cm ising on a river. which flows to the east at 7 miles per hour. As before, the bow of the boat points toward the northeast, and its speed relative to the water is 20 miles per hour. Ln this case. the vector u =

[:~~].which we calculated in the prev ious example, represents the

boat's veloc ity (in miles per hour) relative to the river. To find the velocity of the boat relative to the shore. we must add a vector v. representing the velocity of the river, to the vector u. Since the river flows toward the east at 7 miles per hour, its velocity vector is v =

[~]. We can represent the sum of the vectors u and v by using

the parallelogram law, as shown in Figure 1.4. The veloci ty of the boat relative to the shore (in miles per hour) is the vector u + v= [

3


IOJ2 + 7] lOJ2 .

A justification of the parallelogram law by Heron of Alexandria (first century c.~) appears in his Mechanics.

Page 2 of 10


10 CHAPTER 1 Matrices, Vectors, and Systems of Linear Equations North

boat

velocity u

'

water velocity Figure 1.4

To find the speed of the boat, we use the Pyrhagorean lheorem, which !ells us that the length of a vector with endpoint (p. q) is jp2 + q2. Us ing the fact that the components of u + v are p = 10J2 + 7 and q = JOJ2, respectively, it follows that the speed of the boat is

Jp 2

+ q 2 ""25.44 mph.

SCALAR MULTIPLICATION We can also represenl scalar multiplicarion graph ically, using arrows. If v

= [~J is

a vector and c is a positive scalar, the scalar multiple c v is a vector that pomts in the same direction as v, and whose le ngth is c times the length of v. This is shown in Figure I.S(a). If c is negative, c v points in the opposite direction from v, and has length lei times the length of v. This is shown in Figure I.S (b). We call two vectors parallel if one of them is a scalar mulliple of the other. y

J

cv

(m . cb)

(a. b)

v

(a, b) X

(ca. cb)

(a) c

>0

(b)c< O

Figure 1.5 Scalar m ultiplication of vectors

VECTORS IN R 3 If we identify 1?.3 as the set of all ordered triples, then I he same geomelric ideas thai hold in 1?.2 are also true in 1?.3 . We may depict a vector v

= [~]

in 1?.3 as an arrow

emanating from rhe origin of the xyz -coordinate system, with the point (a, b, c) as its


Page 3 of 10



11

"?' " :;l ~ "2 + ''2 j£=-----i-+-'--+-___,_-=---+--+-=-----= )' ..· /

............................~.,....

a

··

X

(b)

(a)

Figure 1.6 Vectors in R

3

endpoint. (See Figure 1.6(a).) As is the case in R 2 , we can view two nonzero vectors in n 3 as adjacent sides of a parallelogram, and we can represent their addition by using the parallelogram law. (See Figure 1.6(b).) In real life, motion takes place in 3-dimensional space, and we can depict quantjtics such as velocities and forces as vectors in n 3 .

EXERCISES In Erercises 1- 12. compwe rile indicared marrices, where

n

-I

A=[;

4

and

1. 4A

2. -A

4. 3A +28

5. (2 B )r

-2]4 .

0 3

3. 4A -28 6. AT +28T 9. AT

8. (A + 2Bj1'

7. A+B 10. A-8

[~

8 =

-I

13. - A

5

2 -6

-~J

C3 .

North )'

if possible,

·{l -ll

and

14. 38 17. A - 8

15. (- 2)A

18. A - 8 T

16. (28l 19. AT - 8

20. 3A +2BT

22. (4Al

23. B -Ar

2 1. (A + B l 24. (BT - Al

In Exercises 25- 28, as.w me rlwr A = 25. Determine a 12. 27. Determine a 1 •

30. Determi ne

31. Determi ne the first row of C. 32. Determine the second row of C.

12. (-B )T

11. - (BT)

In Exercises 13- 24, compure rile indicared matrices, where

A =[~

29. Determine c1 •

[

2~

= [2;

~~l


-3 12

Figure 1.7 A view of the airplane from above

33. An airplane is Aying with a ground speed of 300 mph
26. Determine a 21. 28. Detennine a 2 •

In Erercises 29- 32, assume rhm C

X

- f - - - - - - - - E ast

0.4]o·

35. A pilot keeps her airplane pointed in a northeastward direction while maintaining an airspeed (speed relative to the surrounding air) of 300 mph. A wind from the west blows eastward at 50 mph .

Page 4 of 10


12

CHAPTER 1 Matrices, Vectors, and Systems of linear Equations

(a) Find the velocity (in mph) of the atrplane relative to the ground. (b) What is the speed (in mph) of the airpl.tne rehuive to the ground? 36. Suppose that in a medical study of 20 people. for each i. I !: i !: 20, the 3 x I vector u, is defined so that its components respectively represent the blood pressure. pulse rate, and cholesterol reading of the i th person. Provide an interpretation of the vector + ll2 + ... + li2Q).

m
~

Exercises 37- 56, detemtille ~ me/1/s are tme or falsi'. /11

11 llnllu

tile state-

37. Matrices must be of the same size for their s um to be dehned. 38. The transpose of a sum of two matrices ;, the sum of the trdnsposcd matrices. 39. Every vector is a matrix. 40. A scalar multiple of the zero matrix is the 7.ero scalar. 4 1. The transpose of a matrix is a matrix of the same size. 42. A submatrix of a matrix may be a vector. 43. If B is a 3 x 4 matrix, then its rows are 4 x I vectors. 44. l11e (3, 4)-entry of a matrix lies in column 3 and row 4. 45. In a zero matrix. every entry is 0. 46. An m x

11

matrix has m

+ 11 entnes.

47. If ' ' and ware vectors such that '' = -3w, then ,. and w are parallel. 48. If A and B are any m x

11

matrices. then

A-B =A+(- 1)8.

49. The (i.j)-entry of AT equals the (j, i)-entry of A. 50. If

A= [~ ~] and B = [~ ~ ~l then A= B.

51. In m1y matrix A. the sum of the e ntries of 3A tXIua ls three times the sum of the entries of A. 52. Matrix addition is commutative. 53. Matrix addition is associative. 54. For any m x 11 matrices A and 8 :md any •calaT>. c and d. (cA +dB)T =cAT+ dBT. 55. If A i' a matrix. then cA i• the same '"e a~ A for e\'ery >calar c.

56. If A is a matrix for which the sum A +A r is defined. then A i' a 'quare matrix.

57. Let A and 8 be matrices of the same size. (a) Prove that the j th column of II + B is

111

+ bi.

(b) Prove that for any scalar c. the jth colunm of cA is

car 58. For any m x 11 matrix A, prove that OA = 0. the m x 11 zero matrix. 59. For any 111 x

11

matrix A. prove thai lA =A.

60. Prove Theorem 1.1 (a). 61. Prove Theorem 1.1 {c). 62. Prove Theorem l.l (d ). 63. Prove Theorem 1.1 (e). 6-1. Prove Theorem l.l(g). 65. Prove Theorem 1.2(b). 66. Prove Theorem 1.2(c). A square mmri.r A is called a diagon al matrix if a,1 = 0 ll'hene•·er i :F j. Exercises 67-70 are co11cemed with dillf(OIIttl 11/lllri·

ces. 67. Prove that u square zero matrix is a diagonal matrix . 68. Prove that if 8 is a diagonal matrix, then cB is a diagonal matrix for any scalar c. 69. Prove that if 8 i' a diagonal matrix, then i> a diagonal matrix. 70. Pro1·e that if 8 and C arc diagonal matrices of the same si1.c, then 8 -1- C i~ a diagonal matrix.

or

A (squa,...)matril A is said to be symmetric ifA= AT. Eurc:is~s 71-78 a,... t'tmtemt'd ll'itlt symmerric matrices. 7 I. Give cx:unple\ of 2 x 2 and 3 x 3 symmetric matricc,. 72. Prove thut the (i .})-entry of a symmetric matnx equals the (j. i)-entry. 73. Prove that a square zero matrix is symmetric. 74. Prove that if B is a symmetric matrix, then so is cB for any scalar c. 75. Prove that if B is a square matrix, then 8 + 8 r i• 'Ymmetric. 76. Prove that if 8 and C are 11 x 11 symmetric matrice,, then so is 8 +C. 77. Is a square submatrix of a symmetric matrix necessarily a •ymmetric matrix? Justify your answer. 78. Prove that a diagonal matrix is symmetric. A (squa,...) mmri.1 A is calletl sk ew-symmetric

if AT= -A.

E.xerci,,e,, 79 -t~ / are <:oncerned with skew·.\'J'mmetric nuurice.f.

79. What must he tnte about the (i, i)-entries of a ,kcwsymmctric matrix? Justify your answer. 80. G ive an example of a nonzero 2 x 2 skew-symmetric m:ttrix 8 . Now show that every 2 x 2 ;kew-symmelric matrix is a scalar multiple of B. 81. Show thai e1ery 3 x 3 matrix can be writlen a' the •um of a symmetric matrix and a skew-symmetric matrix. 82~ The t race of an 11 x 11 matrix A. written trace(A ). is defined to be the sum trace(A) = a 11 +au+···+ a,... Prove that. for any 11 x 11 matrices A and 8 and sc:1lar c, the following Matements are true: (a) tracc(A + 8) = trace(A) +trace(B). (b) trace(cA) = c · tr:tce{A). (c) trace(A 7) = tmce(A). 83. Probability IW'tors are \'CCtors whose comi>Oncnt' are nonnegative and have a sum of I. Show that if p and q arc probability vectors and a and b are nonnegatile scalars with ll + b = I. then a p + bq is a probabtlity vector.

• This exercise is used in Se

Page 5 of 10


1.2 linear Combinations, Matrix-Vector Products, and Specia l Matrices

In the following etudsr. uu tdthrr a calculator with matrix capabilities or computu sajfll'art' .wch a.r MATI.AB to soh·t' tht' 11roblem:

and

B=

84. Constder the matrices

A-

-3.3

2.1

r,

5.2 3.2 08 -1.4

6.0 -I. I 3.4 1.1 -4.0 - 12. 1 5.7 0.7 4.4

2.3

-2.6 -1.3 3.2

l

2.6 2.2

-1.3 -2.6

7.1

1.5 - 1.2

0.7 1.3 -8.3 2.4

-0.9

1.4

r

-0.9 3.3

13

-4.4] -3.2 4.6 . 5.9

6.2

+ 28.

(a) Compute A

(b) Compute A - B. (c) Compute AT + B T.

SOLUTIONS TO THE PRACTICE PROBLEMS I. (a) The (I. 2)-entry of A

IS

-- [23

2.

(b) The (2. 2)-entry of A is 3.

2. (a) A - 8

0 4 I

= [: (b) 2A = 2(; (c) A+ 38

=

-~J- [~

- I

= [;

- I

3. {a)

-~J = [:

0

= [~

~]

-!]

- I

[i

3

-2 0

- I

_;J

9 •] + [3 6 -3 ~~]

-I

0

-2

8

I(')]

-3

j]

AT= [- :

(b) {3Bf = [~ (c) (A+ B)T

=

[~

12

2 - I

0

(!}]

or

9 -3

=

[3 9 -~] 0

~r = [~

12

-n

LINEAR COMBINATIONS, MATRIX- VECTOR PRODUCTS, AND SPECIAL MATRICES

In this ~ction. we explore some applications involving matrix oper..tions and introduce the product of a matrix and a vector. Suppose that 20 students are enrolled in a linear algebra course. in which two

= lltl ;- . where 11, denotes the score II>

tests. a quiz. and a final exam are given. Let u

r

1120

of the ith student on the first test. Likewise, define vectors v, w. and z >imilarly for the second test. quiz. and final exam. respectively. Assume that the instructor computes a student's course average by counting each test score twice as much as a quiz score, and the final exam score three times as much as a test score. Thus the weights for the tests, quiz, and final exam score are. res pectively, 2/11 ,2/1 1, 1/ 11, 6/ 11 (the weights must sum to one). Now consider the vector

The first component Yt represents the first si'Udent 's course average, the second component .\'2 represents the second student's course average. and so on. Notice that y is a sum of scalar multiples of u, v, w, and z. This form of vector sum is so imponanl that it merits its own definition.


Page 6 of 10


14


Definitions A linear combination of vectors u 1, u 2, . . . , u* is a vector of the form

where Ct, c 2•..• , Ck arc scalars. These scalars arc called the coefficients of the linear combination. Note that a linear combination of one vector is simply a scalar multiple of that vector. In the previous example, the vector y of the students' course averages is a linear combination of the vectors u, v, w, and z. The coefficients are the weights. Indeed, any weighted average produces a linear combination of the scores. Notice that

Thus

[~]

is a linear combination of [:].

[~].and

[ _ :]. with coefficients - 3. 4,

and l. We can also write

m = [:J + 2m - ~[-:J. This equation also expresses

[~]

as a linear combination of[:].

but now the coefficients are I, 2. and - I. So the set of coefficients vector as a li near combination of the others need not be unique.

IJ&l,l.!fjl

.

[ 4]. .

(a) Determme whether _ (b) Detennine whether [ (c) Detemline

Solution

IS

1

=~J

whether [~]

~~].and [ _ :]. ~at express one

. . [2] [3]

a linear combmat10n o f

is a linear combination

3

and

of [~] and

is a linear combination of [; ] and

1

.

[7].

[~l

(a) We seek scalars x 1 and x2 such that

[2]

[3] [ - 4] l =x1 3 +x,- l

[3x2] [2xt + 3x2] = [2tt] 3Xt + lx2 = 3Xt + X2 .

That is, we seck a solution of the system of equations

2xt + 3x2 = 4 3xt + x 2 = - I. Because these equations represent nonparallel lines in the one solution, namely,


x1

= - I and

x2

= 2. Therefore

~lane,

[_; J is

there is exactly

a (unique) linear

Page 7 of 10


1.2 Linear Combinations, Matrix-Vector Products, and Special Matrices 15

combination of the vectors [;] and

[~]. name ly,

(Sec Figure 1.8.) y

_../

[~]

2

.. .· /

...

,:

.. /

X

/ [- 4]1

...······························

Figure 1.8 The vector

[_~]is a linear combination of

(b) To determine whether [

=~J

[i]

and

is a linear combination of

[~l

[~]

and

[~l we

pe1form a similar computation and produce the set of equations

6x1 3xl

+ 2x2 = -4 + x2 = - 2.

Since the first equation is twice the second. we need only solve 3xl + x2 = - 2. Thi s equation represents a line in the plane, and the coordinates of any point on the line - 2 and X2 4. In this case. we have give a solution. For example, we can let Xi

=

=

There are infinitely many solutions. (See Figure 1.9.)

[~]

Figure 1.9 The vector [


=~J is a linear combination of[~] and [~].

Page 8 of 10


16 CHAPTER 1 Matrices, Vectors, and Systems of Linear Equations

(c) To determine

if[~] is a linear combination of [~] and

[!l

we must solve

the system of equations

=

3xl + 6x2 3 2x1 +4x2 =4.

-!

If we add times the first equation to the second, we obtain 0 = 2, an equation with no solutions. Lndeed, the two original equations represent parallel lines in the plane, so the original system has no solutions. We conclude that combination of

[~]

is not a linear

and [ : ]. (See Figure 1.10.)

Figure 1.10 The vector[!] isnoralinearcombinationof

Example2

[!]

[;]and[~].

Given vectors u 1, u 2, and u 3, show that the sum of any two linear combinations of these vectors is also a linear combination of these vectors.

Solution

Suppose that w and z are li near combinations of u 1, u2, and u3. Then we

may write

w = a u 1 + bu2 + c u3

and

where a ,b, c, a', b',c' are scalars . So w + z = (a + a ')u 1 + (b + b')u2 + (c + c')u3,

which is also a linear combi nation of U J, u 2, and U3.

STANDARD VECTORS We can write any vector [: and [


J in 1?}

as a linear combination of the two vectors [

~J

~J as follows:

Page 9 of 10


1.2 Linear Combinations, Matrix-Vector Products, and Special Matrices 17

The vectors

[01]

write any vector

[~]

and

[n

[~] in

are called the srandard vecrors of 1<.2 Similarly, we can

n 3 as a linear combination of the vectors

[~l [~land

as follows:

The vectors [

[n=a [~]

l

+b

[~] +c [~]

~ [~l and [~] are called the

sTandard vecTors of

nJ

In general. we define the standard vectors of 'R." by

(S ee Figure I II. ) )'

.,

.,

)'

., .r The standard Vettors of n2

Tile standard vector~ of

n'

Figu re1 .11 y

au

·······

From the preceding equations, it is easy to see that every vector in 'R." is a linear combination of the standard vectors of 'R.". In fact. for any vector v in 7?!' .

.r

(See Figure I. 13.) Now let u and v be nonparallel vectors, and let w be any vector in 2. Begin w ith the endpoint of w and create a parallelogram with sides a u and bv, so that w is its d iagonal. It follows that w =au + bv; that is, w is a linear combinat ion of the vectors u and v. (See Figure 1.1 2.) More generally, the following statement is true:

n

Figure 1.12 The vector w is a lin-

ear combination of the nonparallel vectors u and v.

If u and v are any nonparallel vectors in 1<.2• then every vector in 1<.2 is a linear combination of u and v.


Page 10 of 10



··....

X

The veclor ,, is a linear combination of standard vectors in n 3.

The vector v i~ a linear combination of standard vectors in 'R2 .

Figure 1.13

Practice Problem 1 •

Let

w=

[~~]and S = {[~]. [-~J } -

(a) Without doing any calculations, explain why w can be wrinen as a linear combination of the vectors in S. (b) Express w as a linear combination of the vectors in S. Suppose that a garden supply store sells three mixtures of grass seed. The deluxe mixture is 80% bluegrass and 20% rye. the standard mixture is 60% bluegrass and 40% rye, and the economy mixture is 40% bluegrass and 60% rye. One way to record this information is with the following 2 x 3 matrix: lie luxe

8=

.80 [ .20

' tandanl .60 .40

economy

.40 ] .60

bluegrass rye

A customer wants to purchase a blend of grass seed containing 5 lb of bluegrass and 3 lb of rye. There are two natural questions that arise: 1. Is it possi ble to combine the three mixtures of seed into a blend that has exactly the desired amounts of bluegrass and rye, with no surplus of either? 2. If so, how much of each mixture should the store clerk add to the blend? Let x 1, xz, and x3 denote the number of pounds of deluxe, standard, and economy mixtures. respectively. to be used in the blend. Then we have

.80x, .20x,

+ .60xz + .40x3 = 5 + .40xz + .60x3 = 3.

This is a system of two linear equations in three unknowns. Finding a solution of this system is equ ivalent to answering our second question. The technique for solving general systems is explored in great detail in Sections 1.3 and 1.4. Using matrix notation, we may rew rite these equations in the form

.80x1 + .60x2 + AOx3 J = [5] _ [.20xt + .40,r2 + .60x3 3


Page 1 of 10


1.2 Linear Combinations, Matrix-Vector Products, and Specia l Matrices

19

Now we use matrix operations to rewrite th is matrix equation, using the columns of Bas

x, [ .80] .20 + x2 [.60] .40 + xJ [ .40] .60 Thus we can rephrase the first question as follows: Is

= [5] 3 · [~]

a linear combination of the

. the box on page 17 provtdes . columns [ ..80] , [.60] . , and [.40] .60 o f 8.? The result m an 20 40

affirmative answer. Because no two of the three vectors are parallel,

[~]

is a linear

combination of any pair of these vectors.

MATRIX- VECTOR PRODUCTS

8

- [.80 .20

X-

.60 .40] .40 .60

[XI] ;~

=Xt

[.80] [.60] [.40] .20 + x2 .40 + xJ .60 ·

This defin ition provides another way to state the first question in the preceding example: Does the vector

[~]

equal Bx for some vector x? Notice that for the

matrix-vector product to make sense, the number of columns of 8 must equal the number of components in x. The general definition of a matrix-vector product is given next. Definition Let A be an m x 11 matrix and v be an 11 x I vector. We define the ma tr ix-vector p rod uct of A and v. denoted by Av. to be the linear combination of the columns of A whose coefficients are the corresponding components of v. That is,

As we have noted, for Av to exist, the number of columns of A must equal the number of components of v. For example, suppose that

A=

[i :]

and

Notice that A has two columns and v has two components. Then

Av =


[~

: ]

m [~] + [;I]+[ii] [~n. =

7

8 [ :]

=

Page 2 of 10


20


Returning to the preced ing garden supply store example, suppose that the store has 140 lb of seed in stock: 60 lb of the deluxe mixture, 50 lb of the standard mixture, and 30 lb of the economy mixture. We let v

= [~~]

represent th is information. Now

the matrix-vector product

8v

=

[~~] 30

.80 .60 .40] [ .20 .40 .60 80

60

40

= 60 [·.20] +50 [·.40] + 30 [·.60] seed (lb)

[~~]

bluegrass rye

gives the number of pounds of each type of seed contained in the 140 pounds of seed that the garden supply store has in stock. For example, there are 90 pounds of bluegrass because 90 .80(60) + .60(50) + .40(30). There is another approach to computing the matrix- vector product that relies more on the entries of A than on its col umns. Consider the following example:

=

Av

= [Ot t 02t

Notice that the first component of the vector Av is the sum of products of the corresponding entries of the first row of A and the components of v. Likewise, the second component of Av is the sum of products of the corresponding e ntries of the second row of A and the components of v. With tllis approach to computing a matrix- vector product, we can omit the intermed iate step in the preceding illustration. For example, suppose

A = [~ -~ ~]

and

Then

Av


=

[7

3

l]

-2 3

[-l] [ ~

=

(2)(- l ) + (3)( 1) + ( 1)(3)

J

[4]

(1)(- 1)+(-2)( 1) + (3)(3) = 6 .

Page 3 of 10


1.2 Linear Combinations, Matrix-Vector Products, and Special Matrices

In general, you can use this technique to compute A v when A is an m x v is a vector in R". ln this case, the i th component of A v is

[ail

a;2 ...

a;,.][~:~ ]

11

21

matrix and

=a; ,v, +ai2v2+·· · + a;,.v,.,

VII

which is the matrix - vector product of the ith row of A and v. The computation of all the components of the matrix-vector product Avis given by

all

OJnl [VJ] a2,. v2

02)

[O02ti1VJ\IJ +at2V2+ atn\ln l + anv2 + ····· · ++ az,, v,.

Av=

[

Practice Problem 2 ...

a,~,. - I

Let A= [ ;

0

v:, =

a,2

a,"

OmJVt + Om2V2;+ · ·· +a,"v"

-~J and v

= [ _ : ] .Compute the following vectors:

.

(a) Av

(b) (A v)T

Example3

A sociologist is interested in studying the population changes within a metropolitan area as people move between the city and suburbs. From empirical evidence, she has discovered that in any given year, IS % of those living in the city will move to the suburbs and 3% of those living in the suburbs w ill move to the c ity. For simplic ity, we assume that the metropolitan population remains stable. This information may be represented by the following matrix: Fro m City Suburb~ To

C ity Suburb~

.85 [ .15

.03] =A .97

Notice that the entries of A are nonnegative and that the entries of each column sum to I. Such a matrix is called a stoch astic m atrix. Suppose that there arc now 500 thousand people living in the c ity and 700 thousand people living in the suburbs. The socio logist would like to know how many people will be living in each of the two areas next year. Figure 1. 14 describes the changes of population from one year to the next. It follows that the number of people (in thousands) who will be living in the city next year is (.85)(500) + (.03)(700) = 446 thousand, and the number of people living in the suburbs is (.15)(500) + (.97)(700) = 754 thousand. If we Jet p represent the vector of current populations of the city and suburbs, we have


Page 4 of 10


22


This yeor

City

Suburbs

500 thousand

700 thousand

t5%

.Vr 97%

Next year

c85X500r

(. t5)(500) + (.97)(700) Suburbs

+ 1.03)r70<))

C U}

Figure 1.14 Movement between the city and suburbs

We can find the populations in the next year by computing the matrix-vector product: A

p

= [ 85 .15

.03] [500] = [(.85)(500) + (.03)(700)] = [446] .97 700 (. 15)(500) + (.97)(700) 754

In other words, Ap is the vector of populations in the next year. If we want to detennine the populations in two years, we can repeat this procedure by multiplying A by the vector Ap . That is , in two years, the vector of popu lations is A(Ap).

IDENTITY MATRICES Suppose we let /2

= [~ ~] and v be any vector in 7?} . Then

So multiplication by /2 leaves every vector v in 7?} unchanged. The same property holds in a more general context. Defin ition For each positive integer n, the 11 x 11 identity m atr ix / 11 is the 11 x matrix whose respective columns are the standard vectors et . e2, . . . , e11 in R 11 •

11

For example,

and

/3

= [0I 0

0 0]0 . I

0

I

Because the colu mns of 1, are the standard vectors of R 11 • it follows easil y that 1, v for any v in R".

=v

ROTATION MATRICES

n

Consider a point Po = (xo,yo) in 2 with polar coordinates (r, a ), where r ~ 0 and a is the angle between the segment OPo and the positive x -axis. (See Figure 1.1 5.) Then xo = r cos a and Yo = r s in a. Suppose that OPo is rotated by an angle B to the


Page 5 of 10


1.2 Lin ear Combin ation s, Matrix-Vector Products, and Specia l Matrices

23

y

Po = (xo. .ro} Yo

0

X

Figure 1.15 Rotat ion o f a vector through the angle 0 .

segment OP 1, where P 1 = for P I· and hence

(XJ. )'J).

Then (r ,c:r + B) represents the polar coordinates

x , =rcos(c:r + B} = r (cosc:r cosO - sin c:r sin 0) = (r cosc:r)cosB - (rsin c:r)sin O

= xo cos O -

Yo sin O.

Similarly, y1 = xo sin B +yo cos O. We can express these equations as a matrix equat ion by usi ng a matrix- vector product. If we define Ao by A _ [ cos O - sin O] 0 sin O cosO •

then A 0 [xo] = [cos 0 - sin O] [xo] = yo smB cosO yo

[xocos O- yo sin B] = xo s1n B + yo cosB

[x']. y1

We call A0 the B-rotation matrix, or more si mply, a rotation matrix. For any vector the vector A0u is the vector obtained by rotating u by an ang le 0, whe re the rotation is counterclockwise if B > 0 and clockwise if 0 < 0.

u,

To rotate the vector

[!]

by 30°. we compute A30o

[!}

that is.

J3 cos 30° [ sin 30°

- sin 30: ] [3] = cos30 4

2 [

1

2

Thus when

4] . + -J3 [3]4 is rotated by 30°, the resultjng vector is "2' [3~ 3 4

It is interesting to observe that the 0° -rotation matrix A0o , which leaves a vector unchanged, is given by A0o "' /2. This is quite reasonable because multiplication by / 2 also leaves vectors unchanged.


Page 6 of 10



Besides rotations, other geometric transformations (such as reflections and projections) can be described as matrix- vector products. Examples are found in the exercises.

PROPERTIES OF MATRIX- VECTOR PRODUCTS I! is usefu l to note that the columns o f a matrix can be represented as matrix-vector products of the matrix with the standard vectors. Suppose. for example, that A =

[i :].

Then

and The general result is stated as (d) of TI1eorem 1.3. For any m x 11 matrix A. AO = 0', where 0 is the 11 x 1 zero vector and 0' is the m x I zero vector. This is easily seen since the matrix-vector product AO is a sum of products of columns of A and zeros. Similarly. for them x 11 zero matrix 0. Ov = 0' for any n x I vector v. (See (f) and (g) of Theorem 1.3.)

THEOREM 1.3 (Properties of Matrix- Vector Products)

Let A and B be m x

11

matrices, and

let u and v be vectors in R". Then (a) A(u + v) =A u + Av. (b) A(c u) = c(Au) = (cA)u for every scalar c. (c) (d) (e) (f) (g)

(A+ B)u =Au +B u. Aej a j for j I. 2. . . . , 11 , where Cj is the jth standard vector in R ". If 8 is an 111 X II matrix such that 8w = Aw for all w in n", then 8 =A. AO is the m x I zero vector. If 0 is the 111 x 11 zero matrix, then Ov is them x I zero vector. (h) J, v = v.

=

=

PROOF We prove part (a) and leave the rest for the exercises. (a) Because the ith component of u + vis u; + v;, we have

A(u + v) = (111 + vr)ar + (112 + 112)a 2 + · · · + (u, + v,)a,

=Au + Av.

•

It follows by repeated applications of Theorem 1.3(a) and (b) that the

matrix-vector product of A and a linear combination of u 1• u 2..... Uk yields a linear combination of the vectors Au 1,A u 2, ... ,A uk. That is, For any m x

11

matrix A. any scalars

CJ. c2,

... , Ck. and any vectors U J. u2, .... Uk

inn",


Page 7 of 10


1.2 Linear Combinations, Matrix-Vector Products, and Specia l Matrices 25

EXERCISES In Eurcises 1- 16, compwe the mmrix- vector products.

1.

3.

[! -~ ~J[-i]

[! -~ -l][!]

29. u = [:].s=

4.

9.

[~ ~r [~J

8.

[~ ~ ~] [~]

ll. [

13. [

-~ ~~] [~]

14.

-~]0

- ;

- 1

u

= [_ :]. S= {[:]}

32.

u

= [:]. s = {[~]. [-~Jl

16. ([

36. ..

38. [:]

I

41.

m

=

fl=210°, u = [=~J

22. fi= l35°, u = [-7]

23.

o = no•, u =

24.

27.

(!

(!

= 240· . .. = [ = 300· . .. =

=n

m

26. 28.

= 30°. .. =

[~J

21.

(!

o= (!

(!


330•. u =

[~]

= 150°. .. = [

-~J

= 120· . .. = [

-~l

S=

U=[~J.s= {[:J.[-m

u=[_Hs= {[j].[-l]} [

42. u =

S= {

[!J.[-iJl

S= {

11

=[

44. ..

=[

43.

-n

I =[-Us= { [-!].[-~] · [-:] } [H GJ. [!] '[~] }

C1

(!

[ -;J

20.

U

19

25.

= 60°. II =

18. (/ = 0°.

Cz

S=

40. .. =

In Eurcises 17- 28, an angle 0 and a 1•ector u are given. Write the corresponding rotation matrix, and compwe the vectorfowtd by rotating u by tire angle 0. Draw a sketch and simplify yortr answers.

=

[

l wJ. [-m = [:l mJ. [-n[:JJ m. {[~].[;].[=;]}

37. u =

-~ ~J + [~ -~D [~J

U

S= {

~: s=

-~ ~r + [~ _;f)[~J

17. (/ = 45°.

=

35. .. = [

39. 15. ([

[i].

[~l j]} 34. .. =[:J. s =I[~J. [-n [~J 1

-~ -~] [~]

[-~3

mJ.[~]}

31.

33. u

[jr [-~J

12. [

-~ -~ -~J [~]

3] [ -:]

[~ ~ ~] [~]

10.

If possible.

30. u = [_:]. S = {[ _:]}

[~ -~H~J

6. [2

7.

In Exercises 29-44, a vector u and a setS are given. write u as a linear combination of the \'ectors in S.

~

=n

-n

5= {

S= {

[~], [~] , [~] } [-:J. [-~J . [-iJ J

bt Eurcises 45- 64, determine wlrether the state·

~ mews are true or false.

45. A linear combina1ion of vectors is a sum of scalar mulli ples of the vectors.

46. The coefficienls in a linear combinalion can always be chosen to be positive scalars.

Page 8 of 10


26


n

47. Every vector in 2 can be wriuen as a linear combination of the standard vectors of 2•

n

n

48. E,·ery vector in 2 i; a linear combinalion of any two nonparallel vectors. 49. The zero vector i; a linear combination of any nonempty set of vectoN. 50. The matrix \'ector product of a 2 x 3 matrix and a 3 x I ,·ector i< a 3 x I 'ector. 51. The matrix vector product of a 2 x 3 matrix and a 3 x I vector equals a hnear combination of the rows of the matrix. 52. The product of a matrix and a standard ,·ector equals a standard vector. 53. The rotation matrix A 1 ~0 equals - / 2• 54. The matrix vector product of an m x 11 matrix and " vector yield' a vector in R.". 55. Every vector in 2 i> a linear combination of two parallel vectors. 56. Every vector v in n" can be wriuen as a linear combination of the standard vectors. using the components of v as the coefficients of the linear combination. 57. A vector with exact ly one nonzero component is called a standard vector. 58. If A is an m x 11 matrix, u i~> a vector in n". and c is a scalar, the n A(c u ) = c(A u). 59. If A is an m x 11 matrix. then the only vector u in n• such that Au = 0 is u = 0 . 60. For any \'ector u in 2• A~u is the vector obtained by rotating u by the angle 0. 61. If 0 > 0. then A. u i> the vector obtained by rotating u by a clockwise rotation of the angle(). 62. If A i' iln m x 11 matri' and u and ,. are vectors in n• such that Au = A''· then u = v. 63. The matrix vector product of an m x 11 matrix A and a \"eCtor U in n• equaJ, 11(8 1 + 112 3 1 - • · • + 11n 3 n.

70.

~t

64. A matnx havmg nonncgauve entries such that the sum of the entrie' in each column is I is called a stochastic matrix. 65. Use a matrix vector product to >how that if () Aov = v for all v in 2•

n

= 0 °. then

66. Usc a matrix vector product to show that if 0 = 180 °. - v for :til v in 2• then Aov

=

n

67. Usc matrix- vector products to >how that. for any angles 0 and fJ and any vector v in 2• Ao(A,9 v) = A0+.8"· 68. Compute A~(Ao u) and Ao(A~ u) for any vector u in n 2 and any :mgle 0. 69. Suppose that in a metropolitan area there me 400 thousand people living in the city and 300 thousand people living in the ;uburbs. Use the >tochastic matrix in Example 3 to determine (a) the number of people living in the ci ty and s uburbs after one year; (b) the number of people living in the city and suburbs after two years.

n


~ ~

:] and u =

GJ

Repre>ent Au as a

linear combination of the columns of A. hr £rercises 71- 74,1t't A = [

-~ ~] a11d u = [~

l

71. Show that Au is the reRcction of u about the y·axis. 72. Prove that A(A u) = u . 73. Modify the matrix A to obtain a matrix 8 ..a that B u i> the reRection of u about the ,, -axis.

74.

~t

C denote the rotation matrix that corre,pond> to

() = 180.

(a) Find C . (b) Use the matrix B in Exercise 73 to >how that

n

n

A= [

A(C u) = C(A u) = B u

and

B(C u) = C(B u) = Au . (c) Interpret these equations in terms of renections and rotations. /11

Erercises 75-79. let A

= [~ ~] and u =

(;:

l

75. Show that Au is the projection of tt on the x·axis. 76. Prove that A(Au ) = Au .

77. Show that if v is any vector whose endpoint lies on the x-axis. then Av = ''· 78. Modi fy the matrix A to obtain a matrix 8 so that B u is the proJection of u on the \'-axi>. 79.

~t C denote the rotation matrix that corresponds to 8 = 180 . (See Exerme 74(a).)

(a) Prove that A(C u) = C(Au ). (b) Interpret the result

111

(a) geometncally.

80. ~t u1 and u 2 be vectors rn n•. Pro\"e that the >urn of two linear combination; of these vectors is also a tinear combination of the>e vecton.. 8 1. ~t 111 and 112 be vecton. in n•. Let v and w be linear combination' of u 1 and u 2. Prove that any linear combination of v and w is also a linear combination of u 1 and u 2• 82.

u 1 and u 2 be vectors in n•. Prove that a >c:llar multiple of a linear combinatio n of thc>c vectors is also a linear combination of these vectors. ~t

83. Prove (b) of Theorem 1.3. 84. Prove (c) of Theorem 1.3.

85. Prove (d) of Theorem 1.3. 86. Prove (e) of Theotc m 1.3. 87. Prove (t) of Theore m 1.3.

88. Prove (g) of Theorem 1.3. 89. Prove (h) of Theorem 1.3. /11 Exercises 90 a11d 91. ll.fl! either a wlclll(l(or ll"ith matrh capo·

bilitirs or compwu softwarr suclr as MATLAB to soh·c caclt problem.

Page 9 of 10


1.3 Systems of Linear Equations

90. In reference to Exercise 69. dete rmine the number of pcopie living in the city and suburbs after 10 years.

and the vectors

91. For the matrices

A=

and B

['' ['' 1.3 4.4 0.2

=

1.3

-0.1 4.5 5.7 1.1

-9.9 -2.2

9.8 1.1 4.8 2.4 -2.1

-1.2 1.3 6.0

27

"{l]

'] '] 6.2 2.0

and

-8.5

3.0 2.4 -5.8 -5.3

(a) compute Au: (b) compute B (u

6.0 2.8 8.2

(c) compute (A

+ v);

+ B )v:

(d) compute A(Bv).

SOLUTIONS TO THE PRACTICE PROBLEMS Using elementary algebra. we sec that

I. (a) The vectors inS are nonparallel vectors in 1?.2

x2

(b) To express w as a linear combination of the vectors in S. we must find scalars Xt and x2 such that

[-1] [2] +x [ 3] = [2x1+ 3x2] 10

=Xt

2 -2

I

Xt

-2r2

That is. we must solve the following system: 2.tt

2x2 =

=4

and

3[

2.(a)Av =[~ -~ -~] [_:] = [,~] (b)(Av/ = [~~r =(4 11]

I 10

CD]

XI

[~~J = [~J _ -n 4

.

+ 3x2 = -

Xt-

= - 3. So

SYSTEMS OF LINEAR EQUATIONS

A linear equation in the variables (unknowns) x 1,x2 • . . . ,x,. is an equation that can be written in the fom1

where a 1,a2, ... ,a,., and b are real numbers. The scalars a 1,a2, . .. ,a,. are called the coefficients, and b is called the constant term of the equation. For example, 3x 1 - ?x2 +x3 19 is a linear equation in the variables x 1, x2, and x3, with coefficients 3. -7, and l. and constant tcnn 19. The equation 8x2- 12xs = 4x t - 9x3 + 6 is also a li near equati on because it can be wrinen as

=

On the other hand. the equations

2xt -1x2 +xi= -3,

and

4J,X;' - 3x2 = IS

are 11ot linear equations because they contai n terms in volvi ng a product of variables, a square of a variable. or a square root of a variable. A system of linear equations is a set of m linear equati ons in the same 11 vru·iablcs. where m and 11 arc positive integers. We can write such a system in the


Page 10 of 10


28


form

= b,

a,,x, + a12x2 + · · · + a,,x, a21X1

+ 022X2 + · ·· + a2,X, = b2

0 111 1X1

+ a,2X2 + · ·· + a,,x,

= h111 ,

where a;j denotes the coefficient of Xj in equation i. For example, on page 18 we obtained the following system of 2 linear equations in the variables x,, x2. and x3: .80x,

+ .60x2 + .40x3 =

.20x,

+ .40x2 + .60x3 =

5 3

( l)

A solution of a system of linear equations in the variables x,,xz, ... ,x, is a vector [:;] in 1?!' such that every equation in the system is satisfied when each x;

s, is replaced by s;. For example, .80(2) + .60(5) + .40( 1)

[~] is a solution of system ( I ) because =5

and

.20(2) + .40(5) + .60( 1)

= 3.

The set of all solutions of a system of linear equations is called the solution set of that system.

Practice Problem 1 ""'

Determine whether (a) u =

[ -~~l

" ' (b) •

~ [1] ~l ,.ioM ore

of oho

''"~

of

linear equations

x, + 2x, - xz

Sx3 -

+ 6x3

X4

=7 = 8.

SYSTEMS OF 2 LINEAR EQUATIONS IN 2 VARIABLES A linear equation in two variables x and y has the form ax+ by= c. When at least one of a and b is nonzero, this is the equation of a line in the xy-planc. Thus a system of 2 linear equat ions in the variables x andy consists of a pair of equations, each of which describes a line in the plane.

+ b,y = c, a2x + bzy = c2

a,x

is the equation of line L,. is the equation of line

Lz.

Geometrically, a solution of such a system corresponds to a point lying on both of the lines L 1 and L2. There are three d ifferent situations that can arise. If the lines are different and parallel, then they have no point in common. In this case, the system of equations has no solution. (See Figure 1.16.) If the lines are different but not parallel. then the two lines have a unique point of intersection. In this case, the system of equations has exactly one solution. (See Figure 1.17 .)


Page 1 of 10



29

y

)'

c,

c,

X

X

£ 1 and £ 2 arc diffcrcnl buo nol parallel.

£ 1 and £ 2 arc parallel.

Exactly one solution

No solution

Figure 1.16

Figure 1.17

Finally. if the two lines coincide, then every point on £, and Lz satisfies both of the equations in the system, and so every point on £ 1 and Lz is a solution of the system. ln this case. there arc infinitely many solutions. (Sec Figure 1.18.) As we will soon see, no matter how many equations and variables a system has, there arc exactly three possibilities for its solution set. X

Every system of linea r equations has no solution, exactly one solution, or infinitely many solutions . .C 1and .C.2 are the sa. me. Infinitely many solulions

Figure 1.18

A system of linear equations that has one or more solutions is called consistent; otherw ise, the system is called inconsistent. Figures 1.17 and 1.18 show consistent systems, while Figure 1.16 shows an inconsistent system.

ELEMENTARY ROW OPERATIONS To find the solution set of a system of linear equations or determine that the system is inconsistent, we replace it by one with the same solutions that is more easily solved. Two systems of linear equations that have exactly the same solutions are called equivalent. Now we present a procedure for creating a simpler, equivalent system. It is based on an important technique for solving a system of linear equations taught in high school algebra classes. To illustrate this procedure, we solve the following system of three linear equations in the variables x 1, xz, and x3:

x, - 2xz - X3 = 3 3x, - 6xz - Sx3 = 3 2x1 - Xz + XJ = 0

(2)

x,

We begin the simplification by eliminating from every equation but the first. To do so, we add appropriate mu ltiples of the first equation to the second and third equations so that the coefficient of x 1 becomes 0 in these equations. Adding - 3 times the first equation to the second makes the coefficient of equal 0 in the result.

x,

-3x, + 6xz + 3x3 = -9 {-3 times equation I) 3xl - 6x2 - Sx3 = 3 (equalion 2) - 2x3 = - 6


Page 2 of 10


30


Likewise, add ing - 2 times the fi rst equation to the th ird makes the coeffic ient of x 1 0 in the new third equation.

-2x, + 4x2 2r, - x2 3x2

+ 2x3 = -6 + XJ = 0 + 3x3 = -6

(-2 times equation I) (equalion 3)

=

We now replace equation 2 with - 2r3 - 6, and equation 3 with 3x2 transform system (2) into the following system: X1 -

+ 3x3 = -

6 to

2t2 - X3 = 3 - 2x3 = - 6 3x2 + 3x3 = -6

In this case, the calculation that makes the coefficient of x 1 equal 0 in the new second equation also makes the coefficient of x2 equal 0. (This does not always happen, as you can sec from the new third equation.) If we now interchange the second and third equations in th is system, we obtain the following system:

Xi - 2x2 -

=

X3

3

= -6

3x2 + 3x3

(3)

- 2r3 = - 6 We can now solve the third equation for X3 by mu ltiplying both si des by - ~ (or equivalently, dividing both sides by - 2). This produces X( -

2r2 - X3 3x2 + 3x3

=

3

X3

=

3.

=- 6

By adding appropriate multiples of the third equation to the first and second. we can el iminate X3 from every equation but the th ird. If we add the third equation to the first and add -3 times the third equation to the second , we obtain

6 = - IS X3

=

3.

Now solve for x2 by multiplying the second equation by X( -

2x2

6

X2

= -5 X3

=

!. The result is

3.

Finall y, adding 2 times the second equation to the first produces the very simple system = - 4

X(

X2

=- 5

X3 =

(4)

3,

whose solution is obvious. You should check that replacing x, by -4, and

X3

by 3 makes each equation in system (2) tme. so that [

=~]

x2

by -5,

is a solution of

system (2). Indeed, it is the only solution, as we soon will show.


Page 3 of 10



31

In each step just presented, the names of the variables played no essentia l role. All of the operations that we performed on the system of equations can also be performed on matrices. In fact, we can express the original system Xt - 2r2 X3 = 3 3xt - 6x2 - 5x3 3 2xt - X2 + XJ = 0

=

(2)

as the matrix equation Ax = b , where

A=

[i -I]

-2 -6 -5 ' - I

X =

I

XI] [ X2

and

,

b=

XJ

[il

Note that the columns of A contain the coefficien ts of x 1, x2 , and x3 from system (2) . For this reason. A is called the coefficient matrix (or the matrix of coefficients) of system (2 ). All the information that is needed to find the solution set of this system is contained in the matrix

[i

-2 - I -6 -5 - I

which is called the augmented matrix of the system. This matrix is formed by augmenting the coefficient matrix A to include the vector b. We denote the augmented matrix by [A b ]. If A is an m x 11 matrix, then a vector u in 'R" is a solution of Ax b if and only if Au = b. l11us [

=~] is a solution of system

=

(2) because

-2 -6 - I

Example 1

For the system of linear equations Xt

+

2rt - x2

5XJ -

X4

+ 6x3

=

7

= - 8,

the coefficient matrix and the augmented matrix arc

[~

0 5 - 1 6

and

[~

0 5 - 1 6

- I

0

respectively. Note that the variable x2 is missing from the first equation and x4 is missing from the second equation in the system (that is. the coefficients of x2 in the first equation and X4 in the second equation are 0). As a result, the ( I, 2)- and (2, 4)entries of the coefficient and augmented matrices of the system are 0.

In solving system (2), we performed three types of operations: interchanging the position of two equations in a system, multiplying an equation in the system by a


Page 4 of 10


32


nonzero scalar, and adding a multiple of one equation in the system to another. The analogous operations that can be performed on the augmented matrix of the system are given in the following definition . Defin ition Any one of the following three operations performed on a matrix is called an elementa r y row operation : I. Interchange any two rows of the matrix.

(in ter change operation)

2. Multiply every entry of some row of the matrix by the same nonzero scalar. (scaling operation)

3. Add a mu ltiple of one row of the matrix to another row. (r ow addition oper ation) To denote how an elementary row operation changes a matrix A into a matrix 8, we use the following notation: r;~ r;

l. A - - --

8 indicates that row i and row j are interchanged.

cr; - r,

2. A - - - - 8 indicates that the entries of row i arc multiplied by the scal ar c. 8 indicates that c times row i is added to row j.

3. A

ijifi ..!.!tjM

Let I

2

and

B=

[~

~] .

2 1

-5

-3 -7

The following seq uence of ele mentary row operations transforms A into B:

A=

[r

2

- 2r1+r2- r2

- I I

0

[i

-3r 1+r, . - r., [I0

~]

r1 # r1

2

I

-3

-3 0

2

I

[i

2

1 - I

0

n

-n _;]

-3 -3 0 -5 - 3 -7

- \rJ -+ 1"2

[~

2 I

1

~] = 8.

-5 -3 -7

We may perform several elementary row operations in succession, indicating the operations by stack ing the individual labels above a single arrow. These operations are performed in top- to-bottom order. In the previous example. we cou ld indicate how to transform the second matrix into the fourth matrix of the example by using the following notation: 2 - I

0


I

_;]

-3 -3 -5 -3 -7

Page 5 of 10



33

Every elementary row operation can be reversed. That is. if we perform an e lementary row operation on a matrix A to produce a new matrix 8, then we can perform an e lementary row operation of the same kind on 8 to obtain A. If, for example. we obtain 8 by interchanging two rows of A, then interchanging the same rows of 8 yields A. Also, if we obtain 8 by mu ltiplying some row of A by the nonzero constant c. then multiplying the same row of 8 by ~ yields A. Finally. if we obtain 8 by adding c times row i of A to row j , then adding - c times row i of 8 to row j results in A. Suppose that we perform an elementary row operation on an augmented matrix lA bl to obtain a new matrix lA' b' l. The reversibility of the elementary row operations assures us that the solutions of Ax b are the same as those of A'x b'Thus performing an elementary row operation on the augmell/ed matrix of a system of linear equations does not change the so/wion set. That is, each elemematy row operation produces the augmellled matrix of an equivalent system of linear equations. We assume this result throughout the rest of Chapter I; it is proved in Section 2.3. Thus, because the system of linear equations (2 ) is equivalent to system (4 ). there is only one solution of system (2).

=

=

REDUCED ROW ECHELON FORM We can use elementary row operations to simplify any system of linear equations unti l it is easy to see what the solution is. First. we represent the system by its augmented matrix, and then use elementary row operations to transform the augmented matrix into a matrix having a special form. which we call a reduced row echelon form. The system of linear equations whose augmented matrix has this form is equivalent to the original system and is easily solved. We now define this special form of matrix. In the following discussion, we call a row of a matrix a zero row if all its e ntries are 0 and a nonzero row otherwise. We call the leftmost nonzero entry of a nonzero row its leading entry.

De finitions A matrix is said to be in row ech elon form if it satisfies the foll owing three conditions: I . Each nonzero row I ies above every zero row. 2. The lead ing entry of a nonzero row lies in a column to the right of the column containing the leading entry of any preceding row. 3. lf a column contains the leading entry of some row, then all entries of that column below the leading entry are 0. 5 If a mau·ix also satisfies the following two additional conditions. we say that it is in

reduced row echelon form 6 4. If a column contains the leading entry of some row, then all the other entries of that column are 0. 5. The leading entry of each nonzero row is I.

5 Condition 3 is a direct consequence of condition 2. We include it in this definition for emphasis, as is

usually done when defining the row echelon form. 6 Inexpensive calculators are available that can compute the reduced row echelon form of a matrix. On

such a calculator, or in computer software, the reduced row echelon form is usually obtained by using the command rre f .


Page 6 of 10


34


l

A matrix having e ither of the forms that follow is in reduced row echelon form. In these diagrams. a denotes an arbitrary entry (that may or may not be 0).

*

~ ~ ~ g: 0 0 0 1 *

[0

l

* * * 0 00 0 ** 0 I 01 *

OOOl__!_ 0 *

0

[ 0 0 0 0 ~ 0 0 0 0 0 0 0

0 0 0 0

Notice that the leading entries (which must be I ' s by condition 5) form a pattem suggestive of a fl ight of stairs. Moreover, these lead ing e ntries of I are the on ly nonzero entries in their columns. Also, each nonzero row precedes all of the zero rows.

Example 3

The following matrices are nor in reduced row echelon form:

0 0 6

A= [!

"]

I 5 7 0 0 2 4 0 0 0 0 0 I 0 I

8 =

[I 72 0 0 0 0

0 0 0 0

-3 9

I

4

6

0 0 0

2

3

0 0 0 0

~]

Matrix A fai ls to be in reduced row echelon form because the leading e ntry of the third row does not lie to the right of the leading entry of the second row. Notice, however, that the matrix obta ined by interchanging the second and third rows of A is in reduced row echelon form. Matrix 8 is not in reduced row echelon form for two reasons. The leading entry of the third row is not I. and the leading entries in the second and third rows are not the only nonzero entr ies in their columns . That is, the third column of 8 contains the first nonzero entry in row 2, but the (2, 3)-entry of 8 is not the onl y nonzero entry in column 3. Notice. however, that although 8 is not in reduced row echelon form, 8 is in row eche lon form. A system of linear equations ca n be eas ily solved if its augmented matrix is in reduced row echelon form. For example, the system

=

Xt

X2 X3

-4 = -5 3

=

has a solution that is immediately evident. If a system of equations has in fin ite ly many solutions, then obtaining the solution is somewhat more complicated. Consider, for example, the system of linear equations Xt -

Jx2

x3

+ 2x4 + 6x4

=7 = 9

xs = 2 0=0.

(5 )

The augmented matrix of this system is

[!

-3 0 2 0 0 I 6 0 0 0 0 I 0 0 0 0

which is in reduced row echelon form .


l Page 7 of 10



35

=

Since the equation 0 0 in system (5) provides no useful in formation, we can disregard it. System !5) is consistent. but it is not possible to find a unique value for each variable because the system has infini tely many solutions. Instead, we can solve for some of the variables. called basic variables. in terms of the others, called the free variables. The basic variables corTes pond to the leading entries of the augmented matrix. ln system (5). for example. the basic variables arc x 1 • XJ. and xs because the lead ing entries of the augmented matrix are in columns I, 3, and 5, respectively. The free variables are x2 and X4. We can easily solve for the basic variables in tem1s of the free variables by moving the free variables and their coefficients from the left s ide of each equation to the right. The resulti ng equations xr x2 X3

= 7 + 3x2 =9 -

x4

xs

2x4

free

6x4

free

=2

provide a gener a l sol ution of system (5). Th is means that for every choice of values of the free variables, these equations give the corresponding values of x r, XJ, and xs in one solution of the system, and furthermore, every solution of the system has this 0 and X4 0 form for some values of the free variables. For example, choosing X2

=

gi~

[!J"' oho~ing

l

=

" = -2 " ' " = I yicldnho.olo
=

The general solution can also be written in vector form as

1,

~'" fo~. opp~rn<

"·~••~Y

;, ;,

'"" ""' rolmim< of iho

lio= oombi,.
"oc""

''"= ;, iho ""'of

[~l "'

[!l"' [~11, ~m'''"' whh
being the free variables x2 and x4, respectively.

Example4

Find a general solution of the system of linear equations

+ 2x4 = 7 - Jx4

X3


=&

+ 6x4 = 9.

Page 8 of 10


36


Solution Since the augmented matrix of thi s system is in reduced row echelon form. we can obtain the general solution by solving for the basic variables in terms of the other variables. In this case, the basic variables are x 1, xz, and x 3, and so we solve for X t . x2, and X 3 in terms of x4. The resulting general sol uti on is

=

Xt 7 - 2x4 xz = 8 + 3x4

x3 X4

=9 -

6x4

free.

We can write the general solution in vector fonn as

There is one other case to consider. Suppose that the augmented matrix of a system of linear equations contains a row in which the only nonzero entry is in the last column, for example,

The system of linear equations corresponding to this matrix is Xt

-

)XJ

=5

Xz + 2x3 = 4 Ox t + Ox2 + Ox3 = I Ox, + Oxz + Ox3 = 0. Clearly. there are no values of the variables that satisfy the third equation. Because a solution of the system must satisfy every eq uation in the system, it follows that thi s system of equations is inconsistent. More generally. the following statement is tme: Whenever an augmented matrix contains a row in which the on ly nonzero entry lies in the last column, the corres ponding system of linear equations has no solution. It is not usually obvious whether or not a system of linear equations is consistent. However, th is is apparent after calculating the reduced row echelon form of its augmented matrix. Practice Problem 2 ""

The augmented matrix of a system of linear equations has

[~

I

0 0

-4 0 0

0

3 0]

1 -2 0 0 0 I

as its reduced row echelon form. Determine whether thi s system of linear equations is consistent and. if so. find its general solution. ~


Page 9 of 10



37

SOLVING SYSTEMS OF LINEAR EQUATIONS So far, we have learned the following facts : l. A system of linear equations can be represented by its augmented matrix, and any elementary row operations performed on that matrix do not change the solutions of the system. 2. A system of linear equations whose augmented matrix is in reduced row echelon form is easily solved.

Two questions rema in. Is it always possible to tmnsfonn the augmented matrix of a system of linear equations into a reduced row echelon form by a sequence of elementary row opemtions? Is that form un ique? The first question is answered in Section 1.4, where an algorithm is given that transforms any matrix into one in reduced row echelon form. The second question is also important. If there were different reduced row echelon forms of the same matrix (depending on what sequence of elementary row operations is used), then there cou ld be different solutions of the same system of linear equations. Fortunately. the following important theorem assures us that there is only one reduced row echelon form for any matrix. It is proved in Appendix E.

THEOREM 1.4 Every matrix can be transformed into one and only one matrix in reduced row echelon form by means of a sequence of elementary row operations. In fact, Section 1.4 describes an explici t procedure for perfonning this trans formation. If there is a sequence of elementary row operations that transforms a matrix A into a matrix R in reduced row echelon form, then we call R the reduced row echelon form of A . Using the reduced row echelon form of the augmented matrix of a system of linear equations Ax = b, we can solve the system as follows:

Procedure for Solving a System of Linear Equations l. Write the augmented matrix

lA

b) of the system.

2. Find the reduced row echelon form [R c) of [A b ). 3. If [R c) contains a row in which the on ly nonzero entry lies then Ax = b has no solution. Otherwise, the system has at Write the system of linear equations corresponding to the solve this system for the basic variables in terms of the free a geneml solution of Ax b .

=

Example 5

in the last column, least one solution. matrix [R c), and variables to obtain

Solve the following system of linear equations: X1 + 2x2- XJ + 2t4 + Xs = 2 - x1 - 2x2 + x3 + 2r4 + 3xs = 6 2x1 + 4x2 - 3x3 + 2x4 = 3 - 3xl - 6x2 + 2x3 + 3xs = 9

Solution The augmented matrix of this system is

[-]

2 - I - 2 I 4 -3 2 -3 -6


2 2

I

3

2 0 0 3

!l Page 10 of 10


38


In Section 1.4, we show that the reduced row echelon form of this matrix is

-1-5]

[ 2 0 0 0 0 [ 0 [

0 0

0

0 -3 I 2 .

I

0

0 0 0 0

0

Because there is no row in this matrix in which the only nonzero entry lies in the last column. the original system is consistent. Thi ~ matrix corrc~pond~ to the system of linear equations

+ 2x2

Xt

-

Xs

X3 X4

= -5

= -3

+ Xs = 2.

In this system, the basic variables are X t , XJ, and

X4,

and the free variables are

x2

and

xs. When we solve for the basic variables in tenns of the free variables. we obtain the following general so lution: Xt

=- 5-

X4

=- 3 = 2

xs

free

X3

2x2

+ X5

free

x2

-

XS

This is the general solution of the original system of linear equations.

Practice Problem 3 ..,.

The augmented matrix of a system of linear eq uations has

2 4]

0 [ -3 0 0 [ - 1 5 0 0 [0 0 0 0 0 0

as its reduced row echelon form. Write the corresponding system of linear equations. and determine if it is consistent. If ~o. find ib general ~olution. and write the geneml solution in vector form. ~

EXERCISES f11 £1ercises I -6. ll'rite (ll) the t'o
-.l z

x,

+ .IJ

3.

4.

+ 213 =

0 =- 1

3x2

+ 212 + lt2

XJ

Xl

+ 21.1 +

2.t 2 - Jx1 5. - x 1 + xz + 2q 2.1t

+

-

xz

+ 3.q = 4

x1

-

x1

-

2rz

=

atio11 on

=3 = 2

-xr - 3Xt 1- 4X2 = I

2rr -

2. 2t1

+ .r. + 1.15 = 5 2x2 + IO.rs = 3 2.tr - 4x2 + -lx• + 8xs 7 In Exercises 7- 14. perform the indicmed elememary row oper· 6.

.\.1

,14 X4

= =

=3

=0 4

6

= 0


[-~

- I 6 2

0

2

3

- I 4

4

-3] I

.

2

7. Interchange rows I and 3. 8. Multiply row I by -3.

9. Add 2 time' row I to row 2.

I0. Interchange rows l and 2.

Page 1 of 10



II. Multiply row 3 by

4.

12. Add -3 times row 3 to row 2.

39.

[t

- I

2]

40.

1 [0 0

41.

1 -2 [0 0

~]

42.

[~ -~ ~]

13. Add 4 times row 2 to row 3. 14. Add 2 times row I to row 3.

In Exercises 15- 22. petformthe indicated elementmy row operation on

[

_: 2

-2 -4

-3

2

-~]

[~

-3

43.

[~

-2 0

45.

47.

[~

0 I 0

0 0

[0~

I 0

0 I 0

0

0 I 0

-2

I

15. Multiply row I by - 2. 16. Multiply row 2 by 17. Add -2 times row I to row 3.

t·

18. Add 3 times row I to row 4. 19. Interchange rows 2 and 3. 20. Interchange rows 2 and 4 .

49.

21. Add - 2 times row 2 to row 4. 22. Add 2 times row 2 to row I.

In Exercises 23-30, deten11i11e whether the gire11 vector is a solution of the system X t - 4x2 + 3x4 = 6 X3 - 2x4 -3.

51.

=

[=ll ~ [-ll , m [ll

I

53.

26

23.

"·[-ll ,. [-ll , Ul , [=!l In Exercises 31- 38, deten11i11e whethenl!e give11 vector i.f a solution of tile system Xt - lr2 + X3 + X4 + 7xs = I Xt - lr2 + 2x3 + lOxs = 2 2r1 - 4x2 + 4x4 + 8xs = 0.

3L

[l] [l]

35. [ ]

32

36

[

~ll

33

37

[]

[

~ ~ll [

-ll ~ [l]

In Exercises 39- 54, the reduced row echelon form of the a11gmemed mmrix of a sysrem of li11ear eq11ario11s is given. Determine whether this system of linear equations is consistent and. if so. find its geneml solwio11. /11 addition. in Exercises

47- 54, write the solution in vector form.


[~

54.

[~

3

0 0

~]

I 0

0 0

0

-3 -4

0

46. 0

~] -3] -4 5

0

~]

4

-3

2

0

4

0

0 0

0 0

0

0

-:]

44.

0 0

6 .

39

48. I

50.

52.

[~

-I 2

3 -5

0

0

-2

0

0

I

[~ ~

0 0

0 0 0 I

~] -3] -4 5

3 2

0 0

0

[~ ~ ~ -~ ~

-l ~]

55. Suppose that the general solution of a system of m linear equations in 11 variables contai ns k free variables. How many basic variables does it have? Explain your answer. 56. Suppose that R is a matrix in reduced row echelon fonn. If row 4 o f R is nonzero and has its leading e ntry in column 5, describe column 5. ~

~

In Exercises 57-76. determine whether the following stateme11ts are tme or false.

57. Every system of linear equations has at least one solution. 58. Some systems of linear equations have exactly two solutions. 59. If a matrix A can be transformed into a matrix 8 by an e lementary row operation. then 8 can be transformed into A by an elementary row operation. 60. If a matrix is in row echelon fo n11 , then the leading entry o f each nonzero row must be I. 61. If a matrix is in reduced row echelon form, then the leading entry of each nonzero row is I. 62. Every matrix c an be transfonned into one in reduced row eche lon form by a sequence of elementary row operations. 63. Every matrix can be transformed into a unique mat.rix in row echelon form by a sequence of elementary row operations.

Page 2 of 10


40


64. Every matrix can be transfonned into a unique matrix in

78. Prove that if R is the reduced row echelon fonn of a

reduced row echelon form by a sequence of elementary row operations.

matrix A, then IR 0) is the reduced row echelon fomt of [A OJ.

65. Perfonning an elementary ro\\ operation on the augmented maui< of a sys tem of Ioncar equations produces the augmented matrix of an equivalent ~ystcm of linear

equation,, 66. If the reduced row echelon fomt of the augmented matrix of a sy,tcm of linear equations contains a zero row. then the system is

const~tent.

67. If the only nonzero entry in some row of an augmented matrix of a system of Ioncar equations hcs in the last column. then the •)'stem i> incons i>tcnt.

68. A system of linear equation> is called consistent if it has

79. Pro\·e that for any m x 11 matrix A. the equation Ax is consistent. where 0 is the uro vector in

n· .

form contains no zero rows. Pro\e that Ax tent for every b m R.•.

contains one more column than the coefficient matrix.

72. A system of linear equations Ax = b has the same solu-

=

tions as the .ystcm of linear equations Rx c. where IR cl is the reduced row echelon fomt of lA bl. 73. Multiplying every entry of some row of a matrix by a scalar is an elementary row operation.

74. Every >oluuon of a con>i>tcnt sy>tcm of linear equations can be obtained by substituting appropriate values for the free variables in its general >olution.

[~

I ()

0

76. If A ts an Ax

111

x

11

matnx. then a solution of the system

= b is a vector u in n• >UCh that Au = b .

•

0 0

0 I

0

• •0

0 0

is a possible reduced row ccltelon form for a 3 x 7 matrix. I low many different .uch forms for a reduced row echelon matrix are possible if the matrix is 2 x 3?

82. Repeat Exercise 8 1 for a 2 x 4 matrix. 83. Suppose that 8 is obt:tincd by one elementary row operation performed on matrix A. Prove th:tt the same type of elementary operation (namely, an interchange. scaling. or row addition operation) that tran>forms A into 8 also transforms 8 into A.

84. Show that if an equation in a >y;tem of linear equations is multiplied by 0. the resulting system need not be equivalent to the original one.

85. let S denote the following sy>tem of linear equations: a 11 x 1 t121.\1

75. If a sy,tcm of linear equations has more variables than equattons. then tt must have mfirutely many solutions.

= b is consis-

81. In a matrix in reduced ro\\ echelon fonn. there are three types of entrtes: The lcadmg entnc> of nonzero rows are required to be Is. cenain other cntric> arc required to be 0.. and the remaining entric> arc arbit.rary. Suppose that these arbitrary entric> are denoted by asterisks. For example.

69. If A is the coefficient matrix of :1 ,ystem of m linear equations in 11 variables. then A is an 11 x m matrix.

71. If the reduced row eche lon fomt of the augmented matrix of a consistent syMem of m linear equations in 11 vari:tblcs contains k notwcro rows, then its general solution contains k basic variables.

0

80. let A be an m x 11 matrix whose reduced row echelon

one or more solutions.

70. The :mgmcntcd matrix of a system of linear equations

=

(IJ t•\1

+ Dt2·''2 +lit l X I = b 1 + a ;u.\ 2 + tiJJ.\ _l = b 2 + 0 )2A2 + tl ~t-I J = bj

Show that if the second equation of S i> multiplied by a nonzero scalar c. then the resulting sy>tem is equivalent toS.

86. let S be the •)'stem of linear equations in Exercise 85. 77. 7 let lA bl be the augmented matrix of a system of linear equations. Prove that if it;, reduced row echelon fonn is [R cl. then R is the reduced row echelon fonn of A.

Show that if k times the first equation of S is added to the third equation. then the resultang >yblem is equivalent toS.

SOLUTIONS TO THE PRACTICE PROBLEMS

=

I. (a) Since 2(- 2) - 3 + 6(2) 5. u is not a solution of the second equation in the given system of equations. Therefore u is not a solution of the system. Another method for solving this problem is to represent the given •ystem as a matrix e
=

A = [~

0 5 - 1 6

and

Because

0 -1

5 6

u is not a solution of the given sy;,tem.

7 This exercise Is used In Sect tOn 1.6 (on page 70).


Page 3 of 10


1.4 Gaussian Elimination (b) Since 5+5( 1)-3=7 and 2(5)-8+6(1)=8. v satisfies both of the equations in the given system. lienee v is a solution of the system. Alternatively. using the matrix equation Ax = b. we see that ' ' is a solution bccau>e

Since the gh·en matrix contains no row who>e only nonzero entry lies in the last column. this ;ystem is consistent. The general solution of thi> >yMem is \2

=

·'• =

0 5 6

x~

+ hJ -

5 +

2.\j

.tj

free.

Note that.\ t. which i' not a ba...ic variable. is therefore a free variable. The general soluuon tn vector form .,

2. In the given matrix. the only nonzero entry in the third row lie> in the la>t column. lienee the >)'Stem of linear equations corresponding to this matrix is not consistent.

H-[!l [!l ... [~l

3. The corresponding >y>tcm of linear equations is

+ ..

+ 2.rs = 4 X5

free 4 free

Xt

·'l

X4-

41

= 5.

+» [ - ;].

ITIJ GAUSSIAN ELIMINATION In Section 1.3, we learned how to solve a system of linear equations for wh ich the augmented matrix is in reduced row echelon form. In thi s section. we describe a procedure that can be used to tr.tnsfonn any matrix into this form. Suppose that R is the reduced row echelon form of a matrix A. Recall that the fir>! nonzero entry in a nonzero row o f R is called the leading enlry of that row. The positions that contain the leading entries of the nonzero rows of R are called the pivot positions of A, and a column of A that conta ins some pivot position of A is called a pivol column of A. For example. later in this section we show that the reduced row echelon form of 2 - I 2 I I 2 3 [ - II - 2

A=

2 4 -3 2 0 -3 -6 2 0 3

~]

is

0 0 0 I 0 R= [ 0" 0 00 I 0 0 0 0

-I

-5]

0 -3 2 . I 0 0

Here the first three rows of R are its no nLero row>, and so A has three pivot positions. The fi rst pivot position is row I. column I because the leading entry in the first row of R li es in column I. The second pivot position is row 2, column 3 because the lead ing entry in the second row of R lies in column 3. Finally. the third pivot position is row 3. column 4 because the leadi ng entry in the th ird row of R lies in column 4. Hence the pivot columns of A are columns I , 3. and 4. (Sec Fig me 1.19.) The pi vot posi tions and pivot columns are easi ly determined from the reduced row echelon form of a matrix. However, we need a method to locate the pivot position s so we can compute the reduced row echelon form. The algori thm that we use to obtai n the reduced row echelon form of a matrix is called G aussian climinalion.s 'l11is • This method is named after Carl Friedrich Gauss (1777-1855), whom many consider to be the greatest mathematician of all t ime. Gauss described this procedur
to determine the orbit of the asteroid Pallas. However, a similar method for solving systems of linear equations was known to the Chinese around 250 ac.


Page 4 of 10


42


. fm.t ,•

pi~ot

pm.ition

/ second pivot posi tion third pivot position

·-r:

2

0 0 0

o.. ·· o

- L ··

0 /'0 cb CD. 0 0

0

-1]

I 0

pivot col umns Figure 1.19 The pivot positions of the matrix R

algorithm locates the pivot positions and then makes certain entries of the matrix 0 by means of elementary row operations. We assume that the matrix is nonzero because the reduced row echelon form of a zero matrix is the same zero matrix. Our procedure can be used to find the reduced row echelon form of any nonzero matrix. To illustrate the algorithm. we find the reduced row echelon form of the matrix

I A=[~0 6~ -7 0 -6

-4

- 5 3 5

2

16

Step 1. Detennine the leftmost nonzero column. This is a pivot column, and the topmost position in this column is a pivot position. Since the second column of is the leftmost nonzero column. it is the first pivot column. The topmost position in this column lies in row I , and so the first pivot position is the row I. col tmut 2 position.

pivot po'ition

A

A ~ [:

@

2

-4

-5

I

- I

I

3

I

6

0

-6

5

16

2

-;l

pivot columns S tep 2. In the pivot column, choose any nonzero9 entry in a row that is not above the pivot row. and perform the appropriate row interchange to bring this entry into the pivot posi tion. Because the entry in the pivot position is 0, we must perform a row interchange. We must select a nonzero entry in the pivot column. Suppose that we select the entry I. By interchanging rows I and 2, we bring this entry into the pivot position. 9 When performing calculations by hand, it

_,_,__,2_

[~ ~

- 1 3 2 -4 - 5 0 - 6 5

may be advantageous to choose an entry of the pivot column

that is± 1, if possible, in order to simplify subsequent calculations.


Page 5 of 10


1.4 Gaussian Elimination

43

Step 3. Add an appropriate multiple of the row containing the pivot position to each lower row in order to change each entry below the pivot position into 0. In step 3, we must add multiples of row I of the matrix produced in step 2 to rows 2 and 3 so that the pivot column entries in rows 2 and 3 are changed to 0. In this case, the entry in row 2 is already 0, so we need only change the row 3 entry. Thus we add - 6 times row I to row 3. This calculation is usually done mentally, but we show it here for the sake of completeness.

0 -6 6 0 6 0

- 18 - 6 5 16

-6 -6

0 6 - 12 - 13

b

10

6

(-6 times row I )

(row 3)

7 13

- 1 3 I 2 - 4 - 5 2 6 - 12 - 13 10

The effect of this row operation is to transform the previous matrix into the one shown at the right.

-~] 13

Du ring steps 1-4 of the algorithm, we can ignore certain rows of the matrix. We depict such rows by shad ing them. At the beginning of the algorithm. no rows are ignored. Step 4. Ignore the row containing the pivot position and all rows above it. If there is a nonzero row that is not ignored. repeat steps 1-4 on the submatrix that remains. We are now fin ished with row I, so we repeat steps 1-4 on the submatrix below row I. : pt vot position

[~

0

0

-4 -5 6 -12 -13

2 10

pi vot columns The leftmost nonzero column of this submatrix is column 3, so column 3 becomes the second pivot column. Since the topmost position in the second column of the submatrix lies in row 2 of the entire matrix, the second pivot position is the row 2, column 3 position. Because the entry in the current pivot position is nonzero. no row interchange is required in step 2. So we continue to step 3, where we must add an appropriate multiple of row 2 to row 3 in order to c reate a 0 in row 3, column 3. The addition of - 3 times row 2 to row 3 is shown as follows: 0 0

0

0 0 0

12 -6 6 - 12 0

0

IS - 13 2

-6 10 4

-Jr2+r3- r3

The new matrix shown at the right.


is

- IS 13 - 2

[~

I

0 0

(-3 times row 2) (row 3)

- I 2 0

-4

0

3 I -5 2 2

4

-~] -2

Page 6 of 10


44


We are now fin ished with row 2, so we repeal steps 1-4 on the submatrix below row 2, which consists of a single row. pivot pos ition

pivot columns At this stage, column 5 is the leftmost nonzero column of the submatrix. So column 5 is the third pivot column, and the row 3, column 5 position is the next pivot position. Because the entry in the pivot position is nonzero. no row interchange is needed in step 2. Moreover. because there are no rows below the row containing the present pivot pos ition. no operations are required in step 3. Since there are no nonzero rows below row 3, steps 1-4 are now complete and the matrix is in row echelon form. The next two steps transform a matrix in row echelon form into a matrix in reduced row echelon form. Unlike ste ps 1-4, which started at the top of the matrix and worked down, steps 5 and 6 start at the last nonzero row of the matrix and work up.

Slep 5. If the leading e ntry of the row is not I, perform the appropriate sca ling operation to make it I. Then add an appropriate multiple of this row to every preced ing row to change each entry above the pivot pos ition into 0. We start by applying step 5 to the last nonzero row of the matrix, which is row 3. Since the leading entry (the (3, 5)entry) is not I, we multiply the third row by I 2 to make the lead ing entry I. This produces the matrix at the right. Now we add appropriate mu ltiples of the third row to every preceding row to change each entry above the lead ing entry into 0. The res ulting matrix is shown at the right.

irJ ~ rJ

s, .. + ' 2 -

-3r .1 + r l -

[~

'2 [ rl

I

- I

0 0

2

0

I

3

-~]

I

-4 -5 2

0

- I

2

0

I

- I

0

-5

0 0 0

2

- 4 0

12

0

0

0

2

-~J

Slcp 6. If step 5 was performed on the first row, stop. Otherwise, repeal step 5 on the preceding row.


Page 7 of 10



Since we just performed step 5 using the third row, we now repeat step 5 using the second row. To make the leading entry in this row a I, we must mu ltiply row 2 by~-

- I I

I

0

Now we must change the entry above the leading entry in row 2 to 0. The resulting matrix is shown at the right.

0

-2 0

2

0 - 1 0 I

0

J]

-5 6

0

45

I

-2 0 6 2

0

We just performed step 5 on the second row, so we repeat it on the first row. T his time the lead ing entry in the row is already I, so no scaling operation is needed in step 5. Moreover, because there are no rows above this row, no other operat ions are needed. We see that the preceding matrix is in reduced row echelon form. Steps 1-4 of the preceding algorithm are called the forward pass. The forward pass transforms the original matrix into a matrix in row echelon form. Steps 5 and 6 of the algorithm are called the backward pass. The backward pass further transforms the matrix into reduced row echelon form.

Example 1

Solve the following system of linear equations:

+ 2x2 - X J + 2r4 + XS = 2 + XJ + 2x4 + 3xs = 6 2xt + 4x2 - 3x3 + 2r4 = 3 -3x t - 6x2 + 2xJ + 3xs = 9 Xt

-x1 - 2x2

Solution The augmented matrix of this system is

[-;

2 - 2

- I

4

-3

-3 -6

2

I

2 I 2 3 2 0 0 3

ll

We apply the Gauss ian e limination algorithm to transform th is matrix into one in reduced row echelon form.

Operations The fi rst pivot position is I ll row I. column I. Since this entry is nonzero, we add appropriate multi pies of row I to the other rows to change the entries below the pivot position to 0. The second pivot position is row 2. column 3. Since the entry in this position is presently 0. we inter-

change rows 2 and 3.


Resulting Matrix r 1 + r2 - r2 -2.r 1 + r 3 -. r J

Jr1 + r4 _. r4

[j

f 2Wf _l

[!

2 - I 0 0

2 4

4

0

- I - I

-2 6

-2 6

2

- I

0

- I

2 -2

-2

0

I

0

0

4

4

0

- I

6

6

-l] 15

-:] 15

Page 8 of 10


46


- I

Next we add - 1 times row 2 to row 4.

2

-!]

-2 -2

- 1 0

0

4

4

8

8

16

- I - 1 0

The third pivot postllon is in row 3, column 4. Since this e ntry is nonzero, we add - 2 t imes row 3 to row 4.

0 2

At th is stage, step 4 is complete, so we continue by performing step 5 on the third row. First we multiply row 3 by ~ ·

- 2 I 0

Then we add 2 times row 3 to row 2 and - 2 times row 3 to row I.

- 1 0 - 1 0

rz - ..I

2r:t+ rz - 2r3 + ra

- 2 I 0

0

I

0 0 - 1 0 I 0

Now we must perfonn step 5 using row 2. This requires that we multiply row 2 by -I.

I

- 1 0 I

0 0

0

0

[o~l ~ ~ ~

Then we add row 2 to row I.

0

0

I

0 0 0

-

=~] 2

0

~ =~] I

2

0

0

Performing step 5 with row l produces no changes, so this matrix is the reduced row eche lon form of the augmented matrix of the given system. This matrix corresponds to the system of linear equations Xt

+ 2x2

-

X5

=- 5 = - 3

X3 X4

+ X5 =

(6 )

2.

As we saw in Example 5 of Section 1.3. the general solut ion of this system is Xt

=- 5 -

X3

=- 3

X4

=

xs Practice Problem 1 .,.

2x2

+ X5

free

x2

2

-

X5

free.

Find the general solution of Xt -

X2 -

3X)

+

X4-

XS

= - 2

=

- 2xt + 2x2 + 6x3 - 6xs -6 3xt - 2x2- 8x3 + 3x4 - 5xs = - 7.


Page 9 of 10



47

THE RANK AND NULLITY OF A MATRIX We now associate w ith a matrix two important numbers that are easy to determine from its reduced row echelon form. If the matrix is the augmented matrix of a system of linear equations, these numbers provide significant information about the solutions of the co!Tesponding system of linear equations. Definitions The rank of an m x 11 matrix A. denoted by rank A. is defined to be the number of nonzero rows in the reduced row echelon form of A. The nullity of A, denoted by nullity A, is defined to be 11 - rank A.

i#ifh.!.!fW

For the matrix

[-;

2

- I I

2 2 2

I

3 4 -3 0 -3 -6 2 0 3 - 2

!]

in Example I, the reduced row echelon form is

[!

2 0 0 -I 0 I 0 0 I 0 0 I 0 0 0 0

-5]2 .

-3 0

Since the reduced row echelon form has three nonzero rows, the rank of the matrix is 3. The nu llity of the matrix, found by subtracting its rank from the number of columns. is 6 - 3 3.

=

Example 3

The reduced row echelon form of the matrix

B

is

~

[j

[j

3

5

I 5

3

0 - I

j]

7 - I

-I

-2

I

I

3

0 0

0 0

0 0

-;]

0 . 0

Since the latter matrix has two nonzero rows, the rank of 8 is 2. T he nullity of 8 is

5 - 2= 3. Practice Problem 2 .,..

Find the rank and nullity of the matrix

h[j User: Thomas McVaney

0 - I 0 I 0 I -2 0 6 0 0 0 I 2 0 0 0 0 0 0 0 0 0 0

-ll Page 10 of 10


48


In Examples 2 and 3, note that each nonzero row in the reduced row echelon form of the given matrix contains exactly one pivot pos ition, so the number of nonzero rows equals the number of pivot positions . Consequently, we can restate the defin ition of rank as follows: The rank of a matrix equals the number of pivot columns in the matrix. The nullity of a matrix equals the number of nonpivot columns in the matrix.

It follows that in a matrix in reduced row echelon form with rank k. the standard vectors e., e2, ... , e.( must appear, in order, among the columns of the matrix. In Example 2. for instance. the rank of the matrix is 3. and columns I. 3, and 4 of the reduced row echelon form are

respectively. Thus, if an m x must be precisely [et e2

the

11

x

If an

11

11

11

matrix has rank 11, then its reduced row eche lon form e,). In the special case of an 11 x 11 (square) matrix,

identity matrix; therefore we have the following usefu l result:

x

11

matrix has rank n. then its reduced row echelon form is /, .

Consider the system of linear equations in Example I as a matrix equation Ax = b . Our method for solving the system is to find the reduced row echelon form [R c] of the augmented matrix [A b]. The system R x = c is then equ ivalent to Ax = b. and we can easily solve Rx = c because IR c ) is in reduced row echelon form. Note that each basic variable of the system Ax = b corresponds to the leading e ntry of exactly one nonzero row of [R c), so the number of basic variables equals the number of nonzero rows, which is the rank of A. Also, if 11 is the number of columns of A, then the number of free variables of Ax = b equals 11 minus the number of basic variables. By the previous remark, the number of free variables equals n - rank A, wh ich is the nullity of A. In general.

If Ax= b is the matrix form of a consistent system of linear equations, then (a) the number of basic variables in a general solution of the syste m equals the rank of A; (b) the number of free variables in a general solution of the system equals the nullity of A. Thus a consistent system of linear equations has a unique solution if and only if the nullity of its coefficient matrix equals 0. Equivalently, a consistent system of linear equations has infinitely many solutions if and only if the nullity of its coefficient matrix is positive.


Page 1 of 10



49

The original system of linear equat ions in Example I is a system of 4 equations in 5 variables. However, it is equivalent to system (6 ), the system of 3 equadons in 5 variables corresponding to the reduced row echelon form of its augmented matrix. In Example I , the fourth equation in the original system is redundant because it is a linear combination of the first three equations. Specificall y, it is the sum of - 3 times the first equation. 2 times the second equation. and the third equation. The other three equations are non redundant. In general, the rank of the augmented matrix [A b] tells us the number of nonredundant equations in the system Ax = b .

Example4

Consider the following system of linear equations: Xt Xt Xt

+x2 + X3 = + 3x3 = - 2+s - X2 + I"XJ = 3

(a) For what values of r and s is th is system of linear equations incons istent? (b) For what values of r and s does this system of linear equations have infinitely many solutions? (c) For what values of r and s does this system of linear equations have a unique solution? Solution Apply the Gaussian elimination algorithm to the augmented matrix of the given system to transform the matrix into one in row echelon form:

[:

1

1

]

0 3 - 2 +s - 1 r 3

-- qr 1 + + rr2.1-. rrz .;

-2r2+rJ~ r~

[

[

1

I

0 - I 0 -2

2

I

0 0

r- 1

- 3 + .1' 2

I - I

2

0 r- 5

- 3+s 8 - 2s

] ]

(a) The original system is inconsistent whenever there is a row whose on ly nonzero entry lies in the last column. Only the third row cou ld have th is form. Thus the original 5 and system is inconsistent whenever r - 5 0 and 8 - 2s ::J 0: that is. when r

=

=

s ::/:4. (b) The original system has infinitely many solutions whenever the system is consistent and there is a free variable in the general solution. In order to have a free variable. we must have r - 5 0. and in order for the system also to be consistent. we must have 8 - 2s = 0. Thus the original system has infinitely many solutions if r 5 and s 4. (c) Let A denote the 3 x 3 coefficient matrix of the system. For the system to have a unique solution, there must be three basic variables, and so the rank of A must be 3. Since deleting the last column of the immediately preceding matrix gives a row echelon form of A, the rank of A is 3 precisely when r - 5 ::J 0; that is, r ::J 5.

=

=


=

Consider the following system of linear equations: Xt Xt

+ 3x2 = I + s

+ I'X2 =

5

(a) For what values of r and s is th is system of linear equations incons istent? (b) For what va lues of many solutions?


r and s does this system of linear equations have infinitely

Page 2 of 10


SO

CHAPTER 1 Matrices, Vectors, and Systems of Linear Equations (c) For what values of r and s does this system of linear equations have a unique solution? <1111 The following theorem provides several conditions that are equivalent 10 to the existence of solutions for a system of linear equations.

THEOREM 1.5 (TestforCo nsistency)

The following conditions are equivalent:

=

(a) The matrix equation A x b is consistent. (b) The vector b is a linear combination of the columns of A . (c) The reduced row echelon fonn 11 of the augmented matrix [A b] has no row of the form [0 0 · · ·0 d ). where d # 0. Let A be an 111 x n matrix, and let b be in 7?!". By the definition of a matrix-vector product, there exists a vector

PROOF

in 'R" such that Av = b if and on ly if

Thus Ax = b is consistent if and only if b is a linear combination of the columns of A. So (a) is equ iva lent to (b). Finally, we prove that (a) is equivalent to (c). Let [R cj be the reduced row echelon form of the augmented matrix lA b]. If statement (c) is fa lse, then the syste m of linear equations corresponding to Rx = c contains the equation

Ox1

+ Ox2 + · · · +Ox" =d.

=

where d # 0. Since this equation has no solutions. Rx c is inconsistent. On the other hand , if statement (c) is true, then we can solve every equation in the system of linear equations corresponding to Rx c for some basic variable. This gives a solution of Rx = c, which is also a solution of Ax= b. •

=

TECHNOLOGICAL CONSIDERATIONS Gaussian e limination is the most efficient procedure for reducing a matrix to its reduced row echelon form. Nevertheless, it requi res many tedious computations. ln fact, the number of arithmetic operations required to obtain the reduced row echelon 3 + i11 2 - ~" · We can form of an 11 x (n + 1) matrix is typically on the order of

in

10

Statements are called equivalent (or logically equivalent) if, under every circumstance. they are all true or t hey are all false. Whether any one of the statements in Theorem 1.5 is true or false depends on the particular matrix A and vector b being considered. 11 Theorem 1..5 remains true if (c) is changed as follows: Every row echelon form of [A b ] has no row in which the only nonzero entry lies i n the last column. • The remainder of this section may be omitted without loss of continuity.


Page 3 of 10



51

easily program th is algorithm on a computer or programmable calcu lator and thus obtain the reduced row echelon form of a matrix. However. computers or calculators can store only a fini te number of decimal places, so they can introduce small errors called roundoff errors in their calculations. Usually. these e1Tors are insignificant. But when we use an algorithm with many steps, such as Gaussian elimination, the errors can accumulate and significantly affect the result. The following example illustrates the potential pi tfalls of roundoff error. Although the calculations in the example are perfom1ed on the TI -85 calculator, the same types of issues arise when any computer or calculator is used to solve a system of linear equations. For the matrix

[~

-1]

- 1 2 3 - 1 2 4 -2 4 8

2 ' 6

the TI-85 calculator gives the reduced row echelon form a~

I 0

0 I

[0 0

- I E-14 0 - 2 0 0

-.999999999999 4 2

(The notation aEb represents a x lOb.) However, by exact hand calculations, we find that the third, fifth, and sixth columns should be

[Jl

and

respectively. On the TI-85 calculator, numbers are stored with 14 digits. Thus a number containing more than 14 s ignificant digits is not stored exactly in the calculator. ln subsequent calculations with that number, roundoff errors can accumulate to such a degree that the final resu lt of the calculation is highly inaccurate. In our calculation of the reduced row eche lon form of the matrix A, roundoff errors have affected the ( I, 3)-entry. the (1, 5)-entry. and the (2, 6)-entry. In this instance. none of the affected entries is greatly changed, and it is reasonable to expect that the true entries in these positions should be 0. - I. and 0, respectively. But can we be absolutely sure? (We will team a way of checking if these entries are 0, - 1, and 0 in Section 2.3.) It is not always so obvious that roundoff errors have occurred. Consider the system of linear equations

(k

kx1 + (k + I )x1 +

I )x2

=I

kx2 = 2.

By subtracting the first equation from the second, this system is easi ly solved, and the solution is x1 = 2 - k and x2 = k - I. But for sufficiently large values of k, roundoff 4,935,937, the T I-85 calculator errors can cause problems. For example, with k gives the reduced row echelon fom1 of the augmented matrix

=

4935937 [ 4935938

4935936 4935937

'] 2

as

[~ User: Thomas McVaney

.999999797404 0

Page 4 of 10


52

CHAPTER 1 Matrices, Vectors, and Systems of Linear Eq uations

Since the last row of the reduced row echelon form has its only nonzero entry in the last column. we would incorrectly deduce from this that the original system is incom,istent! The analysis of roundoff errors and related mailers is a serious mathematical ~ubject that is inappropriate for thi~ book. (It is Mudied in the branch of mathematics called llllmerical a11alysis.) We encourage the usc of technology whenever possible to perform the tedious calculations associmed "ith matricc~ (~uch as tho;,e required to obtain the reduced row echelon fonn of a matrix). evertheless. a certain amount of ;,kepticism is healthy when technology i; used. JuM because the calculations are performed with a calculator or computer is no guarantee that the result is correct. In thi; book. however. the examples and exercises u;ually in\olvc ~imple numbers (often one- or two-digit integers) and small matrices; so there is lillie chance of serious errors resulting from the use of technology.

EXERCISES In £..\ercises I /6, dt'll'nniltt' whether the gil'en system ;,,.con·

.sisrem. and if so, find irs ge11eml so/111io11. I. 2x1

+ Ci.rz

=- 4

2

. 3

x 1 2xz = -l.r1 + 3.rz =

5.

211 - 2x2 + 4xl = - 411 + 412- 81 ) = -3

6.

-'1 - 21! - XJ- 3 2.t 1 + 4.tl + 2rt = 6 3.t 1 - 6.r2- J,, 9

7

6 7

4

+

.t 1

+

xz

x.

"'

_ x1

-

2.r1 +

x2

-

x2 -

3x3 3x3

=3 =0

=

x 1 - x 2 +x.l x1 - 2.r2 - r3 = 14 2x1 -X.l = -x 1 - 4xz - X3 = _

16.

_ -'1 - 2.tl - -'l = -3 2.•1 - 4t2 + 2.t, = 2 x1

XI - X2 = 3 -2xl + 2x2 = -6

8 10 2

=

.r1 - 12 + \4 = -4 XI - X! + 2.t4 + 2.t5 = -5 3xl- 3.t·2 + l.r4- 2.1s =- II

/11 E.urcises 17· 26. detem1i11e the ••trlues of r. if lliiJ. for ..-hich the gi••e11 system of lillt'ilf equati1111.r i.r iiiWIUiltell/.

2

2.\z - 3x,- 12.t~ = -3 l J + 6x4 0

3x1 + r.t2 = - 2 6

17.

-x1 + -lxz = 3 3x1 + rxz = 2

18.

19.

x1- 2rz = 0 4xl - 8.rz = r

20.

XI+ 2.q

21.

XI - 3x2 = -2 2.r1 + rx2 = - 4

22.

-2.rl + .Q =5 rx 1 + 4x2 = 3

= -2

23.

-x1 + rx1 = 2 r.r1 - 9.Q = 6

24.

.I I + l'.fl = 2 n, + 16rz = 8

fJ + t4 = - I XJ 5 ·'I+ 112 + 2rJ + 3xo~ = 2

25.

Xt .r2 + 2r3 = 4 3.rl + r.r2- .fJ = 2

26.

X1 + 2.r2 - 4XJ = I -2.r1 - 4x2 + rx3 3

=

,\I Xz - 3.q + .\4 = 0 9. -l.r1 + X2 + 5.n = -4 4.tl - 2xz - IO.tJ + x. = 5

,\I -

J.\2

+

XJ

+

X4

10. -3xl + 9xl - 2rJ- 5.q 2q - 6x2 - .n + sx.

=

0

=

.\ 1 + 3..12 +

I I. -2\1 - 6x2 -

-'1

12.

7

·' 1 - ' J - lr4 - &.15 = -3 15. -2xl + XJ + 2.14 + 9.t5 = 5 3x1 - 2tJ - 3.\4 - 15xs = -9

=

8.

.

X1 + 2x2 + X! = I _ -2.r 1 - 4xz - XJ = 0 13 5x1 + 10x2+3xJ=2 3x1 + 6x2 + Jx3 = 4

+

211+

.\2

+

.\.\=- I

.\!- X.\=

- lr3

=

2 3

=- 1


-

-r1 + 4.12 = f.\2

= -3 =-6

=

/11 E.rercises 27-34, de/ermine the t•alues ofr cmd sfarwhich the gi•·en sysrem af li11ear eqtwtimt< lw< (II) no .rolurio11s. (b) exactly one solwion. 11nd (c) infinirely "'""-"so/Ill ions.

27

.r1 + rxz · 3x1 + 6.<2

=5 =•

28

-.II

+ 4 .\2

=.

· 2.11+n2=6

Page 5 of 10


1.4 Gaussian Eliminatio n

+ 2x2 = S + fX2 = 8

30.

-xt 4xt

Xt 2xt

+ fX2 = -3 + 5x2 = s

32.

Xt -3xt

-x 1 3x1

+ rx2 =

34.

2xt- X2 = 3 4x 1 + rx2 = s

29.

Xt - 4Xt

31.

33.

-

9x2

s

=- 2

+ 3xz = s + rxz = - 8 + fX2 = 5 + 6x2 = s

In Exercise.< 35-42. jind 1/re rank tmd 1wlli1y of 1he given mtllrix.

35.

36.

37.

-I -I

[-j [-j

-2 2

-2

-3 6

- I

-9

2 2

-I

-3 3

4 - 9

[-~

-I

2

u [i 0 -I I

2

39.

40.

41.

42.

3

-I

- I

38.

- I

-2 -2

I 2 0

[-l [-l [-l

4 - 4

0 -I

45. A patient needs to consume exactly 660 mg of magnes ium. 820 IU of vitamin D. and 750 meg of folate per day. Three food supple ments can be mixed to provide these murients. The amounts of the three nutrients provided by each of the supplements is given in the following table:

-3 8

-2

-I

- 6

-2

2 6

3

Food Supplement Magne>ium (mg) Vitamin D (IU) Fo late (meg)

I

~] 0 - I

- I -I

-3

_i]

0

2

-6

0 -I

3 5

- I

-4

-2

2

I

0

0

2

3

15 20 15

36 44 42

(b) Can the three s upplements be mixed to provide exactly 720 mg of magnesium. 800 IU of vitamin D. and 750 meg of folate? If so. how?

-l]

-8 5

-2

I 10 10 15

(a) What is the maximum amount o f supple ment 3 that can be used to provide exactly the required amounts of the three nutrients?

2

-4

44. A company makes three types of fertil izer. The first type contains 10% nitrogen and 3% phosphates by weight, the second contains 8% nitrogen and 6% phosphates, and the third contains 6% nitrogen and I% phosphates.

(b) Can the company mix these three types of fertilizers to supply exactly 600 pounds of fertil izer containing 9 % nitrogen and 3.5% phosphates? If so, how?

-l]

- 4 -3

-2

(b) Can the company supply exactly 40 tons of highgrade, 100 tons of medium-grade, and 80 tons of low-grade ore? If so, how many days should each mine operate to fill this order?

to supply exactly 600 pounds of fertil izer containing 7.5% nitrogen and 5% phosphates? If so, how?

!] I -3 -7

5

low-grade ore? If so. how many days should each mine operate to fill this order0

(a) Can the company mix these three types of fertilizers

-I - I

I

53

-2

46. Three grades of crude o il :ue to be blended to obtai n 100 barrels of oil costing $35 per barrel and containing 50 g m of s ulfur per barrel. The cost and sulfur content of the three grades of oil are given in the following table: Grade Co>t per barrel Sulfur per barrel

-ll

43. A mining company operates three mines that each produce three grades of ore. T he daily yield of e"ch mine is shown in the following table:

lligh· gradc ore \1edium -gratlc ore Low-grade ore

Daily Yield Mme I Mme2 Mine J I ton I ton 2 to ns 1 ton 2 to ns 2 to ns 2 tons I ton 0 tons

B

c

$32 62 gm

$24 94 gm

(a) Find the amounts of each grade to be ble nded that use the least oil of grade C. (b) Find the amounts of each grade to be blended that use the most oil of grade C. 47. Find a polynomial function f(x) = ax 2 + bx + c whose graph passes through the points (- I. 14). ( 1.4). and (3. 10). 48. Find a polynomial function f(x) = a.~ 2 + bx + c whose gmph passes through the points ( - 2, -33). (2. -I), :md (3,-8).

= ttxl + bx 2 +ex+ d whose graph passes through the points (-2, 32). (- 1, 13),

49. Find a polynomial function f(x)

(a) Can the company supply exactly 80 tons of highgrade. 100 tons of medium-grade. and 40 tons of


(2.4), and (3. 17).

Page 6 of 10


54 CHAPTER 1 Matrices, Vectors, and Systems of Linear Equations 50. Find a polynomial function f(x) = ar 3 + bx 2 +ex+ d whose gruph passes through the points (-2. 12). (- 1. - 9 ). (I. -3). and (3. 27). 51. If the thrrd pivot posruon of a matrix A is m column j. what can be sard about column j of the reduced row echelon fonn of A? Explain your answer. 52. Suppo-.c that the founh pivot po
/11 £.\t'rt"iu.< 5.l -72. tf<>tumi11e ll"iletilu tile state· 1111!11/S llYt' /Yilt' ()Y /llfSt'.

53. A column of a matrix A is a pivot column if the corresponding column in the reduced row echelon form of A contains the leading entry of some nonzero row.

7 1. The third pivot position in a matrix lies in column 3. 72. If R is an 11 x 11 matrix in reduced row echelon form that has rank"· then R = 1•. 73. Describe an m x 7~.

11

matrix with rank 0.

What is the smallest possible rank of a 4 x 7 matrix? Explain your answer.

15. What is the largest possible rank of a 4 x 7 matrix? Explain your answer. 76. What is the la.rgest possible rank of a 7 x 4 matrix? Explain your answer. 77. What is the smallest possrble nullrty of a 4 x 7 matrix? Explain your answer. 78. What is the smallest possible nullity of a 7 x 4 matrix? Explain your an>wer.

54. There is a unique sequence of elementary row operations that transforms a matrix into its reduced row echelon form.

79. What is the largest possible rank of an m x Explain your
11

matrix?

55. When the forward pass o f Gaussian e limination is complete. the orig imrl matrix has been transformed into one in mw echelon fonn.

80. What is the s ma llest possible nullity of an m x Explain your
11

matrix?

56. No ;,caling opcrutions arc ret' a 5 x 8 matrix with runk 3 a nd nullity 2. 60. If a ;ystem of m linear equations in 11 variables is equivalent to a ' )'>tem of 11 linear equation.' in q variable>. then "'=fl.

61. If a S)"Stem of m linear equations in II variables is equi\'3· lent to a system or p linear equations in q variables. then II

=q.

62. The equation Ax = b •s consistent if and only if b is a linear combrnatron of the column; of A.

63. If the equation Ax = b ;, inconsi~tent. then the rank of [A bJ is greater than the rank of A. 64. If the reduced row echelon form of lA b ) contains a zero row. then Ax = b mu>t have infinitely many sol utions. 65. If the reduced row echelon form of lA b l contains a zero row. then Ax = h mu>t be con>istent. 66. If some column of m:1trix A is a pivot column. the n the corresponding column in the reduced row echelon form of A is a ;tandurd vector. 67. If A is a matrix with r:u1k k. the n the vectors e 1• e2 ••••• et :rppe;rr as columns of the reduced row echelon form of A. 68. The surn of the rank and nullity of a matrix equals the number of rows in the matrix . 69. Suppose that the pivot •·ows of a matrix A :~re rows I. 2..... k. and row k + I becomes zero when applying the Gaussian e limination algorithm. The n row k + I must

equal borne linear combination of rows I. 2..... k, 70. The third pivot po;ition in a matrix lies in row 3.


8 I. Let A be a 4 x 3 matrix . Is it possible that Ax = b is Explain your an>wcr. consistent for every b in

n•?

82. Let A be an m x 11 matrix and b be a vector in n"'. What must be true ahout the rank of A if Ax = b has a unique solution? Justify your ;mswcr. 83. A >ystern of linear equations is called wrdertletermi11ed if it has fewer equation; than variables. What can be said ahout the number of solut ions of an underdetennined system'! 84. A system of linear equations is called Ol'etrletumill<'d if it has more equations than variables. Gi,·e examples of overdetermined 'Y'tem' that ha\e (a) no >Oiution>. (b) exactly one solution. and

(c) infinitely many solullons. 85. Prove that if A is an m x 11 matrix with rank m. then AX = b b con'i'tent for e•cry b in

n,..

86. Prove that a matrix equation Ax = b is consi>tent if and only if the rank• of A and [A bJ arc equal. 87. Let u be a solution of Ax = 0. where A i' an m x n matrix. Must c u be a solution of Ax = 0 for every scalar c? Justify your answer. 88. Let u and v be solutions of Ax = 0. where A is an m x 11 matrix. Must u + v be a solution of llx = 0? Justi fy your answer.

89. Let u and '' be solution$ o f Ax = b. where A is ~u1 m x 11 matrix and b is a vector in Prove that u - v is a solutio n of Ax = 0 .

nm.

90. Let u be a solution of Ax = b and v be a solution of Ax = 0. where A is an m x 11 matrix :md b i' a vector in n'". Prove that u + v is a solution of Ax = b. 91. Let II be an m x

11

matrix and b be a vector in

n'"

such

that Ax = b i~ con~istenl. Prove that Ax = cb is consistent for e\·ery scalar c.

Page 7 of 10


1.4 Gaussia n Elimination 92. Let A be an m x 11 matrix and b 1 and b2 be vectors in n"' such that both Ax = h t and Ax = b2 are consistent. Prove that Ax = b t + h2 is consistent. 93. Let u and v be solutions of Ax = b, where A is
=

Exercises 94- 99. 11se either a calculator with matrix capabilities or complller software such as MATLAB to solve each pmblem.

4Xt- X2 7xt - 6xz 96. 9xt - 5x2

&-,

97.

/11

1.3xt + 0 .5x2 - I .IXJ + 2.7x, - 2.1xs 94. 2 .2xt - 4.5xz + 3.1XJ - 5. 1x4 + 3.2.xs 1.4xt - 2.1xz + 1.5xJ - 3. Lr4 - 2.5xs

x, 95

_ 2Xt

+

3x1 2r1

-

= = =

12.9 - 29.2 -11 .9

+ 3x 3 - x. + 2x5 = 5 + 4XJ + x 4 - Xs = 7 + 2xJ - 2x, + 2.rs = 3 4x2 - XJ - 4x4 + 5xs = 6

1.7 98. [ '3. 1 1 4. 1 6.2

w

2.3 1.2 4 .1 9 .9

3.1 - 2 .1 3.4

x2

X2 x2

2x, + xs = 0 3x4 + 8xs = 15 + 4xJ - 7.r• + xs = 6 + 9X3 - 12r, - &·s = II -

+

/11 Exercises 97- 99, ji11d the rank and the nul/if)• of the mmrix.

/ 11

Exercises 94- 96. use Gaussia11 e/imi11atio11 011 the augmemed mmrix of the sysrem of linear equmio11s to test for consisTency, cmd to ji11d the general solution.

[ ,,

+ 5XJ

55

[·l

-1.1

1.3

1.6

2 .3 -1.8 - 1.1 -1.7

-1.2 4.2 2.1 3.4

- II I 2 14

2 0 -9 -I I

2

0

'']

1.0 1.4 -1.2

-2.1 2.3 -2.0

2.4 0.5 7. 1

2.2 1.5 2 .1 3. 1 1.2 1.5

-LO] 2.2 1.4

0.0 2 .0

-:]

4

8

-4

3

7

9

16

10

SOLUTIONS TO THE PRACTICE PROBLEMS I. The augmented matrix of the given system is

[-~

I

- I

0

-6 - 5

- I - 3 2

6

-8 3

-2

The final matrix conesponds to the following system of linear equations: Xt - krJ + XS = 2 x2 + XJ - 2rs = - I X4 - 4xs = -5

- 2] . -6 - 7

Apply the Gaussian e limination algorithm to the augmented matrix of the given system to transfom1 the matrix into o ne in row echelon form:

[-~

-I 2 -2

l r 1 + r2 - Jr1 + r.\ -

-3 6

-8 r2 r.\

r 2- r.\

iq- q

- r.3+ r 1- q

r1+ r 1- r 1

[~ [~ [~ [~ [~

-6 -5

3

- I 0

-I I 0

-2] -6

x3

-7

X4

-3 0

-3

-I I 0

- 3 I 0

0


+ krJ - XS - XJ + 2rs

free

=

-5

+ 4xs

free.

-10

-2

-I

2. The matrix A is in reduced row echelon fonn. Furthermore. it has 3 nonzero rows. and hence the rank of A is 3. Si nce A has 7 columns. its nullity is 7- 3 = 4.

-2] -I

3. Apply the Gaussian elimination algorithm to the augmented matrix of the given system to transform the matrix imo one in row echelon form:

0 2

-2

I 0

-I

-8

-2 -4

0 0

3

-2 -4

0 0

2

=-I

I -I -8

-2]

-I

I 0

- 2

xs

2 0

-3 I 0

-I I 0

0 I 0

Xt = X2

-I

I 0

The general solution of this system is

I

-2 - 4

-10 -2] -I -5

-~]

-)

-~]

- )

3

3

r

r-3

I

+s ]

4-s

(a) The original system is inconsistent whenever there is a row whose only nonzero e ntry lies in the last co lum n. Only the second row could have thi s fom1. Thus the o riginal system is inconsistent whenever r - 3 0 and 4 - s =I 0: that is. when r = 3 and s =/4. (b) The origi nal system has infinitely many solutions whenever the system is consistent and there is a free variable in the general solution. In order to have a free variable, we must haver - 3 0. and in order for the

=

=

Page 8 of 10


56


>yMem :1lso to be consi>tent. we must have 4 - s = 0. TI1u< the origmal >y>tem has infinitely many solutions if r 3 and s 4. (c ) Let A denote the coefficient matrix of the system. For the ~ystem to have a unique solution, there must be

=

=

two basic v:uiables. so the rank of A mu>t be 2. Since deleting the la>t column of the preceding matrix gives a row echelon form of A. the mnk of A is 2 precisely when r - 3 ¥= 0 : that is. "hen r ¥= 3.

l1.s• j APPLICATIONS OF SYSTEMS OF LINEAR EQUATIONS Systems of linear equations arise in many applications of mathematics. In this section. we present two such app)jcations.

THE LEONTIEF INPUT- OUTPUT MODEL In a modern industrialized country, there arc hundreds of different industries that supply goods and services needed for production. These industries arc often mutually dependent. The agricultural industry, for instance, requ ires farm machinery to plant and harve~t crops. whereas the makers of farm machi nery need food produced by the agricultural industry. Because of this interdependency. events in one industry, such as a st rike by factory workers. can significantl y a ffect many other ind ustries. To better understand these complex interactions, economic planners usc mathematical models of the economy. the most important of which was developed by the Ru ssian-born economist Wassily Leontief. While a student in Berlin in the 1920s, Leontief developed a mathematical model, called the inpw- owpurmodel, for analyzing an economy. After arriving in the United States in 1931 to be a professor of economics at Harvard University, Leontief began to coUect the data that would enable him to implement his ideas. Finally. after the end of World War II. he succeeded in extracting from government statistics the data necessary to create a model of the U.S. economy. This model proved to be highly accurate in predicting the behavior of the postwar U.S. economy and earned Leontief the 1973 Nobel Prize for Economics. Leontiefs model of the U.S. economy combined approximately 500 industrie. into 42 sectors that provide products and services. such as the electrical m:~chincry sector. To illustrate Leontiefs theory. which c;m be :~pplicd to the economy of any country or region. we next show how to construct a general input- output model. Suppose that an economy is divided into n sectors and that ~ector i produce~ M>me commodity or service S, (i = I. 2 .... , n ). Usually. we measure amounts of commodities and services in common monetary units and hold costs fixed so that we can compare diverse sectors. For example. the output of the steel industry could be measured in mi ll ions of dollars worth of stee l produced. For each i and j. let Cij denote the amount of S, needed to produce one unit of S1 . Then the 11 x n matrix C whose (i .})-entry is c,; is cttllcd the iniJUt - output matrix (or the consumption matrix) for the economy. To illustrate these ideas with a very simple example. consider an economy that is divided into three sectors: agriculture, manufacturing. and services. (Of course. a model of any real economy, such as Leontiefs orig in ttl model, wi ll in volve many more sectors and much larger matrices.) Suppose that each dollar's worth of agricultural output requires inputs of $0.10 from the agricultural sector. $0.20 from the

• This section can be omined without toss of continuity.


Page 9 of 10


1.5 Appl ications of Systems of Linear Equations

57

manufacturing sector, and $0.30 from the services sector; each dollar's worth of manufacturing output requires inputs of $0.20 from the agricultural sector. $0.40 from the manufacturing sector, and $0.10 from the services sector; and each dollar's worth of services output requires inputs of $0.10 from the agricultural sector, $0.20 from the manufacturing sector, and $0.10 from the services sector. From this information, we can form the following input-output matrix: Ag.

C=

Man. .2 .4

Svcs.

['I .I .I] .I .2 .3

.2

Agriculture Manufactunng Services

Note that the (i ,j)-entry of the matrix represents the amount of input from sector i needed to produce a dollar 's worth of output from sector j. Now let x 1, x 2. and x3 denote the total output of the agriculture, manufacturing, and services sectors, respectively. Sinc-e x 1 dollar' s worth of agricultural products arc being produced, the first column of the input-output matrix shows that an input of .lx1 is required from the agriculture sector. an input of .2r1 is required from the manufacturing sector, and an input of .3x1 is requ ired from the services sector. S imilar statements apply to the manufacturing and services sectors. Figure 1.20 shows the total amount of money flowing among the three sectors. Note that in Figure 1.20 the three arcs leaving the agriculture sector give the total amount of agricultural output that is used as inputs for all three sectors. The sum of the labels on the three arcs .. lx 1 + .2r2 + .lx3. represents the amount of agricultural output that is consumed during the production process. Similar statements apply to the other two sectors. So the vector

Figure 1.20 The flow of money among the sectors


Page 10 of 10



59

For an economy with 11 x 11 input-output matrix C. the gross production necessary to satisfy exactly a demand d is a solution of(/, - C)x = d .

Example 2

For the economy in Example I. determine the gross production needed to meet a consumer demand for $90 mill ion of agricu lture, $80 million of manufacturing, and $60 million of services. Solution We must solve the matrix equation (13 - C )x = d. where C is the input-output matrix and

is the demand vector. Since

0 0] [.1 .2 .l] I 0 -

.2

.4 .2

0

.3

.I

I

.l

.9 - .2 - .2

.6

[ - .3 - .I

-.1]

- .2 . .9

the augmented matrix of the system to be solved is

.9 - .2 [ - .3

- .2 - . 1 90] .6 - .2 80 . - .1 .9 60

Thus the solution of (/3 - C)x = d is

170] . [240 ISO so the gross production needed to meet the demand is $170 million of agriculture, $240 million of manufacturing, and $ 150 million of services.

Practice Problem 1

~

An island 's economy is divided into three sectors-tourism, transportation, and services. Suppose that each dollar's wonh of tourism output requires inputs of $0.30 from the tourism sector. $0.10 from the transpo1tation sector, and $0.30 from the services sector; each dollar' s wonh of transportation output requires inputs of $0.20 from the tourism sector, $0.40 from the transpo1tation sector, and $0.20 from the services sector; and each dollar's wonh of services output requires inputs of $0.05 from the tourism sector, $0.05 from the transpo1tation sector, and $0. 15 from the services sector. (a) Write the input-output matrix for th is economy. (b) If the gross production for th is economy is $ 10 million of tourism, S I S million

of transportation. and $20 mlllion of services. how much input from the tourism sector is required by the services sector? (c) If the gross production for this economy is $10 million of tourism, $15 million of transportation, and $20 million of services, what is the total va lue of the inputs consumed by each sector during the production process?


Page 1 of 10


60


(d) If the total outputs of the tou rism, transportation, and services sectors are $70 million. $50 million. and $60 million. respectively. what is the net production of each sector? (e) What gross production is required to satisfy exactly a demand for $30 million of tourism. $50 million of transportation, and $40 million of services? <4

CURRENT FLOW IN ELEGRICAL CIRCUITS When a battery is connected in an electrical circuit. a current flows through the circuit. If the current passes through a res istor (a device that creates resistance to the flow of electricity). a drop in voltage occu rs. These voltage drops obey Ohm's law. 12 which states that V = Rl , where V is the voltage drop across the resistor (measured in volts), R is the resistance (measured in ohms), and I is the current (measured in amperes).

r-------41r-------~ 20 volt.' 2ohms

3 ohms

Figure 1.21 A simple electrical

Figure 1.21 shows a simple e lectrical circui t consisting of a 20-vol t battery (i ndicated by t l ) and two resistors (indicated by /WI. ) with resistances of 3 ohms and 2 ohms. The current flows in the direction of the arrows. If the value of I is positive, then the flow is from the posi ti ve terminal of the ballery (indicated by the longer side of the battery) to the negative terminal (indicated by the shorter side). In order to determine the value of I , we must util ize Kirchhoff's 13 volwge law.

circuit

Kirchhoff's Voltage Law In a closed path within an electrical circuit. the sum of the voltage drops in any one direction equals the sum of the voltage sources in lhe same direction. In the circuit shown in Figure 1.21. there arc two voltage drops. one of 31 and the other of 21. Their sum equals 20, the voltage supplied by the single battery. Hence 31

+ 21 = 20 51= 20 I = 4.

Thus the current flow through the nctv

Determine the current through the following e lectrical circuit:

-H52 volts

3 ohm~

5ohms

20 voll<

12 Georg Simon Ohm (1787- 1854) was a German physicist whose pamphlet

Die galvanische Kette mathematisch bearbeitet greatly influenced the development of the theory of electricity. n Gustav Robert Kirchhoff (1824- 1887) was a German physicist who made significant contributions to the fields of electricity and electromagnetic radiation.


Page 2 of 10



c 1 ohrn

30 volts 4 ohms

I ohm

D

£

A

I,

I,

61

ll 2ohms

12

I,

ll

¥H 8

ll

F

Figure 1.22 An electrical circuit

A more complicated circuit is shown in Figure 1.22. Here the junctions at A and 8 (indicated by the dots) create three branches in the circuit, each with its own current How. Starting at B and applying Kirchhoff' s voltage law to the closed path BDCAB , we obtain 111 + 111 - 412 = 0; that is, (7)

Note that, s ince we are proceeding around the closed path in a clockwise direction, the How from A to 8 is opposite to the direction indicated for h Thus the voltage drop at the 4-ohm resistor is 4( - h ). Moreover, because there is no voltage source in this closed path. the sum of the three voltage drops is 0. Similarly, from the closed path 8AEF8 , we obta in the equat ion (8)

and from the closed path 8DCAEF8. we obtain the equation

(9) Note that, in this case, equation (9) is the sum of equations {7) and (8). Since equation (9) provides no informat ion not already given by equations (7) and (8). we can discard it. A similar situation occurs in all of the networks that we consider, so we may ignore any c losed paths that contai n only currents that are accounted for in other equations obtained from Kirchoff s voltage law. At this point, we have two equations (7) and (8) in three variables, s o another equation is required if we are to obtain a unique solution for 1, , h and l 3. This equation is provided by another of Kirchhoff s laws.

Kirchoff's Current Law The current How into any junction equals the CUITent How out of the junction.

In the context of Figure 1.22, Kirchhoffs current law states that the How into junction A, which is 11 + h equals the ClllTent How out of A. which is h Hence we obtain the equation 11 + h = l3 , or (1 0)

Notice that the current law also applies at junction 8 , where it yields the equation l3 = 11 + h However. since trus equation is equivalent to equation ( 10). we can ignore

it. In general, if the current law is appl ied at each junction, then any one of the resulting equations is redundant and can be ignored.


Page 3 of 10


62


Thus a system of equations that determines the current flows in the circuit in Figure 1.22 is

=

0

4/2 + 2 /3 = 30

It+ h -

/3

=

0.

Solving this system by Gaussian elimination. we see that It = 6. /z = 3. and / 3 = 9. and thus the branch currents arc 6 amperes. 3 amperes. and 9 amperes. respectively. Determine the currents in each branch of the following electrical circuit:


-l 33 \'Oil< 2 ohms

--E '•/,

---

6ohms

l ohm

/J

3ohm, 8 volt'

11

EXERCISES ~

In F.lert·ises I 6. dner111i1w whether the Jtatements ~ are rme or foist'.

I. The (i .j)-entry of the mput- output matrix represents the amount of input from sector i needed to produce one unit of output from sector j. 2. For an economy \lith 11 x 11 mput-output matrix C. the gro\\ production necessary to satisfy exactly a demand d i' a 'olution of (1. - C)x = d . 3. If C is the input -output matrix for an economy with gross production vector x. then C x i' the net production vector. 4 . In any closed path wtthin an electrical circuit. the alge· braic >Um of all the \Oitage drop> in the same direction equals 0. 5. At every junction in an electrical circuit. the current How into the junction equals the current flow out of the junction. 6. The voltage drop :11 each re>i;,tor in an electrical network equals the product of the reststancc and the amount or current through the resistor. In Exerdses 7 16. snppnse that tm econo111y is divided into four M!Ctor.v (agriculwre, numufat'lfll'ing . services, anti enter·

winmem ) witlt the follml'ill8 input o11tp111111atrix:

At;.

c=

["

.20 . 18 .09

Man . II .08 . 16 .07

'in' . 15

.24 .06 .12

l:nt

"]

.07

.22 .05


Agtintlturc MJnu factunng Sen rce' Entcrtammcnt

7. What amount of input from the services sector i> needed for a gross production of $50 million by the entertainment sector? 8. What amount of input from the manufacturing sector is needed for a gross production of S I 00 million by the agriculture sector? 9. Which sector is lea.t dependent on services? I0 . Which 'ector i' rno't dependent on

\CJ'\ ice'?

II. On which sector is agriculture least dependent? 12. On whrch sector rs agriculture most dependent? 13. If the gross production for thi> economy is S30 million or agriculture. $40 million of manufacturing. S30 million of services. and $20 milhon of entertainment. what is the total value of the inputs from each scctor con,umed during the production procc>s? 14. If the gross production for thi s economy i' $20 million of agriculture. $30 million of manuf:rcturi ng. $20 million or services, and $ 10 million of entert:rinmcnt. what is the total value of the inputs fro m each sector con>umcd during the production procc»? 15. If the gross production for this economy is $30 million of agricu lture. $40 mill ion of m
16. If the gross production for thb economy is $20 million of agricu lture, $30 million of manufacturing. $20 million of sel'\•ices. and $ 10 million of cntcrt:ri nmcnt. what is the net production of each ;ector?

Page 4 of 10


1.5 Applications of Systems of Linear Equations

17 The input output matrix for an economy producing trans-

63

(a) What is the net production corresponding to a gross production of S40 mall ion of transportation. S30 million of food. and SJS milhon of oil?

21. Consider an economy that is divided into three sectors: finance. goods. and services. Suppose that each dollar" s worth of financial output requares inputs of $0.10 from the finance sector. $0.20 from the goods sector. and $0.20 from the service; ;,ector: each dollar's worth of goods require' inputs of SO 10 from the finance sector. $0.40 from the good; ;,ector. and $0.20 from the services sector: and each dollar'; worth of services requires inputs of $0.15 from the finance sector. SO.IO from the goods sector. and $0.30 from the ;ervices sector.

(b) What gro.s productaon is required to sati>fy exactly a demand for $32 million of transportation. $48 million of food. and S24 million of oil?

(a) What is the net production corresponding to a gross production of $70 million of finance. SSO million of goods. and $60 nulhon of ;crvaccs"!

Ill The input- output matrix for an economy with sectors of metals. nonmetal>. and service; follows:

(b) What is the gross production corre;ponding to a net production of $40 million of finance. S50 million of goods, and S30 million of ;erviccs?

portation, food. and oil follows: I

I·LO
.2

.20 .30 .25

.4

[ .2

M~t

t\ unm

.2 .4 ["2 .2

.4 .2

Oal

.3]

.I

.3

Tran,port<~hon

Food Oal

S \ C\

Metah . I] Nonme tab .2 . I Scrvace'

(;a) What i~ the net pi"Oduction coaTe;,pondi ng to a gross production of $50 million of metals. $60 million of nonmetals, and S40 million of >ervices? (b) What gross production is required to satisfy exactly a demand for $120 million of metals. Sl80 million of nonmetals. and S ISO million of services? 19. Suppose that a nation's energy production is divided into 1\\0 \CCtor.: electricit) :md oil. Each dollars worth of elccaricity output requires SO.IO of electricity input and $0.30 of oil input. and each dollars wonh of oil output requires $0.40 of electricity input and S0.20 of oil input.

(c) What gross production is needed to satisfy exactly a demand for $40 million of linance. $36 million of goods. and $44 million of ;crviccs? 22. Consider an economy that is divided into three sectors: agriculture, m:mufacturing. :and services. Suppose that each dollar's woa1h ol agricu ltural output requires inputs of$0.10 from the agricu ltural sector. $0.15 from the manufacturing sector. and $0.30 from the services sector: each dollars wonh of manufacturing output requires inputs of $0.20 from the agricultural sector. $0.25 from the manufacturing sector. and $0.10 from the services sector: and each dollar's worth of ;,ervices output requires inputs of $0.20 from the agricultural sector, $0.35 from the manufacturing sector. and SO.IO from the services sector.

(a) Write the input -output matrix for this economy.

(a) What is the net production corrc;ponding to a gross production of $40 milhon of agriculture. S50 million of manuf:acturing. and $30 million of \Crvices?

(b) What as the net production corresponding to a gross production of $60 mall ion of electricity and S50 million of oil?

(b) What gross production is needed to ;atisfy exactly a demand for $90 million of agnculturc. $72 million of manufacturing. and 596 mallion of services?

(c) What gro" production is needed to satisfy exactly a demand for $60 million of electricity and $72 million of oil?

23. Let C be the input- output matrix for an economy. x be the gross production vector. d be the demand vector. and p be the vector whose components arc the unit prices of the products or ;ervicc' produced by each sector. Economists call the vector ''= p - cr p the mlu~·addul a•ecror. Show that pTd = vTx. (The single entry in the I x I matrix pTd represents the gross domestic product of the economy.) /lint: Compute I>Tx in two different way,. First. replace p by v + cr p. and then replace x by

20. Suppose thnt an economy is divided into two sectors: nongovernment and government. Each dollars worth of nongovernment output requires $0.10 in nongovernment input ;md $0.10 in government input. oond each dollars worth of government output requires $0.20 in nongovernment input and $0.70 in government input. (a) Write the input output matrix for thi s economy. (b) What is the net production correspondi ng to a gross production of $20 mil lion in nongovernment and $30 mi Ilion in govemment? (c) What gross production is needed to satisfy exactly a demand for $45 million in nongovernment and $50 million in government?


C x + d. 24. Suppose that the columns of the input output matrix

Af

c = [I.2 .3

\lin

.2 .4 .I

lex

:~]

.I

A~rinolturc \l lll~l .th

Tcxll lc'

Page 5 of 10


64

CHAPTER 1 Matrices, Vectors, and Systems of Linear Eq uations

measure the amount (in tons) of each input needed to produce one ton of output from each sector. Let p 1 , p2. and /)3 denote the prices per ton of agricultural products, mi nerals, and texti les, respectively.

(a) Interpret I he vector CT f), where

f) =

~~·] l;~ .

2ohms

3 ohms

28.

(b) Interpret p- CT f). 6ohms

In Exercises 25-29. determine the crm¥'11/S in each branch of the given circuit.

---:::-1 60 volts

3ohms

,,

lz

2 ohms I ohm

25.

L 29

2ohms 2ohms

,,

v~lts

I

"E

,;J,.,.. 6ohms

-'NV"-

5 volts

'• 26.

2ohm~

lz

,,

..,.,.,.._

3ohms

vvolts

- H-

c=

.05

.II

.03 .I I .21 .18 .1 5 .23

.20 .06 .I I .13 .07 .06

.10

.II .15

.22 .12 .05

.05 .07 .I I .03 .19 .15

.07 .06 .18 .14 .19

wl

and

2ohm~

d

=

150 100] 200 125 . [ 300 180

/3

I ohm

["

.2 1

.16 .07

lz

5ohms

m

31. Let

6ohms

27.

1:

-~ /3

In the following exercise, use either a calculator with matrix capabilities or computer software suclr as MATLAB to solve tire

/J

I ohm

II -

proiJ/em:

30 volts

ri,.

/6

30 volts

30. In the following e lectrical net work. determine the value of v that makes l2 = 0:

/3

I ohm

Is I ohm

4ohms

4ohms

I ohm

29.

lz

,,

I ohm

/}

__,


where C is the input- output matrix for an economy that has been d ivided into six sectors, and d is the net production for this economy (where units are in millions of dollars). Find the gross production vector required to produce d .

Page 6 of 10



65

SOLUTIONS TO THE PRACTICE PROBLEMS I. (a) The input-output matrix is as follows: Tour. C= [ .3I .3

Tran. .2 .4 .2

Svc>. .05] .05 .15

.7

-.2

-.05

- .1

.6 -.2

- .05

[ -.3

Touri>m Tran,pon:u ion Services

.85

30] . 50 40

and so the gross production vector is

(b) Each dollar's wonh of output from the services sector requires an input of $0.05 from the tourism sector. Hence a gross output of $20 million from the services sector requires an input of 20($0.05) $ 1 million from the tourism sector.

=

Thus the gross productions of the tourism. transponation, and services sectors are $80 million, $ 105 mi llion, and $100 million, respectively.

(c) The total value of the inputs consumed by each sector during the production process is given by

2. The algebraic sum of the voltage drops around the cir8/. Since the current flow from the cuit is 51 + 3/ 20-volt battery is in the direction opposite to I. the algebraic sum of the voltage sources around the circuit is 52+ ( -20) = 32. Hence we have the equation 8/ = 32. so I 4 amperes.

=

Hence, during the production process, $7 million in inputs is consumed by the tourism sector. $8 million by the transponation sector. and $9 million by the services sector.

=

3. There are two junctions in the given circuit, namely, A and B in the following figure:

c

(d) The gross production vector is

£

A I

33 volts

I

/3

/21 ~ I ohm

2 ohms

3 ohms

6ohms

and so the net production vector is D

x-Cx

70] [.3 [

= ~ - :~

.2 .4

..05OS] [70] 50

.2

.1 5

60

F

Applying Kirc hoff's current law to junction A or junction B g ives It /2 + /3. Applying Kirchoffs voltage law to the closed path ABDCA yields

=

[~] -[EJ = [~~] .

112 + 61, +2ft

= 33.

Similarly. from the c losed path AEFBA. we obtain Hence the net productions of the tourism. transportation. and services sectors arc S36 million. $20 million. and $20 million, respectively. (e) To meet

a demand

3/J + 1(-/2) = 8. Hence the system of equations describing the current flows

is It 8ft +

the gross production vector must be a solution of the equation (/3 - C)x = d . The augmented matrix of the system is


=

0

+ 3/3 =

8.

/2 -

12 -/2

/3

=

= 33 =

Solving this system gives / 1 4. / 2 = I, and / 3 3. Hence the branch currents are 4 amperes. ampere, and 3 amperes. respectively.

Page 7 of 10


66


~ THE SPAN OF A SET OF VECTORS In Section 1.2, we defined a linear combination of vectors u, , u 2, ... , Uk in 1?." to be a vector of the form c 1u 1 + c2u2 + · · · + q lib where c 1• q, ... , q are scalars. For a given set S = {u ,, u 2.... , Uk} of vectors from 1?.", we often need to find the set of all the linear combinat ions of u 1, u2, . . . , uk. For example, if A is an n x p matrix, then the set of vectors v in 1?." such that Ax = v is consistent is precisely the set of all the linear combinations of the columns of A . We now define a term for such a set of linear combinations. Defin ition For a nonempty set S = {u 1• u2, . ... Uk} of vectors in 1?." . we define the span of S to be t he set of all linear combinations of u 1, u2, .. . , uk in 1?." . This set is denoted by SpanS or Span {u, , u2, . . . , uk}.

A linear combination of a single vector is just a multiple of that vector. So if u is in S. then every multiple of u is in SpanS. Thus the span of {u} consists of all multiples of u. In particular, the span of {0} is {0). Note, however, that if S conta ins even one nonzero vector, then SpanS contains infinitely many vectors. Other examples of the span of a set foil ow.

Example 1

Describe the spans of the following subsets of 1?.2:

and

Solution The span of S 1 consists of all linear combinations of the vectors in S 1 • Since a linear combination of a s ingle vector is just a multiple of that vector, the span

J

of S 1 consists of a ll multiples of [ _ : -that is, all vectors of the form [ some scalar c . These vectors all li e along the line with equation y in Figure 1.23.

-~ J for

= - x , as pictured

The span of S 2 consists of all linear combinations of the vectors [ _ : Jand [ - ; ] . Such vectors have the form

a [_ :]

+b

[-;]=a [_:]- [_:]=(a2b

2b) [ _ : ],

where a and b are arbitrary scalars. Taking c =a - 2b. we see that these are the same vectors as those in the span of S 1. Hence Span S2 Span S 1. (See Figure 1.24.)

=

The span of S3 consists of all linear combinat ions of the vectors [ _ :

[7]. User: Thomas McVaney

Note that the vectors [ _ :] and

[7]

J. [-;J.

and

are not parallel. Hence an arbi trary vector

Page 8 of 10


1.6 The Span of a Set of Vectors

67

Figure 1.23 The span of S 1

Figure 1.24 The span of S 2

v in 1?.2 is a linear combination of these two vectors, as we learned in Section 1.2. Suppose that v =a [ _ : J+ b

[~]

for some scalars a and b. Then

so every vector in 1?.2 is a linear combination of the vectors in S3. It follows that the is R 2 . span of Finally, since every vector in 1?.2 is a linear combination of the nonparallel vectors

s3

[ _:Jand [~] , every vector in 1?.2 is also a linear combination of the vectors in S 4 . Therefore the span of S4 is again R 2.

IJiflrrl,!tjM

For the standard vectors


Page 9 of 10



e, )'

c,

X

Span{ e,. e2 } = .\)··plane

y

Figure 1.26 The span of (u, v}, w here u and v are nonpara llel vectors in 'R. 3

in R 3 • we see that the span of {e 1, e 2) is the set of vectors of the form

Thus Span {e., e 2 ) is the set of vectors in the xy-pllme of R 3 . (See Figure 1.25.) More generally, if u and v are nonpara lle l vectors in R 3• then the span of {u, v} is a plane through the origin. (See Figure 1.26.) Furthermore. Spa n {e3} is the set of vectors that li e along the z-axis in R 3. (See Figure 1.25.)

From the preceding examples, we see that saying "v belongs to the span of S = {u 1, u2. .... ud" means exactly the same as saying ·'v equals some linear combination

of the vectors U J, u2, ... , u k."So our comment at the beginning of the section can be rephrased as follows:

={

u 1• u 2, .. .. Uk ) be a set of vectors from R", and let A be the matrix Let S whose columns are u 1, u 2• . .. , uk. Then a vector v from R" is in the span of S (that is. v is a linear combination of \I J, u2. . .. , Uk) if and on ly if the equation Ax = v is consistent.


Page 10 of 10



e, )'

c,

X

Span{ e,. e2 } = .\)··plane

y

Figure 1.26 The span of (u, v}, w here u and v are nonpara llel vectors in 'R. 3

in R 3 • we see that the span of {e 1, e 2) is the set of vectors of the form

Thus Span {e., e 2 ) is the set of vectors in the xy-pllme of R 3 . (See Figure 1.25.) More generally, if u and v are nonpara lle l vectors in R 3• then the span of {u, v} is a plane through the origin. (See Figure 1.26.) Furthermore. Spa n {e3} is the set of vectors that li e along the z-axis in R 3. (See Figure 1.25.)

From the preceding examples, we see that saying "v belongs to the span of S = {u 1, u2. .... ud" means exactly the same as saying ·'v equals some linear combination

of the vectors U J, u2, ... , u k."So our comment at the beginning of the section can be rephrased as follows:

={

u 1• u 2, .. .. Uk ) be a set of vectors from R", and let A be the matrix Let S whose columns are u 1, u 2• . .. , uk. Then a vector v from R" is in the span of S (that is. v is a linear combination of \I J, u2. . .. , Uk) if and on ly if the equation Ax = v is consistent.


Page 1 of 10



1$if!,j,)§i

69

Is or

·{i]

a vector in the span of

If so, express it as a linear combination of the vectors in S.

Solution Let A be the matrix whose columns are the vectors in S. The vector v belongs to the span of S if and only if Ax = v is consistent. Since the reduced row echelon form of [A v] is

[~ ~ ~ -~] 0 0 0 0 0 0

0 . 0

Ax = v is consistent by Theorem 1.5. Hence v belongs to the span of S. To express v as a linear comb ination of the vectors in S, we need to find the actual solution of Ax= v. Using the reduced row echelon form of [A v]. we sec that the general solution of this equation is Xt

=

X2

=

x3 For example. by tak ing x 3

1 - 3x3

-2-

2x3

free.

= 0, we find that

In the same manner, w belongs to the span of S if and only if Ax = w is consistent. Because the reduced row echelon form of [A w] is

Theorem 1.5 shows that Ax of S.

PracticeProbleml

~

Are u =


[-n

and v=

= w is not cons istent. Thus w does not belong to the span

[~] inthespanofS=

1[-!l[-l])? Page 2 of 10


70


In our examples so far, we stat1ed with a subset S of R" and described the set V =SpanS. In other problems. we might need to do the opposite: Start with a set V and find a set of vectors S for which SpanS V. If V is a set of vectors from R"

=

and SpanS = V. then we say that S is a gener ating set for V or that S generates V . Because every set is contained in its span, a generating set for V is necessarily contained in V.

l§ifi,!.)tji

Let

[~J. [lJ.[!J.[=~J I·

s =I Show that SpanS =

n 3.

Solution Because S is contained in 1?.3, it follows that SpanS is contained in n 3 . Thus, in o rder to show that SpanS n 3 , we need only show that an arbitrary vector v in n 3 belongs to SpanS. Thus we must show that Ax= vis consistent for every v, where

=

A=[~ ~ -~]. - I

Let [R c] be the reduced row echelon form of [A v]. No matter what v is, R is the reduced row echelon form of A by Exercise 77 of Section I .3. Since

R

=

[~

0 0

-~]

l 0 0 - I

has no zero row. there can be no row in [R c] in which the only nonzero entry lies in the last colu mn. Thus Ax v is consistent by Theorem 1.5, and so v belongs to SpanS. Since v is an arbitrary vector in n 3 . it follows that SpanS= 1?.3 .

=

The following theorem guarantees that the technique used in Example 4 can be applied to test whether or not any subset of R"' is a generating set for R"':

THEOREM 1.6 The following statements about an

111

x n matrix A arc equivalent:

(a) The span of the columns of A is R"'. (b) The equation Ax = b has at least one solution (that is, Ax = b is consistent) for each b in n"'. (c) The rank of A ism, the number of rows of A. (d) The reduced row echelon form of A has no zero rows. (e) There is a pi vot posi tion in each row o f A. PROOF Since, by Theorem 1.5. the equation Ax "" b is consistent precisely when b equals a linear combination of the columns of A, statements (a) and


Page 3 of 10


1.6 The Span of a Set of Vectors 71 (b) are equ ivalent. Also, because A is an 111 x 11 matrix , statements (c) and (d) are equivalent. The details of these arguments are left to the reader. We now prove that statements (b) and (c) are equivalent. First, let R denote the reduced row echelon form of A and e, be the standard vector

in 7?:". There is a sequence of elementary row operations that transforms A into R. S ince each of these elementary row operations is reversible. there is also a sequence of elementary row operations that transforms R into A. Apply the latter sequence of operations to [R e,] to obtain a matrix [A d ] for some d in 7?:". Then the system Ax = d is e<1u ivalent to the system Rx = e,. If (b) is tmc. then Ax= d. and hence Rx = e,. must be consistent. But then Theorem 1.5 implies that the las t row of R cannot be a zero row, for otherwise [R e,) wou ld have a row in which the only nonzero entry lies in the last column. Because R is in reduced row echelon form, R must have no nonzero rows. It follows that the rank of A is m. establishing (c). Conversely, assume that (c) is true, and let [R c) denme the reduced row echelon fonn of [A b]. Since A has rank 111, R has no nonzero rows. Hence [R c) has no row whose only nonzero entry is in the last column. Therefore, by Theorem 1.5. Ax b is consistent for every b. •

=

Pra ctice Problem 2 •

Iss=

I[~l [-il [~l [-nI

a generating

s~

3

for R ?

MAKING A GENERATING SET SMALLER In Example 3, we found that v is in the span of S by solving Ax = v, a system of 4 equations in 3 variables. If S contained only 2 vectors, the corresponding system would consist of 4 equations in 2 variables. Ln general. it is easier to check if a vector is in the span of a set wi th fewer vectors than it is to check if the vector is in the span of a set with more vectors. The following theorem establishes a useful property that enables us to reduce the s ize of a generating set in certain cases.

THEOREM 1.7 Let S = {u t, u2, .. . , uk) be a set of vectors from R", and let v be a vector in R". Then Span {u t, U2,···· "*·v} =Span {u,. u 2, .... uk} if and only ifv belongs to the span of S. PROOF Suppose that vis in the span of S. Then v = a 1u 1 + a2u2 + · · · + ak u k for some scalars at.a2, .... ak. lf w is in Sp;m {u t, u2..... uk. v}, then w can be written w = Ct Ut + c2 u2 + · · · + ck u k + bv for some scalars Ct, c2, ... , Cko b. By substituting a, u, + a2u 2 + · · · + ak u k for v in the preceding equation. we can wri te w as a linear combination of the vectors u , , u 2, ... , llk. So the span

of {u, . u2, .. . , llk, v) is contained in the span of {u ,, u2, .. . , Ilk). On the other hand , any vector in Span {u, , u 2, ... , llk} can be written as a linear combination


Page 4 of 10


72


of the vectors u 1, u 2, ... , u t, v in wh ich the coefficient of v is 0; so the span of {u J, u 2, ... , uk} is also contained in the s pan of {u1 . u 2, ... , v). It follows that the two spans are equal. Conversely, suppose that v does not belong to the span of S. Note that v is in the span o f {u 1, u2, .. . , uk, v} because v = Ou 1 + Ou2 + · · · + Ou k + I v. He nce Span {u 1. u2, .. . , uk) f. Span {u 1, u 2.... , Uk. v) because the second set contains v, but the first does not •

"*'

Theorem 1.7 provides a me thod for reduc ing the size of a generating set If one of the vectors in S is a linear combination of the others. it can be removed from S to obtain a smaller set having the same span as S. For instance, for the set S of three vectors in Example 3. we have

Hence the span of S is the same as the span of the smaller set

For the set S in Practice Problem 2, find a subset with the fewest vectors such that SpanS= R 3 . ~


EXERCISES In Exercises 1-8. determine whether the given ••ector is in Span { [

1.

5.

l [-:l [n}.

~

[-~]

[-:J

2.

[~]

3.

6.

[-~J

7.

[-!] [_:J [-n [~]

4.

8.

In Exercises 9- 16 detemrine whether the given vector is in

' '" ![-i]·[-l]· [-!ll 9

[-l] "[l] "[~1] "[-l]


In Exercises 17-20, determine the values of r for which v is in the span of S.

17.

S=

{ [_~J.[-n } . v = [_~]

18.

S=

{[jJ.[=i]}.v= [;]

19.

s=

{[

-~l

[-n l·

20. S= {[- :]. [

v

= [

-~J

-~] }· = [~] v

Page 5 of 10



In Exercises 21- 28, a set of vectors in R." is given Detennine whether this .~et i.~ a generating set for R.". Zl.

23.

26

. {

H- ~l[-;J}

22.

{[_;J. [-~]}

n-!l m. [-~]}

24.

{[-;J. [-~l [-!]}

[-!l[_i]} -~l [-ll[-~] } -n.[-~l [-~l [-n 1 n.Gl [-!l [-:Jl -~l

25. { [

. {[

27 . {[

28

31.

[-~

33.

30 [' 2

-;J

0 0

[ ~ -:] -2

32.

[_:

34.

[j

2

2

35.

[i :] 3

4

36

[~

~

mJ.[~J l

39. { [

40

[ -il [-~l [~] } -n. [~~l [~J l -~l -~l [~] }

. {[

4

1. {[


[!l [i]}

In Exercises 45- 64. determine wlwther the statemenrs are true or false.

45. Let S = {u ~o u2..... uk} be a nonempty set of vectors in R.". A vector v belongs to the span of S if and only if v = c 1u 1 + c2 u 2 + · · · + Ct Uk for some scalars Ct. C2., ••• CJ"

46. The span of {0) is {0 ).

47. If A = [u t u2 ... Ut ) and the matrix equation Ax = v is inconsistent. then \' does not belong to the span of {Ui. U2····· uk}. 48. If A is an m x 11 matrix. then Ax = b is consistent for if and only if the mnk of A is 11. every b in

nm

49. LetS= {u 1• u 2 • • •• , Ut ) be a subset of R.". Then the span ofSisR." if;mdonlyifthcrankof[u 1 u2 ... u,)isiL 50. Every finite s ubset of R." is contai ned in its span .

-2] -4 I

-3

;J

-:]

0 - I 3

-2

-3 0

2

I

3

4

4

i]

Jn Exercises 37- 44, a set S of vectors i11 R." is give11. Find a subset of S with the scm1e span asS that i.s a.s small as pos.sible.

37.

-~l

[~l [ll [:]. [~] }

~

Jn Exercises 29- 36. an m x 11 matrix A is given. Detennine whether the equation Ax = b is consiste/11 for every b in R."'.

[-~ ~]

[~l [ll [:]}

43. {[

44 . {

. {[-

29

42

73

5 1. If 5 1 and 5 2 are finite subsets of R." s uch that 5 1 is contai ned in Span 5 2 • then Span 5 1 is contai ned in Span 5 2 •

52. If 5 1 and S 2 are finite subsets of R." having equal spans, then S, = Sz, 53. If and Sz are finite subsets of R." having equal spans, then St and Sz contain the same number of vectors.

s,

54. Let S be a nonempty set of vectors in R.". and let v be in R.". The spans of Sand S U {v) arc equal if and only if vis in S. 55. The span of a set of two nonparallel vectors in R. 2 is R.2. 56. The span of any finite nonempty subset of R." contains the zero vector. 57. If v belongs to the span of S. so does c v for every scalar c. 58. If u and v belong to the span of S, so

doe.~

u

+ v.

59. The span of {v) cons ists of every multiple of v. 60. If S is a generating set for R."' that contains k vectors, then k 2:: m. 6 1. If A is a n m x 11 matrix whose reduced row echelon form contains no zero rows, then the columns of A form a generating set for R."'.

62. If the columns of an n x n matrix A form a generating set for R.", then the reduced row echelon form of A is 1,.. 63. If A is an m x n matrix such that Ax = b is inconsistent for some b in R."', then rank A < m. 64. If S 1 is contained in a fi nite set S2 and S 1 is a generating

set for 'R'", then S2 is also a generating set for 'R"'.

Page 6 of 10


74

65.


Lel u 1 =[-~]:uld uz= [ ~l

75.

(a) llow many ''eCIOrs are in (u 1, uzJ? (b) llow many vcc1ors are in the span of (u 1, uzl?

76. Let V be the span of a finite subset of n•. Show 1hat either V = (01 or If comains infinitely many vectors.

66. Give three different generating sets for the set of vec1ors that lie in 1he Xl'·plane of 'R.3 .

67. Lei A be an 111 X 11 matrix with 111 > 11. Explain why Ax = b is inconsi>lent for >Orne b in 'R.'". 68. Whal can be said about the number of vectors in a generating sci for 'R."''? Explain your answer. 69. Le1 S 1 and S 2 be fini1e sub:.et> of n• such that S 1 is contamed m Sz. Prove that if i> a genemting set for n•. then so is Sz. 70. Let 11 and v be any vectors m n•. Prove that the spans of (u. vi :md (11 + v. u - vi are equal.

s,

77.

Let S be a finite ~ubset of n•. Prove that if u and v are in the span of S. then so i~ u + cv for any scalar c.

Let B be a matrix ob1ained from Aby performing a single elementary row operauon on A. Pro' e that the span of the rows of B equals the span of the rows of A. Him: Usc Exerci.c.' 71 and 72.

78. Pro'e thai e'ery linear combination of the rows of A can be wrinen a.~ a linear combination of the rows of the reduced row echelon form of A. Hmt: Use Exercise 77.

/11 Exucis~s 79 ·82, liS~ ~ithu" calwill/or ll'ith mlllrix CliJXIbili· ties or comp111er software wch (IS MATLAB to detemli11e u·herher e(lch gh·e11 vuror is ill the span of

UJ. IIz .... , Ut be VCCIOrs in 'R." and CJ.Cz ..... Ct be nonzero sc:1lars. Prove thai Span I u 1. 112 ••••• "*I = Sp:m (c, ll , .l·z uz, .. .,l'l llt ).

71. Let

72. Lei 111. uz..... u1 be vectors in n" "nd c be " scal"r. Prove lhlll 1hc sp:m of ( u 1, u 2, ...• lit I is equal 10 Ihe span of (u 1 + cuz, llz, .... 73. Let R be Ihe reduced row echelon form of an m x 11 ma1rix A. Is Ihe span of the columns of R equal 10 the span of lhe columns of A'! Justify your answer.

-0.1 - 1.7 3.4] 2.3 0.0 [ 3.1 . 2.4 . - 1.1 1.7 2.6 - 1.9

1.2] [

[

"*).

74.

Let s, and Sz be finite :.ubsets of n• such that S 1 is conlained in S 2. Usc only the definition of spa11 to prove lhat SpanS1 is contained in SpanS2 •

79.

0.0 -1.8 3. 2.51] [ - 7.7] 0.2 1.6 . 3.9 3.2 3.8 1.7 -1.0

-::~] [-~:~] [-::~] [-~:~]

[

-4.6 -3 8 3.9

6.4

SO.

7.3 8.7 -5.8

.u

Sl.

2.4 4.0 -1.5

82 .

7I 5.4

-7. 1

4.3

-4.7

SOLUTIONS TO THE PRACTICE PROBLEMS I. Lei A be the mn~rix "hose columns are the vectors of S. Then u is in the span of S if and only if Ax = u is conSIStent. Because the reduced row echelon form of (A u ( is / 3 • this sy>tem 1s inconsi:.lent. lienee u is not in the span of S . On the o1hcr lt
[~

Thus the rank of A IS 3. so S " a generatmg set for 'R.3 by Theorem 1.6. 3. From the reduced ro" echelon form of A in Pracuce Problem 2. we see that the last column of A is a linear cornbirunion of 1he fir>l 1hree columns. Thus the vec1or [-:] c:m be removed from S wilholll changing ils span. So

Thus Ax = v is consis1en1. and so v is in the span of S. ln f:1c1, 1hc reduced mw echelon form of [A v] shows thai

n

2. Let A be lhe m:urix whose columns arc the vectors in S. The reduced row echelon form of A is

[~

0

0

I 0 0


~].

is a subset of S 1hat is a generating set for 3 • Moreover. this set is lhe smallest generating set possible because removing any vcclor from S ' leaves u :.et of only 2 vcclors. Since 1he malrix who...c: columns :tre the vec1ors in S ' is :1 3 x 2 matrix. it cannot have rank 3 and so cann01 be a generatmg set for 3 by Theorem 1.6.

n

-2

Page 7 of 10


1.7 Linear Dependence and Linear Independence

[!I]

75

LINEAR DEPENDENCE AND LINEAR INDEPENDENCE

In Section 1.6, we saw that it is possible to reduce the size of a generating set if some vector in the generating set is a li near combi nation of the others. Ln fact, by Theorem 1.7, this vector can be removed without affecti ng the span. In this section. we consider the problem of recognizing when a generating set cannot be made smaller. Consider. for example. the setS= {u 1, u z. UJ. u 4}, where

In this case, the reader should check that u4 is not a linear combination of the vectors u 1, uz, and UJ. However, this does nor mean that we cannot find a smaller set having the same span asS because it is possibl e that one of u 1. liz. and liJ might be a linear combination of the other vectors in S. Ln fact, this is precisely the situation because u 3 = Sll1 - 3ll2 + 01.14. Thus checking if one of the vectors in a generating set is a linear combination of the others could require us to solve many systems of linear equati ons. Fortunate ly, a better method is available. In the preceding example, in order that we do not have to guess which of ll 1, liz, liJ. and u 4 can be expressed as a linear combination of the others, let us fommlate the problem d ifferently. Note that because liJ = Su 1 - 3uz + Ou4, we must have - Su 1 + 3u 2 + u 3 - 0114

= 0.

Thus, instead of tryi ng to wri te some u; as a linear combination of the others, we can try to write 0 as a linear combin ation of lit. liz. liJ. and 1.14. Of course, this is always possible if we take each coefficient in the linear combination to be 0. But if rhere is a linear combinaTion of li t. liz, liJ. and 1.14 rhat equals 0 in which nor all of tile coefficienTs are 0, then we can express one of the u; 's as a linear combinarion of rhe otl1ers. In this case. the equation -Sli t + 3u z + l13- Ou4 = 0 enables us to express any one of u 1, uz, and liJ (but nor 1.14) as a linear combination of the others. For example. since - Sli t + 3llz + ll3 - 0114 = 0, we have - Su 1 = -3llz- ll3 + Ou4

3

I

s

s

li t = - uz + - liJ + 0lJ4 . We see that at least one of the vectors depends on (is a linear combi nation of) the othe rs. Th is idea motivates the following defi ni tions. Definitions A set of k vectors {u 1. u 2, .. .. uk} in R" is called linearly dependent if there exist scalars c 1, Cz, ... , ck, not all 0, such that

In this case, we also say that the vectors u 1, liz, . ... llk are linearly d ependent .


Page 8 of 10


76


A set of k vectors (u 1, u2, .. . , u*) is called linearly independent if the only scalars c , , c2, ... , Ck such that

are c , = c2 = · · · = ck = 0. In this case, we also say that the vectors u , , u 2, ... , llk are linearly independent. Note that a set is linearly independent if and o nl y if it is not linearly dependent.

l§ifh•l•!tjl

Show that the sets and are linearly dependent.

Solution The equation

=

=

is true with c , 2, c2 - I. and CJ = I. Since not all the coefficients in the preceding linear combination are 0, S 1 is linearly dependent. Because

and at least one of the coefficients in th is linear combination is nonzero, S 2 is also linearly dependent.

As Example I suggests, any finite subset S = {0. u 1, u2, .. . , uk) of 'R" rltar conrains the zero vector is linearly dependelll because

is a linear combination of the vectors in S in which at least one coefficient is nonzero.

0 CAUTION

Whi le the equation

is true, it te lls us nothing about the linear independence or dependence of the set S 1 in Example 1. A similar statement is tnte for any set of vectors {u,. u 2. . . . . llk ):

Ou,

+ Ou2 + · ·· + Ot~k = 0.

For a set of vectors to be linearly dependent, the equation

must be satisfied with at least one nonzero coefficient.


Page 9 of 10



Since the equation c 1u 1 + c2u 2 + · · · + q vector product

uk

=0

77

can be written as a matrix-

we have the followi ng useful observation: The set {u 1, uz, ... , u k} is linearly dependent if and on ly if there exists a nonzero solution of Ax = 0. where A = (u , u 2 . . . uk].

i#iflrrl,!tfW

Determine whether the set

s=

I[n ·[~J ·[~J ·[iJI

is linearly dependent or linearly independent.

Solution

We must determine whether Ax

A=

= 0 has a nonzero solution. where

I

I

I

2

0

4

I

1

[ I

i]

is the matrix whose columns are the vectors in S. The augmented matrix of Ax is

[~

I

I

0

4 2 1 3

=0

~l

and its reduced row echelon form is

[~

0

2

I

- I

0

0

0 0

~l

Hence the general solution of this system is

x, = -2x3 X2

=

X3 X4

=

X3

free

0.

Because the solution of Ax = 0 contains a free variable, this system of linear equations has infinitely many solutions, and we can obtain a nonzero solution by choos ing any 1, for instance, we see that nonzero value of the free variable. Taking XJ

=


Page 10 of 10


78


is a nonzero solution of Ax = 0. Thus S is a linearly dependent subset of n 3 since

is a representation of 0 as a linear combination of the vectors in S.

+;:£1,!.111

Determine whether the set

is linearly dependent or linearly independent.

Solution

As in Exa mple 2, we must check whether Ax= 0 has a nonzero solut ion,

where

There is a way to do this without actually solving Ax= 0 (as we did in Exa mple 2). Note that the system Ax = 0 has nonzero solutions if and on ly if its general solution contains a free vari able . Since the reduced row echelon form of A is

[~

0 0]0 , I

0

I

the rank of A is 3. and the nullity of A is 3 - 3 = 0. Thus the general solution of Ax = 0 has no free variables. So Ax 0 has no nonzero solutions, and hence S is linearly independent.

=

ln Example 3. we s howed that a particular set S is linearly independent without actua ll y solving a system of linear equat ions. Our next theorem shows that a similar technique can be used for any set whatsoever. Note the relationship between this theorem and Theorem 1.6.

THEOREM 1.8 The following s tatements about an

111

x n matrix A are equivalent:

(a) The columns of A are linearly independent. (b) The equation Ax = b has at most one solution for each b in 'R"'. (c) The nullity of A is zero.


Page 1 of 10


1.7 Linear Dependence and Linear Independence 79

(d) The rank of A is n, the number of columns of A. (c) The columns of the reduced row echelon form of A arc distinct standard vectors in

nm.

(f) The only solution of Ax = 0 is 0. (g) There is a pivot position in each column of A.

PROOF We have already noted that (a) and (f) are equivalent, and clearly (f) and (g) are equivale nt. To complete the proof, we show that (b) implies (c), (c) implies (d), (d) implies (c), (c) implies (f). and (f) implies (b). (b) implies (c) Since 0 is a solution of Ax = 0, (b) implies that Ax= 0 has no nonzero solu tions. Thus the general solution of Ax = 0 has no free variables. Since the number of free variables is the nullity of A, we see that the nullity of A is zero. (c) implies (d) Because rankA + nullity A= n, (d) follows immediately from (c). (d) implies (e) Lf the rank of A is n, then every column of A is a pivot column , and therefore the reduced row echelon form of A consists entirely of standard vectors. These are necessarily distinct because each column contains the first nonzero e ntry in some row. (e) implies (f) Let R be the reduced row eche lon form of A. If the columns of R are distinct standard vectors in 'R"'. then R = (e1 !!;! .. . e,.]. Clearl y, the only solution of Rx = 0 is 0. and since Ax = 0 is equivalent to Rx = 0, it follows that the onl y solution of Ax = 0 is 0. (f) implies (b) Let b be any vector in 1?!". To show that Ax = b has at most one solution, we assume that u and v are both solutions of Ax = b and prove that u = v. Since u and v are solutions of Ax = b, we have

A(u - v) =A u - Av = b - b = 0. So u - v is a solution of Ax = 0. Thus (f) implies that u - v = 0: that is. u = v. It follows that Ax = b has at most one solution. • Is some vector in the set


a linear combination of the others? The equation Ax = b is called homogeneous if b = 0 . As Examples 2 and 3 illustrate, in checking if a subset is linearly independent. we are led to a homogeneous equation. Note that, un like an arbitrary equation, a homogeneous equation must be consistent because 0 is a solution of Ax = 0. As a result. the important question concerni ng a homogeneous equation is not !f it has solutions, but whether 0 is the only solution. If not, then the system has infinitely many solutions. For example, the general solution of a homogeneous system of linear equations with more variab les than equati ons must have free variables. Hence a homogeneous system of linear equarions wirtz more variables rhan equations has infinirely many solwions. According


Page 2 of 10


80


=

to Theorem 1.8, the number of solutions of Ax 0 determines the linear dependence or independence of the columns of A. In order to investigate some other properties of the homogeneous equation Ax 0, let us consider this equation for the matrix

=

A=[~

-4 2 - I -8 3 2

Since the reduced row echelon form of [A 0) is

[~ the general solution of Ax

-4 0 0

7

-8

-4

5

= 0 is Xt

= 4xz - 7x4

X2

4x4 - 5xs

x3 = X4

xs

+ 8xs

free free free.

Expressing the solurions of Ax = 0 in vector form yields

Thus the solution of Ax

= 0 is the span of

=

In a similar manner, for a matrix A , we can express any solution o f Ax 0 as a linear combination of vectors in which the coefficients arc the free variables in the general solution. We call such a representation a vector form of the general solution of Ax = 0. The solution set of this equation equals the span of the set of vectors that appear in a vector form of its general solution. For the preceding set S. we sec from equation ( II ) that the only linear combination of vectors in S equal to 0 is the one in which all of the coefficients are zero. So S is linearly independent. More generally. the following result is tmc:

When a vector form of the general solution of Ax = 0 is obtained by the method described in Section 1.3. the vectors that appear in the vector form are linearly independent. Practice Problem 2 .,.

Determine a vector fonn for the general solution of Xt -

3xz -

2xr - 6x2 - 2xr + 6xz


+

X3

+

X3 -

X4 -

3x4 -

XS

=0

9xs = 0

+ 3x3 + 2r4 + I lxs = 0.

Page 3 of 10


1.7 Linear Dependence and Linear Independence 81

LINEARLY DEPENDENT AND LINEARLY INDEPENDENT SETS The following result provides a useful chamcterization of linearly dependent sets. In Section 2.3, we deve lop a simple method for implementing Theorem 1.9 to write one of the vectors in a linearly dependent set as a linear combination of the preceding vectors.

THEOREM 1.9

=

Vectors UJ. u 2, ... , u k in R " are linearly dependent if and only if Ut 0 or there exists an i :::; 2 such that u ; is a linear combination of the preceding vectors li t, ll2 , . . . , Ui - 1·

PROOF Suppose first that the vectors u 1. u 2, . .. , u k in R" are linearly dependent. If u 1 = 0. then we arc finished: so suppose u 1 # 0. There exist scalars c 1, c2, ... , ck, not all zero, such that

Let i denote the largest index such that c; # 0. Note that i :::; 2, for otherw ise the preceding etlttation would reduce to c 1ll 1 0, which is false because c 1 # 0 and U t # 0. Hence the preceding equation becomes

=

where

C;

#

0. Solving this equation for u;, we obtain

C2

- Ct

ll;

= -

ut C;

C;

Ci-1 u 2 - · · · - - - u ;-J. C;

Tims u ; is a linear combination of ll t , ll2, ... , ll;- t. We leave the proof of the converse as an exercise.

•

The followi ng properties relate to linearly dependent and linearly independent sets. Properties of Linearly Dependent and Independent Sets l. A set consisting of a single nonzero vector is linearly independent, but (0} is linearl y dependent.

=

2. A set of two vectors (u 1, u 2} is linearly dependent if and only if u 1 0 or u2 is in the span of ( ll t}: that is, if and only if li t = 0 or u 2 is a multiple of li J. Hence a set of two vectors is linearly dependent if and only if one of the vecron is a multiple of the other.

=

3. Let S (li t, ll2•... , Uk} be a linearly independent subset of R". and v be in R". Then v does not belong to the span of S if and onl y if {li t. ll2, ... , u k. v) is linearly independent. 4. Every subset of R" contai ning more than n vectors must be linearly dependent. 5. If S is a subset of R" and no vector can be re moved from S without changing its span, then S is linearly independent.


Page 4 of 10


82


0 CAUTION

The result ment ioned in item 2 is valid only for sets contain ing two vectors. For example. in n 3, the set (Ct. e2. Ct + e2} is linearly dependent. but no vector in the set is a multiple of another. Properties 1, 2. and 5 follow from Theorem 1.9. For a justification of propetty 3, observe that by Theorem 1.9, u 1 "# 0, and for i 2::. 2. no u; is in the span of (u 1• u2. .. .. u, _ J). If v does not belong to the span of S . the vectors u 1, uz, ... , uk. v are also linearly independent by l11eorem 1.9. Conversely. if the vectors u ,. u 2..... Uk . v are linearly independent, then v is not a linear combination of u 1, u2, ... , u k by Theorem 1.9. So v does not belong to the span of S. (Sec Figure 1.27 for the case that k = 2.)

. k -

0

u,

Span l u 1• u 2 )

Span {u 1• u 2)

( u 1, u 2• v} i~ linearly dependent.

{ua. u2. l'} is linearly independent.

Figure 1.27 Linearly independent and linearly dependent sets of 3 vectors

To justify property 4, consider a set (u ,, u 2.... , uk} of k vectors from R". where 11. The 11 x k matrix [u 1 u 2 . . . uk] cannot have rank k because it has on ly 11 rows. Thus the set (u t. u2..... uk} is linearly dependent by Theorem 1.8. However, the next exa mple shows that subsets of R" containing n or fewer vectors may be either linearl y dependent or linearly independent.

k >

Example 4

Determine by inspection whether the following sets are linearly dependent or linearly independent:

Solution

Since St contains the zero vector. it is linearly dependent. To determine if S2. a set of two vectors, is linearly dependent or linearly independent, we need only check if either of the vectors in S2 is a multiple of the other. Because

5[-4] [-!0] ,o '

2 we see that


12 6

15

s2 is linearly dependent.

Page 5 of 10


1.7 Linear Dependence and Linear Independence 83 To see if $ 3 is linearly independent. consider the subsetS= (u 1, 112} , where and

Because S is a set of two vectors, neither of which is a multiple of the other, S is linearly independent. Vectors in the span of S are linear combi nations of the vectors in S, and therefore must have 0 as their third component. Since

has a nonzero third component, it docs not belong to the span of S. So by propctty 3 in the preceding list, S3 = {Ut, 112, V} is linearly independent. Finally, the set S 4 is linearly dependent by propcny 4 because it is a set of 4 vectors from 1?.3. Practice Problem 3

~

Determine by inspection whether the following sets are linearly dependent or linearly independent:

ln this chapter. we introduced matrices and vectors and learned some of their fundamental propenies. Since we can wri te a system of linear equations as an equation involving a matrix and vectors, we can use these arrays to solve any system of linear equations. It is surprising that the number of solutions of the equation Ax = b is related both to the si mple concept of the rank of a matrix and also to the complex concepts of generating sets and linearly independent sets. Yet this is exactly the case, as Theorems 1.6 and 1.8 show. To conclude this chapter, we present the follow ing table, which summarizes the relationships among the ideas that were established in Sections 1.6 and 1.7. We assume that A is an 111 x n matrix with reduced row echelon fom1 R. Properties listed in the same row of the table are equivalent.


The rank of A

The number of solutions of Ax = b

The columns of A

The reduced row echelon form R of A

rankA =m

Ax = b has at least one solution for every b in n_m.

The columns of A are a generating set for'Rm.

Every row of R contains a pivot position.

rank A = n

Ax = b has at most one solution for every b in n_m.

The columns of A are linearly independent.

R contains a pivot

Every column of position.

Page 6 of 10



EXERCISES In £urcises 1- 12, determine by inspection ll'hether the gil•en sets are linearly depentlent.

. {[- 2] .[o]. [-4!] } 21 0 4 3 0

u:nm '·I[_:n-m "· 1[l [~Wl l -il[-~] } !l[~l [-n } Exe~i:s 23~30. d~tennine

I

4

J. { [

' { [-

5. { [Jl[~l[-~] } 6.![_!].[;] .[-~]} 7 ·

n~;J.[-!Jl

' l[~!l II.

&.

10 0

I

{[~l [~l [~] }

24.

WH:H-:]l

25 1[

subset of S comaining the fewest vectors that has the same span asS.

[ [~l [-:]} , w;n:nm "mnn [iHlll "· 1Hl t:Hm l[-~l [-n ·[_~J. [-~J l -~l ~:] }

n~l

20 . {[

i].[-i],[j]j -2

3

"II[[l [1][-;]! J.-l ,: ,

1] [ 0] [1] [OJ ! ~

27.

14. {

I&.

19.

{[-:]. [-~].[:]}

-I

"·1[-iHil Hl Ull

whether the given set is linearly

23. {[:ll[-~J.[~] }

mJ.[-;J.[~Jl

In Exercises 13- 22, a setS is given. Detennine by impection a

13. { [

In mdependem.

28

1[ilnllll tm

"·I[ll f!lfil Pll =

30

IDH-!].[-!][=ll l

In Exe~ises 31 - 38, a linearly dependem setS is given. Write some vector in S as a linear combintllinn of the others.

[-;J. m J

_iJ. [-~l [~~l [=~]}


Page 7 of 10



85

'· 1[-iH-iH!lDll In Exercises 51- 62, wrire rhe vecror fonn of rile general .wlwion of rile given sysrem of linear equations. x, + Sx3 0 51. x, - 4xz + 2r3 0 52. x2 - 3x, 0

=

53.

+ 2r4 = 0

+ 3xz

Xt

X3 -

x1

6x~

XI

54.

=0

= = + 4x, = 0 x2 - 2r4 = 0

+ 4x3 - 2r4 =0 + 7x4 = 0 + llx• = 0 2X2 - XJ - 4X4 = 0 4xz + 3xJ + 7x• = 0

55. -x1 + x2 - 7x3 2x, + 3.r2 - XJ

Xt 56.

2rt -

+ 4X2 + X3 + 5X4 = 0 + 2x3 - 5x• + xs - X6 = X3 + 3x4 - xs + 2r6 = x 1 + x3 - x4 + xs + 4x6 =

- 2rt

-x, 57. x 1 In Exercises 39- 50, derennine, if possible, a l'tliue of r for which rile given ser is linearly dependenT.

2x3 - x. - 5xs . - x2 + 3x3 + 2.r4 58 - 2.r, + xz + XJ - x. + 8xs 3x1 - xz - 3x3 - x. - 15x5 -x 1

-

0

0 0

=0

=0 =0 =0

X2 + X4 = 0 + 2xz + 4x• = 0

XI +

59.

Xt 2xt

- 4x•

=0

+ XJ + x. + 1xs = 0 + 2x3 + 10xs = 0 + 4x4 + 8xs = 0 x 1 + 2..rz - X3 + 2.rs - X6

x, - 2rz 60. x, - 2x2 2x 1 - 4.r2

6

43.

45.

H7J. m. [~J l

n-~l m. [~]} jJ. L!l [-~J }

47. { [

l . lUHllPll 48

. {[

-~l Gl [n


+ 4xz -

2r3 - .r4 - .r, - 2xz + XJ + x, x,

1. 2x,

x1

-

x2

-

. 2rt - 2rz 62 -XI

+

Xl

X3 -

~

2x, - x 5 7.r• - xs

+ X) + 5X4 + XS X)

~

=0 5x6 0 + 2.rs + 4x6 = 0 + 4x5 + 3x6 = 0

+ 3X4 + XS

-

=

+ 4.r6 = + 5x6 = -

0

0 3X6 = 0 X6 = 0

In Erercises 63-82. derennine wheTher rile state· ments are true or false.

63. If Sis linearly independent, then no vector inS is a lineM combi nation of the others. 64. If the onl y solution of Ax = 0 is 0. then the rows of A are linearly independent.

65. If the nullity of A is 0, then the columns of A are linearly dependent.

66. If the columns of the reduced row echelon form of A are distinct s tandard vectors, then the only solution of A x 0

=

is 0.

67. If A is an 111 x 11 matrix with rank A are linearly independent.

11,

then the columns of

Page 8 of 10


86

CHAPTER 1 Matrices, Vectors, and Systems of linear Equations

68. A homogeneous equation i> alway• consistent. 69. A homogeneous equation always has infinitely many solulion~.

70. If a vector form of the general ;olution of Ax= 0 is obtained by the method described in Section 1.3, then the vectors that appear in the vector fonn are linearly independent. 71. For any vector v. {v i is linearly dependent.

72. A set of vectors in R!' is linearly dependent if and only if one of the vectors is a multo pie of one of the others. 73. lf a subset of n• is linearly dependent. then it must contain at least 11 \'CCtors. 74. If the columns of a 3 x .f matrix arc distinct. then they are linearly dependent. 75. For the ;ystcm of linear equation; Ax = b to be homogeneous, b must equal 0. 76. If a subset of n • contains more than linearly dependent.

11

vector., then it is

77. If every column of an m x 11 mmrix A contains a pivot position, then the matrix equation Ax = b is consistent for every b in n•. 78. If every row of an m x 11 matrix A contains a pivot position. then the matrix equation Ax = b is con;i;tcnt for C\Cry b in n•. 79. 1f <'o ll o+czuz+ .. ·+ct u.t= O for <'t=n= .. ·= <'t = 0. then {u o. uz..... u.t I is linearly independent. 80. Any >Ubset of n• that contain' 0 i\ hncarly dependent. 81. The sCI of Standard vector' in

n•

i' linearly independent.

82. The largest number of linearly independent vectors in

n•

is 11. 83. Find a 2 x 2 matrix A such that () is the only solution of Ax = (). 84. Find a 2 x 2 matrix A such that Ax many solutions.

= 0 has

infinitely

85. Find an example of linearly independent sub>cts {u 0, u 2} and {\'1 of n 3 such that (u o. uz. vi is linearly dependent. 86. Let (u 1 • uz..... u.t} be a linearly independent set of vectors in n•. and Jet V be a \'eCtOr in n• such that \' = co u o + Cz ll z + · .. + <'t Ut for some scala" c 0.c2 ..... Ct. with Ct # 0. Prove that (v. u 2• ••• • Ut} "linearly independent. 87. Let u :ond v be d istinct vecton< in R". Prove that the set {u. v } is linearly independent if and only if the set {u + v, u - ' '} is linearly independent.

90. Complete the proof of Theorem 1.9 by s howing that if u o = 0 or u, is in the span of {u o. uz..... u, t} for some i ~ 2. then {u t. uz..... Ut} is linearly dependent. Him: Separately consider the case in which u 1 = 0 and the case in which vector u, is in the span of (u t. u 2••••• u, _ 1). 4

9 1~ Prove th:ot :ony nonempty subset of a linearly independent subset of n" is linearly independent.

92. Prove th:ot if S o is a linearly dependent subset ofR" that is contained in a fi nite set S 2. then S2 is li nearly dependent. 93. Let S = {u 0• 111 ..... Ut} be a nonempty set of veeton< from n•. Prove that if S is linearly independent. then every vector in Span S can be written as Ct Ut + czu z - .. · + Ct Ut for wuqu~ scalars l't.l'l ... . . c• . 9.f. State and prove the con\'crsc of Exercise 93. 95. LetS = {u ,. u z..... Ut } be a noncmpty sub>ct of R.• and A be an m x 11 matrix. Prove that if S i~ linearly dependent and S' {Au 1.Au 2 , •••• Auk} contains k distinct vectors. then S' is linearly dependent.

=

96. G ive an example to show that the preceding exercise is false if lillf!ttrly dependem is changed to li11early i11depe11· de11t. 97. LetS= (u 1• u 2 •.•.• lit} be a nonempty subset of 'R" and A be an m x 11 matrix with rank 11. Prove that if S is a hnearly mdependent set. then the set (Au 0.Au 2..... Aut} i~ also linearly mdependcnt. 98. Let A and 8 be m x 11 matrices :.uch that 8 can be obtained by performing a single elementary row operation on A. Prove that if the rows of A are linearly independent. then the rows of 8 arc also linearly independent. 99. Prove that if u matrix is in reduced row eche lon form. then its non1ero row' are linearly independent. 100. Prove that the rows of an m x 11 matrix A :ore linearly independent if and only if the rank of A ism . Him: Usc Exercises 98 and 99. In £terl'ises /OJ 104. use t'ither a calculmor ll'ith matri.l capabiliries or comp11tu softll'ai'F s11ch as MATU\8 to dnen11i11~ ll'herhu each 1/11'<'11 St'l is lmt'arl)' dependem. In tit~ cau tltot th~ set is linearly tlependmt. ll'rit~ somt' •·e ctor i11 the set tiS a linel/r Cflmbinatitm t1{ the mhus.

~.~

101.

88. Let u. v. and w be distinct vectors in R". Prove that (u. v. wl is linearly independent if and only if the set (u + v. u + w. v + w} is linearly independent. 89. Pro'c that if (u o. uz, .... ud i~ a linearly independent 'ub>et of n• and co. cz ..... ct arc non1ero scalar-;, then (co u o.c2u 2..... ct u.t} is also linearly mdependent. 14

102.

[

1

2.3 r1.4 2.7 • 3.6 0.0 -

4.2 8.1 2.4 1.7 r-5.7 r-5.0 r -1.1 2 .9 6.2 - 4.3 1. 1 2.6 0.0 . 7.2 • 3.4 . 1.6 10 5 3.3 0.0 1.3 4.0 2.9 6. 1 3.2

1 1 1 1

-5.4 4.2 2.4 r-0.6 4.2 l..f 1.2 r-1.7 r-5.0 r - 2.4 1.1 ' 2.4 . 0 .0 3.7 . 6.2 . -2.6 0.0 3.4 -1.0 5.6 [ 0.3 1.3 3.3 8.3 2.3 l..f -4.0 6.1 -2.2 -1.0

1 1 1 1 1

This exercise is used in Section 7.3 (on page 5 t4).


Page 9 of 10


Chapter 1 Review Exercises

103 ·

[-li]. [ ll]. [Ji]. [-i!]. [-~]· [ ll] 17 10

- 66 II

25 16

-7 22

- 15 22

-15 24

104 ·

87

[-li].[-~]· [~il]. [-:1]. [~l]. [Jl] 17 10

-66 II

0 2

45 15

- 28 -3

34 7


----------------------

I. Let A be the matrix "hose columns are the vectors in S. Since the reduced row echelon form of A is / 4 • the columns of A arc linearly independent by Theorem 1.8. Thus S i> linearly independent. and so no vector in S is a linear combination of the others. 2. The augmented matrix of the given system is

To oblain its vector form. we expre.s the general solution as a linear combination of vector> in which the coefficients are the free variables.

XI] [3] [ 2] ~~ = [3Xl + -lis] 1s = lz ~ +xs -~ [xs -l
[j

3

6 6

- I I

I

- I

-3

-9

3

2

II

.14

0

Since the reduced row eche lo n form of this matrix is

[~

3 0 0 - 2 OJ

0 0

I 0

0 I

l 2

0 . 0

the general solution of the given >ystem is

11

=h2+

~2

free

..l \

.14

1s

CHAPTER 1

=

=

3. By properly I o n page 81. S 1 i< linearly indepe ndent. By property 2 o n page 81. the lir>t two vectors in S2 arc linearly dependent. Therefore S2 is linearly dependent by Theorem 1.9. By properly 2 on page 81. S1 is linearly independent. By property 4 on page 81.

s. is linearly dependent.

l1s - X'\

free.

-l
REVIEW EXERCISES

~ bt E.urcist!s I 17. detemtillt! wltt'tlter tltt! statmumts

~

1

""' tmt! or falu.

I. If 8 is a 3 x 4 matrix. then its columns arc I x 3 vectors.

2. Any scalar multiple of n vector v in bination of v.

n"

is a linear com·

3. If a vector ,. lies in the spru1 of a finite subset S of n", then v is a line:tr combi nation o f the vectors in S. 4. T he matrix vector product of an m x 11 matrix A and a vector in n" is a !incur combination o f the columns of A. 5. 111e rnnk of the coeftlcie nt matrix of a consistent sys· tcm of linear equations is cqtml to the number of basic varillbles in the general ,olution of the sy,tem. 6. The nullity of the coefficient matrix of a consistent system of linear equations i> equal to the number of free variables in the general solution of the system.


7. Every mmrix can be tran;formed into one and only one matrix in reduced row echelon form by means of a sequence of elementary row oper.ltion>. 8. If the last row of the reduced row echelon form of tm aug· mentcd matrix of a system of linear equation' has only one nonzero entry. 1hc n the syMem is inconsistent. 9. If the last row of the reduced row echelon form o f an aug· mented matrix o f a syste m of linear equations has only zero entries, then the system has infi nitely muny solutions. 10. The zem vector ofn" lies in the >pan of any fi nite subset ofn". II. If the rank of an m x 11 matrix A is A are line:trly independent.

111.

then Lhc rows of

12. The set of columns of an 111 x 11 matrix A is a generating set for n"' if and only if the rank of A is m.

Page 10 of 10


88


13. If the columns of an m x 11 matrix are linearly dependent. then the rank of the matrix is less than 111. 14. If S is a linem·ly independent subset of 1?.'' and v is a vector in 'R" such that SU (v} is linearly dependent. then v is in the span of S. 15. A s ubset of 'R" containing more than 11 vectors must be linearly dependent. 16. A subset of 1?!' containing fewer than 11 vectors must be linearly independent.

during January of las t year. Provide an interpretation of the vector (O.I )(vt + v2 + · · · + Vto).

/11 Exercises 30-33. compute the matrix-vector products.

30.

[~

32.

A4s• [-~J

17. A linearly dependem s ubset of'R" must coma in more than 11 vectors.

-l] [~]

[:

3 1.

3 -I

33. A_30o [

-~J

34. Suppose that 18. Determine whether each of the followi ng phrases is a misuse of terminology. If so, explain what is wrong w ith each one:

Vt

=

[!]

and

=[

"2

[

~r -:J

I 4

-!l

Represent 3v t - 4v2 as the product of a 3 x 2 matrix and

(a) an inco nsistent matrix

a vector in 1?. 2•

(b) the solution of a matrix (c) equivalent matrices

111 Exercise.r 35- 38, determine ll'hether the given vecror v is in the span of

(d) the nullity of a system o f linear equations (c) the span of a matrix (f) a generating set for a system of linear equations

(g) a homogeneous matrix If so, write v as a linear combination of the vectors in

(h) a linearl y independent matrix

19. (a) If A is an m x 11 matrix with rank 11 . what can be said about the number of solutions of Ax = b fo r every b

in"R111 '! (b) If A is an m x 11 matrix with rank m , what can be said about the number of solutions of Ax = b fo r every b

in 7?./11 ?

/11 Exercises 20-27. use tire jol/owi11g matrices to compute tire g iven expression, or give a

deji11ed:

A= [

-~ ~l

8

=

retiSOII

why the e..rpression is nor

G-~l

21. A+8

22. 8C

23. AD 7

24. 2A- 38

25. ATDT

26. A 7 -8

27. C 7

-

37.

UJ

V=[l]

= [~]

38. v

=

[~~] Xt

c=

m'

Xt

+ lx2 -

XJ

X1

X1

42.

2tt

+ lx2 + 3XJ + X2 + XJ -

2D

28. A boat is traveling o n a river in a southwesterly direction. parallel to the riverbank. at 10 mph. At the same time. a passenger is walk ing from the southea st side of the boat to the northwest at 2 mph. Find the velocity and the speed of the passenger w ith respect to the riverbank.

29. A s upermm·ket c hai n has 10 stores. For each i such that I :S i :S I0. the 4 x I vector v, is defined so that its respective components represent the total value of sales in produce, meats. dairy, and processed foods at store i

43.

X2

+

X3

=

3

= I = 2

4x2 - 7xJ = 4

+ 3X2 + 2<) + + X2 + X3 -

XI -

+

40. - 2rt + 4x2 + 2<3 = 7 2Xt - X2 - 4XJ = 2

=I

and 41. lr1 x1


36. v

111 Exercises 39-44, determi11e whether the given system is consiste/11, lllld if so. ji11d its ge11eral sollllion.

39.

D =[I - I 2]. 20. A+8 7

35. v =

S.

lx2 -

XJ -

X4 X4

lr4

=2

=3 =4

2:: ! 3;~ ! l;~ ~ ;: : -~ + 4x2 - lr3 + 2r4 = 4 + x2 + 4XJ + 2r4 = I 4x 1 + 6x2 + x 3 + 2r4 = I

2r1 44. 2rt

111 Exercises 45-48.jirulthe m11k and llllllity of tire give11 matrix. 45. [1

2

-3

0

1]

46.

[~

2 0

-3 0 0 0

~]

Page 1 of 10



47. [_~

2 I

I 0

-3

2

-I I 2

48.

[-llll

49. A company that ships fruit has three kinds of fruit packs. The first pack consists of I 0 oranges and 10 grapefruit. the second pack consists of 10 oranges, 15 gnapefruit, and I 0 apples, and the third pack consists of 5 oranges, 10 grapefruit, and 5 apples. How many of each pack can be made fro m a stock of 500 oranges, 750 grapefruit. and 300 apples?

In Exercises 50- 53, a .set of vectors in R" is given. Detem1i11e whether tile set is a generating set for R".

, l[-iH!H-ill 53. {

In Exercises 64-67, a linearly dependent set S is given. Write some vector in S as a linet1r combinati()n of the others.

64

{ [-llGl [~J}

65

.

{[~l [-il[~J }

l[!][l] [lll "·l[-l][lHltill M.

X3 + x. = 0 + 2x2 - X3 = 0 XJ + X2 + X3 = 0 x 1 + 3x2 - 3x3 = 0 2xt + Sx2 - x3 + x 4 = 0 Xt + 3x2 + 2x3 - X4 = 0 3Xt + x 2 x3 + X4 =0 2r, + 2xz + 4x3 - 6x4 = 2r, + x2 + 3x3 - xa = 0

68. x 1 + 2r2 -

[~l [_:J. [D}

[~l

. . . }

63. {[ 6.63 !:~~] [~:~!] 9.27 [~:~~] 1.41 [~~] 41

In Exercises 68- 71, write rhe vectorform ofthe general solwion of the given system of linear equations.

51. {[-:].[-:]. [_:] } 52. {

89

X!

69.

[-:l [:J. [!]}

?O.

In Exercises 54- 59. an m x I! mallix A is given. Determine whether the equation Ax= b is consistelll for every b in R".

71.

0

72. Let A be an m x n matrix. let b be a vector in R'", and 54.

[~ ~]

55.

[~ ~]

56.

[~

57

[-:

58.

[j -~]

- I

0

:]

2

0

59.

suppose that v is a solution of A x

-I

_:]

2

73. S uppose that Wt and w 2 are linear combinations of vectors v 1 and v 2 in R" such that w 1 and w 2 are linearly independent. Prove that ' ' t and v 2 are linearly independent.

[-l ll

WH-lHnl . j[-l][l][!ll

., W lHll User: Thomas McVaney

=

(b) Prove that for any solution u to Ax = b, there is a solution w to Ax = 0 such that u = v + w.

-3

In Exercises 60- 63, determine whether the given set is linearly dependem or/inearly independe/11.

60

= b.

(a} Prove that if w is a solution of Ax = 0. then v + w is a solution of Ax b.

74. Let A be an m x n matrix with reduced row echelon fo rm R. Describe the reduced row echelon form of each of the following matrices: (a} [A 0]

(b) (a 1 a2 · · ·

ad

fork < 11

(c) cA, where c is a nonzero scalar (d)

lim A]

(e)

lA cA I. where c

is any scalar

Page 2 of 10


90


CHAPTER 1 MATLAB EXERCISES For the following exercises, use MATlAB (or comparable software) or a calculator with matrix capabilities. The MATlAB funaions in Tables 0.1, 0.2, 0.3, 0.4, and 0.5 ofAppendix O may be useful.

I. Let 3.2

6. 1 -1.7 1.5 4.3 2.4 . 4.1 2.0 5. 1 4.2 6. 1 -1.4 3.0 -1.3 Use the matrix- vector product o f A and a vector to com· pute each of the following linear combinations of the columns of A: (a) 1.5at - 2.2a z + 2.7a3 + 4:14 (b) 2a t + 2.1 a z - l.la4

-23]

1.3 -2.5 [ 21 A= -1. 2 1.5

(c) 3.3a 2 + 1.2a3 - 34

(a) Compute A20o

(b) Compute A30• ( A2o• (c) Compute A so• (d) Compute

[~]).

l

[~

A_20• (Aw•

[~]) .

(c) Make a conjecture about As1+02 and its relationship 10 Ao1 and Ao2 for any angles (), and 82.

(f) Prove your conjecture.

2. Let A=

[ 6.131 - 2.2 2.2

8 =

[-4.~ 21 2.) -0.7

2.1 2.4 5.1 6.1

-3.3 -1.3 3.2 7.2

41]

(g) Make a conjecture about the relationship between Ao and A_8 for any angle 9.

-3.1 2. 1 . -5.1

-I. I 8.1 5.2 2.8

1.2 9.2 -3.3 -6.3

,. =

[- 4.6 32] 1.8 .

(h) Prove your conjecture. 4. Let

42]

-3.3 4.2 . 4.7

A=

[I I 3.1 7.1 2.2

and

7.1 (a) Compute 3A - 28. (b) Compute A- 4B r. (c) Compute P = f
(e) Compute pT and QT to see that P in (c) is symmetric and Q in (d) is skew-symmetric. Then compute P + Q. What does it equal? (f) Compute Av. (g) Compute B(Av). (h) Compute A(B v). (i) Evaluate the linear combination 3.5a 1 - 1.2a 2 + 4. 1a3 + 2a4,

Gl

Gl

and determine a vector w such that Aw equals this linear combination. Let M be the 4 x 4 matrix whosej th column is Ba1 for I .:;: j .:;: 4. Verify that M e1 = B (Ae;) for all j . and verify that Mv = B(A v). Stale and prove the generalization of this result to all vectors in 4 •

2.0 -1.5 - 8.5 4.0

4.2 4.7 5.7 8.4

2.7 8.3 19.5 6.5

2.3 6.7 . -3.4

(b) Use technology to compute the reduced row echelon form of A directly. (For example, enter rref(A) when using MATLAB.) Compare this to the resu lt obtained in (a). 5. For the matrix A in Exercise 4 and each of the following vectors b. determine whether the system of linear equations Ax = b is consistent. If so, find its general solution:

(a) b

n


0I]

(a) Use elementary row operations to transform A into reduced row echelon form . (Note that the MATLAB function command A ( i, : ) described in Table 0.4 in Appendix 0 is useful for this purpose. For example, to interchange row i and row j of A, enter the three commands temp = A ( i, : ) , A(i,:) = A(j,:), and A( j ,:) =temp. To multiply the entries of row i of A by the scalar c. enter thecommand A( i , :) = c*A( i ,:) . Toadd the scalar multiple c of row i to row j of A, enter the commandA( j , :) = A( j , : ) + c•A(i, :).)

=

[-H]

(b) b

1.1] = ;:~

1.6

3. Let Ao denote the rotation mattix of() degrees. as defined in Section 1.2. For the following computations. it is useful to apply the imponed MATLAB function rotdeg, as described in Table 0.5 of Appendix 0:

1.2 -3.1 - 11.7 2.1

3.8]

(c) b

= ~:~ [

12.0

(d) b =

[- 1.4

[-l] Page 3 of 10


Chapter 1 MATLAB Exercises

6. Let

c=

0.06 0.12 0.21 0.10 0.10

0.05 0.12 0.11 0.20

ro"

0.20 0.14 0.06 0.15 0.20

0.13 0.10 0.14 0.20 0.05

0.20

"1

00. 15 0. 10 0. 17

and 100

d

rw1

= 1~~

.

where C is the input-output matrix for an economy that has been divided into five sectors. and d is the net production for thi s economy. where units are in billions of dollars. Find the gross production vector required to produce d , where each component is in units of billions of dollars. rounded to four places after the decimal point. 7. Determine whether each of the sets that follow is linearly dependent or linearly independent. If the set is linearly dependent, write one of the vectors in the set as a linear combination of the other vectors in the set.


,,, s.

91

~ -!HiH-!nil [il [

~) ~- [-;][_i][][-l]{] 8. Detennine whether each o f the vectors shown is in the span of St as defined in (a) of Exercise 7. If the vector lies in the span of S 1• tl1en represent it as a linear combination of the vectors in S 1 .

,, [ij] ~.[-!]

(+ll

(d) [

-l]

Page 4 of 10




Page 5 of 10


2

INTRODUCTION

Fingerprint recognition was accepted as a valid personal identification method in the early twentieth century and became a standard tool in forensics. The rapid expansion of fingerprint recognition in forensics created enormous collections of fingerprint cards. Large staffs of human fingerprint examiners struggled to provide prompt responses to requests for fingerprint identification. The development of Automatic Fingerprint Identification Systems (AF/5) over the past few decades has improved dramatically the productivity of law enforcement agencies and reduced the cost of hiring and training human fingerprint experts. The t echnology has also made fingerprint recognition a practical tool for unsupervised access control.

Type

Fingerprint identification for law enforcement is Rodge

based on the location and type of m inutiae, which are

Endpoont

peculiarities in t he friction ridges (or simply ridges) on th e fingertips. Analysis usually focuses on a particular pair of minutiae: ridge endpoints where a ridge terminates and bifurcation points where a ridge splits. An example of each is shown in the center of each

Bifurcation

of the figures at the left, where the ridges are black

poonl

against a white background. The identification then hinges on locating the minutiae in the fingerprint,

a task originally performed by human examiners. 93


Page 6 of 10


94

2 lntrod uction

The AFIS analysis uses a digital image, wh ich is usually acquired by a live-scan fingerprint sensor. The original digital image goes through a preprocessing phase that extracts the ridge pixels, converts them to binary va lues (pixels are either 0 or 1), and thins the image to obtain a set of one-pixel width curves that are approximately the centerlines of the ridges, as shown in the figure at the right. A few of the minutiae are labeled with blue squares. Matrices such as

The process of detection involves sliding the matrices over the image and looking for 3x3 blocks of image pixels that match one of the matrices. A match indicates an endpoint. Other uses of matrices whose entries consist of Os and 1s, called (0, 7)-matrices, are examined in Section 2.2.

000 0 I I ] and [00 0 I 01 ] [ 000 000

are used to detect ridge endpoints. Interpreting the entries as 0 =white and I = black, the matrices match the possible arrangements of pixels at a ridge endpoint. Similarly, matrices such as

I 0 [ I

0 0 I

I

0 0

l

I

0

[ 0

0

I

I

0

I ()

l

and

are used to find bifurcation endpoints.


Page 7 of 10


CHAPTER

2

MATRICES AND LINEAR TRANSFORMATIONS n Section 1.2. we used matlix-vector products to perform calculations in examples involving rotations of vectors, shifts of population, and mixtures of seed. For these examples. we presented the data-points in the plane. population distributions. and grass seed mixtures-as vectors and then formulated rules for transforming these vectors by matrix-vector products. In s ituations of this kind in which a process is repeated, it is often usefu l to take the product of a matrix and the vector obtained from a previous matrix-vector product. This calculation leads to an extension of the definition of multiplication to include products of matrices of various sizes. In th is chapter, we examine some of the elementary properties and applications of this extended definition of matrix multiplication (Sections 2.1-2.4). Later in the chapter, we study the matrix-vector product from the functional viewpoint of a rule of correspondence. This leads to the definition of linear transformation (Sections 2.7 and 2.8). Here we see how the functional properties of linear transformations correspond to the prope1ties of matrix multiplication studied earlier in this chapter.

I

III]

MATRIX MULTIPLICATION

In many applications, we need to mu ltiply a matrix-vector product Avon the left by A again to form a new matrix-vector product A(Av). For instance, in Example 3 in Section 1.2, the product Ap represents the population distribution in a metropolitan area after one year. To find the popu lation distribution after two years, we multip ly the product Ap on the left by A to get A(A p). For other problems, we might need to multiply a matrix-vector product Bv by a different matrix C to obtain the product C(Bv). The following example illustrates such a product:

Example 1

In the seed example from Section 1.2, recall that deluxe

8 = [ .80 .20

s tandard

economy

.60

.40 ]

.60

.40

bluegrass rye

gives the prop01tions of bluegrass and rye in the deluxe, standard, and economy mixtures of grass seed, and that

v

=

[~~] 95


Page 8 of 10


96

CHAPTER 2 Matrices and Linear Transformations

gives the number of pounds of each mixture in stock. Then Bv is the vector whose components are the amounts of bluegrass and rye seed, respectively, in a blend obtained by combining the number of pounds of each of the three mixtures in stock. Since

.80

= [ .20

Bv

.60 .40

.40] [ 60] .60 ;~

= [90] 50 ,

we conclude that there are 90 pounds of bluegrass seed and 50 pounds of rye seed in the blend. Next suppose that we have a seed manual with a table that gives us the germination rates of bluegrass and rye seeds under both wet and dry conditions. The table, given in the form of a matrix A, is as foll ows: bluegraS!.

rye

.70] .40

A -- [.80 .60

w et

dry

The ( 1, I)-entry. which is .80, signifies that 80% of the bluegrass seed germ inates under wet conditions, whi le the ( 1,2)-entry, which is .70, signifies that 70% of the rye seed germi nates under wet conditions. Suppose also that we have a mixture of Yt pounds of bluegrass seed and Y2 pounds of rye seed. Then .80y, + .70y2 is the total weight (in pounds) of the seed that germinates under wet conditions. S imilarly, .60y 1 + .40)'2 is the total weight of the seed that germinates under dry conditions. Notice that these two expressions are the entries of the matrix-vector product Av, where v

= [~~l

Let's combine this with our previous calculation. Since Bv is the vector whose components are the amounts of bluegrass and rye seed in a blend, the components of the matrix- vector product lbs . o f >eed ~emunated

A(Bv)

= [:~~

.70] [90] .40 50

=

- []~~]

wet dry

are the amounts of seed that can be expected to germinate under each of the two types of weather condit ions. T hus 107 pounds of seed can be expected to germinate under wet conditions, and 74 pounds under dry conditions.

ln the preceding example. a matrix-vector product is mu ltiplied on the left by another matrix. An exam ination of this process leads to an extended definition of matrix multipl ication. Let A be an 111 x 11 matrix and B be an 11 x p matrix. Then for any p x I vector v. the product Bv is an n x I vector, and hence the new product A(Bv) is an 111 x I vector. This raises the following question: Is there is an m x p matrix C such that A(B v) C v for every p x I vector v? By the definition of a matrix-vector product and Theorem 1.3, we have

=

A(Bv)

= A(v, b , + v2b2 + ·· · + v1, bp) = A(v, b , )


+ A(v2b 2) + · ·· + A(vpbp)

Page 9 of 10


2.1 Matrix Multiplication

= v,Ab ,

97

+ vzAbz + · ·· + vpAb,

= [Abt Ab2 . . . Abp]v. Let C be them x p matrix [Ab , Ab2 . . . Abp)-that is. the matrix whose j th column is Cj = Abj. Then A(Bv) = Cv for all v in 1?1'. Funhennore, by Theorem 1.3(e), C is the only matrix with this propetty. This conven ient method of combining the matrices A and 8 leads to the following defini tion. Definition Let A be an m x 11 matrix and 8 be an 11 x p matrix. We define the (matrix) p roduct AB to be the 111 x p matrix whose j th column is Abj. That is. C

= [A b ,

Ab 2

...

Abp].

In particular. if A is an m x 11 matrix and 8 is an n x I column vector, then the matrix product AB is defi ned and is the same as the matrix-vector product defined in Sect ion I.2. In view of this defi nition and the preceding di scussion, we have an associative law for the product of two matrices and a vector. (See Figure 2. 1.)

For any m x

11

matrix A, any

11

x p matrix 8 , and any p x I vector v, (A B )v

= A(Bv) .

1?."

v ••~--------------------------------------------~·•

multiply

(A8),. = A(B' ')

byAB

Figure 2.1 The associative law of multiplication

Later in this section, we extend this associative law to the product of any three matrices of compatible sizes. (See Theorem 2 . 1(b).) Notice that when the sizes of A and 8 arc written side by side in the same order as the product, that is, (m x 11)(11 x p), the inner dimensions must be equal, and the outer dimensions give the size of the product AB. Symbolically, (m X 11 )(11 X p )


= (111 X p) .

Suppose that A is a 2 x 4 matrix and 8 is a 2 x 3 matrix. (a) Is the product BAT defined? If so, what is its s ize? (b) Is the product AT8 defined? Lf so, what is its size?


Page 10 of 10


98 CHAPTER 2 Matrices and Linear Transformations

IJi£1,!.)§4

Let

= [ - 31

8

and

I]

2 .

Notice that A has 2 columns and B has 2 rows. Then AB is the 3 x 2 matrix with first and second column

respectively. Thus

AB


For

A=[~

= [Ab t Ab2l = [

4

I 17

~]- compute AB .

0 -3

- 1

~ ~l -

13

-2

AB _ [ -80 - .60

.70] [ 80 .40 .20

.60 .40

.40] _ [.78 .60 - .56

.76 .52

.74] .48 ,

and hence

(AB)w =

[:~~

.76 .52

.74] .48

[~] = 30

[107]. 74

Th is is the same result we obtained in Example I , where we computed A(B w).

Example4

Recall the rotation matrix Ao described in Section 1.2. Let v be a nonzero vector, and Jet a and {3 be angles. Then the expression Ap(Ac:r v) = (ApA.. )v is the result of rotating v by a followed by {3. From Figure 2.2, we see that this is the same as rotating v by {3 followed by a. which is also the same as rotating v by the angle a+ {3. Thus

Since this equation is va lid for every vector v in R 2 , we may apply Theorem 1.3(e)

- --1'"""'=------------"x

to obtain the result

Figure 2.2 Rotating a vector v in

n 2 twice


Page 1 of 10



99

=

Although ApAa AaA_s is true for rotation matrices, as shown in Example 4, the matrix products AB and BA are seldom equal.

Matrix Multiplication Is Not Commutative For arbitmry matrices A and B , AB need not equal BA.

In fact, if A is an m x 11 matrix and B is a n x p matrix, then BA is undefi ned unless p m . If p m , then AB is an m x m matrix and BA is ann x n matrix. Thus. even if both products are defined, AB and BA need not be of the same size. But 11 , AB mi ght not equal BA , as the next example shows. even if m

=

=

=

i#if! ..!.!tjW

Let

and Then

and hence AB :j; BA . (See Figure 2.3.) muhiply by A

.,

multiply byB

"2 = (AB)e 1

e1 = (8A) e 1

Figure 2.3 Matrix multiplica tion is not commutative.

Our defin iti on of the matrix product tells how to compute a product AB by finding its columns, as in Example 2 . However, there is also a method for computing an individual entry of the product, withou t calculating the enti re column that contains it. This is often called the row-column rule and is use fu l when we need to find only some specific entries of the product. Observe that the (i .j)-entry of AB is the i th componenr of its j th column, Abj. This entry equals

[an a;2 . ..

a;nl[:~l

= ailbtj+a;2b2j+ ·· · +a;n bnj ·

bnJ

which is the product of row i of A and column j of 8 . We can describe this formul a by means of the following diagram:


Page 2 of 10



column } o f 8

!

I

n )w i of A

Row-Column Rule for the (i, Jl-Entry of a Matrix Product To compute the (i ,))-entry of the matrix product AB. locate the ith row of A and the jth column of 8 as in the preceding diagram. Moving across the ith row of A and down the j th column of B, multiply each entry of the row by the corresponding entry of the column. Then sum these products to obtain the (i.j)-entry of A8. ln symbols. the (i.j)-entry of AB is

To illustrate this procedure, we compute the (2. I) -entry of A8 in Example 2 . In this case, when we multiply each entry of the second row of A by the co1Tesponding entry of the first column of 8 and sum the results, we get [3 4] [

-~] = (3)(- 1) + (4)(3) = 9,

which is the (2, I )·entry of A8 . Identity matrices, introduced in Section 1.2 , leave matrices unchanged under matrix multiplication. (See Theorem 2. 1(e) and Exercise 56.) Suppose, for example, that A =

[4I 25 3]6 . Then

and

A/3

= [4I

25 36]

[I OOJ

[I 2 3] =A.

~ ~ ~ = 4 5 6

Also, the product of a matrix and a zero matrix is a zero matrix because each column of the product is a zero vector. (Sec Theorem 1.3(1).) The next theorem summarizes various properties of matrix multiplication and illustrates the interplay between matrix multiplication and the other matrix operations.

THEOREM 2.1 Let A and 8 be k x m matrices, C be an m x 11 matrix, and P and Q be n x p matrices. Then the following statements are tn1e:


Page 3 of 10



(a) (b) (c) (d) (e) (f)

101

s(AC) = (sA)C = A(sC) for any scalars. A(CP) = (AC)P.

(associative law of matrix multiplication) (A + 8)C AC + 8C. (right distributive law) C(P + Q) CP + CQ. (left distribu tive law) /tA=A=A I, . The product of any matrix and a zero matrix is a zero matrix. (g) (AC)T = CTAT.

= =

We prove (b). (c). and (g). The rest arc left as exercises. (b) First observe that both A(CP) and (AC)P are k x p matrices. Let ui denote column j of CP. Since Uj = C pj. column j of A(CP) is Aui = A(C pj). Furthermore, col umn j of (AC)P is (AC)pj A(C pj) by the boxed result on page 97. It follows that the corresponding columns of A(CP) and (AC)P are equa l. Therefore A(CP) (AC)P. (c) Both (A+ 8 )C and AC + 8C arc k x 11 matrices. so we compare the corresponding columns of each matrix. For any j, Aci and 8 Cj are the j th columns of AC and 8C . respectively. But thcjth column of (A+ 8 )C is PROOF

=

=

by Theorem 1.3(c), and this is the jth column of AC + 8C. This establishes (c). (g) Both (AC)T and cT AT are 11 x k matrices, so we compare the corresponding entries of each matrix. The (i ,j)-entry of (ACf is the (i, i )-entry of AC, which is

(ajl Oj2

e, · eli]

~~.

... Ojm]

=

Ojleli

+

Oj2e2i

+ ···+

Ojme,; .

[ Cm,

Also, the (i .j)-entry of C rAT is the product of row i of cT and column j of AT, wh ich is

(eli

eu

...

e,;(

a·, ail] =et;Ojt [ I-

+euaj2+· ··+e,;ajm·

Op,

Since the two displayed expressions arc equal, the (i,j)-cntry of (ACl is equal to the (i ,j)-entry of CT A r_ This establishes (g). • The associative law of matrix multiplication, Theorem 2.1 (b). allows us to omit parentheses when writing products of matrices. For this reason, we usually write ABC for a product of the matrices A, 8. and C. If A is an 11 x 11 matrix, we can form products of A with itself any number of times. As with real numbers. we use the ex ponential notation Ak to denote the product of A with itself k times. By convention, A 1 =A and A0 = !,.


Page 4 of 10



. matnx . A Recall that we can use the stochasllc

[.85 .03] .tn Example 3 o f .l .

5 97 Section 1.2 to study population shi fts between the c ity and suburbs. In that example, if the components of p are the current populations of the city and suburbs. then the components of A p are the populations of the city and suburbs for the next year. Extending A(Ap ) is the the argument of that example to subsequent years, we see that A 2p vector whose components are the populations of the ci ty and suburbs after two years. In general. for any positive integer m. A111 p is the vector whose components are the

=

populations of the city and suburbs after m years. For example, if p

= [~~~leas in popt~ations

Example 3 of Section 1.2), then the vector representing the city and suburb in ten years is given by AIO

~ [ 241.2]

p ~ 958.8 .

( Here the entries are rounded.) A sociologist may need to determine the long-range trend that occurs in populations as a result of these an nua l shifts. In terms of matrix multiplication , thi s problem reduces to the study of the vectors A"' p as m increases. A deeper understanding of matrices. which we will acqui re in Chapter 5, provides us with tools to understand the long-term behavior of this population bener.

If A and 8 are matrices with the same number of rows. then we denote by [A 8] the matrix whose columns are the columns of A followed by the columns of 8, in order. We cal l [A 8 I an augmented m a t r ix. For example, if

A= / 2 = [~ ~] then

LA

8 = [

and

0

2

- I

3

:].

8] is the 2 x 5 matrix

LA Bl = [~

0

2

0

- I

3

:].

Occasionally. it is usefu l to include a vettical line when displayi ng an augmented matrix LA 8] in order to visually separate the colum ns of A from those of 8 . 1lms, for these matrices A and B. we may write [A B] = [ I 0

0 I

I - 21

0 3

I ] I .

Because of the way that matrix multiplication is defined, it is easy to see that if P is an m x 11 matrix and A and 8 are matrices with 11 rows, then P LA 8 J LPA P8].

=

liifluj.)Q

Let

A=h=[b ~].

2 0 8 = [- 1 3

and

P =

[i -fl

Use the equation P LA 8 j = (PA P8 j to compute the product P LA 8 ].


Page 5 of 10



Solution

103

Observe that PA = Ph = P and

6

-3 3 Therefore

PIA Bl=IPA PBI= [

Practice Problem 3 ~

~

21

- 1 1

05

- 1

-~ 3]·

Let

] [I 3 -1 4·

A=25

8

Find the fotuth column of A[B

[ 2-2]~·

=_~

and

C =

[

3 0 -1] I

5

-6 0

2

2

Cl without computing the entire matrix.

.

~

SPECIAL MATRICES We briefly examine matrices with spec ial properties that will be of interest later in this text. We beg in with diagonal matrices, which are important because of their simplicity. The (i .j)·entry of a matrix A is called a diagonal entry if i = j. The diagonal entries form the diagonal of A. A square matrix A is called a diagonal m a trix if all its nondiagonal entries arc zeros. For example. identity matrices and square zero matrices are diagonal matrices. If A and B are 11 x 11 diagonal matrices. then AB is also an 11 x 11 diagonal matrix. Moreover, the diagonal entries of AB are the products of the corresponding diagonal entries of A and 8. (See Exercise 60.) For example. suppose that

A=

1 0 0] [00 02 03

and

B=

3 0 0] [00 - 01 02 .

Then

AB

3 0 0] [0 0 6

= 0 -2 0 .

The relationship between diagonal and nondiagonal square matrices will be studied in depth in Chapter 5, where we will see that in many c ircumstances, an ordinary square matrix can be replaced by a diagonal matrix. thus simplifying theoretical arguments as well as computations. Observe that any diagonal matrix is equal to its transpose. In general, a (square) matrix A is called symmetric if AT= A. For example, diagonal matr ices are symmetric. As another example, let


Page 6 of 10


104 CHAPTER 2 Matrices and Li near Transformations

Then

2

4]T

3

- I

- I

5

-n

[~ j

= A.

So A is symmetric. ln general. a square matrix A is symmetric if and only if a,1 for all i andj.

=a"

EXERCISES In Exuci a 2 x 3 matrix. 4. G ive an example of malrices A and 8 such that BA is defined, but AB is not.

In Exercises 5-20. 11.<1! the jollowi11g malrices to compute each expressio11, or gh•e a l't!OSOII why the expl't!ssion is 1101 deji11t!d:

[~ -~]

X=

8 =

[i]

5. Cy

y= [

j]

6. Bx

II. AB

12. AC

14. BA

15. c8r

16. CB

17. A3

18. A 2

19. C 2

us~ til~

~2.

21. Verify that I2C- Ch =C.

= A(BC).

23. Verify that (A8) 7

= BIA'.

24. Verify thtll z(AC)

=

11

x p matrix, then

= BA.

For any matrices A and 8 for which the product AB is defined. the (i .j)-cntry of AB equals the ;um of the prod· uct' of corrc,ponding cntrie' from the llh column of A and thejth row of B.

43. If A. 8. and C arc matnccs for which the product A(BC) is defined, then A(BC) (1\8 )C.

=

44. If A and 8 are 111 x 11 then (A+ B)C A8

=

(7A )C.

111 Exercise.< 25 -32. 11se til~ following matricl!s to comp111e each req11esred ewry or column of the motrix prod11c1, wirllo11t computing rhe entire matri~:

~ --~2

=

36. If A is a square matrix. chen Az is defined.

41. For any matrices A and 8 for "hich the product AB is defined, the (i .j)-entry of AB equal> a,1 b,1 •

llllllrices A. B. C. and z from

22. Verify that (AB)C

35. For any matrices A and B. if the pmducts AB und BA Me both defined, then A8 BA.

40. For any matrices A and 8 for "hich the product A8 is defined. the jth column of AB cquah the matrix- vector product of A and the jth column of 8 .

8. By

13. BC

In Exerc:is~s 21 U. E.rucift'\ 5 -20.

matrice> is defined.

39. There exbtt10n1ero mat rice' A and 8 for which AB

10. Azr

9. ACx

11

34. For any matrices A and 8. if the product AB is defined. then the product 8A is also defined .

38. If A is an m x 11 matrix and 8 is '"' (ABl =AT8r.

z = [7 -I]

7. n

/11 E.urc:ises 33 -50, dt'lermint' whether the stmeme/1/s ore trrte or false.

37. If A and 8 are matrices. then both A8 :md BA arc defined if :md only if A and 8 arc :.quare matrices.

80 4I]

~]

[i

~

33. The product of two m x

2. A is a 2 x 4 matrix, and 8 is a 4 x 6 matrix .

A=

~

!]. [-! ~] ,

m;~trices

and C is an

11

x p matrix,

+ BC .

45. If A and 8 ;~re 11 x 11 matrices. then the diagon"l entries of the product matrix A8 are attblt.ll22bzz .... ,ll,,b., . 46. If the product AB is defined :mel either A or 8 is a zero matrix, then A8 is a zero matrix.

-I] -2

47. If the product AB b defined and AB is u zero matrix , then either A or 8 is a zero matrix.

25. the (3, 2)-entry of AB

26. the (2. I )-entry of BC

48. If Aa and A~ are both 2 x 2 rotation matrices. then A 0 Ap is a 2 x 2 rotation matrix.

27. the (2. 3)-entry of CA

28. the (I. I) entry of C8

49. The product of two diagonal matrices is a diagonal matrix.

29. column 2 of AB

30. column 3 of BC

50. In a symmetric 11 x 11 matrix. the (i ,})- :md (j. i)-entries are equal for all i I. 2... . 11 and j = I. 2..... 11 .

31. column I of CA

32. column 2 of CB

A = [

-3

8

=

0

3

-2


C

= [;

I

3

=

Page 7 of 10


1OS


(b) Suppose that 111 = 100 pupils buy a hot lunch and 112 = 200 pupils bring a bag lunch on the fi~t

51. L..ct

da].

Fro C'lt) Cny To

Suhurth

['85 . 15

of school. Compute A["']· A2 ["•]. and A

S 1l>urt"

1·1

people live (a) Find a 2 x 2 matrix B such that if in the city and l'z people live 111 the suburbs, then

B[

= [

11 '].

where

112,

11 1

people live in
houses and u 2 people live in multiple-unit houses.

52. Of those vehicle owners who live in the city, 60% drive cars, 30% drive vans, and 10% drive recreational vehicles. Of those vehicle owners who live in the suburbs. 30% drive cars. 50% drive vans. and 20% drive recreational vehicles. Of all ''chicles (in the city and ;,uburbs), 60% of the cars. 40% of the vans. and 50% of the recreational ''chicles are dark in color. (a) Find a matrix 8 such that 1f I'J vehicle owners live in the city and 1· 2 vehicle owners live in the suburbs,

[~:]=[::~]-where u1 people drive cars,

112

Ill

people drive recreational

113

(b) Find a matrix A :.uch that A

[:~] = [::::].where u

1,

and 113 are as in (a), "'' is the number of people who drive dark vehicle;,. and 11'2 is the number of people who drive light (not dark) vehicles.

[1'1 ]

(c) Find a matrix C such that w -. = C , ... . where and •·2 are a~ in (a) and w 1 an 11· 2 are as m (b).

1'1

53. In a certain elementary school. it has been found that of those pupils who buy a hot lunch on a particular school day. 30% buy a hot lunch and 70% bring a bag lunch on the next ;,chool d:~y. Furthermore, o f those pupils who bring a bag lu nch on a particular school day. 40% buy a hot lunch and 60% bring a bag lu nch on the next school day. (a) Find a matrix A such that if 111 pupils buy a hot lunch and u 2 pupils bring a bag lunch on a particular day, then A["'] = [''• ]. where ••1 pupils buy a hot lunch 112

''2

and ··~ pupils bring a bag lunch on 1he nex1 school day.


pute A IOO ["t] ll2

= [IV']. Explain the significance U!'2

thi' rc•ult. Now compute A result with

["''] • 11 2

•

[~~'•]. U'l

of

and compare thi<

Explain.

5-I. Prove (a) of Theorem 2.1.

55. Prove (d) of Theorem 2.1. 56. Prove (e) of Theorem 2.1.

n 2 aboul the

~-ax is; !hal is.

Compute BA. and describe geometrically how a •ector ,. is affected by multiplication by BA. 59. A square matrix A is called lower triangula r if the (i .J )entry of 1\ is zero whenever i < j. Prove that if A and B arc both 11 x 11 lower triangular matrices. then AB is also a lower triangular matrix. 60. L..ct A be an

11

x

11

matrix.

(a) Prove thai A is a diagonal matrix if and only if its jlh column equals ai1 e1 .

112.

[II'J

(c) To do this problem, you will need a calcu lator with matrix capabilities or access to computer soflware s uch as MATLAB. Using the notation o f (b), com-

58. L..ct A = A uul'' , and let 8 be the matrix that reflects

1'2

people drive vans. and vehicles.

u2

57. Prove (I) of Theorem 2.1.

(b) Explain the signi ficance of 8A [ "'].

then B

["' .

Explain the significance of each result.

be the stochastic matrix used in Example 6 to predict popuhltion movement between the city anu its suburbs. Suppose that 70% of city residents live in single-unit houses (as opposed to multiple-unit or apartment housing) and that 95% of suburb re;,idents live in single-unit hou•es.

1 '' ] l'J

uz

"2

.03] = A .97

1

(b)

u,c

(a} to prove that if A and B are 11 x 11 diagon:•l matrices. then AB is a diagonal matrix who;e jth column is aJJbJJ e1 •

61. A Mjuarc matrix A is called upper triangular if the (i .j)entry of A is zero whenever i > j. Prove that if A and B are both 11 x 11 upper triangular matrices. then AB is also an upper triangular matrix.

62. L..ct A-

[j ~: -! -i]. -5

3 -4

Find a nonLero 4 x 2

7

mtllrix IJ wil h mnk 2 such that AB = 0. 63. Find an example of 11 x AB = 0, bul BA '# 0 .

11

matrices A and B such that

64. L..ct A and B be 11 x 11 matrices. Prove or di>prove that the ranks of AB and 8A are equal. 65. Recall the definition of the troce of a matrix. given in Exerci>e 82 of Section 1.1. Prove that if A i> an m x 11 matrix and 8 is an 11 x m matrix. then trace(AB) = tmce(BA).

Page 8 of 10


106 CHAPTER 2 Matrices and Li near Transformations

66. Let I < r. s < 11 be integers. and let £ be the 11 x

(c) Make a conjecture about an expre»ionthat is equal to (A+ 8)2 for arbitrary 11 x 11 m:ttrices A and B. Justify your conjecture.

67. Prove that if A i< a~ x m m:llrix. 8 i< an m x 11 matrix. and C i< an 11 x p malrix. then (ABC )T c TsJAT.

(d) Make a conjecture about the relationship that must hold between AB and BA for the equation in (b ) to hold in general.

11 malrix wilh I a< the (r ..r )-entry and Os elsewhere. Let 8 be any 11 x 11 matrix . Describe £8 in tenm of the entries of 8.

=

68. (a) Let A and 8 be S),nrnetric matrices of the same size. Pro\'e that AB is >ymmetnc if and only AB BA. (b) Find symmetric 2 x 2 matrices A and 8 such that

=

AB

#

BA .

(e) Pro'e your conjecture in (d). 71. Let A be the stochastic matrix used in the population appli-

cation in Example 6.

/11 £urr:ts~s

69 72. uu t!llhu a calculator with matrix capabilitit'S or compmu .roft11'lt~ such as MATLAB to soll•e each pmblmL

(a) Verify that

(c) Determine the populations of the ctty and suburbs after 50 years.

= 1r / 2, compute A~ by hand. For 0 = 1r/3, compute AJ .

(a) For 0

(d) Make a conjecture ahout the eventual populations of the city and suburbs.

=

(c) For 0 1r /8. compute A~. (d) Usc the previous re>uhs to make :1 conjecture about where 0 1t /k . (e) Draw n sketch 10 support your conjecture in (d).

=

At.

72. Let A and 8 be 4 x 4 random matrice>. (a) Com1Jute A8 and it• rank hy finding the reduced row echelon form of A8.

70. Let A. 8. and C be 4 x 4 mndom matricc.o;. (a) Ill ustrate the di•tributive law A(B +C)= AB

b as given in the example.

(b) Determine the population• of the city and suburbs after 20 years.

69. Let A11 be the 0 -rot:llion matrix . (b)

A 10p

(b) Compute 8A and its rank by fi nding the reduced row echelon form of BA .

+ AC.

(b) Check the validily of the equation

(c) Compare your answers Exercise 64.

(A + 8) 2 = A 2 + 2AB + 8 2.

with

your solution

to


---------------------and its third column is

I. (a) Since 8 is a 2 x 3 malrix and A T is a 4 x 2 malrix. the product 8A 7 i' not defined. (b) Since AT IS a x 2 matnx and 8 ts a 2 x 3 malrix. the product A 7 8 is defined. hs >ize is 4 x 3.

-1

- I 4

2. The matrix AB is a 2 x 3 matnx. hs fin.t column is

[~

- I 4

][-1]

3 - 2

its second column

[~

- I 4

0 - -2+0 + 9 7 3 -[-1+0-6] - [-7].

Hence AB

3.

ll>

2 3] [-~] = [-12 +- 164+- 6]4 = [-6] 22 .

-2

= [-~

6

-1-1

Since the fourth column of 18 C I il> e2. the fourth column of AlB CJ is

3] [-~]I -- [00 +- 123+- 23] -- [- 14. 6]

-2

12.2•1APPLICATIONS OF MATRIX MULTIPLICATION In this section. we present four applications of matrix multiplication.

THE LESLIE MATRIX AND POPULATION CHANGE The popu lation of a colony of animals depends on the birth and mortality rates for the various age groups of the colony. For example. suppose that Lhe members of • This section can be omined without loss of continuity.


Page 9 of 10


2.2 Applications of Matrix Multiplication

107

a colony of mammals have a life span of less than 3 years. To study the bitth rates of the colony, we divide the females into three age groups: those with ages less than I , those with ages between I and 2, and those of age 2. From the mortality rates of the colony. we know that 40% of newborn females survive to age I and that 50% of females of age I surv ive to age 2. We need to observe only the rates at which females in each age group give birth to female offspring since there is usually a known relationship between the number of male and female offspring in the colony. Suppose that the females under I year of age do not give birth ; those with ages between I and 2 have, on average, two female offspring; and those of age 2 have. on average. one female offspring. Let Xt. x 2 , and X 3 be the numbers of females in the first, second, and th ird age groups, respectively, at the present time, and let Yl· Y2 · and Y3 be the numbers of females in the corresponding groups for the next year. The changes from this year to next year are depicted in Table 2.1.

Table 2.1

The vec.tor x =

[~~]

is the population distribution for the female population

X3

of the colony in the present ye
= [f~].

Note that Yt . the number of females under age I in next year's popu lation, is simply equal to the number of female offspring born during the current year. Since there are currently x2 females of age 1-2, each of which has, on average. 2 female offspring, and X3 females of age 2-3, each of which has, on average. I female offspring, we have the following f01mula for )'I: )' I

=

2x2

+ X3

The number Y2 is the total number of females in the second age group for next year. Because these females are in the first age group this year. and because only 40% of them will survive to the next year, we have that Y2 = 0.4xt. Similarly, )'3 = O.Sx2. Collecting these three equations. we have

Yt Y2 )'3

+

2.0x2

l.Ox3

0.4Xt O.Sx2 .

These three equations can be represented by the single matrix equation y = Ax, where x and y arc the population distributions as previously defined and A is the 3 x 3 matrix

0.0 2.0

1.0]

A= 0.4 0.0 0.0 . [ 0.0 0 .5


0.0

Page 10 of 10



For example, suppose that x

= [ 1000] 1000 ; that is , there are cu rrently

1000 females in

1000

each age group. Then

y =A x =

0.0 2.0 1.0] [ 1000] [ 3000] 0.4 0.0 0.0 1000 = 400 . [ 0.0 0.5 0.0 1000 500

So one year later there are 3000 females under I year of age, 400 females who are between I and 2 years old. and 500 females who are 2 years old. For each positive integer k, let Pt denote the population distribution k years after a given initial population distribution Po· In the preceding example.

PO= X=[::]

and

1000

Then, for any positive integer k , we have that PJ. = APt- I· Thus

In this way, we may predict population trends over the long tenn. For example, to predict the population distribution after 10 years. we compute PIO = A10 po. Thus

10

Pt o =A po =

1987] 851 , [ 387

where each entry is rounded off to the nearest whole number. If we continue this process in increments of 10 years, we find that (rounding to whole numbers)

2043]

Pzo= 819

and

[ 408

P3o = P40 =

[

2045] ~~ .

It appears that the population stabilizes after 30 years. In fact, for the vector

z=

2045] 818 [ 409

.

=

we have that Az z precisely. Under this circumstance, the population distribution z is stable; that is, it does not change from year to year. In general. whether or not the distribution of an animal population stabil izes for a colony depends on the survival and birth rates of the age groups. (See, for example, Exercises 12- 15.) Exercise 10 gives an example of a population for which no nonzero stable popu lation distribution exists. We can generalize this situation to an arbitrary colony of animals. Suppose that we divide the females of the colony into n age groups, where x; is the number of members in the ith group. The duration of time in an individual age group need not


Page 1 of 10


2.2 Applicat ions of Matrix Multiplication

be a year, but the various durations should be equal. Let x

= [;;]

109

be the population

x, distribution o f the females of the colony, p; be the portion of females in the i th group who survive to the (i + l )st group. and b; be the average numbe r of female offspring

of a member of the i th age group. If y =

[~~]

is the population for th e next time

y, period , then

Yt Yz

btXJ

+

b2X2

+

+

bnXn

PtXJ

pzxz

.\'3

y,

Pn-IXTl-1·

Therefore, for

[b, Pt A~

bz 0 pz

:

0

...

we have

P11 - t

n

y =A x. The matrix A is called the Leslie matrix for the population. The name is due to P. H . Leslie, who introduced this matrix in the 1940s. So if xo is the in itial population distribution. then the distribution after k time intervals is


The life span of a cen ain species of mammal is at most 2 years, but only 25% of the females of this species survive to age I. Suppose that, on average. the females under l year of age give birth to 0.5 females, and the females between l and 2 years of age give birth to 2 females. (a) Write the Leslie matri x for the popul ation. (b) Suppose that thi s year there is a population of 200 females under I year of age and 200 females between l and 2 years of age. Find the population dis tribution for next year and for 2 years from now. (c) Suppose this year's population di stri bution of females is given by the vector 400] . dtstn . .butJons . ?. LOO . What can you say about all f uture popul ation [ (d) Suppose that the total po pulation of females this year is 600. What should be the number of females in each age group so that the population distribution remain s unchanged from year to year? ~


Page 2 of 10


110 CHAPTER 2 Matrices and Linear Transformations P,

,.,

03

:,

0.7

P,

0.5

0.3 p2

0.2

Yz

z2

I

Ps

0.8

0.2 0.5

0.6 Y3

I

:3

0.4

l\.·'2

Ps

z 3

""'J

IV4

Figure 2.4 Traffic flow along one-way streets

ANALYSISOF TRAFFIC FLOW Figure 2.4 represents the flow of traffic through a network of one-way streets, with arrows indicating the direction of traffic flow. The number on any street beyond an intersection is the portion of the traffic entering the street from that intersection. For example. 30% of the traffic leaving intersection P, goes to P4. and the other 70% goes to P2 . Notice that all the traffic leaving Ps goes to Ps. Suppose that on a particular day. cars enter the network from the left of P1, and x 2 cars enter from the left of P 3. Let w1, 1112, 1113 , and 1114 represent the number of cars leaving the network along the exits to the right. We wish to determine the values of the w; •s. At first glance, this problem seems overwhe lming since there are so many routes for the traffic. However, if we decompose the problem into several simpler ones, we can first solve the simpler ones ind ividually and then combine their solutions to obtain the values of the w;'s. We begin with only the portion of the network involving intersections P 1, P2. and P 3 . Let )'J, Y2· and YJ each be the number of cars that ex it along each of the three eastward routes, respectively. To find y 1, notice that 30% of all cars entering P 1 continue on to P4 . Therefore y 1 0.3Q.r 1. Also, 0.7x1 of the cars tum ri ght at P,. and of these, 20% tum left at P2. Because these are the on ly cars to do so, it follows that Y2 = (0.2)(0.7)x1 = 0.14x 1• Furthermore, since 80% of the cars entering P2 continue on to P 3, the number of such cars is (0.8)(0. 7)x, = 0.56x,. Finally, aU the cars entering P 3 from the left use the street between P 3 and P6 , so )'3 0.56x, + x2. Summarizing, we have

x,

=

=

)' J

Y2 )'3

0.30x, 0. 14x, 0.56x,

+

x2 .

We can express this system of equations by the single matrix equation y =Ax, where 0.30 OJ A= 0.14 0 and y [ 0.56 I Y3

= [~~]

Now cons ider the next set of intersect ions P4 , Ps. and P6. If we let Zl , Z2, and ZJ represent the numbers of cars that exit from the right of P4, Ps. and P6. respectively,


Page 3 of 10



11 1

then by a similar analysis, we have Zl Z2

0.5y) 0.5yl

+

Yz

Z3

+

0.6)'3 0.4)'3.

or z = By, where

z=

[~~J

and

8

=

0

[ 0.5 0.~

I

o.~J.

0 0.4

Finally, if we set

W=

["'] IV2

IVJ

and

c=

W4

[!

.~]

0.30 0.20 0.35 0.15 0.3

then by a similar argument, we have w = Cz. It follows that

w = Cz

= C(By) = (CB)Ax = (CBA)x.

Let M = CBA. Then I 0.30 0OJ [ 0 .5 0 OJ [0.30 OJ 0 0.20 0.5 I 0.6 0.14 0 - 0 0.35 0.7 [ 0 0 0.4 0.56 I 0 0.15 0.3

M _

=

12 [ 0.1252 0.3378 0. 18] 0.3759 0.49 . 0.161 1 0.21

For example, if 1000 cars enter the traffic pattern at P 1 and 2000 enter at P 3, then, for x =

[~:].we have 0.3378 0.18] 0.1252 0.12 w = M x = 0.3759 0.49 [ 0.1611 0.21

1000

[zooo]

=

[ 697.8] 365.2 1355.9 · 581.1

Naturally, the actual number of cars traveling on any path is a whole number, unlike the entries of w . Since these calculations are based on percentages, we cannot expect the answers to be exact. For example, approximately 698 cars exit the traffic pattern at P1, and 365 cars exit the pattem at Ps. We can apply the same analysis if the quantities studied represent rates of traffic How-for example, the number of cars per hour-rather than the total numbe r of cars. Finally, we can apply this kind of analysis to other contexts, such as the Aow of a Auid through a system of pipes or the movement of money in an economy. For other examples. see the exercises. Practice Problem 2

~

A midwestern supermarket chain imports soy sauce from Japan and South Korea. Of the soy sauce from Japan, 50% is shipped to Seattle and the rest is shipped to San

Francisco. Of the soy sauce from South Korea, 60% is shipped to San Francisco and the rest is shipped to Los Angeles. All of the soy sauce shipped to Seattle is sent to


Page 4 of 10


112


Chicago; 30% of the soy sauce shipped to San Francisco is sent to Chicago and 70% to St. Louis; and all of the soy sauce sh ipped to Los Angeles is sent to St. Louis. Suppose that soy sauce is sh ipped from Japan and South Korea at the rates of 10,000 and 5,000 barrels a year. respectively. Find the number of barrels of soy sauce that are sent to Chicago and to St. Lou is each year.
(0, 1)-MATRICES Matrices can be used to study certain relationships between objects. For example, suppose that there are five countries. each of which maintains diplomatic relations with some of the others. To organize these relationships, we use a 5 x 5 matrix A defined as follows. For I ~ i ~ 5, we let a;; = 0, and for i i= j. a;·

= {I

~

0

if eounlly i maintains diplomatic relations with counlly j otherwise.

Note that all the entries of A are zeros and ones. Matrices whose only entries are zeros and ones are called (0, I)-matrices, and they are worthy of study in their own right. For purposes of illustration, suppose that

In this case. A =AT; that is, A is symmetric. The symmetry occurs here because the underlying relationship is symmetric. (That is, if country i maintains diplomatic relations with country j. then also country j maintains diplomatic relations with country i .) Such symmetry is true of many relationships of interest. Figure 2.5 gives us a visual guide to the relationship, where country i is shown to maintain diplomatic relations with country j if the dots representing the two countries are joined by a line segment. (The diagram in Figure 2.5 is called an undirected graph, and the relationships defined in Exercises 21 and 26 lead to directed graphs.) Let us consider the significance of an entry of the matrix 8 A 2 ; for example,

=

country I

coumry 2

Figure 2.5 Diplomatic relations among countries


Page 5 of 10



113

A typical term on the right-side of the equat ion has the form 02J:Ok3 · This term is I if and on ly if both factors are ! - that is, if and on ly if country 2 maintains diplomatic relations wi th country k and country k maintains diplomatic relations wi th country 3. Thus b23 gives the number of countries that link country 2 and country 3. To see all of these entries, we compute

Since b23 = I , there is exactly one country that links countries 2 and 3. A careful examination of the entries o f A reveals that azs a53 I, and hence it is country 5 that serves as the link. (Other deductions arc left for the exercises.) We can visualize the (i ,j)-entry of A 2 by counting the number of ways to go from country i to country j in Figure 2.5 that use two line segments. By looking at other powers of A, additional information may be obtained. For example. it can be shown that if A is an n x 11 (0. I)-matrix and the (i .j)-entry of A+ A2 + · · · + A"- 1 is nonzero, then there is a sequence of countries beginning with country i. ending with country j. and such that every pair of consecutive countries in the sequence maintains diplomatic relati ons. By means of such a sequence, coun tries i and j can communicate by passing a message only between countries that maintain diplomatic relations. Conversely, if the (i,j) -entry of A + A2 + · · · + A"- 1 is zero, then such communication between countries i and j is impossible.

=

Example 1

=

Consider a set of three countries, such that country 3 maintains d iplomatic relations with both countries I and 2, and countries I and 2 do not maintain diplomatic relations with each other. These relationships can be described by the 3 x 3 (0, I)matrix

In this case, we have

and so countries I and 2 can communicate. even though they do not have diplomatic relations. Here the sequence li nking them consists of countries I , 3, and 2 .

A (0, I)-matrix can also be used to resolve problems involving schedul ing. Suppose, for example, that the adm inistration of a small college w ith m students wants to plan the times for its n courses. l11e goal of such planning is to avoid scheduling popular courses at the same time. To minimize the number of time conAicts, the students are surveyed. Each student is asked wh ich courses he or she would like to take during the follow ing semester. The result s of this survey may be put in mat rix form .


Page 6 of 10



Define the m x

11

matrix A as follows: a;·

={I

~

if st\1dent i wants to take course j

0 otherwise.

In this case, the matrix product A 1A provides important information regarding the scheduling of course times. We begin with an interpretation of the entries of thi s matrix. Let 8 =AT and C =ATA= BA. T hen, for example,

=

I A typical term on the right s ide of the equation has the form OkJOk2. Now, OkJOk2 if and on ly if ak 1 = I and ak2 = I ; that is, student k wants to take cou rse I and course 2. So c12 represents the number of students who want to take both courses I and 2. In general, for i # j, C;j is the number of students w ho want to take both course i and course j. In addition, c;; represents the number of students w ho desire c lass i.

Example 2

Suppose that we have a group of 10 students and five courses. The resu lts of the survey concerni ng course preferences are as follows:

Course Number Student

1

0

2

3

0

0

0

0

0

0

0

0

0

0

5

0

0

7

0

0

8

0

0

0

0

0 0

0

0

0

6

10

S

0

4

9

4

0

0

0

-

0

Let A be the 10 x 5 matrix w ith entries from the previous table. T hen

I

3 A1il

=

[l

2 I 4

0 I

0 0 3 4

l

From this matrix, we see that there are four students who want both course 3 and course 5. All other pairs of courses are wanted by at most two students. Furthern1ore,


Page 7 of 10



115

we see that four students want course I, three students des ire course 2, and so on. Thus. the trace {see Exercise 82 of Section 1.1) of A rA equals the total demand for these five courses {counting students as o ften as the number of courses they wish to take) if the courses are offered at different times.

Notice that although A is not symmetric, the matrix A 'll is sy mmetric. Hence we may save computational effort by computing only one of the {i ,j)- and (j, i)-entries. As a final comment , it should be pointed out that many of these facts about (0, I )-matrices can be adapted to apply to nonsymmetric relationships. Practice Problem 3 .,.

Suppose that we have four ci ties with airports. We define the 4 x 4 matrix A by

a;· ~

={I

if there is a nonstop commerc ial flight from ci ty i to city j

0 if not.

{a) Prove that A is a {0, I )-matrix. {b) Interpret the {2, 3)-entry of A2 . {c) If

compute A2 . {d) How many flights are there with one " layover" from city 2 to city 3? {e) How many fl ights are there with two layovers from city I to city 2? {f) Is it possible to fly between each pair of ci ties?

AN APPLICATION TO ANTHROPOLOGY In this application, 1 we see a fascinating use of matrix operations for the study of the marriage laws of the Natchez Indians. Everyone in this tribe was a member of one of four classes: the Suns, the Nobles, the Honoreds, and the michy-miche-quipy (MMQ). There were well-defined ru les that determined class membership. The ru les depe nded exclusively on the classes of the parents and required that at least one of the parents be a me mber of the MMQ. Furthennore, the c lass of the child depended on the class of the other parent, according to Tabl e 2.2. We are interested in determining the long-range distributions of these classes that is, what the relative sizes are of these classes in future generat ions. It is clear that there will be a problem of survival if the size of the MMQ becomes too small. To simplify matters, we make the following three assumptions: l. ln every generation. each class is d ivided equally between males and females.

2. Each adult marries exactly once. 3. Each pair of parents has exactly one son and one daughter. 1

This example is taken from Samuel Goldberg, Introduction to Difference Equations (with Illustrative

Examples from Economics, Psychology, and Sociology), Dover Publications, Inc., New York. 1986, pp. 238- 241.


Page 8 of 10



Table2.2

Mother is in MMQ Father

Child

Father is in MMQ Mother

Child

Noble

Honored

Noble

Noble

Honored

MMQ

Honored

Honored

MMQ

MMQ

MMQ

MMQ

Because of assumption I, we need keep track only of the number of males in each class for every generation. To do this, we introduce the following notation: SA

= number of male Suns in the kth generation

nk

= number of male Nobles in the kth generation

Ilk = number of male Honoreds in the kth generation mk =

number of male MMQ in the kth generation

Our immediate goal is to find a relationship between the numbers of members in each class of the kth and the (k - I )st generations. Since every Sun ma le must have a Sun mother (and vice versa). we obtain the equation {I )

The fact that every Noble male must have a Sun fathe r or a Noble mother yields

In addition, every Honored male must have a Noble father or an Honored mother. Thus

(3) Finall y, assumption 3 guarantees that the total number of males (and females) remains the same for each generation. (4)

Substituting the right sides of equations ( I ) , (2), and (3) into (4), we obtain Sk-I +(sk- I + nk-ll+(nk-1 + hk- l) + mk =Sk- I + llk-1

+ hk-1 + lllk-1,

which simplifies to

(5) Equations ( I ), (:!), (3) , and (5) relate the numbers of males in different classes of the kth generation to the numbers of males of the previous generation. If we let


Page 9 of 10


2.2 Applications of Matrix Multipl ication

and

A= [

l

- I

117

0I 00 0]0 I I - 1 0

0 I

'

then we may represent all our relationships by the matrix equation

Because this equation must hold for all k. we have

=A-

To evaluate Ak. let B positive integer k .

/4. We leave it to the exercises to show that. for any

At= I +k8

k(k - I)

+ - - -8 2, 2

fork ~ 2.

(Sec Exercise 24.) Thus, carrying out the matrix mu ltip licati on. we obtain

so

Xt

no

+

kso

ho

+

kno

= mo

+

kno

(6)

~~\- 11 so

''~!''so

It is easy to see from equation (6 ) that if there arc initially no Suns or Nobles (i.e .. no= so= 0). the number of members in each c lass will remain the same from generation to generation. On the other hand. con~idcr the last entry of x~. We can conclude that. unless no =so = 0. the size of the MMQ will decrease to the point where there are not enough of them to allow the other members to marry. At this point the social order ceases to exist.

EXERCISES

~

In Exerr:i.res I 9. determine whether the statemems tnte or falte.

lll't'

I. If A b a

Lc~lic matrix and ,. is a population di,tribution. then each entry of Av must be greater than the correspond· ing e ntry of v.

2. For any population disuibution v, if tl is the Leslie matrix for the population. then as 11 grows. A" v approaches a >IJCcilic population di>tcibution. Lc~lie matrix. the (i .j)·entry equals the average num· ber of female off,pri ng of a member of the i th age group.

3. In a

4. In a l..c>lie matrix. the (i

of females

tn

+ I. i)-entry cquab the portion

the ith age group who survive to the next

age group.


5. The applicalion in I his sec1ion on traffic flow relies on the associative law of malrix multiplic:ction. 6. If A and 8 are matrices and x, y. and z are vectors such that y =Ax and z = By. then z (AB)x. 7. A (0. 1)-malrix is a matrix with Os and Is as its only e nlries. 8. A (0, I)-matrix is a square matrix with Os and Is as its only entries. 9. Every (0, I )·matrix is a symmetric malrix.

=

Exercises /0-/6 are

COIICU11ed

with

Le<1/ie

matrices.

I 0. By observing a certain colony of mice. re•earchcrs found that all animals die within 3 year>. or tho•e off,pring that are females, 601! live for at least I year. Of these. 20% reach their second birthday. The females who are under

Page 10 of 10


118 CHAPTER 2 Matrices and Li near Transformations I year of :~ge ha\e, on average. three female offspring. Those females between I and 2 years of age have, on average. two female off:.pring while they are in this age group. None of the females of age 2 give birth. (a) Construct the Lc.he matrix that describes this situa-

tion. (b) Suppose that the current population distribution for females h given b) the \ector

[I~J Find the pop-

ulation dbtribution for next year. Also. find the population dis tribution 4 years from now. (c) Show th:u there is no nonzero stable population distribution for the colony of mice. Him: Let A be the Leslie matrix. and suppose that z is a stable population distribution. Then Az = z. This is equivalent to (A - /3)7. 0 . Solve this homogeneous system of linear equations.

=

I I. S uppose that the females of a certain colony of animals arc divided into two age groups. and suppose that the Leslie matrix for this popu lation is

(e) For the value of q fowtd on (d). what happens to an initial population of 200 females of age less than I. 360 of age I. and 280 of age 2?

(f) For what value of q does (A - I 3 )x nonzero solution?

=0

have a

(g) For the value of q found in (f). find the general solution of (A - / 1 )x 0. llow doe> thts relate to the stable distributions in (d) and (er/

=

13. A certain colony of bat> has a life span of less than 3 years. Suppose that the females are divided into three age groups: those under nge I. tho>c of age I. and those of age 2. Suppose further that the proportion of newborn females that >urvi ves until age I i> q and that the proportion of one-year-old females that survives until age 2 is .5. Assume also that female> under age I do not give birth, those of age I have. on average, 2 female offspring. and those of age 2 have. on uverage. I female offspring. Suppose there are initially 300 females of age less than I. 180 of age I. and 130 of :1gc 2. (u) Write a Leslie matrix A for this colony of bats. (b) If q

= .8, what will happen to the popu lation in 50

years'! (c) If q

= .2. what wi ll

happen to the population in 50

years? (a) What proportion of the female> of the first age group survive to the ;econd age group? (b) I low many female offspring do females of each age group average? (c ) If x

= [;;]

1>

the current population distribution for

the females of the colony. descnbe all future population dtstributions.

(d) Find by trial and error a value of q for which the bat pol>ulmion reaches a nontcro ;table di>tribution. What is this stable distribution? (e) For the value of q found on (d). what happens to an initial population of 210 females of age less than I. 240 of age I. and 180 of age 2'1 (f) For what value of q does (A - t 1)x nonzero solutaon?

=0

have a

Ia E.urci
(g) For the value of q found in (f). find the general solution of (A - /1)x 0. llow does thi' relate to the

12. A certain colony of lizards has a life span of less than 3 years. Suppose that the females arc divided into three age group': tho>c under :age I. tho>e of age I. and those of age 2. Suppose further that the proportion of newborn females that survives until age I is .5 and that the proportion of one-year-old females that survives until age 2 is q. A»ume also that females under age I do not give birth, those of uge I have, on average, 1.2 femule offspring. and those of age 2 have, on average, I female offspring. S uppose there nrc initially 450 females of age less than I. 220 of age I. and 70 of age 2.

14. A certain colony of voles has a life span of less than 3 years. Suppose that the females arc davided into three age groups: those under age I. tho>e of age I. and those of age 2. Suppose further th:ll the proportion of newborn females that survives unti I age I is . l and that the proportion of one-year-old female" that survives until age 2 is .2. Assume also that females under age I do not give birth. those of age I have, on :tverngc. b female offspring. and those of age 2 h:tvc, on avcr:tgc, I0 female offspring. Suppose there are initially 150 females of age less than I, 300 of age I, and 180 of age 2.

(a) Write u Le>li e matrix A for this colony of lizards.

(a) Write a Lesl ie matrix A for this colony of voles.

=

stable distribution< in (d) and (e)?

(b) If q .3. what will happen to the population in 50 years?

=

(b) If b = 10. what will happen to the popultuion in 50 years?

= .9. what will h:tl>pen to the population in 50

(c) If b = 4, what will happen to the population in 50 years?

(c) If q

years? (d) Find by trial und error a value of q for which the li1.ard populltion reache~ a nonrero stable distribution. What is this stable dis tributJon?


(d) Find by trial and error a value of b for which the vole population reaches a nonzero stable distribution. What is this stable distribution?

Page 1 of 10


119

2.2 Applications of Matrix Multiplication (c) For the value of b found in (d). what happens 10 an initial population of 80 females of age les.~ than I, 200 of age I. and 397 of age 2? (f) For what value of b does (A - !))x

=0

have a

nonzero solution? (g) Let p

= ~;]

be an arbttrary population vector. and

let b have the value found in (f). Over time. p appro:tche' a 'table population vector q. Express q in term~ of PI· I'!· and l'l· 15. A cenain colony of squirrels has a life span of less than 3 years. Suppose that the females arc divided into three age group>: those under age I. those of age I. and those of age 2. Suppo'e funhcr that the proponion of newborn females that •urvivcs until age I is .2 and that the proponion of one-year-old females th:u survives until age 2 is .5. Assume also that females under age I do not give birth. thO>C of age I have, on average. 2 female offspring. and tho'e of age 2 have, on averuge. h female offspri ng. Suppose there arc initi:t lly 240 females of age less than one, 400 of age one. :md 320 o f age two. (a) Write a Leslie mmrix II for this colony of «1u irrcls. (b) If b = 3. what will happen to the population in 50 years? (c) If b 9. what will happen to the population in 50 year'? (d) Find by trial and error a value of b for which the >quirrel popul:ttion rcachc> a nontero >table di>tribution. What ts this stable distribution?

=

(e) For the value of b found in (d). what happens to an initial population of 100 females of age less than one. 280 of age one. and 400 of age two? (f) For what value of b does (A - / 3)x

=0

Exercises 17 aud 18 use the teclmique de1•e/oped in the traffic flow applicatiou. 17. A cenain medical foundation rccetves money from two sources: donattons and mtercst earned on endowments. Of the donations rccci,cd. 30'.l i' used to defray the costs of raising funds: only IO<;t of the interest is u~ed to defray the cost of managing the endowment funcb. Of the rest of the money (the net mcome). 40<:l ts ul. then M [

~] = [;].

where m and/ are the material and personnel costs of the foundation. respectively. Table2.3

personnel costs

20%

30%

SO'!Io

18. Water is pumped into a system of pipes at points P1 and Pz shown in Figut'C 2.6. AI each of the junctions P;. 1'4. Ps. 1'6. P1. and P~. the pipes are split and water flows according to the portions indicated in the diagram. Suppose that w:tter flow; into P 1 and 1'2 at p and q gallons per minute. respectively. and flows out of />9 • P 10 • and P11 at the rates of a. b. and c gallons per minute. respectively. Find a matrix M such that the vector of outputs is given by

have a

nonzero solutton·' (g) Let P =

~~l be an arbitrary population vector. and

let b ha\c 1 c value found in (f). O'er time, p approaches a stable population vector q . Express q in terms of fl!· and fiJ.

p}~

p9

I 0.4

I'•·

16. The maximum membership term for each member of the Service Club is 3 years. Each first-year and seoond-ycar member recruit< one new I>Cr>On who begins the membership term in the following year. Of those in their first year of membership. 50~ of the members resign. and of those in their second year of membership. 70% resign. (a) Write a 3 x 3 matrix A so that if x, is c urrentl y the number of Service Club members in their i th year of membership und y, is the number in their i th year of membership a year from now, then y = ilx. (b) Suppose that there arc 60 Service Club members in their lin.t year. 20 members in their second year. and 40 member> in their third year of membership. Find the di>tribution of members for next year and for 2 years from now.


Po 0.8

P,

10.2 P,

p.

Ps

1o.5

"·

1'2

0.5

~ PIO

1o.3 J'll

0.7

Flgure2.6

Exercises 19- 23 ttre COIIcemed ll'itlt (0, I)-matrices. 19. With the interpretation of a (0. I )-matrix found in Prnctice Problem 3. ~uppo;e that we have five cities with the

Page 2 of 10


120

CHAPTER 2 Matrices and Li near Transformations associated matrix A given in the block fonn (sec Section 2.1)

Exerr:ises 22 and 23 rt'fer to till' .vcltedulirrg example.

22. Suppo-c that student preference for a ;.et of courses is given in the following table: \\'here 8

i~

a 3 x 3 matrix. 0 1 is the 3 x 2 zero matrix.

Oz is the 2 x 3 zero matrix. and C is a 2 x 2 matrix.

Coun• Numbltt'

Student

1

l

4

J

5

(a) What docs the matn~ A tellm. about Hight connections bet\\CCn the citie\? (b) U.c block multiplication to obtain A 2• A 3• and At for any positive integer k. (c) Interpret your result the citie,,

tn

(b) in terms of Rights between

20. Recall the (0. 1)-matrix

A ~ [l

0

0 0

0

l

0 0

0 0 0

I

l1

in which the entri es describe countries that maintain diplomatic relations with one anotheL (a) Which pairs of countries maintain diplomatic relations'/ (b) I low many countries link country I with country 3? (c) Give an imcrpretation of the (1, 4)-entry of A 3 • 21. Suppose that there is a group of four people and an associated -' x 4 matrix A dehned by

a,,

1 if i '# j and person i likes person j = { 0 otherwise.

We say that persons 1 and j are frimds if they like each I. Suppose that A is given by other: that i~. if a,1 a"

= =

(a) G ive all pairs of courses that arc desired by the most

students. (b) Give a ll pairs of courses that arc desired by the fewest students.

(c) Construct a m:llrix whose diagonal entrie> determine the number of student;. who prefer e:tch course.

23. Let A be the matrix in E'ample 2. (a) Justify the following interpretation: For i # j. the (i .j )-entry of AA r ts the number of classes that are desired by both students i and j. (b) Show that the (I. 2)-cntry of AA r i> I and the (9. I)entry of AA T IS 3. (c) Interpret the answen. to (b) in the context of (a) and the data in the scheduhng example. (d) Give an interpretation of the diagonal entries of AA r. £urcises 24 arrd 25 art' corrum~d wrtlr tilt' amhropology application.

(it) List

:111 pail" of friends.

(b) Give an interpretation of the entries of A 2 . (c) Let 8 be the 4 x 4 matrix defined by

1>,,

l = {0

if persons i and j are friends otherwise.

Determine the matrix B. Is 8 a symmetric matrix? (d) A clique is a set of three or more people, each of whom is friendly with al l the other members of the set. Show that person i belongs to a clique if and only if the (i. i)-entry of 8 1 is positive. (e) Usc computer software or a calculator that performs matrix arithmetic to coumlhe cliques that exist among the four friends.


24. Recall the binomial formula for 'caJars positive integer~ : (a+ 1>)1

= a1 + ka1

tl

and b and any

1> + ...

1

+ -*-'-o1 -'b' +··. + li . i!(k - i)! where i! (i factorial) is given by i! = I· 2 · · · (i- I) . i. (a) Sup1X>se that A and 8 arc m x m matrices that commute: that is. AB = BA . Prove that the binomial formula holds for A and 8 when k = 2 and k 3. Specifically. prove that

=

and

Page 3 of 10


2.2 Applications of Matrix Multiplication (b) Usc mathematical induction to extend the results in pan (a) to any positive integer k.

=

(c) For the matricc' A and 8 on page 117. >how 8 3 0 and that A :md 8 commute. Then use (b) to prove

A4 = 1,

k(k

I) ,

+ tB + - -- s2

fork :: 2.

Ia E.urrisrr 25 and 26. list' rithu tt mlculawr \lith marri.r £'tll](l· bilitil's or computu sofnl'lm' well a.r MATU\8 to soh·<' l'ach problem. 25. In reference to the apphcallon in the text involving the N:1tchel Indians. s uppose that initially there arc 100 Sun males, 200 Noble males. 300 Honored males, and 8000 MMQ males. (a) I low m:my males will there be in each class ink 2. and 3 gcncrntions?

(a) Show that A is a (0. I )-matrix. (b) Give an interpretatiOn of what it means for the term (IJ2(/2J to equal one. (c) Show that the (3. I )-entry of A 2 represents the number of ways that person 3 can send a mc~sage to person I in two stag<'s that i>. the number of people to whom person 3 can ;end a mes.agc and "ho in tum can send a message to person I. fl im: Con~ider the number of terms that arc not equal to zero in the expression

(d) GencraliLc your rc,ult in (c) to the (i.j)-cntry of A 2 • (c) Generalize your rc'ult in (d) to the (i.j)·entry of A'". Now 'uppo'e

= I,

0 I 0

(b) Usc a computer to dctcnni ne how many males there 9. 10. and II generations. will be in each c lass ink

=

(c) What do your answers to (b) suggest for the future of the N:11che1 Indian< if they hold to thei r cu•·re nt marriage laws? (d) Produce an algebraic proof that for some k there will not be enough MMQ to a llow the other members to marry.

26. Suppose that we have a group of six people. each of whom own< a communication device. We define a 6 x 6 matrix 0: and for i -# j.

A a; follows: For I !5 i !5 6. let a,

=

if person i can send a message to person j othen,i •e.

I

a,, = { 0

121

0 I I 0

0

0 0

0 () 0

l

(f) Is there uny person who c:~nnot receive ll mc.-sagc from anyone else in one >luge? Justify your :~nswer. (g) How many way' can pcr..on I ;end a mt»:tgc to per· son 4 in I, 2. 3, and 4 s l:lges? (h) The (i.j)-entry of A +1\ 2 + .. ·+ A"' can be shown to equal the number of ways in which person i can send a message to person J in at most m stages. Use this result to dctcrmmc the number of ways in which person 3 can send a message to person 4 in at most 4 stage.~.

SOLUTIONS TO THE PRACTICE PROBLEMS I. (a) A

= [~:;~

~l

the population di;lribut•on docs not change from year to year.

(b) The population d1stnbution of females for this year is gh·cn by the vector

[~:]. and hence the population

distributio n of females for next year is 200] A [ 200

= [0.50 0.25

2] [ 200] [500] 0 200 = 50 .

and the popu lation dist•·ibution of females 2 years from now is 500] = [0.50 0.25

A [ 50

2]0 [500] 50 = [350] 125 .

(d) Let x 1 and .1 2 be the numbers of females in the first and second age group>. rc,pectivcly. Since next year's population dimibution is the s:1mc liS this year's. Ax x. and hence

=

0.50 [ 0.25

2] [' '] 0 x2

=

4xz. But .It + xz = 600. :u1d therefore 4xz + xz = Sxz = 600, from which it follows that x 2 = 120. Finally. x 1 = 4x2 = 480.

Thus -~•

2 . Let Zt and ;:2 be the mtes at which soy sauce is s hipped to Chicago and St. Louis. respectively. Then

(c) Since A

[400] _[0.50 100 - 0.25


= [0.501t + 2xl] = ['']. 0.25x, .r2

0.3 0.7

OJ I

[10000] [7400] 0.5 o.o] ~:~ 5000 = 7600 .

[0.5

0

Page 4 of 10


122

CHAPTER 2 Matrices and Li near Transformations

3. (a) Clearly. every entry of A is either 0 or I. so A is a (0. I )-mauix. (b) Tite (2. 3)-cntry of A2 i<

(d) Because the (2, 3)-entry of A2 i> I. there is one Hight with one layover from city 2 to city 3. (e) We compute A·1 to find the number of nights with two layovers from city I to city 2. We have

A typical tenn ha' the form ti!J.OH. which equals or 0. This tenn equals I if and only if au I and tto I. Con;cqucntly. thb tenn equals I if and only tf there •~ a nonstop fltght between city 2 and city k. as well as a nonstop fltght between city k and city 3. Th:1t i;,, O:!J;Otl I means that there is a flight with one ltt)'OI>er (the plane stops at city k) from city 2 to city 3. Therefore we may interpret the (2. 3)-entry of A 2 as the number of floghts woth one layover from city 2 to city 3.

=

=

=

3 0

0

2

Because the (I. 2)-entry of A1 os 3. "e see that there are three night> -.ith two layover. from city I to city 2.

=

, (c) A·

0 3 2 0] [

'

A=2001' 0 2 I 0

(I) From the entries of A. we see that there are nonstop ftighLs between cities I and 2. citie> I :md 3. and cities 2 and 4. From A2 • we see that there are flights between cities l and 4. as well as between cities 2 and 3. Finally. fro m A 1, we di >cover that there is a night between c ities 3 and 4. We conclude that there are flights between all pairs of cities.

20 02 0I 0I] [0I 0I 0I 0I '

I 2.3 I INVERTIBILITY AND ELEMENTARY MATRICES In this section, we introduce the concept of invenible matrix and exami ne special invertible matrices that a re intimately associated wit h elementary row operations, the

elementary matrices. For any real number a -::ft 0. there is a unique real number b. called the multiplicatil·e im·erse of a, with the property that ab = ba = I. For ex:unple, if a = 2, then b = I / 2. In the context of matrices. the identity matrix In is a multiplicative identity: so it is natural to ask for what matrices A does there exist a mauix 8 such that A8 = 8A = ln. Notice that this last equation is possible only if both A and 8 are 11 x 11 matrices. This discussion motivates the following definitions: Definitions Ann x 11 matrix A is called invertible if there exists an such that AB = 8A = 1•. ln this case, 8 is call ed an inverse of A.

11

x

11

matrix 8

If A is an invertible matrix. then its inverse is unique. For if both B and C are inverses of A. then AB = BA = 1. and AC = CA = ln. Hence

8

= 81, = 8 (AC) = (BA)C =

I,C =C.

Whe n A is invertible, we denote the unique inverse o f A by 11A - 1A = /11 • Notice the similarity of this statement and 2 · 2 - t

1,

so that AA - 1 =

= 2 - t · 2 = 1. where

2- t is the multiplicati ve inverse of the real number 2.

8$!ii .. i.)fj8 _ . ,..... ·--

Let A=


[t3 52]

and 8

= [ 3s

_

2]1 . Then

Page 5 of 10


2.3 lnvertibility and Elementary Matrices

123

and

So A is invettible, and 8 is the inverse of A; that is, A-

If A = [ - :


~ -~] and 8 =

1 =

8.

[;

2 - 1 - 1

5

Because the roles of I he matrices A and 8 are the same in the preced ing definition, it follows that if 8 is the inverse of A, then A is also the inverse of 8. Thus, in Example I , we also have

Just as the real number 0 has no multiplicative inverse. the 11 x 11 zero matrix 0 has no inverse because 08 0 =/= I,. for any 11 x n matrix 8 . But there are also other

=

square matrices that are not invettible; for example, A

l

= [~ ~

For if B

= [~ ~]

is any 2 x 2 matrix, then

A8

= [21

1] 2

[ac

b]

d

=

[

a+ c + 2c

2a

b

2b

+ + d2d ] ·

Since the second row of the matrix on the right equals twice its first row, it cannot be the identity matrix

[~ ~l SoB

cannot be the in verse of A, and hence A is not

invertible. In the next section, we learn which m atrices are invettible and how to compute their inverses. In this section, we discuss some elementary properties of invertible matrices. The inverse of a real number can be used to solve certain equations. For example, the equati on 2x 14 can be solved by multiplying both sides of the equation by the inverse of 2:

=

=T 1 (2- 2)x = 7 T 1(2x)

1

( 14)

lx = 7 X

=7

In a similar manner, if A is an invertible 11 x 11 matrix , then we can use A - • to solve matrix equations in which an unknown matrix is mu ltiplied by A. For example, if A is invertible, then we can solve the matrix equation Ax = b as follows: 2 Ax = b 1

A- (Ax) 2 Although

= A- 1b

matrix inverses can be used to solve systems whose coefficient matrices are invertible, the

method of solving systems presented in Chapter 1 is far more efficient.


Page 6 of 10


124


If A is an invertible n x solution A- 11>.

11

matrix, then for every b in R". Ax = b has the unique

In solving a system of linear equations by using the inverse of a matrix A, we observe that A -I ··reverses" the action of A: that is. if A is an inve1tible n x 11 matrix and u is a vector in 'R!', then A- 1(A u) = u . (See Figure 2.7.) multiply by A

1?." • Au

muhiplr by A

Figure 2.7 Multiplication by a matrix and its inverse

Example 2

Use a matrix inverse to so lve the system of linear equations

x,

3x, Solution

+ 2xz =4 + Sxz = 7.

Th is system is the same as the matri x equation Ax= b , where

X=[X'], xz

b=

and

[~].

We saw in Example I that A is inve1tible. Hence we can solve this equation for x by multiplying both sides of the equation on the left by

r • = [-~

_7]

as follows:

T herefore x 1 Practice Problem 2 II>

= - 6 and x2 = 5 is the un ique solution of the system.

Use the answer to Practice Problem I to solve the following system of linear equations:

- x, XJ

2x1 -


+

+ 2.Y2 -

X2 -

x3

=

I

2.Y3

=

2

X3

=

- I

Page 7 of 10



1$if!,j,)§i

125

Reca ll the rotation matrix A _ [ cosO 0 -

sin O

- sinO] cosfJ

considered in Section 1.2 and Example 4 of Section 2.1. Notice that for fJ = 0°, = / z. Funhermore, for any ang le a .

Ao

Similarly. A_.,A., = h. Hence A., satisfies the definition of an invenible matrix with inverse A_, . Therefore (Aa)- 1 =A _, . Another way of viewing A_.,A., is that it represents a rotation by a, followed by a rotation by - ex , which results in a net rotation of 0°. This is the same as multiplying by the identity matrix.

The following theorem states some useful properties of matrix inverses:

THEOREM 2.2 Let A and 8 be n x

11 matrices . (a) If A is invertible. then A- 1 is invertible and (A- 1)- 1 =A. (b) If A and 8 are invertib le, then AB is invertible and (AB)- 1 (c) If A is inve1tible, then AT is inve1tible and (AT)- 1 (A-I)T

=

= s- 1A- 1.

The proof of (a) is a simple consequence of the definition of matrix inverse. (b) Suppose that A and 8 arc invertible. Then

PROOF

Similarly, (B - 1A- 1)(AB) = /,. . He nce AB satisfies the definition of an inve1tible matrix with inverse s- 1A- 1; that is, (AB)- 1 s- 1A- 1. (c) Suppose that A is in verti ble. Then A- 1A = /,.. Usi ng Theorem 2. l (g). we obtain

=

=

Si mi larly, (A -If AT I,. . Hence AT satisfies the definition of an in vertible matri x with the inverse (A - 1f; that is, (AT)- 1 = (A- 1/ . • Part (b) of Theorem 2.2 can be easily extended to products of more than two matrices. Let A 1, A2 , .. . ,Ak be invc1tiblc. and


11

x

11

invertible matrices. Then the product A 1Az · · · Ak is

Page 8 of 10



ELEMENTARY MATRICES It is interesting that every elementary row operation can be performed by matrix multiplication. For example, we can multiply row 2 of the matrix

A= [~~] by the scalar k by means of the matrix product

Also. we can interchange rows I and 2 by means of the matrix product

We can add k times row I of A to row 2 by means of the matrix product

The matrices

[~ ~]. [~ ~l

and

[!

~]

are examples of e/ememary maTrices. In general, an 11 x n matrix E is called an elementary matrix if we can obtain E from 1, by a s ingle elementary row operation. For example, the matrix

E =

I

0

[2

0 0] I 0

0

I

is an elementary matrix because we can obtain E from /3 by the elementary row operation of adding two times the first row of /3 to the third row of h Note that if

then

~].

10

Hence we can obtain EA from A by adding two times the first row of A to the third row. This is the same elementary row operation that we applied to /3 to produce E. A simi lar result holds for each of the three elementary row operations.

Let A be an 111 x n matrix. and let E be an m x m elementary matrix resulting from an elementary row operation on/,. Then the product EA can be obtained from A by the identical elementary row operation on A .


Page 9 of 10



Practice Problem 3 Ill>

Find an elementary matrix E such that EA

A= [~

-4

= B,

where

and

5

127

-4 13

As was noted in Section 1.3. any elementary row operation can be reversed. For example, a matrix obtained by adding two times the fi rst row to the third row of A can be changed back into A by adding -2 times the first row to the third row of the new matrix. The concept of a reverse operation gives us a way of obtaining the inverse of an elementary matrix. To sec how this is done, consider the preceding matrix E. Let

F

J = [ 0I 0I O 0 -2 0 I

be the elementary matrix obtained from /3 by adding -2 times the first row of / 3 to the third row of /3. If this elementary row operation is applied to E , then the result is / 3 . and therefore FE= h Similarly, EF =h. Hence E is invertible. and £- 1 =F. We can apply the same argument to any elementary matrix to establish the following result: Every elementary matrix is invertible. Furthermore, the inverse of an elementary matrix is also an elementary matrix.

The value of elementary matrices is that they allow us to analyze the theoretical properties of elementary row operations using our knowledge about matrix multiplication. Since we can put a matrix into reduced row echelon form by means of ele mentary row operations. we can carry out this transformation by multiplying the matrix on the left by a sequence of elementary matrices, one for each row operation. Consequently, if A is an 111 x 11 matrix with reduced row echelon form R. there exist elementary matrices E 1, £ 2, ... , Ek such that

=

EkEk - l ·· · £,.Then Pis a product of elementary matrices, so P is an invettLet P ible matrix, by the boxed res ult on page 125. Furthermore, PA = R. Thus we have established the following result:

THEOREM 2.3 Let A be an m x 11 matrix with reduced row echelon form R. Then there exists an invertible m x m matrix P such that PA = R.

This theorem implies the following impo1tant result that justifies the method of row reduction (described in Section 1.4) for solving matrix equations:

=

Corollary The matrix equation Ax h has the same solutions as Rx is the reduced row echelon form of the augmented matrix [A b ]. PROOF

= c, where [R

c)

There is an invertible matrix P such that PIA bj = [R c) by Theorem 2.3.

Therefore [PA P b j = P[A b j = [R c j,


Page 10 of 10



and hence PA = R and P b = c. Because P is inveltible, it follows that A = p - JR and b = p-•c. Suppose that v is a solution of Ax = b. Then Av = b, so Rv = (PA)v = P(Av) = P b =c. and therefore v is a solution of Rx = c. Conversely. suppose that v is a solution of Rx = c. Then Rv = c, and hence Av = (P - 1R)v = p - 1(Rv) =p- Ic= b. Therefore v is a solution of Ax = b. Thus the equations Ax = b and Rx = c have • the same solutions. As a special case of the preceding corollary, we note that if b = 0, then c = 0. Therefore Ax = 0 and Rx = 0 are equivalent.

THE COLUMN CORRESPONDENCE PROPERTY In Section 1.7. we found that if a subsetS of 'R:' contains more than 11 vectors, then the set is linearly dependent, and hence one of the vectors inS is a linear combination of the others. However. nothing we have leamed tells us which vector in the set is a linear combination of other vectors. Of course, we could solve systems of linear equations to find out how to do this, but this method could be extremely inefficient. To illustrate another approach, consider the set

s~ ~ll' [~~] ·[~ll' m'm· ml' [[

n

Because S is a subset of 4 and has five vectors, we know that at least one vector in this set is a linear combination of the others. TI1us Ax = 0 has nonzero solutions, where

A=

_:

-2

2

4

[

2

-3 -6

- 1 2

I 2 -3

2

2 0

Hl

is the matrix whose columns are the vectors in the set. But as we have just proved. the solutions of Ax = 0 are the same as those of Rx = 0. where

R

=

I 2 0 0 -1 -5] [ 0 0 I 0 0 0 0 1 0 0 0 0

0 I 0

- 3 2 0

is the reduced row echelon form of A (obtained in Section 1.4). Since any solution of Ax = 0 gives the coefficients of a linear combination of columns of A that equals 0,


Page 1 of 10



129

it follows thai the same coefficients yield the equ ivalent linear combinat ion of the columns

of R, and vice versa. As a consequence, if column j of A is a linear combination of the other columns of A. then column j of R is a linear combination of the other columns of R with the same coefficients, and vice versa. For example, in S' it is obvious that r 2 2 r1 and rs - r 1 + r 4. Although it is less obvious, similar relationships hold in S : a2 = 2a1 and as = - a 1 + 34. The preceding observations may be summarized as the cohmm correspondence properry.

=

=

Column Correspondence Property Let A be a matrix and R its reduced row echelon form. If columnj of R is a linear combination of other columns of R , then column j of A is a linear combination of the correspond ing columns of A using the same coefficients, and vice versa. Practice Problem 4 ....

For the preced ing matrix A, lL~e the column correspondence property to express a linear combination of the other columns.

<16

as -
The column COJTespondence property and the following theorem are proved in Appendix E:

THEOREM 2.4 111e following statements are true for any matrix A: (a) The pivot columns of A are linearly independent. (b) Each nonpivot column of A is a linear combination of the previous pivot columns of A, where the coefficients of the linear combination arc the entries of the corresponding column of the reduced row echelon form of A.

For example. in the previous matrices A and R. the first. third, and fourth columns are the pivot columns, so these columns are linearly indepe ndent. Furthennore, the fitih column of R is a linear combination of the preceding pivot columns. In fact, rs = ( - l )r 1 + Or 3 + I r 4 . Hence Theorem 2.4(b) guarantees that a simi lar equation holds for the corresponding columns of A; that is.

Example4

The reduced row echelon fom1 of a matrix A is

R


=

[~

2 0 0 1

0 0

-I]

1 . 0

Page 2 of 10



Determine A, g iven that the first and third columns of A are

and

Solution Since the first and third columns of R are the pivot columns, these are also the pivot columns of A. Now observe that the second column of R is 2r ,, and hence, by the column con-espondence prope11y, the second column of A is

Furthermore. the fourth column of R is r 4 con-espondence propeny,

= (-

+ r 3, and so, again by the column

l )r ,

~=(-!)a, + a 3 = (- 1) [U+

[n

=

[~].

Thus

2 2 4

2

2 3 ~

Practice Problem 5

Suppose that the reduced row echelon form of A is

-3 0

0

5 2

I

3]

- 2 .

0 0 0

0

Detennine A if the first and third columns of A are a ,= [

-l]

and a3 = [

respectively.

-~l .,.

EXERCISES For each ofrhe mmrices A and Bin Exercises 1- 8, derennine whether B = A 1• I. A= [:

-n

2. A=[~ ;] 2 3. A=

[i

I 3

and

and

n

8

=[: - 0.5] I

5 A" [

8= [-53 -~] and

8

=[-I~


4.A= [~

I I

0

i

-I 5

-2 -I

-:]

B

[-'

n

and

-2 -2

I

-I

0

0

-I

- I

= ~

-3

-I

2

0 I

B

-1]

=[-I:

-I 2 0

_;]

and

-I

-7]

2 -2 -2

I

Page 3 of 10



6.

A~ [j B=

2

2

I

I

3 5

2 7

[-:

-2

-2

[I

3 2 -4 -2

-I

2 2

-I

_

-4

-I I

-I

7. A=

l

4

2

-2

8 =

8.

A~ B=

u [j

- I

5

-4

I

-I

-2 2

I

I

I

2

2

2

-I

[-i

-4

2

-2

-:] -;]

26. A= [ _:

-3 4

27. A= [

1]

29. A =

I

A-· =

n _:]

32. A= [

I

8-• =

and

[~

12. (BA)- 1

- I 0 -2

~l

II. (A B)-1

10. (BT)-l 13. (AB 'I )-1

9. (AT)-1

14. (AT B1)-l

In Exercises 15- 22, find file inverse of each elemelllaty malrix.

15.

17.

19.

21.

[: i]

[-i

[j [j

16.

~]

0 I

0 0 4 0 0 0 0 0

0 0 I

0 0 0 I

0

18.

[~

-n

A = [~ ~]

!]

20.

i]

22.

8 =

8 =

~]

and

8 =

I -3

~

-~]

2 -I

8 = [

[~ ~] 0 I

[-1

3


-~]

~

33. 34. 35. 36.

-l -~

~] 2 I -I

2

I -5

8 =

and

and

8 =

[-1 [i - 5

-~]

0 3 2

~]

-3

[-li

8 =

[~

- I -3 19 2 -I

3

rJ

-~]

[~ ~] 2 8

5

3 3 0

!]

3 3 -6

and

j]

In Exercises 33-52. determine wiletl!er fil e slate· menls are true or false.

Every square matrix is invenible. Invertible matrices arc square. Elementary matrices are invertible. If A and 8 are matrices such that AB then both A and 8 arc invertible.

= !,.

for some

11,

37. If 8 and C are inverses of a matrix A, then 8 =C. 38. If A and 8 are invenible n x 11 matrices, then AB T is

8 is the inverse of A.

0

41. For any matrice.~ A and 8 , if A is the inverse of Br. then A is the transpose of 8 - •. 42. If A and 8 are invertible 11 x 11 matrices, then AB is also

0

43. If A and 8 are invertible 11 x n matrices, then (AB)- 1

[~ ~]

[j

~

25 8

and

-3] -5

invertible.

0

and

and

2 -4

8 = [:

39. An invertible matrix may have more than one inverse. 40. For any matrices A and 8, if A is the inverse of 8 , then

0

0 0

0

I

0

0

inve1tible.

=

A- 18 - 1 •

44. An e lementary matrix is a matrix that can be obtai ned by

!]

In Exercises 23- 32, find tm elementary nwtri.x E such film EA = B.

23.

-~]

-3 4

31. A= ['4 7

For Exercises 9- 14, find file value of each malrix expression. where A and 8 are life inverlible 3 x 3 mal rices sucil1ila1

2 0

and

-I

-I]

-I

-3

-;]

[i !]

30. A= [

and

-2

0

[~

8 =

10]

-I

j]

2 3 -2

-~

28. A= [-2I

and

and

2 4 2

~~J

8= [- 01

and

- 3] 10

25. A=[;

- I

I

~]

24. A= [- 1 2

and

131

a sequence of elementary row operations on an identity matrix .

45. An elementary 11 x

11

matrix has at most

11

+I

nonzero

entries.

46. The product of two e lementary

mentary 11 x

11

11

x

11

matrices is an e le-

matrix.

47. Every elementary matrix is invertible.

Page 4 of 10



48. If A and 8 are m x 11 matrices and 8 can be obtained from A by an e lememary row operation on A, then there is an elementary m x m matrix E such that 8 = EA.

49. If R is the reduced row echelon form of a matrix A, then there exists an invertible matrix P s uch that PA = R. 50. Let R be the reduced row echelon form of a matrix A. If colunm j of R is a linear combination of the previous columns of R, then column j of A is a linear combination of the previous columns of A.

51. The pivot columns of a matrix are linearly dependent.

52. Every column of a matrL' is a linear combination of its pivot columns.

65. Prove that if an 11 x 11 matrix has rank 11, then it is invertible. Him: Use Exercises 63 and 64.

o,J

66. Let M = [ A , where A and 8 are square and Ot 02 8 ;md 0 2 arc zero matrices. Prove that M is invct1iblc if and only if A and 8 are both invertible. Him: Think about the reduced row echelon form of M. /11 Exercises 67- 74, find the matrix A, given the retluced row echelon jon11 R of A a11d i11jormatio11 abow cenai11 co/wm1s of A.

67. R

= [~

53. 3 Let A., be the a-rotation matrix, Prove that (A,)T = (A,)-1. 54. Let A = [:

~

l

(a) Suppose ad - be

68.

#

0. and

8 - -I - [ d - ad- be -c Show that A8 = 8A = /, and hence A is invertible and B =A- 1.

-

(b) Prove the converse of (a): If A is invertible, then tul- be# 0.

69. R

=

[~

-I 0

55. Prove that the product of e le mentary matrices is invertible. 56. (a) Let A be an invertible 11 x 11 matrix, and let u and v be vectors in n" such that u # v. Prove that

0

Au # Av, (b) Find a 2 x 2 matrix A and distinct vectors u and v in n 2 such that Au = Av.

57. Let Q be an invertible 11. x

11 matrix, Prove that the subset {u 1 , u 2, ... , uk} of n" is linearly independent if and only if {Q u ~o Q u2, .. . , Qud is linearly independent.

70.

58. Prove Theorem 2.2(a).

59. Prove that if A, 8, and C are invertible 11 x 11 matrices, then ABC is invettible and (ABC )- 1 = c - •n - tA - 1 . 60. Let A and 8 be n x 11 matrices such that both A and AB are invertible. Prove that 8 is invettible, by writing it as the product of two invertible matrices, 61. Let A and 8 be 11 x 11 matrices such that AB = 1,.. Prove that the rank of A is 11. Him: Theorem 1.6 can be useful.

62. Prove that if A is an

matrix and 8 is an 11 x p matrix, then rank A8 ::; rank 8. Hi11t: Prove that if the kth column of 8 is not a pivot column of 8. then the kth column of AB is not a pivot column of AB. 111

x

7l.R = [~ ~ 72. R

=

[~ -~

11

and as= [

n

O

-n

m -~ -~ -Ha=[=il 32

0 I 0

= [;].and a3 =

2

63. Prove that if 8 is an 11 x 11 matrix with rank 11 , then there exists an 11 x 11 matrix C such that BC = 1,.. Him: Theorem 1.6 can be useful. 64. Prove that if A and 8 arc 11 x 11 matrices such that A8 = 1,., then B is invertible and A= s - t . Him: Use Exercises 62 and 63.

3

The result of this exercise is used in Section 4.4 (page 269).


Page 5 of 10



133

88. Let A be an m x n matrix with reduced row echelon form R. Then there exists an invcnible matrix P \Uch that PA = R and an invertible matrix Q ;,uch that QRT is the reduced row echelon form of RT. De;,cribe the matrix PAQT in terms of A. Justify your answer. 89. Let A and 8 be 111 x 11 matrices. Prove that the following ccmdit ion' arc cquiv:~lcnt. (a) A tmd 8 have the same reduced row echelon form.

In Exercises 75-78. write the indicated colwnn of

A=[~

-2 I

-1 I

-6 0

6

-4

(b) There is an invertible m x m matrix fJ such that

8 =

=

91. Let A be a 2

x 3 matrix, and let E be an elementary matrix obtained by an elementary row operation on /2. Pro\·e that the product £A can be obtained from A by the tdenllcal elementary row operation on A. Him: Prove thi;, for each of the three kind' of elementary row opemtions.

as tl linear combination of the pn·m co/um11r of A. 75.

a~

76. a 3

/11 E.1erciw•,, 79-82, u·rite the indicmed wi1111111 of

8

~

[

; -I

0 - I I

0

I

-3

3 -2

!!

- I

2

- I

92. Let A and 8 be m x

11 matrices. Prove that the following conditions are equivalent:

:]

-6 -3

5

(a) A and 8 have the same reduced row echelon form. (b) The system of equations Ax = 0 is equivalent to 8x 0.

=

as a lint'flr combination of tilt' pivot columns of 8.

80.

b~

81. bs

83. Suppose that u and v are linearly independent vectors in 'R3 . Find the reduced row echelon form of A = la1 3 2 3 J a.). given that a , = u.

32

= 2u ,

33

=

~1.

90. Let A be an 11 x 11 matrix. Find a property of A that i' equivalent to the 'tatemcnt A8 AC if and only if 8 = C. Justify your answer.

u + v.

and

a.= v.

84. Let A be an " x n invenible matrix, and let CJ be the jth standard vector of 'R". 1

(a) Prove that the j th column of A is a solution of Ax = e1 . (b) Why does the result in (a) imply that ;\ 1 is unique? (c) Why doe.- the rCMih in (:~) imply th:tt r:mk A =

11?

85. Let A be a matrix with reduced row echelon form R. Usc the column correspondence propeny to prove the following: (a) A column of A is 0 if and only if the corresponding column of R is 0 .

(b) A set of columns of A is linearly independent if and only if the corresponding 'et of columns of R is line:~rl y independent.

86. Let R be an 111 x 11 matrix in reduced row echelon form. Find a relationship between the columns of Rr and the columns of Rr R. Justify your answer. 87. Let R be an 111 x 11 matrix in reduced row echelon form with rank R = r. Prove the following:

\\?can tf~jiJtt' tm e/ementllry ,-nbmm operation in ll mttmrt'rmwlogous to tht> dejimtion in Section 1.3 of011 elemmtan· roll' OJH'T· atimr. Et1cl1 of the folloll'ing operations on a matrix 1s called an elementary column operation: imerchanging a11y tii'O colw11ns of the matri.•. mulriplying Oil)' column by a 11011~ero swlar. a11d adding tl multiplt' of o11e co/um11 ofthe matri.t 1t1 tmotltu mlrmm of the matrix.

93. Let f:.' be an 11 x 11 matrix. Prove that E is an elementary matrix if and o nly if£ can be obtained fro m / 11 by a s ingle e lement ary column operut ion.

94. Prove that if tt matrix E i> oblinglc e lementary column operation, then for any m x 11 matrix A. the product A£ can be obtained from A by the identical elementary column operation on A.

In E.•urift'f 95 99. liSt' eithu a calculator with lllfltrit rapa· bilities or compmer sofn•a~ such as MATU\8 to solt·e elicit problmr.

95. Let

A- [

~

2 - 1

0

I -)

0

-~]

- 3 . I

Let 8 be the matrix obtained by interchanging row• I and 3 of A. and let C be the matrix obtained by interchanging rows 2 and 4 of A.

(a) The reduced row echelon form of Rr is the 11 x m matrix [e, Cl ... e, 0 ••. \\here e, i' the jth standard vector of n• for I ~ j ~ r.

(b) Sho\\ that 8 and C are im·cnible.

(b) rank RT =rank R.

(c) Compare 8

or.


(a) Show that A is invenible. 1

and C

1

with A

1

Page 6 of 10


134 CHAPTER 2 Matrices and Linear Transformations (d) Now let A be any invertible 11 x 11 malrix, and lei 8 be I he matrix oblained by inlerchanging rows i and j of A, where I ::: i < j ::: 11 . Make a conjecture about lhe relationship belween B - I and A- t.

(c) State and verify a si milar result to (b) for A 3. (d) Generalize and prove a res ult analogous to (c) for lhe 11th power of any invertible matrix. 98. Consider the system Ax

(e) Prove your conjecture in (d).

96. Let

A=

Xt Xt

15

30

17

30 17 [ 31

66 36

36 20

61 3 1] 35 .

61

35

65

2tt

Xt

(b) Show that A- t is symmetric and invertible. (c) Prove that the inverse of any symmetric and invertible matrix is also symmetric and invertible. 97. Let

0

5

-I

6

(b) Compute lhe inverse of (A - 1)2.

99. The purpose of this exercise is to illustrate the method of finding an inverse described in Exercise 84. In Exercise 97, you found the inverse of

A- [~ -

(a) Show that A and A 2 are invertible by using their reduced row eche lon forms. A2 ,

4 -3 -I 2

(b) Show that your solution is correct by verifying that it salisfies the matrix equation Ax = b .

l

4

+ 3X2 + 2.f3 + .q = + 2x2 + 4X3 = + 6x2 + 5x3 + 2r4 = + 3x2 + 2t3 + 2t4 =

(a) Compute the inverse of A and use it to solve the system.

(a) Show that A is symmetric and invertible.

2

= b.

and show that it equals

2 3

2

5

0 -I

4 6

!l

(a) Solve the system Ax = e 1• and compare your solution w ith the first column of A-t. (b) Repeal (a) for lhe vectors e 2 • e3. and e4 •

SOLUTIONS TO THE PRACTICE PROBLEMS I. Since

AB =

and

BA =

[-l

2 -I

-~] [~

-1

n[-l

[~

B = A-

0

!] [~ ~] 0 I 0

5

0 2 -I

-~] [~0

-1

=

0 I

0

~l

1.

2. This system of linear equations can be wrillcn as the maltix equation Ax= b . where A is 1he matrix given in

-~~~ POOOI"" I, ~ - [

j ]·

Since the matrix 8 given in Practice Problem I is equal to A-t. we have

3. The matrix B is obtained from A by adding -2 times row I of A to row 2. Adding - 2 times row I of h to row 2 pro-

.

duces the elementary matnx E =

[_ I OJ . Thts. matrtx. 2

1

has the propetty that EA = 8 . 4. From the malrix R. it is clear that r 6 = - 5r 1 - 3r 3 + 2r 4. So by the column correspondence property. it follows that a6

= -5a 1 -

3aJ

+ 2a•.

5. Let r " r 2 , r 3. r 4 • and r 5 denote the columns of R. The pivot columns of R are columns I and 3, so the pivot columns of A are columns I and 3. Every other column of R is a linear combination of its pi vot columns. In fact. r2 = - 3rt . r4 5r t + 2r3. and rs = 3r t - 2r J. Thus the column correspondence propetty implies that column 2 of A is

=

a2 = - 3a,

=

-3 [

-i]

= [

~n.

Simi larly. the fourth and fifth columns of A are

l11erefore Xt = 4,

x2

= 4, and .1'3 = 5.


Page 7 of 10


24 The Inverse of a Matrix

and

135

respectively. I fence

~ -~] .

8

8

12.4 1THE INVERSE OF A MATRIX In this section. we learn which matrices are invenible and how to find their inverses. To do this. we apply what we have learned about invenible and elementary matrices. The next theorem tells us when a matrix is invenible.

THEOREM 2.5 Let A be an 11 x 11 matrix. Then A is invenible if and only if the reduced row echelon form of A is 111 • PROOF Fi rst, suppose that A is invertible. Consider any vector v in 'R" such that Av = 0. Then. by the boxed result on page 124. v = A 10 = 0. Thus the on ly solution of A x 0 is 0, ;md so rank A 11 by Theorem 1.8. However, by the second boxed statement on page 48, the reduced row echelon form of A must equal/11 • Conversely. suppose that the reduced row echelon form of A equals 111 • Then. by Theorem 2.3, there ex ists an invenible 11 x 11 matrix P such that PA = /11 • So

=

=

A= InA= (P - 1P)A

= P - 1(PA) =

= p - 1•

1

P

111

But. by Theorem 2.2. p-I is an invenible matrix. and therefore A is invenible .

•

Theorem 2.5 can be used as a test for matrix invenibility. as follows : To detennine if an 11 x 11 matrix is invertible. compute its reduced row echelon fonn R. Lf R = / 11 • then the matrix is invertible: otherwise. if R 'l in. then the matrix is not invenible.

Example 1

Usc Theorem 2.5 to test the following matrices for invenibility:

I 2 3] [

A= 2 5 6 3 4 8

and

8 =

I I 2] [ 2

l

I

0

l

- 1

You shou ld check that the reduced row echelon form o f A is /3. The refore A is inve rtible by Theorem 2.5. On the other hand. the reduced row echelon form of B is

I

0

0 I [0 0

-I]

3 . 0

So. by Theorem 2.5. 8 is not invertible.


Page 8 of 10



AN ALGORITHM FOR MATRIX INVERSION Theorem 2.5 provides not only a method for determining whether a matrix is invertible. but also a method for actually calculating the inverse when it exists. We know that we can transfonn any 11 x 11 matrix A into a matrix R in reduced row echelon form by means of elementary row operations. Applying the same operations to the n x 2n matrix [A 111 ] transforms th is matrix into an 11 x 2n matrix [R B] for some n x n matrix B. Hence there is an invertible mau·ix P such that P[A /11 ] = LR Bl. Thus [R B]

= P[A

= [PA

111 ]

P/11 ]

= [PA

P].

It follows that PA = R, and P = B. If R '# !,,then we know that A is not invertible by Theorem 2.5. On the other hand. if R 111 , then A is invenible. again by Theorem 2.5. Funhermore, since PA = /11 and P = B , it follows that B = A- I. Thus we have the following algorithm for computing the inverse of a matrix:

=

Algorithm for Matrix Inversion Let A be an 11 x n matrix . Use e lementary row operations to transform [A /,] into the form IR 8], where R is a matrix in reduced row echelon form. Then either (a) R = !,, in which case A is invertible and B =A - I ; or (b) R '# !,, in which case A is not inve11ible.

Example 2

We use the algorithm for matrix inversion to compute A - 1 for the invertible matrix A of Example I . This algorithm requires us to transform [A /3] into a matrix of the form l/3 B J by means of elementary row operations. For this purpose. we use the Gaussian el im ination algorithm in Section 1.4 to transform A into its reduced row echelon form / 3. while applying each row operation to the entire row of the 3 x 6 matrix.

[A / 3]

=[

~

5

311 6 0

4

8 0 0

2

21"2 +r.\~ rJ

0I 00

I

2

•• + •2- rz [

-Jr,

+ •J- •J

I

[~ [~ [~ ~ I 2

l-

-3rJ+r l-"' r a

31 0 - 2I

-2r2+r, - r, [


I 2 3 I 0 -2 0 7

2 I 0

~

0 0 I 0 0 0

2 I

0 -2 0

0 - 1 - 7 2

-r.t~ rJ

01

0 I -2

0 031- 2I I - 1 - 3 0

~l

~]

J] l ~l

- 20 6 3 -2 0 7 -2 - I

- 16

4

-2 I 7 - 2 - I

= [/) 8]

Page 9 of 10


2.4 The Inverse of a Matrix

13 7

Thus

- 16 A-

1

= 8=

[

~]·

4

I

- ;

-2

- I

In the context of the preceding discussion, if [R 8 J is in reduced row echelon form, then so is R. (See Exercise 73.) This observati on is usefu l if there is a ca lculator or computer available that can produce the reduced row echelon form of a matrix. In such a case, we simply find the reduced row echelon form [R 8 J of [A /11 ] . Then , as before, ei ther (a) R = !,.. in which case A is invenible and B = A-I : or (b) R 'f /11 , in which case A is not in vertible.

Example3

To illustrate the previous paragraph , we test

for invertibility. If we are performing calculations by hand, we transform [A /2l into a matrix [R B] s uch that R is in reduced row echelon form: I 1I I lA 121 = [ 2 2 0

0 1

J

-2r ,+r2- r1 [

I

I I

0

0

I

- 2

0

I

J _- [R B ]

Since R 'f / 2, A is not invertible. Note that in th is case [R B] is not in reduced row echelon form because of the -2 in the (2, 3)-entry. If you use a computer or calculator to put lA /zl into reduced row echelon form, some additional steps are performed, resulting in the matrix

IR

cJ =

[ 1

I

I 0 0 0 I

0.5 - 0.5

J.

Again , we see that A is not invertible because R 'f h It is not necessa ry to pe rform these add itional steps w hen solving the problem by hand-you can stop as soon as it becomes clear that R 'f h


For each of the matrices

determ ine whether the matrix is invertible. If so, find its in verse. Practice Problem 2 .,..

Consider the system of linear equations

x, -

x2

x,+ 2r2

+ 2r3 =

2 3

- x2+ x3= - l.


Page 10 of 10



(a) Write the system in the form of a matrix equation Ax (b) Show that A is invertible, and find A - I . (c) Solve the syste m by using the answer to (b).

= b.

Although the next theorem includes a very long list of state ments, most of it follows easily from previous results. The key idea is that when A is an 11 x 11 matrix, the reduced row echelon form of A equals /11 if and only if A has a pi vot position in each row and if and only if A ha~ a pi vot pos ition in each column. Thus, in the special case of an 11 x 11 matrix, each statement in Theorems 1.6 and 1.8 is true if and only if the reduced row echelon form of A equa ls /11 • So all these s tatements are equivalent to each other.

THEOREM 2.6 (Invertible Matrix Theorem)

Let A be an

11

x

11

matrix. The following state-

ments are equivalent: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)

A is invertible.

The reduced row echelon form of A is / 11 • The rank of A equals 11 . The span of the co lumns of A is n". The equation Ax b is consistent for every b in n". The nullity of A equals ze ro. The columns of A are linearly independent. The on ly solution of Ax = 0 is 0. There exists an 11 x 11 matrix B such that BA = 111 • There ex.ists an 11 x 11 matrix C such that AC = ! , . A is a product of e leme ntary matrices.

=

Statements (a) and (b) are equivalent by Theorem 2.5. Because A is an x 11 matrix , statement (b) is equi valent to each of (c), (d), and (e) by Theorem 1.6 on page 70. Similarly, state me nt (b) is equivalent to each of (f), (g), and (h) by Theorem 1.8 on page 78. Therefore statement (a) is equivalent to each of stateme nts (b) through (h). Proof tltat (a) implies (k): The assumption that A is inverti ble implies that the reduced row echelon form of A is 111 • Then, as in the proof of Theorem 2.5, there exists an invertible 11 x 11 matrix P such that PA 1, (Theorem 2 .3). Therefore A =p - I , and the discussion preceding Theorem 2.3 shows that P is a product of invertib le matrices. Thus p -I is the product of the in verses of these e lementary matrices (in reverse order), each of which is an elementary matrix. It follows that A is a product of elementary matrices. and so (a) implies (k). Proof that (k) implies (a) : Suppose that A is a product of ele me ntary matrices. Since e lementary matrices are invert ible, A is the product of inverti ble matrices and hence A is invertible, establishing (a). Thus statements (a) and (k) are equivalent. It re mains to show that (a) is equivalent to (i) and (j). Clearly, (a) implies (i), with B = A- 1• Conversely. assume (i). Let v be any vector in n" such that A v 0. Then PROOF

11

=

=

v

= /, v = (BA)v = B (Av) = BO= 0.

It follows that (i) implies (h). But (h) implies (a); so (i) implies (a). Thus statements (a) and (i) are equi valent.


Page 1 of 10



139

Clearly, (a) implies Q), wi th C =A - •. Conversely. assume Q). Let b be any vector in 7?./' and v = C b. Then Av = A(C b) = (AC)b = l, b = b.

Ul implies (e). Since (e) implies (a), it follows that Ul impl ies (a). Hence (a) and Gl are equivalent. Therefore all of the statements in the Invertible Matrix Theorem are equ iva• lent.

It follows that

Statements ( i) and Gl of the Invertible Matrix Theorem are the two conditions in the definit ion of invertibility. It follows from the In vertible Matrix Theorem that we need verify only one of these conditions to show that a square matrix is invertible. For example, suppose that for a given 11 x 11 matrix A , there is an 11 x 11 matrix C such that AC = !, . The n A is invertible by the lnvenible Matrix Theorem, and so we may multiply both s ides of thi s equation on the left by A -L to obtain the equation A - L(AC) =A -•t,. which reduces to C =A - L. Similarly. if BA = 1, for some matrix 8, then A is invertible by the lnveLtible Matrix Theorem, and

Note that the matrices B and C in statements (i) and U) of the in vertible Matrix Theorem must be square. There are nonsquare matrices A and C for which the product AC is an identity matrix. For instance, let 1

A = [:

2

~]

Then AC = [:

1

2

and

~] [-~

C = [-~

-n [~ ~] =

-n=h

Of course, A and C are not invertible.

COMPUTING* A- 18 When A is an invertible 11 x 11 matrix and 8 is any 11 x p matrix, we can extend the algorithm for matrix inversion to compute A - L8. Consider the 11 x (11 + p) matrix [A B ]. Suppose we transform this matrix by means of elementary row operations into the matrix [/, C ]. As in the discussion of the algorithm for matrix inversion, there is an 11 x 11 inve1tible matrix P such that [/, C)= P[A 8] = [PA PB], from which it follows that PA = 1, and C = PB. Therefore P = A C =A - L8 . Thus we have the following a lgorithm:

I•

and hence

Algorithm for Computing A_, B Let A be an invertible 11 x 11 matrix and B be an 11 x p matrix. Suppose that the 11 x (11 + p) matrix [A B) is transformed by means of elementary row operations into the matrix I/, C J in reduced row echelon fonn. Then C =A- LB .

• The remainder of this section may be omitted without loss of continuity.


Page 2 of 10



Example4

Use the preceding algorithm to compute A -• 8 for

A=

I 2

[2

2 5

and

4

Solution

We apply elementary row operations to transform (A 8 J into its reduced row echelon fo1m, [!3 A- 18].

[A 8]

=[

1

~~

I 0

2~ - I ] 3

-2q -2r1

+ rl -~ + r.l

2 I

)I

2

- I ]

I

4

- 4

- 1 -3

[~ ~ ~I !

5

-4

0

I

I 0

I

2 I

- 1I - 32

0 0

2

[ 0 0

0

r2 [ rJ

- 1

3 ++rr2 1- ._r r1 2 -rr,

.

[

- 51 ]

-4

4

2 0 -2 ) 0 I 0 I 0

0

4

3] I

- 4

] -4

Therefore

A- 18 =

[-4 I] I 4

I - 4

.

One of the appl ications of the algorithm for computing A - I B is to solve several systems of linear equations shari ng a common invertible coefficient matrix. Suppose that Ax = b,, I ::;: i ::;: k, is such a collection of systems, and suppose that X; is the solution of the ith system. Let

Then AX = 8. and therefore X =A - •s. So if A and B are as in Example 4 and b 1 and b2 are the columns of B , then the solutions of Ax = b1 and Ax = b2 arc

and

X2

=[

_!]-

AN INTERPRETATION OF THE INVERSE MATRIX

=

Consider the system Ax b , where A is an in vertible 11 x 11 matrix. Often, we need to know how a change to the constant b affects a solution of the system. To examine what happens, let P A-._ Then

=

[e 1 e2 ... e,.j


= 1,. = AP = A[p 1

P2 . . . p"]

=

[A p 1 Ap2 .. . Ap,. j.

Page 3 of 10



=

141

=

It foll ows that for each j, A pi ei. Suppose that u is a solution of Ax b, and assume that we change the kth component of b from bk to bk + d , where dis some scalar. Thus we replace b by b + d ek to obtain the new syste m Ax b + dek . Notjce that

=

so u + d pk is a solution of Ax = b + dek . This solution differs from the original solution by d pk; that is, d pk measures the change in a solution of Ax= b when the kth component of b is increased by d . An example in which the vector b and the resulting solution are changed in thi s way is provided by the Leontief input-output model (discussed in Section 1.5). When an economy is being planned, the gross production vectors cotTcsponding to several different demand vectors may need to be calculated. For example, we may want to compare the effect of increasing the demand vector from

to

If C is the input-output matrix for the economy and

/ 3-

C is invertible, 4 then

=

such comparisons are easi ly made. For in this case, the solution of (/3 - C)x d 1 is (/3- C)- 1d t. as we saw in Section 2.3. Hence the gross production vector needed to meet the demand d z is

where p 1 is the first column of (/3 - C)- 1 . For the economy in Example 2 of Section 1.5, we have

(13 -

c)-' =

1.3 0.4 75 0.25] o 6 t.950 o.5o [ 0.5 0.375 1.25

1 (13 - C)- dt

and

=

170] 240 . [ 150

So the gross production vector needed to meet the demand d 2 is

1 (13 - C)- d 1

Example 5

+ 10p 1 =

170] 240 [ 150

+ 10

[1.3] 0.6 0.5

=

[183] 246 . 155

For the input-output matrix and demand in Example 2 of Sectjon 1.5. determine the additional inputs needed to increase the demand for services from $60 million to $70 million.

Solution The addi tional inputs needed to increase the demand for services by one unit are given by the third column of the preceding matrix (/3 - c)- 1 . Hence an increase of $10 mmion in the demand for services requi res additional inputs of

10 0.25] 0.50 [ 1.25 4

=[

2.5] ' 5.0 12.5

In practice, the sums of the entries in each column of Care less than t, because each dollar's worth of

output normally requires inputs whose total value is less than S1. In this situation. it can be shown that In- C is invertible and has nonnegative entries.


Page 4 of 10


142


that is, additional inputs of $2.5 million of agriculture, $5 million of manufacturing, and $12.5 million of services. Practice Problem 3 ~

Let A be an inve11 ible 11 x 11 matrix and b be a vector in 1?..". Suppose that x 1 is the solution of the equation Ax = b . (a) Prove that for any vector c in 1?..", the vector x 1 + A- 1c is the solution of Ax= b + c. (b) For the input-output matrix and demand vector in Example 2 of Section 1.5. apply the result in (a) to determine the increase in gross production necessary if the three demands are increased by $5 million for agriculture. $4 million for manufacturing, and $2 million for services. ..,.

EXERCISES In Eurcises 1- 18, detennine whether each matrix is inrertible. If so. find its im•erse.

1.

[: ;]

2.

3.

[-~ -~]

4.

[; ;]

6.

5.

7.

9.

-2 0 - 1

[: :] [i !]

8.

- 1

10.

3

j]

- 1 11.

13.

15.

[:

-3

[~

2

7

-~]

-I

-I

;] [_; -:] [~

[-~

[~ [~

-I

12.

14.

0

16.

I

1 1 0

1

2

I

- I

0

3

18.

-2 -5

-I]

8

- 1 2 -1 1 - 1 -2 -3

19.

A = [~

i]

and

8 -- ['1


0

7]

0

I 0 0

~

= [ -I

and

-~] -I

8 =

and

8 =

n

and

~] ~] _:]

8 =

0

2 6 I 0 2 5 0 0 0

-I I

-I I

l

-I -I I

[i

-~]

2 8

4 -2

8 =

and 8 =

I

[~ -~]

[~

and

0

[! [!

-2] - 1 2 0

-n

1

2 -2

-~]

-I 0 -2 -I 2 4 - I

and

8 --

[i

I I 0

3

-j]

and

-3

I

-8

3

5 2

-2

_:]

- I -3

In Erercises 27- 34, a matrix A is given. Determine (a) the reduced row echelon form R ofA and (b) an invertible matrix P such thai ~\ = R.

27.

[-~

In £rercises 19- 26, use rl1e algt•rithm for comiHifing A_ , 8 . -I

I

2

[l

-2

2 5 3 4

-:

0 ~

u '=

_:]

2 4

3 1

[ -2

['

2~. A =

-2 -3

[t

-2

2

[l ~] -2

-1 - 1

24. A = [ :

!]

0

A = [~

23. A=

n

3

[j j] [j [j l [-j 0 1 0

22.

- I

0

17.

3 5

~]

21. A= [;

-~]

_;]

20. A = [-1 2

29.

[- 1

~

-I

-~]

0 I

2 I

3

-I

-:] -5

28.

[: -n

30.

[~

- 1 0

-2 -4

-I I

-n Page 5 of 10


2.4 The Inve rse of a Matrix

0 3 1.

32.

33.

34.

- I

[-1

-2

- I I

[-l

[j

2

3 0 I -3

-5

I 2 - I - I

0

- 2 3

[-l

7 -3

0

- I

-I I

- I

143

52. If the reduced row echelon fo rm of [A /, ]is [R 8]. then B is an invertible matrix.

-!]

53. If the reduced row echelon form of [A 1, ] is [R 8], then BA equals the reduced row echelon form of A. 54. S uppose that A is an invertible matrix and u is a solution

-I -2 4

:]

of A< =

m

m

TI" •ol,foo of A< =

dfff= rn>o •

by 2p 3, where P3 is the third column o f A-

_j]

1

•

55. Prove directly that statement (a) in the Invertible Matrix Theorem implies statements (e) and (h).

-3 -8

I 3

5

-2

_:]

2

-I

-3

In Exercises 56- 63, a sysrem of linear equarions is given. (a) Wrire each sysrem as ammrix equarion Ax = b. (b) Show rlwr A is invertible. and find A-

1

•

1

(c) Use A- ro solve each sysrem. ~

In Exercises 35- 54, determine wherlter rile sraremenrs are rrue or false.

~

35. A matrix is invertible if and o nly if its reduced row echelon form is an identity matrix. 36. For any two matrices A and B, if A8 itive integer 11 , then A is invertible. 37. For any two BA 1,.

=

11

=1,

for some pos-

x n matrices A and B. if AB = 1,, then

38. For any two 11 x 11 matrices A and B, if AB is invettible and A- t = B.

= 1,, then A

39. If an

11

x

40. If an

11

x n matrix is invertible. then it has rank

11

matrix has rank

11 .

11.

42. lf A is an 11 x 11 mattix such that the onl y solution of Ax = 0. then A is in vertible. 43. An 11 x 11 matrix is invettible if and o nly if its columns are linearly independent. 44. An 11 x 11 matrix is invertible if :md only if its rows arc linearly independent. 45. If a square matrix has a column consisting of all zeros. then it is not invertible. ;.-~

row consisting of all zeros. 1hcn

47. Any invertible matrix can be wrillen as a product o f e lementary matrices. 48. If A and 8 a re invertible invertible.

11 x

=9 =3

=

=

50. If A is an invertible 11 x 11 matrix and the reduced row eche lon fo m1 of lA B ] is [/, C ], then C = s - tA.

51. If the reduced row echelon fonn of [A /,] is [R B]. then 1•


-x 1 - 3x2 2Xt + 5X2

59.

-XJ + X3 -4 3 XJ + 2Y2 - 2x3 2rt - X2 + X3 =

6 1.

-X t -

Xt

-

2xt · - Xt +

X3

X2

+

X3

X2

+ X3 + X,J = -}

Xt -

63.

Xt

+

-x1 -3xt +

2x2 -

X,J

X) +

X2

x2 X2

X,J

-

+

x3 + 2r3

X,J

+ xo~

(a) Verify that A2

-

=

4

= -2 = I =-I

a

64. Let A = [ :

-6 2Yt + 3x2- 4x3 x 2 + 2r,~ = 5 -X2 + XJ = 3

= 3 = -2 = 4

XJ + x.J

X2 -

+

= =

=

=

-5 x 1 +x2 +x3 60. 2rt + xz + x.1 = -3 3xt 2 + x3 =

62

= -6 = 4

57.

=

3A + l2 = 0.

(b) Let B = 3/2 -A. Use 8 to prove that A is in vertible and 8 = A- 1• 65. Let A = [ _

~ ~ -

_

n matrices. then A + 8 is

49. If A is :m n x " matrix s uch that Ax b is consistent for ever)' b in 'R." , then Ax b has a unique solution for every b in n•.

8 =A -

Xt + 2x2 2Xt + 3X2

4 XJ + X2 + X3 58. 2XJ + X2+4XJ= 7 3xt + 2xz + 6x3 = - I

then it is invertible.

41. A square matrix is invertible if and only if its reduced row echelon fonn has no zero row.

46. If a square matrix has it is not invertible.

56.

n

=

(a) Verify that A 3 - 5A 2 + 9A- 413 0. I (b) Let 8 = ;j(A 2 - 5A + 9/j). Use 8 to prove that A is invertible and 8 =A -

I.

(c) Explain how 8 in (b) can be obtained from the equation in (a).

66. Let A be an

11

x 11 matrix such that A2 "" 1, . Prove that A

is invertible and A-

1

=A.

Page 6 of 10


144 CHAPTER 2 Matrices and Linear Transformations 67. Let A be an n x 11 matrix such that AA = 1. for some posttl\·e integer k. (a) Prove that A is invertible. (b) Express A -

I

as a power of A.

68. Prove that if A is an m x 11 matrix and P is an invertible 111 x m matrix. then rank PA = r:mk A. /lim: Apply Exercise 62 of Section 2.3 to PA and to P - I (PA). 69. Let 8 be an

11

x p matrix . Prove the fol lowing:

78. In Exerci\e 20(c) of Section 1.5. how much is required in addition:tl inputs from each sector to increase net production in the nongovernment sector by S I million? 79. In Exercise 21(b) of Section 1.5. how much is required in additional inputs from each sector to increase the net production of scrvtces by $40 million? 80. In Exercise 22(b) of Section 1.5. how much is required in udditional inputs from each sector to increase the net production of manufacturing by $24 mill ion?

(a) For any m x 11 matrix R in reduced row echelon fonn. rank R8 !:: rank R. Him: Usc the definition of the reduced row echelon fonn. (b) If A is any m x 11 matrix. then rank A8 !:: rank A. /lim: Use Exercises 68 and 69(a).

8 I. Suppo;.e that the input- output matrix C for an economy i> ;.uch that 1. - C is invenible and every entry of (1. -C) t is positive. If the net production of one panicular sector of the economy must be increased. how does this affect the gross production of the economy?

70. Prove that if A is an m x 11 matrix and Q is an im·ertible 11 x 11 matrix. then rank AQ = rank A. /lim: Apply the re•ult of Exercise 69(b) to AQ and (AQ)Q - 1•

82. U;c matnx transposes to modify the algonthm for computing A- 18 to devise an algorithm for computing A8 - t. and JU'tify your method.

71. Prove that for any matrix A. rank AT= rank A. Him: u,e Exercise 70. Exerci'e 87 of Section 2.3, and Theorems 2.2 and 2.3.

83. Let A be un m x n matrix with reduced row echelon form R .

72. Usc the In vertible Matrix Theorem to prove that for any subsetS o f 11 vectors in R.". the setS h. linearly independent if and only if S is a generating set for

nn.

7 3. Let R and S be matrices with the same number of rows. and suppose that the matrix [R S I is in reduced row echelon fonn. Pro,·e that R is in reduced row echelon fonn. 74. Consider the system of linear equations Ax

A=

[i

2 and

3 4

b

= b. where

20] .

= [!~

(a) Solve this matrix equation by using Gaussian eliminatlon.

(b) On a TJ-85 calculator. the value of A_, b is given as

=

(u) Prove that if rank A m, then there is u unique m x m matrix P such that PA = R. Furthermore. I' is invertible. /lim: For each j. let u; denote the jth pivot column of A. Prove that the m x m matrix U (u 1 u 1 . . Um] is invertible. Now let PA = R. and show that P u-t.

=

=

(b) Pro' e that tf rank A < m. Lben there is more than one invcnible m x m matrix P such that PA = R. flint: There is an elementary m x m matrix E. distinct from 1,.. ;.uch that ER = R.

Let A and 8 ~ 11 x n matric~s. \\'1> say that A is similar ro 8 if 8 = P 1AP for Willi! im•t'rtible matrix P. Exercises 84 88 are co11cemed with this relation. 84. Let A, fl. :tnd C be statements:

11

x

11

matrices. Pwvc the following

(a) II is similar to A. (b) If II is similar to 8, then 8 is similar to A. (c) If II is similar to 8 and 8 is similar to C. then A is Mmilar to C. But this is

11ot

a solution of Ax

= b . Why?

75. Repeat Exercise 74 with

matrix.

=1•. 11

t.ero matrix.

(c) Suppose thut 8 = cl., for some •calar 1· . (l11e mutrix 8 is called u ~calor matri..r.) Whut can you suy about A if A is simihtr to 8 ?

76. Repc:tt Exercise 74 with

[i

II X II

(a) Prove that if A is similar to 1•. then A (b) Pro'e that if A is similar to 0. the 11 x then A= 0.

2 3 7

A=

85. Let A be an

2

5 8

77. In Exercise 19(c) of Section 1.5. ho" much is required in additional inputs from each sector to increa.~e the net production of oil by $3 million?


86. Suppose that A and 8 are 11 x 11 mutrices such that A is similar to 8 . Prove that if A is inveniblc. then 8 is invertible. and 11 - t is similar to 8 - 1 . 87 . Suppo-c that A and 8 arc 11 x n matrice' such th:ll ;\ h similar to 8 . Prove that AT is similar to 8 T. 88. Suppo-c that A and 8 are n x 11 matrices 'uch that A is ~imtlar to B. Pro\·e that rank A = rank B. /lim: Use Exerci>es 68 and 70.

Page 7 of 10


2.4 ThelnverseofaMatrix

In Exercises 89- 92. use either a calcularor with marrix capabiliries or compwer softwarr! such os MATI.A8 ro solve each problem.

145

91. Show that A is invertible by computing its rank and using the Invertible Matrix Theorem.

92. Show that the matrix

Exercises 89- 9 I refer to rhe matrix 5 8 6 9

l

6

9

5 7

89. Show that A is invertible by c-omputing its reduced row echelon form and using Theorem 2.5. 90. Show that A is invertible by solving the system Ax = 0 and using the Invertible Matrix Theorem.

2 3 4

-1 2 -1

6

-2

l

is inve11ible. lllustrate Exercise 68 by creating several random 4 x 4 matrices A and showing that rank PA = rank A.


I. The reduced row echelon f01m of A is

G~

ll

=

Since this matrix is not / 3 , Theorem 2.5 implies that A is not invertible. The reduced row echelon form of 8 is / 3, so as Theorem 2.5 implies, 8 is invertible. To compute 8 - t, we find the reduced row echelon form of [8 / 3 ].

0 4

I

0

0

I

r1 + r-3

~

r.\

[~I I

5 - 5r2+r , - r 1

[

~

-I I

0

I

I 4

-3 0

I

-I

-3 16

I

0

_ -_•J_-_•.,_ [~

0

I I

0 0

3. (a) First observe that Xt = A-t b . If x1 is the solution of Ax = b + c. then

I

13

-16

0 0 I

T hus the unique solution of the given system of linear equations is Xt = 5, x 2 = -1 , and x3 = -2.

-3 -16

I

0

[~ [~

(c) The solution of Ax = b is

~]

4 4 0 0 - Jq + •2- rz

( b) Because the reduced row echelon of the 3 x 3 matrix in (a) is 13• the matrix is invertible.

0 0

-12 13

- 16

0

(b) In the context of (a). let A = l3 - C. where C is the input- output matrix in Example 2 of Section 1.5 and b = d , the demand vector used in that example. The increase in the demands is given by the vector c=

Gl

and therefore the increase in the gross pro-

duction is given by

11ms8-t= [-:~

_:

- 16

5

-:].

- 1

2. (a) The matrix form of the given system of linear equations is


=

1.3 0.6 [ 0.5

0.475 1.950 0.375

0.25] [5] = [ 8.9] 0.50

4

11.8

1.25

2

6.5

.

Page 8 of 10



12.5* I PARTITIONED MATRICES AND BLOCK

MULTIPLICATION Suppose that we wish to compute A 3 . where

A=

I

0

0

I

00 00]

6 8 5 0 . [- 7 9 0 5

This is a laborious process because A is a 4 x 4 matrix. However. there is another approach to matrix multipl ication that simplifies thi s calculation. We start by writing A as an array of 2 x 2 submatrices.

A=

[-~;,- -,!. -t-~:;,- -,~'" "l -7 9 0 5

We can then write A more compactly as

A= [/82 Sh 0] ' where

h= [~ ~]

and

8=[-~ ~].

Next. we compu te A2 by the row-column rule. treat ing / 2, 0, B, and they were scalar entries of A. A2

= [/2 8

Sh as if

0] [/28 5h0] = [8!2/2/z++(S/2)8 OB

S/2

Finally,

/2

= [ 318 This method for computing A3 requ ires only that we multiply the 2 x 2 matrix B by the scalar 3 1 and / 2 by 53. We can break up any matrix by drawing horizontal and vettical lines within the matrix , which divides the matrix into an array of submatrices called blocks. The resulting array is called a partition of the matrix, and the process of forming these blocks is ca lled partitioning. • This section can be omitted without loss of continuity.


Page 9 of 10


2.5 Partitioned Matrices and Block Multiplication

147

While there are many ways to partition any g iven matrix, there is often a natural prutition that si mplifies matrix multiplication. For example, the matrix

A= [2 0 I

02

2 0

3

-i]

can be written as

A=

[ 2 0

0 2

I

3

-1]

I

2 0

3

.

0

The horizontal and vertical lines prutition A into an array of four blocks. The first row of the pa11it ion consists of the 2 x 2 matrices

2h

= [~ ~]

[I -1]

and

2

3 •

and the second row consists of the I x 2 matrices 11 31 and 0 = 10 0]. We can also pm1ition A as

2 0 11 0 2 2 [ I 3 0

-1] 3 0

.

In this case, there is only one row and two columns. The blocks of this row are the 3 x 2 matrices

[I -1]

and

2

3 .

0

0

The first pmtition of A contains the submatrices 2/z and 0 , which are easily multiplied by other matrices, so this partition is usually more desirable. As shown in the preceding example, prutit ioning matrices appropriate ly can simplify matrix multiplication. Two pa11itioned matrices can be multiplied by treating the blocks as if they were scalars, provided that the products of the individual blocks are de fined.

l§ifi,!.!tjl

Let

A =

[_;~1:_~0::o+---:-3~-_;~ :._] - 1

and

We can use the given partition to find the entries in the upper left block of AB by computing

J [ 4 62] [20 -1 ]_[4 [0l 3]5 [I O 2 + 3 - 5 1


- 1

Page 10 of 10



Similarly, we can find the upper right block of AB by computing

We obtain the lower left block of AB by computing

[I 01 [:

~] + [3 - 11[~ -~]=[1

0) + [6 - 6)=[7 - 6].

Finally, we obtain the lower right block of AB by computing

I

I OJ[~]+ [3

- I][~] =

[3J+ [SJ = [81.

Putting these blocks together, we have

AB =

12

8

;

29 -6

[

In general, we have the following rule:

Block Multiplication Suppose two matrices A and 8 are partitioned into blocks so that the number of blocks in each row of A is the same as the number of blocks in each column of B. Then the matrices can be multiplied according to the usual rules for matrix mul tiplication, treating the blocks as if they were scalars, provided that the individua l products arc defined.

TWO ADDITIONAL METHODS FOR COMPUTING A MATRIX PRODUCT Given two matrices A and B such that the product AB is defined, we have seen how to compute the product by using blocks obtained from partitions of A and B. ln this subsection, we look at two specific ways of partitioning A and B that lead to two new methods of computing their product. By rows Given an m x n matrix A and an n x p matrix 8, we partition A into an m x I array of row vectors a;, a~, ... , a;,. and regard B as a single block in a I x I array. In this case, AB=

1

a;l [aa;s'Bl [a:.. a;,.s a~

B=

.

(7)

Thus the rows of AB are the products of the rows of A with B. More specifically. the ith row of AB is the matrix product of the ith row of A with B.

M;:f! .. l.!t!fM

Let A= [


I

- I

2

and

B

-2

I

=[ :

-3 - I

~].

- I

Page 1 of 10



S ince

a;B=[I

2

- 1] [-~ -~ ~]=[- 1 I

- 1

149

- 4 9]

- 1

and I

a;B = [- 1

-3 - I

we have AB =

[

-I 6

~] =[6

- I

-7I] '

-4 -7

It is interesting to compare the method of computing a product by rows with the definition of a matrix product, which can be thought of as the method of computing a product by columns.

By outer products Another method to compute a matrix product is to partition A into columns and B into rows. Suppose that a1, a2... . , a, are the columns of an m x n matrix A, and b;, b2.. .. . b;, arc the rows of an n x p matrix B. Then block mult iplication gives

AB

= [a 1 a2

(8)

Thus AB is the sum of matrix products of each column of A wi th the corresponding row of B. The terms a; b; in equation (8) arc matrix products of two vectors. namely. column i of A and row i of B. Such products have an especially simple form. In order to present this result in a more standard notation, we consider the matrix product of v and wT, where

v'] [v~, vz

V=

and

It follows from equation (7) that

Thus the rows of the m x 11 matrix vwT are all mu ltiples of wr. If follows (see Exercise 52) that the rank of the matrix vwT is I if both v and w are nonzero vectors. Products of the form vwr , where \' is in R"' (regarded as an m x I matrix) and w is in 1?/' (regarded as ann x I matrix), are called outer products. In this terminology,


Page 2 of 10



equation (8) states that the product of an m x 11 matrix A and an 11 x p matrix 8 is the sum of n matrices of rank at most I, namely, the outer products of the columns of A with the corresponding rows of B. ln the special case that A is a I x n matrix. so that A = (a 1 a2 ... a,] is a row vector, the product in equation (8) is the linear combination

of the rows of 8 with the corresponding entries of A as the coefficients. For example,

[2

Example 3

3)[-~ ~] =2[- 1

4]+3[5 0]=[13 8].

Use ou ter products to express the product AB in Example 2 as a sum of matrices of rank I.

Solution

First form the outer products in equation (8) to obtain three matrices of

rank I,

-6 8]

-3 4 .

and

[-'] 3

(I

- I

- IJ

=

[-' 3

I

-3

Then I - I

-3

as guaranteed by equation (8).

We summarize the two new methods for computing the matrix product AB.

Two Methods of Computing a Matrix Product AB We assume here that A is an m x n matrix with rows a'1• a; ..... a;,. and 8 is an x p matrix with rows b'1 , b~ • . . . , b;, .

11

I. By rows The ith row of AB is obtained by multiplying the ith row of A by

B-that is, a;s. 2. By outer products The matrix AB is a sum of matrix products of each column of A with the corresponding row of 8. Symbolically,


Page 3 of 10



151

EXERCISES In Exercises 1- 12, compwe the product of each partitioned marrix using block nmltiplication.

-~I:J

1. [-1 3 I] [

2. [

3. [

4.

I 0

-I I

0 2

A=

[~

[ ~ -~ !] , = [-! ~] , B

-3

c --

J[i]

-2

-1 1

0

3

- 2

[24 ~ =~J

13. row I of AB 15. row 2 of CA 17. row 3 of BC 19. row 2 of A2

I 1n [~ J

b

In Exert:i.fe.s 13- 20, compute the indicated row of the given product without compwing rite entire matrix.

14. row I of CA 16. row 2 of BC 18. row 2 of B T A 20. row 3 of A 2

In Exercises 21- 28, use the mmrices A. B, and C from Exercises 13-20.

-: 1n [+J

21. Use outer products to represent AB as the s um of 3 matrices of rank I.

22. Usc outer products to represent 8C as the sum of 2 matri5.

6.

[-lll [ [-lll [

- 1 2 2 2

3

0

-1

2

ces of rank I. 23. Use outer products to represent CB as the s um of 3 matrices of rank I.

24. Use outer products to represent CA as the sum of 3 matri-

ces of rank I.

- 1 2 2 2

3

-1

1n

25. Use outer products to represent 8 r A as the sum of 3 matrices of rank I.

26. Usc outer products to represent ACT as the sum of 3 matrices of rank I.

7.

8.

[-lll [

- 1 2

un[

I~

3

0

-1

2

2

0

0

0

0 0

0

1 -~ J

12. [

0

~~~:

2 0 0

] ( A40•

~

~

In Exercises 29- 34. determine wlwther the stmemems are true or false.

29. The definition of the matrix product AB on page 97 can be regarded as a special case of block multiplication .

-n [#]

[i i J[! I

11.

2 -2

matrices of rank I.

28. Usc outer products to represent CAT as the sum of 3 matrices of rank I.

·un[: ;] 10. [

27. Usc outer products to represent Ar 8 as the sum of 3

0 I

0 0

I Aw


-I I 0

30. Let A and B be matrices such that AB is defined, and let A and B be partitioned illlo blocks so that the number of blocks in each row of A is the Siimc as the number of b locks in each column of B. T hen the matrices can be multiplied according to the usual ntle for matrix multiplication by treating the blocks as if they were scalars. 31. The outer product vw7 is defined only if v and w are both

in'R.". 32. For any vectors v and win 'R."' and 'R.", outer product \ ' WT is an m x 11 matrix.

re.~pectively,

the

33. For any vectors v and w in 'R."' and 'R.", respectively, the outer product \ ' WT is an m x 11 matrix with rank I. 34. The product of an m x 11 nonzero matrix and an 11 x p nonzero matrix can be written as the sum of at most 11 matrices of rank I.

Page 4 of 10


15 2


In Exercises 35-40, assume thm A. 8. C, and D are 11 x nmmrices, 0 is the 11 x 11 zem mmrix. ami, in Exen·ises 35- 36, A is invenible. Use block multiplicmio11 to ji11d each pmducr.

35. [A- 1 1,) 37.

39.

[~]

36. [AI, ' ] [A 1, )

[~ ~r

-~P

~pgPB

=[ I, invertible and P = (!,. - 8C)- 1.

48. Let A and 8 be

11

x

11

l

where 1, - BC is

matrices and 0 be the

38.

[~ ~] [~ ~]

~r[~ ~] 40. [I" ;][~ ~] 1,

41. Show that if A. B , C. and D are 11 x A is invettible.

11

matrices such that

for any positive integer k. 49. Let A and 8 be 11 x 11 matrices and 0 be the

for any positive integer k. 50. Let A and 8 be invertible 11 x [~

~]

x

11

zero

x

11

zero

~ ~r ~ ~r

matrices. Prove that

11

is invertible if and only if A - BA - I 8 is

11

x

11

matrices. then

[~ ~] is invertible. Find the inverse in terms of A- 1,

= [~ In this context, the matrix D - CA - t 8 is called the Schur complement of A. In Exercises 42- 47, assume that A. 8 , C, and Dare 11 x 11 matrices. 0 is the 11 x 11 zerv IIWirix. and A and D are invertible. Use block mu/riplicario11 10 verify each equation.

or'-

42.

[~

D

- [A0- I

43.

[~

Ar 0

=

o-' r'--

45.

[g

Ar 0

In Exercise 53, use either a calculator with matrix capabililies or computer sojtll'are such as MATLA8 to solve each problem.

53. Suppose that A is a 4 x 4 matrix in the block form.

A= [~

(a) Use a random matrix for A to illustrate that A 2 =

[~

2

[A -1

[0

A- 1

o

g].

where the blocks are all 2 x 2 matrices.

A- 1

[~

o-'. and 0. 52. Suppose a and b are nonzero vectors in n'" and n•. respectively. Prove that the outer product ab7' has rank I.

D~']

[0 o;'J

44.

=

11

matrix. Use block multiplication to compute [

invert ible. 51. Prove that if A and 8 arc invertible

8

11

matrix. Use block multiplication to compu1e [

[~ ~] [~ ~] [~

47.

-A - '8g]

; 2 ].

where

* represents

some 2 x 2 matrix.

(b) Use a random matrix for A to illustrate that A3

J

[

0 -1 -A - 1CD - 1

~

; 3 ].

where

* represents

=

some 2 x 2 matrix.

(c) Make a conjecture about the block form of Ak, where k is a positive integer. (d) Prove your conjecture for k 3.

=

46. [ OD

I

12.6* THE LU DECOMPOSITION OF A MATRIX In many applications , it is necessary to solve multiple systems of linear equations with the same coefficient mau·ix. In these situations. using Gaussian eli mination on each system involves a great deal of duplication of effort since the augmented matrices for these systems arc al most identical. In this section, we examine a method that avoids this duplication. For now, suppose that an m x 11 matrix A can be transformed into a matrix U in row echelon fonn withom the use of row interchanges. The n U can be w ritten as

• This section can be omitted without loss of continuity.


Page 5 of 10


2.6 The LV Decomposition of a Matrix

153

where £ 1, .• . , Ek - t, Ek are the elementary matrices corresponding to the elementary row operations that transfonn A into U. Solving this equation for A, we obtain

where

(9) Observe that U is an m x 11 matrix and L, which is a product of invertible m x m matrices. is an invertible m x m matrix. The matrices L and U have special forms, which we now describe. Since each elementary row operation used in the process of transforming A into U is the result of adding a multiple of a row to a lower row of a matrix, the corresponding elementary matrix £" and its inverse£;; ' are of the forms

0

row j

~

row i

~

Ep= c

0

t

columnj and

0

rowj

~

£ fl- 1 = rowi ~

-c

0

t

columnj where c is the multiple and j < i are the rows. Notice that £1, and £1; 1 can be obtained from !, by changing the (i ,j)-entry from 0 to c. and from 0 to - c. respectively. Since U is in row echelon form. the entries of U below and to the left of the diagonal entries are all zeros. Any matrix with this description is called an upper trian gular matrix. Notice that the entries above and to the right of the diagonal entries of each £;; ' are zeros. Any matrix with th is description is called a lower tria ngular matrix. Furthermore. the diagonal entries of each E fl- l are ones. A lower triangu lar matrix whose diagonal entries are all ones is called a unit lowe r triangular matrix. Since the product of unit lower triangu lar matrices is a unit lower triangu lar matrix (see Exerc ise 44), L is also a unit lower triangu lar matrix. Thus we can factor

A "" LU into the product of a unit lower triangular matrix L and an upper triangu lar matrix U.


Page 6 of 10


15 4


IJifi,!.!tjl

Let

A=

[~ ~l 0 2

8 =

4

c=

[~

0 1 0 3 0 3

-] 4

[! ~l

and

.

0

Both A and B are lower triangular matrices because the entries above and to the right of the diagonal entries are zeros. Both diagonal entries of B are ones, and hence B is a unit lower triangular matrix. whereas A is not. The entries below and to the left of the diagonal entries of C are zeros, and hence C is an upper triangular matrix.

Defin ition For any matrix A, a factorization A =LV. where L is a unit lower triangu lar matrix and V is an upper triangular matrix, is called an L V decomposition of A. If a matrix has an LV decomposition and is also invertible. then the LV decomposition is unique. (See Exercise 46.) Not every matrix can be transformed into a matrix in row echelon form without the use of row interchanges. For example, to put the matrix [

~ ~]

into row echelon

form. you must interchange its rows. If a matrix cannot be transformed into a matrix in row echelon form without the use of row interchanges, then the matrix has no LV decomposition.

COMPUTING THE LU DECOMPOSITION For the present, we consider a matrix A that has an LV decomposition. We describe a method for finding the matrices L and V, and show how to use them to solve a system of linear equations Ax b. Given the LV decomposition of A, the number of steps required to solve several systems of linear equations with coefficient matrix A is significant ly less than the total number of steps requ ired to use Gauss ian e limination on each system separate! y. We begin the process of finding an LV decomposition of A by using elementary row operations to transform A into an upper triangular matrix V. TI1en we use equation (9) to compute L from 1111 by applying the e lementary row operations corresponding to the E;; 1 's, starting with the last operation and working our way back to the first. We illustrate this process in the followi ng example :

=

Example 2

Find the LV decomposition of the matrix - 1

- I

-4 Solution First, we use Gaussian elimination to transform A into an upper triangu lar matrix V in row echelon form without the use of row interchanges. This process consists of three e lementary row operations performed in success ion: adding - 3 times row 1 to row 2 of A , adding -2 times row I to row 3 of the resulting matrix , and


Page 7 of 10



155

adding I times row 2 to row 3 of the previous matrix to obtain the final result. The details are as follows:

A=[~

- I - I

-4

n

- 3 r, + r2~r2

rz+ r;\-r.l

[~

- I 2

!]

-4

- 2 rl +r.l~ r .l

[~

- I

2 -2

u.

n

[i n - I 2 0

=U

The reverse of the last operation, adding - I times row 2 to row 3 of a matrix, is the first operation used in the transfonnation of /3 into L. We continue to apply the re verse row operations in the opposite order to complete the transformation of /3 into L

/3

=

[~

~ ~]-r2+r3~r3 0 I 3r 1+r2- fl

[~ [~

0 - I

0 I - I

~]

2r , + rJ- r.l

[~

0 - I

~]

()] ~ = L

We can obtain the entries of L below the diagonal directly from the row operations used to transform A imo U. In particular, the (i ,))-entry of L is - c, where c times row j is added to row i in one of the elememary row operations used to transform A into U. For example, in the first of the three elementary row operations used to transform A into U, - 3 times row I is added to row 2. Thus the (2, I )-entry of L is 3. Since - 2 times row I is added to row 3 in the second operation, 2 is the (3, I )-entry of L. Finally, I times row 2 is added to row 3 to complete the transformation of A to U, and hence - I is the (3, 2)-entry of L. These entries below the diagonal of L are called multipliers. We summarize the process of obtaining the LU decomposition of a matrix.

The LU Decomposition of an m x n Matrix A (a) Use steps I, 3, and 4 of Gaussian elimination (as described in Sec tion 1.4) to transform A into a matdx U in row echelon fonn by means of elementary row operations. If thi s is impossible, then A has no LU decomposition. (b) Whi le performing (a), create an m x m matrix Las follows: (i) Each d iagonal ent ry of L is I. (ii) If some ele mentary row operation in (a) adds c times row j of a matrix to row i, then l ;j = -c: otherwise. lij = 0.

Example 3

Find an LU decomposition of the matrix

A= [


-~

-2 2

4 2 -2

4

-~l 14

Page 8 of 10


15 6


Fi rst, we transform A to U.

Solution

A= [

-~

-2 2 4 2 -2 4

-~]

rt + rl-.Q

14

l-31rl+r.t -r.l [~

-2 2

0

4

2 4

[~

-2 2 2 4 -2 4

~ll-21rJ+r.l-•.l [~

-2 2

0

~~l -2 2 0

2 4

~] = u

- 10 -4

We arc now prepared to obtain L. Of course. the diagonal entries of L arc Is . and the entries above the d iagonal are Os. The entries below the diagonal, the multipl iers, can be obtained directly from the labels above the arrows in the tr
L= [


-l 0 0] I

0 .

2

I

Find an LU decomposition of

A=[-i

-l

-2 2 2

-~]I ·

USING AN LU DECOMPOSITION TO SOLVE A SYSTEM OF LINEAR EQUATIONS

=

G iven a system of linear equations of the form Ax b, where A has an LU decomposition A = L U. we can take advantage of this decomposition to reduce the number of steps required to solve the system. Since

Ax = LUx= L(U x) = b, we can set

Ux = y.

and hence

Ly = b.

The second system of equations is eas ily solved for y because L is a unit lower triangular matrix. Once y is obta ined, the first system can then be easily solved for x because U is upper triangular. (See Figure 2.8.) To illustrate this procedure, consider the system

x, 3x, -

x2 x2 2x, - 4x2 with coefficient matrix

+ 2r3 =

2

+ Sx3 =

10 4

+ ?x3 =

-12]

- 1 7 . - 4

5

The L U decomposition of A. which we obtained in Example 2, is given by

L=


[i _: ~]

and

u=

2] [~ -1 0 2 2

l

.

Page 9 of 10


2.6 The LV Decomposition of a Matrix 1 57

w• • Ux

=y

· ~----------------------~------------------_. multiply

Ax = Ly = b

byA = LU

Figure 2 .8

Solving a system of linear equations using an LU decomposition

The system Ax= b can be rewrinen as LU x = b . Set y = Ux so that the system becomes Ly b. where

=

y=

[~~]

and

.. 3

Thus we have the system

Yt 3y, + )'2 2y, - Y2

=

2

+ Y3 =

4.

= 10

The first equation gives us the value of y1 • Substituting this into the second equation, we solve for Yl to obtain )'2 = 10 - 3(2) = 4. Substituting the values for Yl and Yl into the third equation, we obtain )'3 = 4 - 2(2) + 4 = 4. Tims

y=

We now solve the system U x

[1l

= y, which we can write as Xt -

X2

lx2

+ 2x3 = 2 +

x3 = 4 4.

2x3

=

Solving the third equation, we obtain X3 = 2. Substituting this into the sc.c ond equation and solving for x2, we obtain x2 = (4 - 2)/2 = I. Finally, substituting the values for XJ and x2 into the first equation and solving for Xt gives Xt = 2 + 1 - 2(2) = - 1. Thus

This method of solving Ux = y is called back s ubs titution. If the coefficient matrix of a system of linear equations is not invertible-for example, if the matrix is not square-we can still solve the system by using an LV decomposition. In this case, the process of back substitution is compl icated by the presence of free variables. The following example illustrates this situation:


Page 10 of 10



Example4

Use LV decomposition to solve the system

2..r1 - 2x2 + 2.x3 + 4..r4 = 6 - 2..rt + 4x2 + 2r3 X4 4 6..r1 - 2r2 + 4x3 + 14x4 = 20.

=

Solution

The coefficient matrix of this system is

2 A=

2 4]

- 2

-2 4 2 [ 6 - 2 4

- 1 . 14

An LV decomposition of A was obtained in Example 3; it is given by

L=

[-t

0I 0] 0 2 I

and

v

2 ~l [~ -2 2 4

=

0 - 10 - 4

Solve the system Ly = b, where

y

=

[f~]

and

b

=

[2~l

As before, the unique solutjon of this system can be shown to be

y= [

~~]·

- 18

Next, we use back substitution to solve the system Vx = y, which we can write as 2rt - 2rz

2r2

+ +

2t3 4xJ

+ 4x4 = + 3x4 =

6

10

- IO..r3- 4x4 = - 18. We begin with the last equation. In this equation, we solve for the first variable ..r3, treating X4 as a free variable. This yields

Working our way upwards, we substitute thi s solution into the second equation and solve for the first variable in this equation, xz.

Hence


Page 1 of 10



159

Finally, we solve for x 1 in lhe fi rs! equalion, substituting the expressions we have already obtained in the previous equations. In this case, there are no new variables other !han x 1, and hence no additional free variables. Thus we have

2x,

= 6 + 2x2 =6+2 26

2x3 - 4x4

(2 -

!_x4) - 2 10

5

(~5 - 5~x4) - 4x4

23

= S - SX4. and hence

Pract ice Problem 2 .,..

Use your answer to Practice Problem I to solve Ax Practice Problem I and

= b . where A

is the malrix in

WHAT IF A MATRIX HAS NO LU DECOMPOSITION? We have seen that not every matrix has an LU decompos ition. Suppose that A is such a matrix. Then, by means of Gaussian elimination. A can be transfonned into an upper lriangular matrix U by elementary row operations lhat include row inlerchanges. II can be shown that if these row interchanges are applied to A initially, then the resulting malrix C can be trdnsformed into U by means of e lementary row operations that do not include row interchanges. Consequently. there is a unit lower triangular matrix L s uch LU. The matrix C has the same rows as A, but in a different sequence. Thus thai C there is a sequence of row interchanges that transforms A into C. Perfonning this same sequence of row inlerchanges on lhe appropriate identity malrix produces a matrix P such that C = PA. Any matrix P obtained by permuting the rows of an identity matrix is called a permutation matrix. So if A does not have an LU decomposit ion, there is a permutation matrix P such that PA has an LU decomposition. In lhe nexl example, we ill ustrale how 10 find such a permutation matrix for a matrix having no LU decompos ition.

=

i#ifi ..!.!tJW

Let

0 0 2 2 2 2 4] 2 A=

[

I

2

2

I

.

2 6 7 5

Find a permulation matrix P and an upper triangular matrix U such !hat PA an LU decomposition of PA for some unit lower triangular matrix L.


= LU is

Page 2 of 10



Solution

We begin by transforming A into a matrix V in row echelon form, keeping track of the elementary row operations. as in Example 2.

A~ [l

2 2 2

2 2 2

6 7

l [j r1 - r,

- rz+r., - r.l

2 2 2

2

6

2 2 7

j]

[j l 2 2

[!

2

0 2

2 0 3

2 2

- 2r, + r4- r4

- r2+r4-r4

[j

2

2 2 2

2

3

2 2

2 2

0 0 0

l i]

In this computation, two row interchanges were pe1formed. If we apply these directly to A, we obtain

A-[: -

2

2

2 2

I

2

2

2

6

7

l

rl '"r3

2 2 2 6

2

[i j] 2 2

r3++ r4

7

[j

2 2

2 2

6 7 2 2

l ~c

To find P, simply perform the preceding row interchanges on / 4 in the same order:

[ Then C

= PA

~ ~

0]

0I 0 - p

0 0 0 I I 0 0 0

-

has an LV decomposition.

To complete the process of obtaining an LV decomposition for the matrix PA given in Example 5, we could apply the methods described earlier in this section to PA. However, it is more e fficient to take advantage of the work we have already done. The next example illustrates how.

Example 6

For the matrices A, P , and V in Example 5, find a unit lower triangular matrix L such that PA = LU.

Solution The method used here is simi lar to the method in Example 3, but w ith one complication. u·. in then the multipliers So. for example, in - 2 times row I to

the process of transforming A to U, two rows are interchanged, that have a lready been computed are switched in the same way. the process of transforming A into V in Example 5, we added row 4. This gave us a multiplier 2 in the (4, I )-position. Then

several steps later. we interchanged rows 3 and 4, which moved th is multiplier to the (3, ! )-position. Since there are no more row interchanges, it follows that /31 = 2.


Page 3 of 10



161

A simple way of keeping track of these row interchanges is to temporari ly place multipliers, as they are created, in the appropriate positions of the intermediate matrices. These replace the actual entries, wh ich, of course. are zeros . Then, when two rows are interchanged, the multi pliers in these rows are also interchanged. So as not to confuse these mu ltipl iers with the actual zero e ntries, we place each multiplier in parentheses. Thus. enhancing the process used in Example S. we obtain the following sequence of matrices:

[' "]

[j j] 2 2 2 2 2 6 7

2

A=

0 I

2 2

2

2

2

I

2 6 7

s

2

rt - rJ

- r1 +ry~ r.,

ul [(~)

r.\~ r4

2 2 2

2 2 0 3

2 2

2 2

( I) ( I)

0

(I)

- 2ra+r.4- q

[i

j]

2 2 2 2 2 2 2 3

l

2 2

2

-r2 + r4 ~ r4

[i

( I) ( I)

2 0

j]

j]

I

Notice that if the entries in parentheses arc replaced by zeros in the last matrix of the previous sequence, we obta in V. Finally, we obtain L by us ing the entries in parentheses in the last matrix of the sequence. The other nondiagonal entries of L arc zeros. Thus

L=

[!

0 0 0 I 0

!]

u

and

[j j]

~

2 2 2 2 0 I 0 0

It is a simple matter to verify that LV = PA.


Find an LV decomposition of PA, where P is a pemllltation matrix and

A

= [-~I _;I 2

!]

- 1

-~

-1

2

- 2

- 1

.

We now use these results to solve a system of linear equations.

Example 7

Use the results of Example 6 to solve the following system of linear equations: ~·2

2xz

Xi

2r,


+ Zx3 + 4x4 = + 2xJ + 2x4 =

+ ~·2 + 2q +

X4

=

+ 6x2 + 7x3 + Sx4 =

-6 -2

3 2

Page 4 of 10


162 CHAPTER 2 Matrices and Linear Transformations Solution

This system can be wrinen as the matrix equation Ax

A =

00 22 22 42] [2I 62 72 5I

= b , where

and

In Example 6, we found a permutation matrix P such that PA has an LV decomposition. Multiplying both sides of the equation Ax= b on the left by P. we obtain the equivalent equation PAx = P b . Then PA = LV for the matrices L and V obtained in Example 6. Setting b' = Pb, we have reduced the problem to solving the system LV = b', where

0 0 I 0] [-6] [ 3] --0001 [ 0 0 0 3 - 2' 6

b' _ Pb _

0

I

0

-2

0

I

-2

2

As in Example 4, we set y = V x and solve the system Ly = b', which has the form

3

.\'I 2 )'1

Y2

= - 2

Y2

+ .\'4 = - 6.

+ Y2 + )'3

2

We solve this sys tem to obtain

J' --

[;:~]- [-~] )'3

-2 .

-

-4

)'4

Finally, to obtain the solution of the original system. we solve Vx = y, which has the form

X1

+ 2x2 + 2x3 + X4 = 2.x2 + 2x3 + 2x4 = X3 + X4 = -

3 2 2 2x4 = - 4.

Using back substitution, we solve this system to obtain the desired solution,

Practice Problem 4

~

Use your answer to Practice Problem 3 to solve Ax = b . where A is the matrix in Practice Problem 3 and

b --

[-~~] I

.

8


Page 5 of 10



16 3

THE RELATIVE EFFICIENCIES OF METHODS FOR SOLVING SYSTEMS OF LINEAR EQUATIONS Let Ax = b denote a system of n linear equations in 11 variables. Suppose that A is invenible and has an LU decomposition. Then we have seen three different methods for solving this system. I. Use Gauss ian e limination to transform the augmented matrix [A bj to reduced row echelon form. 2. Apply elementary row operations to the augmented matrix [A / 11 ] to compute A-t , and then calcu late A- 1b. 3. Compute the LU decomposition of A, and then use the methods described in this section to solve the system Ax = b. We can compare the relative efficiencies of these methods by estimating the number of arithmetic operations (add itions, subtractions, multiplications, and divis ions) used for each method. In calculations petformcd by computers, which arc required for matrices of substantial size, any arithmetic operation is called a flop (floating point operation). The total number of flops used to perform a matrix computation by a patticu lar method is called the flop count for that method. Typically. a flop count for a computation involving an 11 x n matrix is a polynomial in 11 . Since these counts are usually rough estimates and are sign ificant only for large va lues of n, the terms in the polynomial of lower degree are usuall y ignored, and hence a flop count for a method is usually approx imated as a multiple of a power of11. l11e table that follows lists approximate flop counts for various matrix computations that can be used in solving a system of 11 equations in 11 unknowns. Note that computing A- 1 to solve Ax= b is considerably less efficient than using Gaussian el imination or the LU decomposition.

Flop Counts for Various Procedures In each case, we have a system Ax

= b , where A is an 11 x 11 invertible matrix. A pproxim ate F lop Coun t

Procedur e computing the LU decomposition of A solving the system using Gaussian el imination solving the system given the LU decomposition of A calculating the inverse of A

If several systems with the same coefficient matri x are to be solved, then the 3 flops to solve each system, cost of using Gaussian elimination is approximately whereas the cost of using an LU decomposi tion involves an initial investment of approx imately ~n 3 flops for the LU decomposition followed by a much lower cost of approximately 211 2 flops to solve each system. For example, the cost of solving 11 different systems by Gaussian elimination is approximately

fn


Page 6 of 10


164 CHAPTER 2 M atrices and Linear Transformat ions

flops , whereas the cost for solving the same n systems by an LV decompositi on is approx imately

flops .

EXERCISES In Exercises 1-8, find an LV decomposition of each matrix. l.

2.

3.

4.

5.

6.

7.

8.

~~]

3 8 -4

[J

- I - I I

~]

[j

- I

-3 2

2 5

[~

- I

2

-3

5

[-l

-I 2 - I

[-l

4 - I 5 0 -I

[-l

[-: -2

- I

i]

-4

2 - 4

6 - 4

~] 2

I

0

-2

7

- I

-I -I

-~]

0 -3

- I -I

-8 5 2

13.

-x• + 2xz

5 -5

0 -I 4

II

-2 -5

j] -l] -2

=

9

2..r, -

X2 -

- x1

+ x2

8x3 - X4 + 5x3 + x.

x2

+ 2x3 + X4

-x• 16

.

17.

18.

19.

21.

2xJ- X2 4XJ - X2

+ 2x3 + x• = I 3x2 + 5x3 + 4x4 = 8

+ 2..rz +

5x5 + 4xs + 3xs -

.t-4

XJ -

+

+ +

I

8 2x6 = -5 4x6 = -2

+ 3xs =

7

=-7 -2x1 + 6x2 - XJ - 5x4 + 7x5 = 6 -x1 - 4x2 + 4x3 + l lx4 - 2..rs = I I 4x2

x1 -

=5


[j [~ [j [-~

[-l

['

- 2

X2

2x1 -3x 1 + 2..r2 - 4x3

2xs

. 15

+ X3 = -I + 4XJ = - 2 +x2 + 2xJ = -2

I I.

+ 3xs = -4

14.

2x1 + Jx2 + 9. 6x1 + 8x2 + l0x3 = 4 -2x•- 4x2- 3x3 = 0

XJ -

x4

2x4 -

+ 7X3- X4 + XS = -2 Jx1 + X2 X3 + X4 = 0 6x1 + 4xz - XJ + 4x• = 15 -3x 1 - x 2 + 2..r3 - x4 = I 3x1 + 5x2 + 3x• = 21 x1 - 3x3 - XJ - 2x5 + x6 =

20.

4x3 =I

-2x•

-

X2

In Exercises 9-16. use the results of Exercises 1-8 to sol1·e each sysTem tif linear equations.

10.

+ 2x3 +

2XJ -

4 3

-I

x2

-

5.q - 5xs

In Exercises 17- 24, for eac/1 matrix A, find (a) a permwmion matrix P such that PA has a11 LV decompositio11 a11d (b) a11 LV decomposition of PA.

-l]

2

3:~: =3;~ ! ~!! ~;:: ! x1

-3

[_~

12.

22.

-~J

- I -2 2

-~]

2 6 3

- I I 2 - 2

-3

- 2

-I

4

-3

2 - I

I 2

_,]

~]

-I

-4 -3 4 I

-- I] I I

-I

3 2

-6 3

~ 9 -9 -3

l Page 7 of 10


165

2.6 The LU Decomposition of a Matrix

23.

[j

2 4 2

- I

5

3

[_j

37. An LU decompo,ition of every matrix is unique. 38. The process for solving U x tion.

- 2

0 2

4

2 2

I

I

2

-2

0

- 3

2

24.

-:]

row j. In an LU decomposition of A. the (i .j)-entry of Lis c .

-10. Suppose that. in transfom1ing A mto a matrix in row ech-

/11 £urr:ist!s 25 ·32. liSt! the rrsults of £terr:ises 17- 24 to soll·e each system of linMr t!qua11o11s.

25.

2.t, -

+ 212 -

-x,

I

.1 .1 -

+ Jr2 -!

Xi

27.

X2

+ 2r2 -

2rt

- Xi -

lr2 -

11 upper triangular matrices. Prove that AB is an upper triangular matrix and that its ith diagonal entry is a;; b1r.

=- 2

26. 2r 1 + 6.r2 Xi

42. U!t A and 8 be 11 x

2

2.\ 2 - X.\ = ,q

=

I

43. U!t U be an invertible upper tri:mgular matrix. Prove

=

21.1

,q

3x1 -

.14

U

I

+ Jx-1 = + 2r.1 = .r, + x2 - .n + .1.1 = 12- 211 .t 1

lrt - 412 + 3x 1 J12 + 2.n

.r, -

=

0

=

2 5

=-

=

44. U!t A and 8 be

- I 2 0

31.

+ 2.12 +

3xt 2.rt

+

lrt Xt

t-1

IJ X.\

+

+ lr2 - X\ + .it2 + 3.rt

It

32.

4X2

.1.1

lr-1

=

= = =

-17. U!t

3.r.~

-

5x~

8 12

5

= -8

using any row interchanges.

35. An upper triangular matrix i> one in which the entries above and to the right of the diagonal entries are all zeros.

decomposition of A. all the diagonal entries of


tn

-

I addi-

C b i, 211 2 •

34. If :1 matrix A has an LU decomi>O,ition. then A c:m be transformed into a matrix in row echelon form without

LU

c be an II X II matnx and b be a vector n·.

(b) Show that the approximate flop count ror computing

f11 Ext•rri.fe.< 3.1 41 , determine ll'hether tire state~ mems are tme orfal.
U are Is.

=

(a) Show that it requires 11 multiplications and 11 tions to compute each component of Cb.

3 2 - 4

~

36. In an

lower triangular mal rices.

46. Suppose that LU and L'U' are two LU decompositions for an inverttble matrix. Pro,·e that L L' and U U '. Thus an LU decomposition for an in' ertible matrix is unique. ffim: Use the result> of Exerci>es 42-45.

7

-

11

45. Prove that a square unit lower tri:mgular ma1rix Lis invertible and that L - • is also a unit lower triangular matrix.

+ 2.\2 + 2.\ 1 + 2.1• + rs = + 4Xl + ltJ + .1.1 + Xl + ,\) + 2.1.1 + 2ts =

- J.q - 2.\2

x

(b) Prove that if both A and 8 :1re unit lower triangular matrices. then AB is also a unit lower triangular matrix.

+ 412 - 611 =2 + ,\ 2 + ).rj + l\J = 7 2rt + 9.12 - 9xJ + X.! = II -lrt + Jr2 - 3.\1 =7

I1

11

(a) Prove that AB is a lso a lower triangular matrix.

2.\, 30. -2.\,

2rt +

th<~t

1

is an upper triungular matrix and that its ith diagonal entry is I /uu.

= 5 X.t + ,I 1 = - I

-.r2 + 4.tJ 28. - 2r, - J.r2 1 lr1

-x, + l12 29.

elon fonn. c time' rO\\ ' of a matrix " added to row j. In an LU decomposition of A. the (i.j)-entry of Lis -c.

-II. For every matrix A. there i< a permutation matrix P such that PA has an LU decomposition.

+ J1 .1 = 6 212 + .il .l = 9 X2

.\t -

is called back substitu-

39. Suppose that. in transforming A into a matrix in row echelon form. c time' row i of a matrix i< added to

j]

I

=y

48. Suppose we are given

11 system> of 11 linear equations in variables. all of which have as their coefficient matrix the :.amc invertible matrix A. E'timate the total flop count for solving all of thc:.e systems by first computing A - •, and then computing the product or A 1 with the constant vector of each system.

11

49. Suppose that A is an

111 x 11 matrix and 8 is an 11 x p matrix. Find the exact flop count ror computing the product AB .

50. Suppose that A is an 111 x

11 matrix. B is an 11 x p matrix. :md C is a I' x q matrix. 1l1e product ABC can be com-

puled in two way•: (a) First compute AB. and then mu ltiply this (on the right) by C. (b) Fin>t compute BC, and then multiply this (on the left) by A. U>e Exercise 49 to devise a strategy that compare, the two way, so that the more efficient one c:m be cho,en.

Page 8 of 10


166 CHAPTER 2 Matrices and Linear Transformations In Exercises 51- 54. use either a calculmor with matrix capabilities or complller software such as MATI.AB 10 solve each problem. 5 In Exercises 51 and 52, find 011 LU decomposition of the given nwtrix. -I 2 3 2 -I I 51. I 15 12

[-;

[ -6 -3 52.

-4

I

9 9 0

0

I

3

7

4

-4

2

-6 -2

-1~

2 2

-6

·ll

In Exetdses 53 and 54.for each matrix A.jind (a) a permlllation mmrLr P such thm PA lws 011 LU dew mpositia11 a11d (b) 011 LU decomposition of PA.

53.

2

-I

3

I

2

-I

0

3 - 1

0 2

[-l

4

" [i

.j]

I -2

-I

3

- I

2

- 3

6

-5 -3

2

4

4 - 4

3 2 2

i]

4

l1

SOLUTIONS TO THE PRACTICE PROBLEMS I. First we apply Gaussian elimination to transform A into an upper triangular matrix U in row echelon form without using row interchanges:

A= [

-~

- I I 0

2r1+r2- r2

-3r 1+ r3- r3

-2 2 2

-~]

[i

-I -I 0

-2 -2 2

[~

-I -I

-2 -2

3

8

[~

-I

We so lve this system to obtain

Next we use back substitution to solve the system U x = y, which can be written

-8]

Xt -

-7

2 <3 + 4x4

-8]

3rz+r.1- r.l

-8]

-~ =

L= [

-~

I -3

~l

2. Substituting the LU decomposition fo r A obtained in Practice Problem I. we write the system as the matrix equation LU x = b . Setting y = U x. the sys tem becomes Ly = b. which can be written

=- 3

)'I

-2yt + )'2 3yt - 3)'2

+ )'J

=

5 -8.

x2 XJ

u

Then L is the unit lower triangular matrix whose (i .j)entry, i > j , is - c, where cr1 + r; -> r; is a label over an arrow in the preceding reduction process. Thus

0

=- 3

= -I = -2.

Treating X4 as a free variable , we obtain the following general solution:

-7 25

Xt

-2 - 1 -2 0 2

2<3 - 8x4

X2 -

-x2 - 2<3 - 7x.

I

X4

=

-2 + X4 3- 3x•

= =

- I - 2f4 free

3. Using the method o f Example 6. we have

A=

[-~I

-2

2

-I

t)

3

-6 2

-I

•-+r.l

2r 1+ r2 - r2

2

_i] -2

[-j

['-'l

I - I 2 -6 2

-2 3

-I

I -I 0 3

-I

0 -6 2

-:] -2

-:] -2

s Caution! The MATLAB function lu does not compute an LU decomposition of a matrix as defined on page 154. (See page 568.)


Page 9 of 10


2.7 Linear Transformations and Matrices

.,_~_··-

/4 _ _

167

[! ; ~ !]

I (2)

0

~

[

[

Thus L

= [

( - 2)

I

I

(2)

-3

(-2~

(-1)

4. Using the permutation matrix P and the LU decomposi· uon of P,\ obtained in Practice Problem 3. we transform the system of equation~ Ax = b into

0

~

LUx= Pr\x

=

-63]I = [-183I] . Pb= [0I 00 00I 00]I [-1 0

-2 and

u-

[!

- I

-3

4

0 0

-2 0

-~] I

I

0

0

8

-6

This system can now be solved by the method used in Prac tice Problem 2 to ob1ain the unique solut ion .

4

X=

Finally. "e interchange the fir.t and third rows and the >econd and founh rows of / 4 to obtain P:

,;~.,] [-~I] [ =

.

j2.1 1LINEAR TRANSFORMATIONS AND MATRICES In Section 1.2. we defined the matrix-vector product Av. where A is an m x 11 matrix and ,, is in nn. The correspondence that associates to each vector v in nn the vector Av in R"' is an example of a function from nn toR"'. We define a function as follows: Definitions Let St and S2 be subsets of nn and nrn. respectively. A function / from St to S2. written f : St -+ S2. is a rule that as;ign; to each vector v in St a unique vector f(v) in S 2 • The vector f(v) is called the image of ,, (under /). The set St is called the domain of a function f. and the set Sz b called the codomain of f. The range off is defined to be the set of images f(v) for all ' ' in S t.

w

In Figure 2.9, we see that u and v both have w as their image. So w =f(u) and =f(v).

s,

C:

I

sl

•w ran Jet

)

•

domain

codomain

Figure 2 .9 The domain, codomain, and range of a function


Page 10 of 10



IJifi,!.!tjl

Define/:

n 3 ~ n 2 by the

rule

Then/ is a function whose domain is

f

So

([!]) m [!] =

[~] 1s the image of both

an image of a vector in com ponent.

i#if!,i.!tfW

n3

n 3 and codomain is n 2

Notice that

and

and [ _

~]

However, not every vector in

n2

1s

because every image must have a nonnegative second

Let A be the 3 x 2 matrix

~].

-I

n3by

Define the function TA: n2 ~

Notice that, because A is a 3 x 2 matrix and x is a 2 x I vector, the vector Ax has size 3 x 1. Also, observe that there is a reversal in the order of the size 3 x 2 of A and the "sizes" of the domain 2 and the codomain n 3 of T11 . We can easi ly obtain a formula for TA by computing

n

A definition similar to that in Example 2 can be given for any 111 x 11 matrix A, in which case we obtain a function TA with domain 'R" and codomain 'R"'.

nm

Defin ition Let A be an Ill X II matrix. The function TA: 'R" ~ defined by TA(X) =A x for all X in n" is called the matrix transformation induced by A .


-2] LetA=[~ - I 4 I

.

(a) What is the domain of TA?


Page 1 of 10


2.7 linear Transformations and Matrices

169

(b) What is the codomain of TA? (c) Compute TA

([~]).

v

We have already seen an important example of a matrix transformation in Section 1.2 usi ng the rotation matri x A =Au. Here. TA: 1?.2 -+ 1?.2 represents the function that rotates a vector counterclockwise by e. To show that the range of TA is all of 1?.2, suppose that w is any vector in 1?.2. If we let v A_ow (see Figure 2.10), then. as in Example 4 of Section 2.1 , we have

=

Figure 2.10 Every vector in 1?. 2 is an image.

So every vector w in 1?.2 is in the range of TA.

+;:f!,!.!ii

Let A be the matrix

So T,1: 1?.3 -+ 1?.3 is defined by

I 0

0 I

[0 0 We can see from Figure 2. 11 that TA(u) is the orthogonal projection of u on the xy·plane. The range of TA is the xy-plane in 1?.3.

["']

;: T_.(u) = ·~

Figure 2.11 The orthogonal projection of u on the xy-plane

So far, we have seen that rotations and projections are matrix transformations. In the exercises, we discover that other geometric transformations, namely, reflections, contractions, and dilations , are also matrix transformations. The next example introduces yet another geometric transformation.


Page 2 of 10



)'

)'

(- 1.6)

( 13.6)

T,,(u) X

X

(b)

(a)

Figure 2.12 A shear transformation

The next resu lt foll ows immed iately from Theore m 1.3.

THEOREM 2.7 For any m x are true:

11

matrix A and any vectors u and v in R". the following statements

=

(a) TA(u + v) TA(u) + TA(v). (b) TA(cu) cT11(u) for every sca lar c .

=

We see from (a) and (b) that TA preserves the two vector operations; that is, the image of a su m of two vectors is the s um of the images, and the image of a sca lar multiple of a vector is the same scalar multiple of the image. On the other hand, the function f of Example I does not satisfy e ither of these properties. For example,

f but

f

([~] + [~]) =! ([~]) = [:].

([~]) +! ([~]) m+ [~J [~J. =

=

So (a) is not satisfied. Also,

f

(2[~]) =f ([~])

= [ :]'

but

So (b) is not satisfied. Functions that do satisfy (a) and (b) of Theorem 2.7 merit their own definition.


Page 3 of 10



171

Definition A function T from R" to R"', wrillen T: R" -+ R"', is called a linear transformation (or simply linear) if, for all vectors u and v in 1?:' and all scalars c. both of the following cond itions hold: (i) T(u + v) = T(u) + T(v). (In this case. we say that T p reser ves vector ad dition.) (ii) T(cu) = cT (u). (In this case, we say that T preserves scala r multiplication.) By Theorem 2.7, every matrix transfonnation is linear. There are two linear transformations that deserve special attentjon. The first is the identity t r.msforma tion I : R" -+ R" , which is defined by I (x) = x for all x in R". It is easy to show that I is linear and its range is all of R". The second transformation is the zero transformation To: R" -+ R"', which is defined by To(x) = 0 for all x in R". Like the identity transformation. it is easy to show that To is linear. The range of To consists precisely of the zero vector. The next theorem presents some bas ic properties of linear transformations.

THEOREM 2.8 For any linear transformat ion T: R" -+ R"', the following statements are true: (a) T(O) = 0. (b) T(- u) = - T(u) for all vectors u in R". (c) T(u - v) = T(u) - T(v) for all vectors u and v in R". (d) T(a u + bv) = aT(u) + bT(v) for all vectors u and v in R" and all scalars a and b. PROOF

(a) Because T preserves vector addition, we have T (O) = T (O + 0) = T(O) + T (O).

Subtracting T(O) from both sides yields 0 = T (O). (b) Let u be a vector in R". Because T preserves scalar multiplication, we have T(- u) = T((- l)u) = (- I)T(u) = - T(u). (c) Combining the fact that T preserves vector addition with part (b), we have, for any vectors u and v in R", T (u - v)

= T(u + (-v)) = T(u) + T(- v) = T(u) + (- T (v)) = T( u)- T (v).

(d) Because T preserves vector addition and scalar multip]jcation, we have, for any vectors u and v in R" and scalars a and b, that T (a u

+ bv) =

T (a u )

+ T (bv) =

aT(u)

+ bT(v).

•

We can generalize Theorem 2.8(d) to show that T preserves arbitrary linear combinations. LetT: R" -+ R"' be a linear transformation. If Ut, u 2 , . . . , u k are vectors in R" and a 1, az, ... , ak are scalars, then


Page 4 of 10


172


Example 5

Suppose that T : R 2 ~ R 2 is a linear transformation such that

r([:]) m =

(a) Find T

([~]).

(b) Find T

([~])and T ([-~]).and use the results to determ ine T ([;~])-

Solution (a) Since

[~] =

3 [: ]. it follows that

T

(b) Observe that [

([~]) =T (3 [:]) =3T ([:])=3m=[~]. ~J= ~ [:] + ~ [ _: T

J

Hence

G[:] + ~ [ _;])

([~]) = T

= ~T ([:]) + ~T ([ _ :]) =

~

m~ -~J m. + [

=

Similarl y, T

([~]) = T

G[:] -~ [_:])

= ~T ([:]) - ~T ([ _ :])

=HiJ - H-~J=[-~J. Finally, T

([:~~]) =

T

(xl m+

X2

[~])

= x1T ([~]) +x2T ([~])

=x1(n +x2[-~J =


[3XJ -X2]. x1 + 2rz

Page 5 of 10




173

Suppose that T: 7<2 --+ 7<3 is a linear transformation such that

and

Determine T

([~~]).

Theorem 2.8(a) can sometimes be used to show that a function is not linear. For example, the function T: 'R--+ 'R defined by T(x ) 2x + 3 is not linear because T(O) = 3 # 0. Note. however. that afimctionf may satisfy the condition that f(O) = 0, yet not be linear. For example, the function/ in Example I is not linear, even though

=

/(0) = 0.

The next example illustrates how to verify that a function is linear.

Define T: 'R2

.....

v be vectors in 7<2 . Then u

T(u

+ v) = T

1

x~

'R2 by T

x

2

1

.

To verify that T is linear, let u and

111 = [u•J. v = [v• J . and u+ v = [ + v•J. So 112 v2 112 + v2 ([111 112

+ Vt]) = [2(111 + V t) - (112 + v2)] .

+ V2

IIJ

+ VJ

On the other hand.

So, T (u + v) = T(u) + T(v). Now suppose that c is any scalar. Then T(cu)

= T ([CIIJ]) - Cll2]. C/12 = [2cu1Cllt

Also, cT (u) =

C

[2UtIIJ- 112] = [2CIItCIIJ - C!/2].

Hence T(c u) = cT(u). Therefore T is linear.

Another way to verify that the transformation T in Example 6 is linear is to find a matrix A such that T = TA, and then appea l to Theorem 2.7. Suppose we let


Page 6 of 10


17 4


Then

SoT = TA. Now we show that every linear trdnsformation with domain 'R" and codomain '1<.111 is a matrix transformation. This means that if a transformation T is linear. we can produce a corresponding matrix A such that T TA.

=

THEOREM 2.9 Let T: R" -+ R"' be li near. Then there is a unique

A= [T(et) T(e2)

...

111

x 11 matrix

T(e11 ) ] ,

whose columns are the images under T of the standard vectors for 1<.", such that T (v) =Av for all v in 'R". PROOF

Le t A

= [T(ei) T (e2)

Vi]

T (e,)j. We show that T

= TA. Notice that

V?

v=

[

:-

= v1 e1 + v2e2 + · · · + v,e,

VII

for any v in 'R"; so we have T (v) = T(v,el + v2e2 + · · · + v, e,)

= VtTCet ) + v2T Ce2) + · · · +

v, T (c")

=A v

=

Therefore T TA. To prove unique ness. suppose that TA To for some m x 11 matrix 8. Then Av = Bv for every vector v in R", and therefore A = 8 by Theorem 1.3(e). •

=

Let T : R!' -+ R 111 be a linear transformation. We call the m x

11

matrix

A= [T(et ) T(e2) ... T(e" )] the sta ndard matrix ofT. Note that, by Theorem 2.9. the standard matrix A ofT has the property that T (v) = Av for every v in 'R".

'ifh,!.!M

3

Let T: R -+ R

2

be defined by T ( [

;~])

- [

~:: + 4::~] . It is straightforward

to show that T is linear. To find the standard matrix of T , we compute its columns


Page 7 of 10



T(e 1), T(e2). and T(e3). We have T(e 1) =

175

[~]. T(e2) = [-~].and T(e3) = [~].So

the standard malrix of T is

OJ 0 I . [~ -4

Figure 2.13 Reflection of 'R. 2 about the x-axis

Determine the standard matrix of the linear t.rnnsformation T: 'R.1

Practice Proble m 3 ~

T

-+

'R. 2 defined by

([::~]) = [ -Jx, 2tt - Sx1]. + 4r XJ

-

3

In the next section. we illustrate the close relationship between a linear transformation and its standard matrix.

EXERCISES Exe1d.ft!.f 1- 20 refn ro the follolt'ill8 mmrices:

A- [2 -

c=

4

-3 0

-~l

8

=

[i

[~ -~]


5 -I

4

I. Give the domai n and codomai n of the matrix transformation induced by A.

~l

-2

and

2. Give the domain and codomain of the matrix transformation induced by 8.

3. Give the domain and codomain of the matrix tr.msformation induced by C .

Page 8 of 10


176 CHAPTER 2 Matrices and Linear Transformations 4. G ive the domai n and codomain of the matrix transforma· lion induced by A r 5. G ive the domain and codomain o f the matrix trans formation induced by B T

6. Give the domain and codomain of the matrix transformation induced by C T

7. Compute TA ( [

-!]).

9. Compute Tc ([;]).

I I. Compute Tn ( [

13. Compute TA ( [

-~]).

j]).

15. Compute Tc ([ _;])

17. Compute Ta ( [

~~]).

8. Compute TB

([i]) -n}

In Exercises 25- 34. linear transformmions are given. Compllle their standard mmrices. 25. T: 'R.2 -> 'R. 2 defined by T

2

2

26. T : 'R -> 'R. defined by T

27. T:

n 3 -> n2 defined by T 2

28. T : n ....

n 3 defined by T

10. compute rA ( [

!]).

([~]).

16. Compute TA ( [

=iJ}

18. Compute Tc ([

=~D·

.X2

Xt

Xz .

3

([;~]) = [•• +{:,+xJJ ([:~]) = [2x;x:_ x2 ] x1 +x2

~

30. T:

31

n 3 -> n 3 defined

r "'

by T

~\2

x2

29. r "' "' "'"" by r ([;;]) = [ ,;;

12. Compute Tc ([-

14. Compute Ta

([x']) = [ + J ([x']) = [2xt4x 1++ 5x2x2]

i; ,]

([;~]) = [~ix~:;x

2]

~ "'"''"'by r ([~]) ["~,"] x'] ) [lr1 -x,++3x~] ([ =

:~

32. T: 'R4 -> 'R.3 defined by T

=

x.

33. T : 'R.3 ->

- x1

2x4

3x2- XJ

n 3 defined by T(v) = ,, for all v in n 3 = 0 for all v in n 3

34. T : n 3 -> 'R. 2 defined by T(v)

~

In Exercises 35- 54. determine whether the stale· ~ mems are true or false. 35. Every function from 'R" to 'R"' has a s tandard matrix. 36. Every matrix transformation is linear.

37. A function from 'R" to 'R"' that preserves scalar multipli-

In Exercises 21- 24, idemify the values of n and m for each linear transformation T : n• -> 'R!". 21. Tis defined by T

22. T is defined by T

23. n

([;~]) = [x, 2~'xJ ([;~]) = [;: -~ :~].

"'""by r ([;;]) =

[ ;;,

~~';,]

catio n is linear.

38. The image of the zero vector under any linear transformation is the zero vector.

39. If T: 'R.3 -> 'R.2 is linear. then its s tandard matrix has size

3

X

2.

40. The zero transformat ion is linear. 41. A function is uniquely determined by the images of the standard vectors in its domain. 42. The first column of the standard matrix of a linear transformation is the image of the first standard vector under the transformation. 43. The domain of a function[ is the set of all images f( x).

44. The codomain of any function is contained in its range. 45. Iff is a function and f(u)

=f( v), then u = v.

46. The matrix transformation induced by a matrix A is a linear transformation.


Page 9 of 10


2.7 linear Transformations and Matrices

n" --> nm is

47. Every linear transformation T: transformation.

a matrix

62. Suppose that Tis linear, such that T

48. Every linear transfonnation T: n" - n"' is the matrix transfom1ation induced by its standard matrix.

::~~e:t::::~:::~~:· i·:d::;·:~he[T"r~~Jin. n

49.

:e

3

T

is

0 0 0 50. Every linea r transformation preserves linear combinations. 51. If T (u + v) = T(u) + T(\') for all vectors u and v in the domain ofT, then T is said to preserve vector addition.

n" -

52. Every function[: tion.

'R!" preserves scalar multiplica-

53. Iff: n" - n"' and 1:: n" - 'R!" are functions such that f(e,) = g(e;) for every standard vector e;. then f(v) = g(v) for every v in n".

answer.

63. Suppose that T is linear. s uch that T T ([

-~]) =

[-n

mine T ([

-~])

and T ([

_!]).

57. Suppose that T is linear and T

Determine T

([ 1 ~])

-~D = [ ~n Deter-

and T ([

=~]).

that T(et ) =

m

and T(ez)

=[~].Determine T ([~]).

n 2 is a linear transformation such and T(~) = [~l Determine T ([;;])

n 2 -+

for any

=

m

[x'J in n X2

66. Suppose that T :

2 • Justify your answer.

n 2 ....,. n 2

= [_~]

T ([::]) for any

is a linear transfonnation

and T(e2 )

= [- ~l

Determine

[~:] in n 2• Justify your answer.

n 3 - n 3 is a linear transformation such

67. Suppose that T: Justify your

Justify your

Jus tify your answer.

such that T(et)

([~]) = [ -~].

([-~])

n 2 --> n 2 is a linear transformation such

64. Suppose that T:

that T(e 1)

56. Suppose that T is linear and T ([

De termine T

([~]) = [~ Jand

answer.

65. Suppose that T:

55. If T is the identity transfomullion. what is true about the domain and the codomain ofT?

([~]) = [-nand

([~]) = [ -~] . Determine T ([ -;]). Justify your

54. If T and U are linear transformations whose st;mdard nuttrit-es are equal. then T and U a re equal.

177

that

answers.

58. Suppose that Tis linear and T ( [

=iD

mine T ( [

s9.

Suppose that

and T ( [

r

is linear and

Detennine T ( [

-~~])

-~]) = [-~]. Deter-

~~]} r

Determine T

and T

([~]) = [

answer.

-n

Determine T

T

([~]) = [

jJD~ermine

answer.


T(et) = [

([~]) = [ -:]

([~]).Justify your

6 1. Suppose that Tis linear, such that T ([

T ([

-~]) =

[n

Justify your

n 3 - n 2 is a linear transformation such

68. Suppose that T: that

(UJ}

60. Suppose that T is linear. such that T

3 •

answer.

([~]) = [ ~n

and T

([;~]) for any [;~] in n

-7],

Determine T (

T(ez) = [

-~J ,

and

[;~]) for any [;~] in n

T(e3) =

3 •

[~] .

Justify your

answer.

69. Suppose that T: that

n 2 -+ n 2 is a linem transformation such

and

-~]). Justify your

and

Determine T

([::~])

for any

[;~]

in

n 2 • Justify your

answer.

Page 10 of 10


178 CHAPTER 2 Matrices and Linear Transformations 70. Suppose that T : R.2 --> R.3 is a linear transfonnation such that and

Defi nitions Let T. U: R." -+ R."' be functions and c be a scalar. Define (T + U): R."--> R."' and cT: R."--> R."' by (T + U)(x) = T(x) + U(x) and (cT)(x) = cT(x) for all x in R.". The preceding definitions are used in Exercises 83-86:

83. Prove that if T is li near and c is a scalar, then cT is linear.

Determine T ([;:]) for any [;:] in R.2 Justify your

84. Prove that if T a nd U are linear, then T

answer. 7 I. Suppose that T: R.3 --> R.3 is a linear transfom1at ion such

that

Gl

T ( [-:])

=

T ( [_ : ] )

= GJ.

Determine 7'

T ( [-: ] )

n.

= [-

([;~]) fonmy [:~] in R.

and

.

Justify your

111 Exercises 72-80, a function T: R." -+ R."' is given. Either prove that 7' is linear, or explain why T is not linear. 72. T:

n

2

.....

n

2

defined by T ([;:]) =

73. T: R.2 -+ R.2 defined by T ([;:]) = 74. T: R.2 -+ R.2 defined by T ([;:]) =

3

75. T: R. -+ R. defined by T

([~~]) =

86. Usc Exercise 84 to prove that if 7' and U are linear with standard matrices A and B. respectively. then the standard matrix of T + U is A + 8. 87. Let T : R.2 -+ R.2 be a linear tmnsformation. Prove that there exist unique scalars a, b, c, and d such that x2

:mswcr.

= [ll~t + bx2 ] C..l t + dxz

77. T:

n defined by T

n 2 -+ n 2 defined by T 2

([~:])

sents the orthogonal projection

[~ ]

79. T: R.--. R.2 defined by T(x)

3

3

90. Define T: R. --> R. by T (

[~. ]

l

T repre-

on the x-axis.

[~!])

=

[2J

T repre-

(a) Prove that T is linear. Xt

+xz +x3 - I

(b) Find the st:mdard matrix of 7'. 91. Define the

+xz +x3

T

by

represents the reflection of R.2

(a) Show that T is a matrix tnmsformation.

92.

~e(fi'['~~Jth)c =li[nea~~Jtra;s::~:~:~::: r::c:o~:f~: t:e:

([:~~]) = [~;~].

82. Prove that the zero transformation To: R."

linear transformation T: R.2-+ R.2

([:~:]) = [ - ;:]. T

(b) Determine the range ofT.

[ 1:~~ 1 ]

81. Prove that the identity transformation I : R." T1• and hence is linear.

= T (v) for every v in R. 3.

aboltl rite y-axis the reflection ofR.2 aboUitite y-axis.

~: ~ ;! J

-XJ

a bow rite .1)' -plane. where a

and b are scalars


[-~

sents the orthogonal projection ofR. 3 on the yz-p/ane.

= [si.~x]

To and hence is linear.

=

(c) Prove that T(T(v)) = T (v) for every " in R.2 .

XJ

2 2 80. T: R. ---> R. defined by T

ofR.2

(a) Prove that T is linear. (b) Find the standard matrix ofT.

[2~ 1 ]

([;:]) = [

2

x2

89. Define T: R.2 -+ R. 2 by 7' ([::])

1

=X)

78. T: R. --> R. defined by T ([;:]) =

for every vector [Xt] in R.2 .

Him: Use Theorem 2.9. 88. State and prove a generalization of Exercise 87.

(c) Prove that T(T(v)) 76. T: R.3 -+

is

85. Suppose that c is a scalar. Use Exercise 83 to prove that if T is linear and has standard matrix A, then the standard matrix of cT is cA.

T ([Xt]) 3

+U

linear.

(a) Show that T is a matrix transformation. (b) Determine the range ofT.

-+

--'>

R." equals R.'" equals

93. A linear transformation T: R." -> R." defined by 7'(x) = kx. where 0 < k < I. is called a comraction (a) Show that T is a matrix transformation. (b) Determine the range ofT.

Page 1 of 10


2.7 linear Transformations and Mat rices

94. A li near transformation T: n" --> n" defined by T(x) = k x , where k > 1, is called a dilation.

102. Suppose that T: n

-->

n"'

n

n

n

n

2 _,. 2 and g : 2 _,. 2 such that /(e t) = g{et) and/(~)= g(e2). butf(v) of g(v) for some v in 'R2 .

97. Let A be an invertible

x

matrix. Determine TA_,('J:,(v)) and TA(TA _,(v )) for all v inn". 11

=

T ([

be a linear transformation. Prove that

T(u) = T(v) if and only if T(u- v) = 0.

96. Find functions I:

11

n• is a linear transformation such

j]) [l] r([i])=[j]

(b) Dctcm1inc the range ofT.

n•

->

that

(a) Show that T is a matrix transformation.

95. Let T:

4

179

T ([

_!]) -l]· = [

and

98. Let A be an m x

11 matrix and 8 an 11 x p matrix. Prove that TAo(v) = TA(To(v )) for all v in nP.

99. Let T: n" --> nm be a linear transformation with stan· dard matrix A. Prove that the columns of A form a generati ng set for the range ofT.

n• _,. n"', prove that its if and only if the rank of its standard mattix

100. For a linear transformation T: range is

n"'

ism . 10 1. Let T :

n" _,. n"'

(a) Find a rule for T.

(b) Is T uniquely determined by these four images? Why o r why not? 103. Suppose that T:

n 4 _,. n 4

is the linear transformation

defined by the rule

be a linear transformation and S =

n". Prove that if the set {T(vt). T(v2) , ••• , T(' '*)) is a linearly independent subset of n"'. then s is a linearly independent subset of n". {v 1• v2, .... vkJ be a subset of

!11 Exercises 102 and 103, use either a calculator with matrix capabilities or compurer software such as MATLAB to solve each problem.

SOLUTIONS TO THE PRACTICE PROBLEMS 1. (a) Because A is a 3 x 2 matrix, the domain of T11 is

(b) The codomain of T11 is (c) We have

2. Since et = (-1)

T

n 2•

n >.

[-~]and ez =~ [~]. it follows that

([~~]) = T(xt e t + x2~) =T

(x,(-1)[-~] +x2 G) [i])

=-XtT ( [ -~]) + ~x2T ([~])


3. Since T(e t)

= [~] .

T(e 3 )

= [ -~] .

and

the standard matrix of T is

[~

0 -3

-5]

4 .

Page 2 of 10



12.8 1COMPOSITION AND INVERTIBILITY OF LINEAR TRANSFORMATIONS In this section. we use the standard matrix to study some basic prope11ies of a linear transformation. We begin by determining whether a transformation is onto and one-toone. which is closely related to the existence and uniqueness of solutions of systems of linear equations.

ONTO AND ONE-TO-ONE FUNCTIONS Once we have found the standard matrix of a linear transformation T, we can use it to find a generating set for the range ofT . For example. suppose that T: n 3 -+ n 2 is defined by

In Example 7 of Section 2.7, we saw that the standard matrix ofT is

-4 OJ 0 I . Now w is in the range of T if and only if w = T (v) for some v in V = VtC t + V2C2 + V3 C3. we see that

n 2•

Writing

which is a linear combination of the columns of A . Likewise, it is clear from the same compu tation that every linear combination of the columns of A is in the range ofT. We conclude that the range of T equals the span of

This argument can be generalized to prove the following result:

The range of a linear transformation equals the span of the columns of its standard matrix. ln what follows. we obtain some additional properties about a linear transfonnation from its standard matrix. First, however, we recall some properties of functions. Defin ition A function f: n" -+ n"' is said to be onto if its range is all of that is, if evety vector in n"' is an image.

n"';

From the preceding boxed statement, it follows that a linear transformation is onto if and only if the columns of its standard matrix form a generating set for its codomain. Thus we can illustrate that the reflection of n 2 about the x-axis in Example 8 of Section 2.7 is onto by showing that the columns e1, - e2 of its standard matrix form a generating set for n 2 . Because e 1 and - e2 are nonparallel vectors in


Page 3 of 10


2.8 Composition and lnvertibility of linear Transformations

181

R 2 , every vector in R 2 is a linear combination of e 1 and - e2 • So the columns of the standard matrix of form a generating set for 2. We can also use the rank of the standard matrix A to determine if T is onto. If A is an 111 x 11 matrix, then, by Theorem 1.6, the columns of A form a generating set for 1?!" if and only if rank A = m.

u

Example 1

n

Determine if the linear transfonnation T : R 3 .....

n 3 defined

by

is onto.

Solution For T to be onto, the rank of its standard matrix must equal 3. But the standard matrix A of T and its reduced row echelon form R are

A= So rank A = 2

[l ~ ,~]

=I 3. and therefore T

0 0]

1 2 .

and

0 0 is not onto.

We now state the first of several theorems that relate a li near transformation to its standard matrix. Its proof follows from our preceding observations and from Theorem 1.6.

THEOREM 2.10 LetT: R" --> R"' be a linear transformation with standard matrix A. The following condi tions are equivalent: (a) T is onto; that is, the range of T is R"' . (b) The columns of A form a generating set for

nm.

(c) rank A = m .

T here is a close relati onship between the range of a matrix transformation and the consistency of a system of linear equations. For example. consider the system

x1

2x1

=I

+ X2 = 3

X1 -

X2 =I.

The system is equivalent to the matrix equation Ax

~] '

- I

X =

[Xi] X2

'

= b. where

and

b =

[~l

Because we can write the matrix equation as TA(x) "" b. the system has a solution if and on ly if b is in the range of TA.


Page 4 of 10


182 CHAPTER 2 Matrices and Linear Transformations Now suppose that we define T: 1?.3 --+ 1?.2 by T

:;~ ) ( [ Xt]

=

Xt + X2 +x3 [ Xt + 3xz - XJ ]

~

It is easy to sec that T = Ts ,where 8 = [ :

and

_:

w=

[~].

J

so T is linear. Suppose that we

want to determine if w is in the range of T. This question is equivalent to asking if there exists a vector x such that T(x) = w. or, in other words. if the following system is consistent:

Xt + X2 + x3 =2 Xt + 3xz - x 3 = 8 Us ing Gaussian elim ination, we obta in the general solution

Xt = - 2x3X2

=

x3

X3

I

+3

free.

We conclude that the system is consistent and there are infinitely many vectors whose image is w. For example, for the cases x3 = 0 and x3 = I, we obtain the vectors

and

respectively. Alternatively, we could have observed that A has rank 2 and then appealed to Theorem 2.10 to conclude that every vector in 1?.2 , including w, is in the range ofT. Another impo11ant property of a function is that of being one-to-one. Definition A function!: 1?." --+ 1?."' is said to be one-to-on e if every pair of distinct vectors in 1?." has distinct images. That is, if u and v are distinct vectors in 1?." , then f(u) and j(v) are distinct vectors in 1?."'. In Figure 2. 14(a), we see that distinct vectors u and v have distinct images, which is necessary for f to be one-to-one. In Figure 2. 14(b), f is not one-to-one because there exist distinct vectors u and v that have the same image w. Suppose that T: 1?." --+ 1?."' is a one-to-one linear transformat ion. If w is a nonzero vector in 1?.", then T(w) "f T (O) = 0, and hence 0 is the only vector in 1?." whose image under T is the zero vector of 1?."' .

(a) Vectors with distinct ima~es

(b) Vectors with identical ima~es

Figure 2.14


Page 5 of 10



183

Conversely, if 0 is the only vector whose image under T is the zero vector, then T must be one-to-one. For suppose that u and v are vectors with T (u ) = T (v). Then T ( u - v) T(u) - T (v) 0. So u - v 0, or u v, and hence T is one-to-one.

=

=

=

=

Defi nition Let T: 7?.!' -+ 1?."' be linear. The null space of T is the set of all v in 1?." such that T (v) = 0. The discussion preceding the definition proves the foll owing result:

A linear transformation is one-to-one if and only if its null space contains only 0.

Note also that if A is the standard matrix of a linear transformation T , then the null s pace of T is the set of solutions of Ax 0.

=

i#iflrrl,!fM

Suppose that T: 1?.3 -+ 1?.2 is defined by T

[ ~~ = ([Xt])

Xt -

- Xt

X2

+ 2x] 3

+ X2 -

3X3

•

Find a generating set for the null s pace of T.

Solution The standard matrix of T is A= [

- I

I - I

Because the null space of T is the set of solutions of Ax = 0, we must compute the reduced row echelon f01m of A, which is

[~

- I

0

This matrix corresponds to the system Xt -

=

X2 X3

0

= 0.

So every solution has the form

Thus a generating set for the null space of T is

I[l] )·

Using Theorem 1.8, we can give addit ional statements that are equ ivalent to the statement that a linear transformation is one-to-one.


Page 6 of 10



THEOREM 2.11 Let T : 1?.." ~ 1?..111 be a li near transformation with standard matrix A. Then the following statements are equivalent: (a) (b) (c) (d)

T is one-to-one.

The nu ll space of T consists on ly of the zero vector. The columns of A are linearly independent. rank A= n.

Determine whether the linear transformation T: 1?..3 -+ 1?..3 in Example I,

Example 3

is one-to-one. For T to be one-to-one, the rank of its standard matrix A must be 3. However, we saw in Example 1 that rank A= 2. Thus Theorem 2 . ll(d) does not hold, and so T is not one-to-one.

Solution

Finally, we relate all th ree topics: linear transformations, matrices, and systems of linear equations. Suppose that we begin with the system Ax = b , where A is an m x 11 matrix , x is a vector in 1?..", and b is a vector in 1?..111 • We may write the system in the equivalent form TA(x) = b. The following list compares the nature of solutions of Ax b and properties of TA :

=

(a) Ax = b has a so lution if and only if b is in the range of TA . (b) Ax b has a solution for every b if and only if TA is onto. (c) Ax b has at most one solution for every b if and only if TA is one-to-one.

= =


Let T: 1?..2 ~ 1?..3 be the linear transformation defined by

T {a) (b) (c) (d)

MS:£1 rrl .Jtjl

3Xt - Xz] ([::~]) = [-x~~ 2x2 .

Determine a generating set for the range of T. Determine a generating set for the null space ofT. Is T onto? Is T one-to-one?

Let

A- [~ ~ -

4 6

4 6


Page 7 of 10



185

Is the system Ax = b consistent for every b? If Ax= b t is cons istent for some b 1, is the solution unique? Solution Although these questions can be answered by computing the rank of A and us ing ll1eorems 1.6 and 1.8, we give here an alternative solution using the matrix transformation 7A: R 5 --+ R 4 • which we denote by T. First, note that the reduced row echelon form of A is 1.5

0

0

0 0

0

0 0 0

0

I

0]

1 3 0 I

.

Because rank A = 3 "#4. we see that T is not onto by Theorem 2.10. So there ex ists a vector bo in R 4 that is not in the range ofT. lt follows that Ax = bo is not consistent. Also, Theorem 2.11 shows T is not one-to-one. so there exists a nonzero solution u of Ax= 0. Therefore if for some b 1 we have Av = b 1, then we also have A(v + u)=A v + Au = b 1 + 0 = b 1. So the solution of Ax = b 1 is 11ever unique.

COMPOSITION OF LINEAR TRANSFORMATIONS Recall that iff: 5, --+ 5 2 and g: 5 2 --+ 5 3. then the composition g • f: 5, --+ 5 3 is defined by (8 o f)(u) = 8 (f(u)) for all u in S 1. (See Figure 2. 15.)

Fig ure 2.15 The composition of functions

ExampleS

Suppose that/: R 2

_,.

"R. 3 and g: "R.3 --+ R 2 are the functions defined by

and

Then 8


0

I: n 2 --+ n 2

is defined by

Page 8 of 10



In linear algebra, it is customary to drop the "circle" notation and write the composition of linear transformations T: 'R!' --+ 'R!" and U : 1?!" --. 'RJ' as UT rather than U o T . In th is case, UT has domain 'R!' and codomain 'RJ'. Suppose that we have an m x 11 matrix A and a p x m matrix 8. so that 8A is a p x 11 matrix. The corresponding matrix transformations are T11 : 1?.." --+ 1?.."', To : 1?.."' --+ 'R!', and ToA : 1?.." --+ 'R!'. For any v in 1?..". we have

It follows that the matrix transformation TnA is the composition of To and TA; that is,

This result can be restated as the following theorem:

THEOREM 2.12 If T: 1?.." --+ 1?.."' and U : 1?.."' --+ R.P arc linear transformations with standard matrices A and 8. respectively, then the composition UT: 1?.." --+ 1(.1' is also linear. and its standard matrix is 8A. PROOF Theorem 2.9 implies that T = TA and U =To. So, by our previous observation, we have UT = To T11 = To11. which is a matrix transformation and hence is linear. Furthermore. since UT = ToA· the matrix 8A is the standard • matrix of UT .


Let U: 1?..3 --+ 1?..2 be the linear transformation defined by

Determine UT

Example 6

([;~]). where T

is as in Practice Problem I.

In 1?..2 , show that a rotation by 180° followed by a reflection about the x-axis is a reflection about they-axis. Let T and U denote the given rotation and reflection, respectively. We want to show that UT is a reflection about the y -axis. Let A and 8 be the standard matrices ofT and U , respectively. Then A = A 180o. and B is computed in Example 8 of Section 2.7. So Solution

-I

and

A= [ 0 Then T = T11 , U = T 8 , and 8A =

UT

-1 [

O

8 =

[~

OJ1 . Thus, for any u = [ttt] 112 , we have

([::~]) =ToT;~([::~]) = ToA ([::~]) = BA

[111] = 112

[- I 0

0] I

[111] = [- 111]. 112 112

which represents a reflection about they -axis. (See Figure 2.16.)


Page 9 of 10



187

u = ["'] "2

Figure 2.16 The rotation of u by 1so• and its reflection about they-axis

We can use the preceding result to obtain an interesting relationship between an invertible matrix and the correspond ing matrix transformation. First. recall that a function f : S 1 ~ S z is invenible if there ex ists a function g: S z ~ S 1 such that gif(v)) v for all v in S t and f(g(v)) v for all v in S2. If f is invcniblc. then the function g is unique and is called the inverse off; it is denoted by f- 1• (Note that the symbo l f- 1 should not be confused with the reciprocal l // . which is usually undefined.) It can be shown that a function is invertible if and on ly if it is one-to-one and onto. Now suppose that A is an 11 x 11 invertible matrix. Then for all v in 1?..", we have

=

=

Likewise, TA- t TA (v)

=

v. We conclude that TA is invertible and

It is easy to see. for example. that if TA represents rotation in 1?..2 by 0 . then TA- t represents rotation by - 9. Using the previous result, we can show this also by computing the rotation matrix A-o. as we did in Example 3 of Section 2.3. If T: 1?.." ~ 1?.." is linear and invertible, then it is also one-to-one. By Theorem 2. 1 I, its standard matrix A has rank 11 and hence is also invettible. Thus we have T - 1 =TA- t. In particular, T - 1 is a matrix transformation, and hence linear. The following theorem summarizes this discussion:

THEOREM 2 .13 Let T: 1?.." ~ 1?.." be a linear transformation with standard matrix A. Then T is invertible if and only if A is invertible, in wh ich case T- 1 TA •· Thus T- 1 is linear, and its standard matrix is A- 1•

=

Suppose that A

.

[~

of Seclton 2.3 that A-t

_

Vt+2vz ] . [ 3 VJ + 5 V2 . We saw m

Example

= [-5 ~l By Theorem 2. 13, we have 3

2] [VJvz] = [- 5Vt +2Vz ] • 3v, - vz

- I


Page 10 of 10


188

CHAPTER 2 M atrices and Linear Transformat ions


Let T: 'R2 ~ 'R2 be the li near transformation defi ned by T

([X'])= [ Xt++ 4X2 ]. x2 7xz 2~,

Show that Tis invertible, and determine

r -t ([::~]).

We list the highlights from th is section in the next table . Let T: 'R" ~ 'R"' be a linear tmnsformation with standard matrix A, which has size m x 11. Properties listed in the same row of the table are equivalent.

Property of T

The number of solutions of Ax = b

Property of the columns of A

Property of the rank of A

T is onto.

Ax = b has at least one solution for every b in

The columns of A are a generating set for n m.

rank A =m

Ax = b has at most one solution for every b in

The columns of A are linearly independent.

rank A= n

Ax = b has a unique solution for every b in

The columns of A are a linearly independent generati ng set for

rank A = m = n

nm. T is one-to-one.

n m.

T is invertible.

n m.

n m.

EXERCISES !11 Exerci;·es 1- 12. find a genemting set for the range of each linear tramformmion T. I T: 'R.2 ·

4

R.2 defined b T Y

([xxz

1 ])

6.

~:(n[~~4J)n: d[e~;:~~lx;x3]

= [2.ft + 3Xz] 4xt + 5xz

X3

7. T:

3. T:

n

2

-+

n3 defined by

~ 1 ]) = [2x~~ .r 2]

T([ 2

XJ

n24n2 ~

defined by T

8 T 1<'

+x2

9. T:

XJ +x3

1<' dofi.W by T

n 3 4 n 3 defined by T

10. T:

n3 4

II . T:

nJ4

12. T: R.3

4

(~:]) = ['~ J

([~])

?l

= [ ;;,

([:~]) = [~]

= v for all v in n 3 = 0 for all v in 'R.3 n 3 defined by T(v) = 4v for al l v in n 3 R. 3 defined by T(v)

R. 2 defined by T(v)

In Exercises 13- 23, find tt genertlling set for 1he null space of

each linear trall.iformatioll T, a11d use your a11swer To deTermine whether T is one-to-one.


Page 1 of 10


2.8 Composi tion a nd lnvertib ility of line a r Transformations

13. T: 'R2 --> 'R2 defined by T

14. T: 'R2 -->

'R

2

defi ned by T

189

([x']) = [XI xz ] X2 +x2

([x']) = [ X2

+ ~X2] + )X2

2

1 ' 4XJ

5

31.

T:(n[''~ll'R defined [x' +2rzby+2rz+2<3 T +X3+x4+6xs+&xsl ;: = + + +2<4 +5xs xs 3x, +2<2 +Sx4 +8xs 4

X2

x1

XI

X3

X2

In Exercises 32-40, find the sTandard matrix ofeach linear Transforma/ion T, and use it to de/ermine wheTher T is olllo. 32. T: n z-+ n 2 defined by T ([;:])

3xz ] 'R3 --> 'R3 defined by T ([x']) x2 = [ 2xx, + - XJ 1

18. T:

X3

Xi

21. T:

3 -->

3

3 -+

2

3

'R 23. T: 'R

22. T:

-->

4 ---.

'R defined by T (

'R

3

34. T : nl ....

[;!])

35.

=XI + 2<3

36. T :

24. T:

n

25. T:

'R

2

--.

n

-->

'R

2

([;~]) = [2r~~ x2] x +x2

'R3 --> 'R defi ned by T ([;~]) = 2<1- Sxz + 4xJ

defi ned by

38.

2

r: n 2 .... n 3 defined by T

1

In £rercises 24- 3 / ,find The swndard mmrix ofeach linear 1/'tmsfomuuion T , tmd use i110 delennine wherher T is one-To-one. 2

n 2defined by T ([;~]) = ['' +{:,+xJ]

3

by

n 2 .... n 2 defined by r ([:~~]) = [x~ J 3

1 + ~xz ] 'R2 __,. 'R2 defined by T ([x'xz]) = [2< 4 x1 + :>xz

+ X2 + 4X3

n n defined by T(v)= vfor all,, in n 20. T: 'R 'R defined T(v)= 0 for all vin 'R 19. T:

33. T:

= [.,, ·~ x2 ]

defi ned by T

T:( n[~~lln•=d[efi:~d -b~ +2, T -2<, +x2 ~2

XJ

([~~]) = [x, ~ x2 ]

defi ned by T ([x']) x2

X4

3 ]

7XJ

Xi -

X2

+ 2<3

-x, +2rz+x3

3 = [24x,' 1 ++ Sx2 xz]

40.

T:(n[:~lln•=d[e~:~~x, ~;+ U+2 +X.!+x•l + +X )+2q x• 3x, + 2rz+ Sx4 2 r3

T

X2

XJ

Xi

X2

~ In Exerci:.·es 41- 60. determine wlw1her The sTale~ mel/Is are /rue or false.

41. A linear transformation with codomain 'R"' is onto if a nd only if the rank of its standard matrix eqmtls


111 .

Page 2 of 10


190 CHAPTER 2 Matrices and Linear Transformations 42. A linear transformation is onto if and o nly if the columns of its standard matrix form a generating set for its range.

63. Define T : n 2

-+ n 2by T ([::~]) = [~l which is the

43. A linear transform
projection of [ ~] on they-axis.

44. A linear transfo rmation is one-to-one if and only if every vector in its range is the image of a unique vector in its domain.

(a) What is the null space of T ?

45. A linear transformation is one-to-one if and only if its null space consists only of the zero vector. 46. A linear transformation is invertible if and only if its standard matrix is invertible. 47. The system Ax = b is consistent for all b if and only if the transfom1ation TA is one-to-one. 48. Let A be an m x 11 matrix. The system Ax = b is consistent for all b in n"' if
nm.

49. A function is onto if its mnge equals its domain. 50. A function is onto if its mnge equals its codomain. 51. The set {T(et ). T(e2) . .. . , T(e,.) } is a generating set for

n•- n"'. n" - n"' is onto if and only

(b) Is T one-to·one? (c) What is the range ofT? (d) Is T onto?

n 3 ..... n 3 by

64. Define T:

the projection of

n• -> nm

54. A function T: n• -> nmis one-to-one if the only vector v in n" whose image is 0 is v = 0. 55. A linear transformation T: n" -> n"' is one-to-one if and only if the rank of its standard matrix is 111 . 56. If the composition UT of two linear transformations T: n"-+ nmand U: nP-+ nqis defined.thcnm = p. 57. The composition of linear transformations is a linear transformation.

n" - nm

and U : 'RJ' - n" are linear transformations with standard matrices A and 8. respectively. the standard matrix of TU equals BA.

58. If T:

59. For every invertible linear transformation T. the function

r -l

is a linear transformation.

X)

(b) Is T one-to·one'l (c) What is the range of T? (d) Is T onto?

r-

n

6 J. Suppose that T: 2 -+ 2 is the linear transformation that rotates a vector by 90°. (a) What is the null space of T? (b) Is T one-to-one? (c) What is the range ofT? (d) Is T onto?

62.

Suppose that T: n 2-+ n 2 is the reflection of n 2about the x-axis. (See Exerc ise 73 of Section 1.2.) (a) What is the null space of T?

n 3 -. n 3

65. Define T:

([;~]) [;~

by T

J

which is

the projection of [ ;:] o n the ,ty-plane. XJ

(a) What is the null space of T ? (b) Is T one-to-one? (c) What is the range of T? (d) Is T onto?

66. Define T:

n3

~n

3

by T

([;~])

[ _::]

(See

Exercise 92 in Section 2.7.) (a) What is the null space of T? (b) Is T one-to·one? (c) What is the range ofT? (d) Is T onto?

60. If A is the standard matrix of an invertible linear transfor1 is A - 1• mation T, then the s tandard matrix of

n

[;~] o n the z -axis.

(a) What is the null space of T ?

the range of any function T :

52. A linear transformation T: if the rank of its standard matrix is 11 . 53. The null space of a linear transformation T: is the set of vectors in n" whose image is 0 .

([~:]) [~J which is

T

that T(et)

=

n 2 -> n 2

m

67. Suppose that T:

is linear and has the property

= [~].

(a) Determine whether T is one-to·one. (b) Determine whether T is onto. 68. Suppose that T: that T(e 1) =

n 2 -> n 2

is linear and has the property

[~] and T(e2) =

(a) Determine whether

[a

T is one-to·one.

(b) Determine whether T is onto.

Exercises 69-75 are co11cemed with the li11ear tra11sj'ormations T: n 2- n 3and U: n 3-+ n 2defined as

(b) Is Tone-to-one? (~)

What is the range of T?

(d) ls T onto?


Page 3 of 10


2.8 Composition and lnvertibility of Linear Transformations

191

:md

= [XI-Xl A!+ 4.\J]. + 3Al

U ( [ ;: ] ) XJ

69. Dclermine I he domain, I he codom:.tin , and lhe rule for UT. 70. U•c the rule for UT obtained in Exercise 69 10 lind lhe >l:tndard malrix of UT. 71. Dclermine the standard matrices A and 8 of T and U. re•pet1ively.

72. Compute the produc1 BA of 1he matrices found in Exercise 7 1. and illus1ra1e Theorem 2.12 by comparing your answer with the resuh obtained m Exercise 70. 73. Delermine the domain. the codomain. and the rule for TU. 74. U•e the rule for TU ob1ained in Exercise 73 to find the Mandard matrix of TU . 75. Compute the product AB of lhc malrices found in Exercise 7 1. and illustrale Theorem 2. 12 by comparing your answer with lhe resuh oblained in Exercise 74.

n n

n

Exerdses 76 82 are concemed ll'itlr tire linear transformations T : 2 -+ 2 and U: 2 ~ 2 defined as

T

n

([:~]) = [ 3:~: =~~]

and

u ([ :~]) = [2x~~ Xl].

76. De1ermine lhe domain. the codomain. and the rule for UT. 77. U•e 1he rule for UT ob1ained in Exerci'e 76 to find 1he Mandard matrix of UT.

78. De1em1ine 1he standard matrice' A and 8 of T and U.

90.

::(n[(])n~ d['~~:;.~~;~~~·l:] x,

13xa - 5xz + Sx3- 6x,

9 1. Prove lhm the composition of two one-to-one linear trans· formation• is one-to-one. ls the result true if I he lrunsfor· rnations are not linear? Justify your answer.

92. Prove th;rl if 1wo linear transformations are onto. then I heir composition is onto. Is the result true if the lmnsformations are not linear? Jus tify your answer.

n

93. In 2 • sho" that the composition of two reflectton:. about the A -axis is the identity transformation.

n

94. In 2 • show that a rellcction about the y-axi\ follo\\ed by :t rotation by 180 is equal to a reflection about the x-axis.

n

respeclively. 79. Compu1c the producl BA of 1he m:urices found in Exercise 78. and illuma1e Theorem 2. 12 by comparing your answer with the resuh obtained in Exercise 77.

80. Detennine the domain, lhe codom:tin. and lhe rule for TU. 81. U>e lhe rule for TU obwincd in Exercise 80 10 find the Mandard matrix of TU. 82. Compute lhe product AB of the matrices found in Exercise 78. and illustrate Theorem 2.12 by comparing your answer with the resuh obtatned m Exerctse 8 I.

In Exurius 83-90. an illl't'rtibl~ lin~ar trtmsfonnation T is d~fined. o~t~mrine a similar definition for tire im·erse of em·/t lmeur trcmsfonuation.

r-t

95 . In 2 • s how that the composition of the proJection on the .\-axis followed by the reflection about lhc y-axis is equal to the composition of the reflection about the )'·axis followed by the projeclion on the x-axis. 96. Prove that the composition of two shear trans forrmuions is a s hear transfommtion. (See Example 4 of Seclion 2.7.)

97. Suppose that T : n•

-+ n'" is linear and one-lo-one. Let (v,, ' '!· .... " • I he a linearly independent >Ub,ct of

n·.

(a) Prove that the set (T(v 1). T( \•2 ) •...• T ( v1 >1 is a lin·

early mdependent subset of

n•.

(b) Show by example: that (a) is false if T is not o ne-to· one.

83. T:

n 2 ~ n 2 defined by T ([x ']) = [2.rt

-xz]

98. Use Theorem 2.12 to prove that matrix multiplication is associative.

84. T:

n 2--+ n 2defined by T ([rr ]) = [ ..x,~·" ++ 3xzXz] .rz

In E.rerr:ises 99 and 100. u.re eitlrer a <·alculator witlr mtllrix capabilities orcomplller software suclr as MATLAB to soll'e eaclr problem.

85. :

12

:(n[~::])n:d[e~~~~Y~2-~ .r1

-.r,

3...3]

Xt +x2

99. The linear tran;,formalions T. U: R as follows:

4

--+

n4

are defined

+ 2r2 + 5x,

86.


Page 4 of 10


192 CHAPTER 2 Matrices and Linear Transformations 100. Define the linear transformation T: R.4 -> R.4 by the rule

and

(a) Compute the standard matrices A and B of T and U. respect ively. (b) Compute the product AB .

(b) Show that A is in vertible and find its in verse.

(c) Use your answer to (b) to write a rule for TU.

(c) Use your answer to (b) to fi nd the rule for r-t.

(a) Find the standard matrix A ofT.


I. The standard matrix ofT is A = [

-! -~]·

2. The standard matrices of T and U are

(a) Since the columns of A form a generati ng set for the range of T , the desired generating set is

3 -1] [-~ ~

A=

and

B

= [~

I

0

-4]

3 .

respectively. Hence. by Theorem 2.12(a). the standard matrix of UT is

!l

(b) The null s pace of A is the solution set of Ax = 0. Since the reduced row echelon fom1 of A is [ we see that the general solution of Ax XJ

=0

X2

= 0.

~

BA

=[

-9 12

Therefore

= 0 is UT

([x1]) = [-9 X2

12

3. The standard matrix of T is A = Hence the null s pace of T is {0 }, so a generating set for the null space ofT is 10}. (c) From (b). we see that the rank of A is 2. Since Theore m 2.10 implies that T is onto if and only if the rank of A is 3. we sec that T is not onto. (d) From (b), we see that the null s pace ofT is 10}. Hence T is one-to-one by Theorem 2.11.

CHAPTER 2 ~

~

[~ ~].

rank of A is 2, A is invertible. In fact, A -

I

Because the

= [- ~

_ ~].

He nce, by Theorem 2.13, T is invertible. and

r -t

([;~]) = [-~

4] [XI]=

-1

X2

[ -7xt + 2Xt -

4xz] . X2

REVIEW EXERCISES

/11 Exercises 1-2 1. determine whether tile statements are true or false.

I. A symmetric matrix equals its transpose. 2. If a symmetric matrix is written in block fom1, then the blocks are also symmetric matrices.

3. The product of square matrices is always defined. 4. T he transpose of an invertible matrix is invertible.


5. It is possible for an invertible matrix to have two distinct inverses. 6. The sum of an invertible matrix and its inverse is the zero

matrix. 7. The columns of an invert ible matrix are linearly independent. 8. If a matrix is invertible, then its rank equa ls the number of its rows.

Page 5 of 10



9. A matrix is invertible if and only if tt> reduced ro" echelon form is an identity matrix.

10. If A i> an n x 11 matrix and the 'Y'tem Ax = b is consistent for some b . then A is invertible. II. The range of a linear tmnsformation i> contained in the codomai n of the linear transfomlation.

In E.urrisu 36 and 37. compflle tlte pro
36[j

12. l11e null s pace of a linear transformat ion is contained in the codomai n of the linear transformation.

13. Linear tr:msformations preserve linear combinations.

37. [ /z

14. Linear transformations preserve linearly independent sets.

15. Every linear transformation has a >tandard matrix. 16. The zero transformation is the only linear transformation whose standard matrix i:. the zero matrtx. 17. If a linear transformation is one-to-one. then it is inverttble. IS. If a linear transformation is onto. then it' range equals its codomain .

19. If a linear transformation is one-to-one. then its range consists exactly of the zero vector. 20. If a linear transformation is onto. then the rows of its standard matrix form a generating set for its codomain. 21. If a linear transformation is one-to-one. then the columns of its standard matrix form a linearly independent set.

193

0

3

I

2

0 0

- I

2

i][

I

-I I 0

2

0 0

0

!]

0 0 2 0

1-htll

In E.utt"itef 38 and 39, dnermine whether elldtmlltrix i.r iiii'UIiblt'. If so. jiud its llll'l!rse; if not, t'Xplain "'')' nm.

u ~] 0

38.

-I

39.

[!

-I 2

3

40. Let A and 8 be square matrices of the same siLC. Prove that if the first row of A is zero. then the lirst row of AB is zero. 41. Let A nnd 8 be square matrices of the same size. Prove that if the first column of 8 is zero, then the first column of A 8 i> zero. 42. Give examples of 2 x 2 matrices A and 8 such that A and 8 are invenible. and (A+ 8)- 1 #A-t+ 8 t.

22. Determine whether each phrase i; a misuse of terminology. If so. explain what is wrong. (a) the range of a matrix (b) the standard matrix of a function f: R"

-+

R•

(c) a generating set for the range of a linear transforma-

/11 £.un1tt't 43 tmd 44. systems ofet/lllltimu art' gn·m. Fir
43. 2.1,

.r,

tion (d) the null space of a system of linear cqumions (e) a one-to-one matrix

23. Let A be an m x

11

a p x q matrix.

matrix and /J be

+.1!

columns of A

R

defined'! (b) If BA is defined. what size is it?

In Exercius 24- 35, uu the git·en matrias to com1mte t'aclt or gi1•e a reason wlty tltt' t>xpreuwn is 1101 dejiJtt'd.

eX/J/l'S~imt.

[24

[2

8-- 4

v=(l

-2 2).

u -n ·

c= and

24. Aw

25. ABA 28.

30. ATB

31. A-18T 3-1. 8 3

33. ACT u

,.c


29. vA 32.

s -'"

35.

u!

:lfC

given by

[~ ~ ! ~

-n.

Jl

~

and

a,

=

[~l

!l

= [-

Detennine the matrix A.

E.rercist's 46-49 refer to Ilte following matrices:

w =[!]

26. Au

=

a1 = [

A

27. C w

=5

3

x 1 + 3x2 + 4xJ = - I 2x1 + 4xz + x, = 2

44.

45. Suppose that the reduced row echelon form R and three

(a) Under what conditions is the matrix product BA

A --

x, + xz + '' =

+ ·'2= 3

= [!

-~ -~]

and

8

=

u -n

46. Find the range and codomain of the matrix tran,formation

TA. 47. Find the range and codomain of the matrix tram.formation

TB.

Page 6 of 10



48. compute

([n}

rA

In Exercises 62 and 63. find rite standard marrLr of each linecrr rran.ifomwrion T . and trse irro derermine wherher T is one-ro-

one.

49. Compute TR ( [ ~]).

62. T:

its xttmdtutl matri.x.

50.

51. T:

n

3~ n

defined by T ([;:])

2

defined by

r

3

= [ x~;. X2]

([~~]) = [z.r~-~ x3]

n 2 defined by T(v) = 6v for all v in 'R2 2 T: n --+ n 2 defined by T(,•) = 2v + U(v) for all

63. T:

u:

in n 2 , where defined by

u

v n 2 ..... n 2 is the linear transformation

([;~]) = [2.t~~ x2]

~n

[x 55. T: n--> n-?defined by T ([XI]) x = [2.t?]

54. T: n

2

defined by

r ([;:]) =

?

64. T:

nJ ~ n2 defined by r ([;~])

56. T:

n ~n

57. T: 'RJ

~ n2

defined by

n 2 ->

'R3 defined by T ([;:]) =

-x3

J

[ x\~ x

2 ]

Erercises 66- 72 are concerned wirh rhe linear Transformations T: n 3 - . n 2 and U: n 2 _,. n 3 defined by T

1 _: ' ]

r ([;:]) = [ x;. J

defined by T

2rt +xl

x 1 +x2

3

65. T:

I]) = [ ;~ ([X

2;
and

X l-

2

= [

XJ

2

2

[:1 ~.r:J

n 2 -. n 3defined by T ([;:]) =

X t +x2

In Exercises 54- 57. a function T: n" --+ n"' is given. Either prove that T is linear or explain why T is not linear. 2

X2 - XJ

In Exercises 64 and 65, find the standard matrix of each linear transformaTion T, and use it to determine whether T is omo.

52. T: 'R2 ~ 53.

[.r1 + 2r2]

=

XJ

In Exercises 50- 53. a linear transformmion is given. Compllle

2 2 T: 'R -> 'R

nJ --+ n2 defined by r ( [;~])

U

([;~]) = [3XtX~X2].

2

([~~]) = [X1 -~.r2]

In Exercises 58 and 59. find a generating set for the range of each linear trcmsfomwtion T.

Xt +x2

66. Detennine the domain, codomain, and the rule for UT. 67. Use the rule for UT obtained in Exercise 66 to find the standard matri x of UT. 68. Detennine the standard matrices A and 8 of T and U, respectively. 69. Compute the product BA of the matrices found in Exercise 68, and illustrate Theorem 2.12 by comparing thi s answer w ith the result of Exercise 67. 70. Detennine the domain. codomain. and the mle for TU. 7 1. Use the mle fo r TU obtained in Exercise 70 to find the standard matrix of TU.

In Exercises 60 and 61. find a generating set for the null space of each linear lrtmsformatitJJr T . Use your answer 10 determine wheTher T is one-to-one.

72. Compute the product AB of the matrices found in Exercise 68, and illustrdte Theorem 2.12 by comparing thi s answer w ith the result of Exercise 71.

In Exercises 73 and 74, an invertible linear Transformation T is 1• defined. Determine a similar definition for

73.

r2 T : n -> n 2 defined by T ([x•]) = [ + t z] X2 - X l + 3x2 2

74. T : n3 .....


Xt

n 3 definedbyT

([XI]) = [u x2 .r3

X1 +X? + X J ]

x 1 +3x 2 + 4x3

1 +4x2 +x3

Page 7 of 10



195

CHAPTER 2 MATLAB EXERCISES For the following exercises, use MATLAB (or comparable software) or a calculator with matrix capabilities. The MATLAB funaions in Tables D. 1, 0.2, 0.3, 0.4, and 0.5 ofAppendix D may be useful.

(ii) Assume that

I . Let

A= [

-I

j

I

0

0

-I

0

I

0

-2

2 2

I

-3

-I

=[ :

0 0

- I

c=

[!

-I -I

I

2

2

I

1

2

3

xo =

2 .

(iii) Prove thai, for any positive integer"· 1 Xn =A" XO +(A - /6)- (A" - /6)b,

and

• = [

3

and use this formula to recompute x5 , requested in (ii). H im for the proof" Apply either the formula for the sum of terms of a geometric series or the principle of mathematical induction.

-ll

(iv) Suppose thai after 11 years. the population is stable. that is. x,,+ 1 = x,.. Prove thai

Compute each of the following matrices or vectors.

(d) A(BT C)

D(B- 2C)

(e)

(g) C(Av)

Xn

DB

(b)

(a) AD

(h)

(I)

3. Let

(CA)v

['! .7

A=

.9

0 0 0

.5 0 0

.8 0 0

.6 0

(a) Compute A 10 . A 100 , and A500 . On the basis of your results. what do you predict w ill be the ultimate fate o f the colony? (b) Now suppose that each year. a particular population o f the species immigrates into the region from the outside, and that the distribution of females in this immigration is described by the vector b . Assume that the original population of females in the region is given by the population distribution xo. and after 11 years. the population distribution is x,. (i) Prove that, for any positive integer 11, =Ax, _ , + b.


0

1

1

0

0

0 0 0

0 0 1

A=

0 0 .3 0

is the Lesl ie matrix for a colony of an animal species living in an isolated region. where each time period is equal to one year.

Xn

= (16 -A) - 1b.

Usc this result to predict the s table distribution of fe males for the vector b given in (ii).

Av

2. Suppose that .3 0

b=

0.0

;l

I

and

where each unit represents 1000 females. Compute the population distributions x 1 • x2 • X3. x4 • and x5 .

-I

-i -:]

'j[ 0] . 1.1

4.3 2.4 1.8

[

- I

I

2

3.1]

2.2

~l

3

0 2

3 0 0

D = [

l

I 2

-I

8

I

2

0 0

0 0 0

0 0 0 I

0 0 0

0 0

0 0

I

I

0 0 0

0 0 0

I

0

0

0

I 0 0 0 0 0 0 0 0

0 0

0 0 0 0

0 0 0 0 1

0 0

be the matrix that describes flights between a set o f 8 airports. For each i and j, a 1i = I if there is a regularly scheduled Hight from airport i to ai.rpot1 j. and a;i = 0 otherwise. Notice that if there is a regularly scheduled flight from airport i to airport j, then there is a regularly scheduled flight in the opposite direction. and hence A is symmetric. Usc the method described on page 10 divide the airports into s ubsets so that one can ny. with possible connecting flights, between any two airpot1s w ithin a subset, but not to an airport outside of the subset.

113

4. Lei

A= [

j -I

2 4 2

2 -2

3 I 2

2

-1

2

-~]

I

- 1

0

1

-1

2

2

-2

2 .

(a) Use the MATLAB function rref(A) to obtain the reduced row echelon form R of A.

Page 8 of 10


196 CHAPTER 2 M atrices and Linear Transformat ions (b) Use the function null(A, ' r'), described in Table 0 .2 of Appendix D , to obtain a matrix S whose columns form a basis for Null A. (c) CompareS and R, and outline a procedure that can be applied in general to obtai n S from R without solving the equation Rx = 0. Apply your procedure to the matrix R in (a) to obtain the matrix S in (b).

5. Read Example 3 of Appendix 0. which describe~ a method for obtaining an invertible matrix P for a given matrix A such that PA is the reduced row echelon form of A. (a) Justify the method. (b) Apply the method to the matrix A in Exercise 6. 6. Let

0 -1

0

(b) Compute A- 's d irectly by comput ing A product of A- t and B.

4 2 6 6 (a) Compute M. the matrix of pivot colum ns of A listed in the same order. For example, you can use the imported MATLAB function pvtcol(A), described in Table 0.5 of Appendix D. to obtain M. (b) Compute R, the reduced row eche lon form o f A, and obtain the matrix S from R by deleting the zero rows of R. (c) Prove that the matrix product MS is defined. (d) Verify that MS =A.

and the

(c) Compare your results in (a) and (b).

A - [~

- 1 I

2

-I

-

I

2

0

-2

0 - 1

3

~r ~] 2

and

b

=[

3

~]

II

(a) Obtain an LV decomposition of A. (b) Use the LU decomposition obtained in (a) to solve the system A x = b .

n 6 --> n 6 be the linear tnons formation defined by

the rule

3

1

8. Let

9. Let T:

3 02] .

- 1

(a) Compute A- t B by using the algorithm in Section 2.4.

T

l :;

([~]] ~:: £;~,~~;,-_2:. ] . =

3x1 + 2n

xs

4x,

x6

+ 2xJ + X4 - 2.rs - 4x6 + 4x2 + 2x3 + 2x4 - 5xs + 3x6

(a) Find the s tandard matrix A ofT. (b) Show that A is invertible and fi nd its inverse. (c) Use your answer to (b) to find the rule for

n

10. Let U: 6 --> by the rule

n4

r - 1•

be the linear tnmsfonn ation defined

(e) Show that for any matrix A, if M and S are obtained from A by the methods described in (a) and (b). then MS =A.

7. Let

I

A+i

2

I

-I

-I

3

0 2

2

2

3

2

4

0

2

-2 3

4

- I 3

2

-ll

let T be the linear transformation in Exercise 9, and let A be the standard matrix of T. (a) Find the s tandard matrix B of U.

and

8= [

-j

-I I

0

2

0

2

I

-I -I

0

2

-I


4

_:] 2 2

3

(b) Use A and B to find the standard matrix of UT. .

(c) Find the rule for UT. (d) Find the rule for

ur- 1•

Page 9 of 10


3

INTRODUCTION

Geographic information systems (GIS) have fundamentally changed the way cartographers and geographical ana lysts do their jobs. A GIS is a computer system capable of capturing, storing, analyzing, and displaying geographic data and geographically referenced information (data identified according to location). Calculations and analyses that wou ld take hours or days if done by hand can be done in minutes with a GIS. The figure at the left, which shows the boundary of a county (heavy line) and the boundary of a state forest (light line), illustrates a relatively simple GIS analysis. In this analysis, the goal is to answer the question: How much of the forest is in the county? In particular, we need to compute the area of the shaded region, whose boundary is a polygon (a figure composed of connected line segments). This problem can be solved using determinants. As explained in Section 3.1, the area of the

L

parallelogram having the vectors u and vas adjacent sides is I det[u v] l. Because the area of the triangle having u and vas adjacent sides is half that of the parallelogram with u and v as adjacent sides, the area of this triangle is ~ 1 det [u v] l. Thus if a triangle has vertices at the points (0, 0), (a, b), and (c, d)

as shown in the figure at the top left of page 198, then its area is where A

= [ : ; ] . The value of del

± ~ det

A,

A is positive if the vertices (0, 0),

197


Page 10 of 10


198 3 Introduction (a, b), and {c, d) are in a counterclockwise order, and negative if the positions of (a, b) and (c , d) are reversed.

det [

-~ ·.

-'l + l ]

Y-+ 1 (<·. tl)

This geometric interpretation of the determinant can be used to find the areas of complex polygonal regions, such as the intersection of the forest and the county. A simple example is shown in the figure at the right. Translate the polygon so that one of its vertices Po is at the origin, and label its successive vertices in a counter-clockwise order around the boundary. Then half the sum of the determinants


for i

= I. 2. .... 6

gives the area of the polygon.

p~

(>,. y~)

P• •

<••· Y•l

I'•- (X.. Y•l = (x,.y,)

Page 1 of 10


CHAPTER

3

DETERMINANTS

he de/erminan/ 1 of a square matrix is a scalar that provides information about the matrix. such as whether or not the matrix is invertible. Determinants were first cons idered in the late seventeenth century. For over one hundred years thereafter, they were studied principally because of their connection to systems of linear equations. The best-known resul! involving determinants and systems of linear equations is Cramer's rule, presented in Section 3.2. In recent years, the use of determinants as a computational tool has greatly d iminished. This is primarily because the size of the systems of linear equations that arise in applications has increased greatly, requiring the use of high-speed computers and efficient computational algorithms to obtain solu tions. Since calculations with determinants are usually inefficient, they are nonnally avoided. Al!hough determinants can be used to compute the areas and volumes of geometric objects, our principal use of determ inants in this book is to detennine the eigenvalues of a square matrix (to be discussed in Chapter 5).

T

[!Ij

COFACTOR EXPANSION

We begin this section by showing that we can assign a scalar to a 2 x 2 matrix that tells us whether or not the matrix is invet1ible. We then generalize th is to n x 11 matrices. Consider the 2 x 2 matrices

A=[~ ~]

and

c

= [

d -b ]. a

-c

We see that

AC

= [~

b] [-cd

d

and si milarly,

1

Although work with determinants can be found in ancient China and in the writings of Gabriel Cramer in 1750, the study of determinants dates mainly from the early nineteenth century. In an 84-page paper presented to the French lnstitut in 1812, Augustin-louis Cauchy (1789- 1857) introduced the term determinant and proved many of the well ·known results about determinants. He also used

modern double-subscript notation for matrices and snowed how to evaluate a determinant by cofactor expansion.

199


Page 2 of 10


200 CHAPTER 3 Determinants I

Thus if ad - be ::/; 0, the matrix - - - C is the inverse of A, and so A is inad - be vettible. Conversely, suppose that ad - be= 0. The previous calculations show that AC and CA equal 0 , the 2 x 2 zero matrix. We show by contradiction that A is not inve rtibl e. For if A were invertible, then

and so all the entries of C arc equal to 0. It follows that all the entries of A equal 0; that is, A= 0 , contradicting that A is invettible. We summarize these results.

.

The matnx A =

[ae

!]

is invertible if and only if ad - be ::/; 0, in which case

A1 - - - I- - [ d

-ad - be -e

Thus the scalar ad - be determines whether or not the precedi ng matrix A is invertible. We call the scalar ad - be the determinant of A and denote it by det A or IAI.

1Ji£1 ..!.!tjl

For A

= [~

~]

and

8 =

[~ ~]'

the detem1inants are detA = l · 4-2 · 3 = -2

and

det8

=

I · 6-2 · 3 = 0.

Thus A is invertible, but 8 is not. Since A is invertible. we can compute its inverse

r' = _..!.._ [

4

- 2 -3


Example 2

Evaluate the determinant of [

-2]- [-2 - !I] . I

-

~

-~ ~]. Is this matrix invertible? If so, compute A- t.

<1111

Determine the scalars e for which A - ch is not in vertible , where

Solution

The matrix A - cl2 has the form A


12] _ c [J0 OJI = [J I- -8 c

JI 8 -9

- ch = [ _

Page 3 of 10


3.1 Cofactor Expansion 201 AI! hough we can use elementa ry row operations to determine the values of c for which this matrix is not invertible, the presence of the unknown scalar c makes the calculations difficult. By using the determinant instead, we obta in an eas ier computation. Since det (A - cl2)

= det [ II_-8 c

12

-9-c

J

= ( 11 - c)(- 9 - c) - 12(- 8)

= (c 2 = c

2

-

2c - 99)

+ 96

2c- 3

= (c + l)(c -

3),

we see that det (A - c/2) = 0 if and only if c = - 1 or c = 3. Thus A - ch is not invertible when c - I or c 3. Calcu late A - (- I)/2 and A + 3/2 to verify that these matrices are not invertible.

=

Practice Problem 2 ....

=

Determine the scalars c for which A - ch is not invertible. where

Our principal use of determinants in this book is to calculate the scalars c for which a matrix A - cln is not invertible, as in Example 2. In order to have s uch a test for n x 11 matrices, we must extend the definition of determinants to square matrices of any size so that a nonzero determinant is equi valent to invertibi lity. For I x I matrices. the appropriate definition is not hard to discover. Since the product of 1 x I matrices satisfies [a J[b] = [ab]. we see that [a) is invertible if and on ly if a -# 0. Hence for a I x 1 matrix [a], we define the determinant of [a J by det [a J = a. Unfortu nately, for n ?; 3, the determinant of an 11 x 11 matrix A is more complicated to define. To begin, we need additional notation. First, we define the (n - 1) x (11 - I) matrix Aii to be the matrix obta ined from A by deleting row i and column}.

ar 1 A;i

=

atf

atn

..

art ..... ~ ..... ~;~ -- .. •···• ... ~;,..

a, ,

an}

~ rO\\' i

a,,

l

col umn ) Thus, for example, if

A=[~

2 5 9

~l

then


Page 4 of 10


202

CHAPTER 3 Determinants

We can express the determ inant of a 2 x 2 matrix by using these matrices. For if

A=[~ ~l then A11

= [d]

and A12 = [c]. Thus detA =ad - be= a· detAtt - b · det At2·

( I)

Notice that th is representation expresses the determinant of the 2 x 2 matrix A in terms of the determinants of the 1 x I matrices A;j. Using equation ( I ) as a motivation, we define the detenninant of an 11 x 11 matrix A for 11 ~ 3 by detA =O tt · detA tt - 012 · detA12

+ · ·· + ( -

1

l) +"at,. · detAtn·

!2)

We denote the determinant of A by det A or IAI. Note that the expression on the right side of equation (2) is an alternating sum of products of entries from the first row of A multiplied by the determinant of the corresponding matrix A tj. lf we let c;i = ( - ti +i · det Aij. then our definition of the detem1inant of A can be wrinen as

(3) The number cii is called the (i, J)-cofactor of A, and equation (3) is called the cofactor expansion of A along the first row. Equations ( I ) and (2) define the detenninant of an 11 x 11 matrix recursively. That is, the determ inant of a matrix is defined in tenus of determinants of smaller matrices. For example, if we want to compute the determinant of a 4 x 4 matrix A, equation (2) enables us to express the determinant of A in terms of determinan ts of 3 x 3 matrices. The determinants of these 3 x 3 matrices can then be expressed by equation (2) in terms of determinants of 2 x 2 matrices, and the determ inants of these 2 x 2 matrices can finally be evaluated by equation ( I ).

Example 3

Evaluate the determinant of A by using the cofactor expansion along the first row, if

I 2 3] [7 95 68 .

A= 4

Solution The cofactor expansion of A along the first row yields detA

= Gt tCJ 1 + Ot 2C J2 + Ot3Ct3 = 1(- 1) 1+ 1 detAtt +2(- 1) 1+2 detAt 2 + 3(- 1) 1+3 detAn

= 1(1)(5. 8 - 6. 9) + 2( - 1)(4. 8 - 6. 7)+ 3(1)(4. 9 - 5. 7) = 1(1)(- 14)+(2)(- 1)(- 10)+3(1)(1)

=-

14+20 + 3

=9.


Page 5 of 10


3.1 Cofactor Expansion

203

As we illustrate in Example 5, it is often more efficient to evaluate a determinant by cofactor expansion along a row other than the first row. The important result presented next enables us to do this. (For a proof of th is theorem, see 14].)

THEOREM 3.1 For any i

= I, 2, ... , 11, we have detA

= a;tC; t + a;2ci2 + · ·· + a;,c;,,

where c;i denotes the (i ,j)-cofactor of A.

The expression ailc; 1 + a ;2c i2 + · ·· + a;,c;, in Theorem 3. 1 is called the cofactor expans ion of A a long row i. Thus the determinant of an 11 x 11 matrix can be evaluated using a cofactor expansion along any row.

Example4

To illustrate Theorem 3. 1, we compute the determinant of the matrix

A=

) 2 3] [7 9 8 4

5

6

in Example 3 by using the cofactor expans ion a long the second row. Solution

Using the cofactor expansion of A a long the second row, we have

= 4(-

1)(2. 8 - 3. 9)+ 5(1)(1 . 8 - 3. 7) + 6( - 1) (1 . 9-2. 7)

= 4(- 1)( - II) + 5(1)(- 13) + (6)( - 1)(- 5) = 44-65

+ 30

=9. Note that we obtained the same value for detA as in Example 3.

Practice Problem 3

~

Evaluate the determinant of

[-~ -4

3 -5 4

-;] -6

by using the cofactor expansion along the second row.


Page 6 of 10


204 CHAPTER 3 Determinants

IJifi,!.)fJW

Let

I 2 3 8 5]

4 5 6 9 I M=79847.

[00

0 0 I 0 0 0 0 I

Since the last row of M has only one nonzero entry. the cofactor expansion of M along the last row has only one nonzero term. Thus the cofactor expansion of M along the last row involves only one-fifth the work of the cofactor expansion along the first row. Using the last row, we obta in

Once again, we use the cofactor expansion along the last row to compute dctMss-

~] = detA Here, A is the matrix in Example 3. llms detM = detA = 9. Note that M has the form

where

and

More generally. the approach used in the preceding paragraph can be used to show that for any m x m matrix A and 111 x 11 matrix 8 ,

Evaluating the determinant of an arbitrary matrix by cofactor expansion is extremely inefficient. In fact. it can be shown that the cofactor expansion of an arbitrary n x 11 matrix requires approximately e · n! arithmetic operations, where e is the base of the natural logarithm. Suppose that we have a computer capable of performing I billion arithmetic operations per second and we use it to evaluate the cofactor

expansion of a 20 x 20 matrix (which in applied problems is a relatively small matrix).


Page 7 of 10



205

For such a computer, the amount of time required to perform this calculation is approximately e · 20! 109 seconds > 6.613 · 109 seconds

> 1.837 · 106 hours > 76,542 days > 209 years. If determinants are to be of practical va lue, we must have an efficient method for evaluating them. The key to developing such a method is to observe that we can easily evaluate the determinant of a matrix such as

-~]

-4 -7 8 -2 0 9 - 1 . 4 0 0 If we repeatedly evaluate the determinant by a cofactor expansion along the last row. we obtain

-4 -7 8 -2 0 9 0 0

-4 8 -2 0 9

-7]

9

= 4 · ( - 1) 3+3 · det

[30 -:]

= 4 · 9 · 8( - I )2+2 • det [3]

= 4 . 9. 8. 3. In Section 2.6, we defined a matrix to be upper tria ngular if all its entries to the left and below the diagonal entries are zero, and to be lower triangular if all its entries to the right and above the diagonal entries are zero. The previous matrix 8 is an upper triangular 4 x 4 matrix, and its determinant equals the product of its diagonal entries. The determ inant of any such matrix can be computed in a s imilar fashion.

THEOREM 3.2 The determinant of an upper triangular 11 x 11 matrix or a lower triangular 11 x matrix equals the product of its diagonal entries.

An impolt
ijif!.,j.jd

11

= I.

Compute the detenninants of

[-2 A=


- :

0 7 - I 3

0

0 -3 9

[~ ~] 3

!]

and

8=

5 0

Page 8 of 10


206


Solution Since A is a lower triangu lar 4 x 4 matrix and B is an upper triangu lar 3 x 3 matrix, we have dctA

Practice Problem 4 ""

= (-2)(7)(-3)(5) = 210

dct 8

and

= 2 · 5 · 7 = 70.

Evaluate the determinant of the matrix

[

-~ -~ 0 0 0] 0

7 -2 0 .

8

9 -5

6

3

GEOMETRIC APPLICATIONS OF THE DETERMINANT*

n

Two vectors u and v in 2 determine a parallelogram having u and v as adjacent sides. (See Figure 3. 1.) We ca ll this the parallelogram determined b y 11 and v . Note that if this parallelogram is rotated through the angle (} in Figure 3. 1, we obtain the parall elogram determined by 11' and v' in Figure 3.2.

.r

u'

= ["'] "2

X

v'

Figure 3.1 The parallelogram determined by u and v

= ~~]

Figure 3.2 A rotation of the parallelogram determined by u and v

Suppose that

u'

= ["1121]

and

' = [VJ] 0 .

v

Then the parallelogram determined by u' and v' has base v, and height 112, so its area is v 1uz

=

Ill

det I [ll2

';;]I= Idet (u'

v']J.

Because a rotation maps a parallelogram into a congment parallelogram, the parallelogram determ ined by 11 and v has the same area as that determined by u' and v'. Recall that multi plying by the rotation matrix Ao rotates a vector by the angle fJ. Using the facts that detAB (detA)(det B) for any 2 x 2 matrices A and 8 (Exercise 71) and dct Ao = I (Exercise 65). we sec that the area of the parallelogram

=



Page 9 of 10



207

determ ined by u and v is

I del [u' v'JI

= I det [Aou

Aov)l

= I det (Aolu

vDI

= l(detAo)(det[u v))l = l(l)(det [u v])l

= I dct [u

vJI .

The area of the parallelogram determined by u and v is I det[u v)l. Moreover, a corresponding result can be proved for 1?./' by using the appropriate n-dimensional analog of area.

Example 7

The area of the parallelogram in 1?} determined by the vectors and is ldet [ - ;

Example 8

~J I =

1(-2)(5)- (1)(3)1 = 1- 131 = 13.

The volume of the parallelepiped in 1?.3 determined by the vectors

[j]

and is det [:

-2

~]

=

6.

- I

Note that the o!1ect in question is a rectangular parallelepiped with sides of length and ,J2. (See F_!$ure 3.3.) Hence, by the fan1iliar formula for volume, its volume should be .J3 · ,)6 · J2 = 6, as the determinant calculation shows.

.JJ, ../6,

Figure 3.3 The parallelepiped determined by 3 vectors in 'R.3


Page 10 of 10


208 CHAPTER 3 Determinants Practice Problem 5 .,.

What is the area of the parallelogram in R 2 determined by the vectors and The points on or within the parallelogram in R 2 determined by u and v can be written in the form a u + bv. where a and b are scalars such that 0 < a < I and 0 :::; b :::; I. (See Figme 3.4.) If T: R 2 -> R 2 is a linear transformation, thenT(a u

+ bv) =

aT(u) + bT(v).

)'

-·· -··

-··

X

Figure 3.4 Points within the parallelogram determined by u and v

Hence the image under T of the parallelogram detennined by u and v is the parallelogram detennined by T(u ) and T(v). The area of this paralle logram is I det [T(u) T(v)JI = I det [A u AvJI =I detA[u vJI = I detA I · I det [u vJI. where A is the standard matrix ofT. Thus the area of the image parallelogram is I detAI times larger than that of the parall elogram determined by u and v. (If T is not invertible, then det A = 0, and the parallelogran1 determined by T(u) and T(v) is degenerate.) More generally, the area of any "sufficiently nice" region R in R 2 can be approximated as the sum of the areas of rectangles. In fact, as the lengths of the sides of these rectangles approach zero, the sum of the areas of the rectangles approaches the area of R. Hence the area of the image of R under T equals ldetA I times the area of R. A corresponding theorem can be proved about the volume of a region in R 3 with sim ilar properties. In fact, by the appropriate n-d imensional analog of volume, the following result is true: If R is a "sufficiently nice'' region in R" and T : R" -> R" is an invertible linear transformation with standard matrix A, then the n -dimensional volume of the image of R under T equals I detAI times the 11-dimensional volume of R.

This result plays a cmcial role when we make a change of variables in calculus.


Page 1 of 10



209

EXERCISES In E
4.

7.

[-~ -~J [~ -~]

-n

[-~

2. [~ -~J

3.

5. [-5 10

6.

[~

8.

-6] 12

[-~ ~] [-~

27.

-~J

-!

-2] - I

~].

-2 6

8

-I

~

0 -2

[

=

[_: -~ -;]

31. u

= [~J v

33.

I0. the (2, 3)-cofactor 12. the (3, 3)-cofactor

along the first row

[i =~

15.

_n

[_:

along the thi rd row

17.

-~]

4 0 0 - I along the second row

001 19.

[4

~

_; -3 2 3 0

-:] -1 -2

-~ -~]

along the third row

[~ ~

18.

-n

[-~I -2 -~ 2~ 0

along the second row

[~

23.

[-~

25.

[~

- I

3 0 0 -3 9 3

6 0

n !]

n

24.

26.


[-! [~ [~

43.

-5

0

-;J =m

v =[

-n

32. u

= [-

u =[~Jv= [-~]

34. u

= [ -~l v = [ -~]

= [~J

36. u

= [ -~]. v = [

=[

-n

v

v =[

-~J

=[

n =m v

_;J

[~ ~] [~ -~] [-~ ~]

38.

[~

4 1.

[-~ -~]

44.

[:

-18] c

39.

[~

_;J

42.

[:

-~]

~]

~ In Erercises 45-64, determine wlwther the stare~ mellls are true or false.

46.

!]

3 - 2

0

along the folll1h row

22.

0

0 -2 5 I

-3 0 I

2 -4

~] _n

~]

det [~

!]

=ad+bc.

47. If the determinant of a 2 x 2 matrix etjuals zero. then the matrix is invertible.

In Exercises 21-28, compwe each determina/11 by any legitimate method.

21.

I

-3

45. The determinant of a matrix is a matrix of the same size.

along the first row

20.

37. 40.

along the second row

16.

j]

In Exercises 37-44, find each value of c for which the matrix is not invertible.

[~ =: -~J

14.

0 6

2 2 -1

30. u

- )

In Exercises 13-20, compute the determinam of each matrix A by a cofactor expamion along the indicated row.

13.

m-

29. u

35. u 9. the ( I, 2)-cofactor I I. the (3, I )-cofactor

-5 0

In Exercises 29-36. compute the area of each parallelogram determined by u and v.

In Exercises 9- 12. compute each indicated cofactor ofthe matrix

A= [

-2

48. If a 2 x 2 matrix is invertible. then its determinant equals zero. 49. If 8 is a matrix obtained by multiplying each entry o f some row of a 2 x 2 matrix A by the scalar k, then det B =kdetA. 50. For n ;:: 2, the (i ,j)-cofactor of an 11 x 11 matrix A is the determinant of the (n - I) x (n - I) matrix obtained by deleting row i and column j from A. 51. For 11 2: 2. the (i .j)-cofactor of an n x n matrix A equals (- 1)1+1 ti mes the determinant of the (n - I) x (n- I) matrix obtained by deleting row i and column ) from A. 52. The dete1minant of an n x n matrix can be evaluated by a cofactor expansion a long any row. 53. Cofactor expansion is an efficient method for evaluati ng the determinant of a matrix. 54. The determinant of a matrix with integer entries must be ;m integer.

Page 2 of 10


210 CHAPTER 3 Determinants 55 . The detenninant of a matrix with po.otive entries must be

77. Prove that

positive. 56. If •ome row of a 'quare matrix con'i't~ only of Lero entrie,, then the determinant of the matri' equals Lero. 57. An upper triangular matrix must be square. 58. A matrix in which a ll the e ntries to the left :md below the di:•gonal e ntries equal zero is called n lower triangular matrix.

det[

c

61. The determinant of an upper triangular 11 x 11 matrix or a lo"er triangular 11 x 11 matrix equal~ the sum of its diagonal entries. 62. The detenninant of In equals I.

63. 1l1e :trea of the parallelogram determined by u and ,, is

kt/

r" J=det[a,. d~>] +k·detU'

78. The T l-85 calculator gives

del[~

59. A 4 x 4 upper triangular matrix has at mo't 10 nonzero entries. 60. The tran,pose of a lower triangular matrix i' an upper triangular matrix.

"+b

a

+~p

n=

2

3 4

det [u ' '].

n,2,

10- 13•

Why must this answer be wrong? Him: State a general fact about the determinant of a square matrix in which all the entries are integers. 79. U-e a determmant to express the area of the triangle 'R.2 having vertices 0. u . and v. 80. Calculate the determinant of

[~

111

; .] if 0 and 0 ' are

In E~en:i.•e.l 8/ ·84, tt-~e either ll (;(1/l'LIIllltJr with matri1 t·apabilirie.< or computer softwilre .welt as MATLAB 10

.wl••e eaclt

problem.

65. Show that the determinant of the rotation matrix

A~

is 1.

66. Show that if A is a 2 x 2 matrix

111 "hich every entry is 0 or I. then the determinant of A equal> 0. I. or -I. 67 . Show that the conclusion of Exercise 66 is false for 3 x 3 matrices by calculating

81. (a) Gcncr:uc random 4 x 4 matrices A and 8 . Evaluate detA, det8. and del (A+ 8). (b) Repeat (a) with random 5 x 5 matrices. (c) Docs dct (A + 8) = detA + det 8 appear to be true for alln x n matrices A and 8 ?

82. (a) Generate r.tndom 4 x -l matrices A and 8 . Evaluate detA, del B. and del (AB).

0

(b) Repeal (:t) with random 5 x 5 matrices.

det(l 68. Prove that if a 2 x 2 matrix has identical rows, then its determinant is zero.

69. Prove that, for any 2 x 2 matrix A, dctAr =
72. What is the determinant of an

11

x

11

(c) Does dct (;\8) = (detA)(det B) appear to be tntc for all 11 x 11 matrices A and B?

83. (a) Gcncmtc a random 4 x 4 matrix A. Evaluate dct A and dctA'.

(b) Repeat (a) with a mndom 5 x 5 matrix. (c) Do you ;uspect that detA = detA r might be true for all t1 x 11 matrices? 84. (a) Let

matrix with a zero

row? Justify your answer.

For t'ach t'lmu•maf')· matrix E in E.urr:ISt'S 73 76. and for tht' mt11rix

A= [

!

-1

b]

A = [~

d

74.

[~ ~]

75.


-I -I

2

0 0 2

-~] I

.

-I

Show that A is invertible. and evaluate detA and dctA 1•

0

••erify that del EA = (det E)(det A).

[~ ~]

X

zero matrice>.

64. If T: 'R.2 - 'R.2 is a linear transformation, then del [T(u) T(v)] = det [u vi for any vectors u and v in

73.

-4

[I

~]

76.

[I k] 0

I

(b) Repeat (a) with a random invertible 5 x 5 matrix. (c) M ake a conjecture about detA and detA - 1 for any

invertible matrix A.

Page 3 of 10


3 2 Properties of Determinants 211


~] = 8(5)- 3(-6) = 40 +

I. We have det [ _ :

= -3(-1 )2

18 = 58.

Becau'e it' detcnninant i' no1ll.ero. the matrix is iR\ertible. Furthermore.

A

1

2 . The matrix A -

A -ch = [

ch

~

8 =

6

3 - ,. 0

I

- I

- 3- c .

Thus

= dct(A - d 2)

= (4 = c2 -

c)( - 3 - c) - 6(-1 )

c-

6

= (c -

3)(c

=~]

-3] -6

I

-4

!]

= -3(- 1)[3(-6)-(-3)(4)]

6]

c

[!

+ 2(- 1)2+-'. det [ -41

8 .

has the form

6] [I 0] = [4 -

- det

+(-5 )(-1)2+2 · dct [

I [5 -3] SSI [5 -3]

= 8 . 5 -3. (- 6) 6

1

+ (-5)(1 )11(-6)- (-3)(-4)) + 2(- 1)[1(4)- 3(- 4)] 3(-6)+ (- 5)( - 18) + (- 2)(16)

= 40

+ 2).

4. Because this matrix is lower triangular. its determinant is 4( -I)( -2)(3) = 24, the pro{lucl of its diagonal e ntries.

Since A - c/2 is not invertible if and o nly if we have dct(A - c/2) 0. we sec thai A - c/2 is not invertible when c - 3 = 0 or r + 2 = 0, 1hat is. when c = 3 or c - 2.

5. The area of the pnrallc logram in 'R.2 determined by the

=

vectors

=

and

3. Let <'•i denote the (i ,})-cofactor of A. The cofactor expan sion along the second row gh•es the following value for del A: detA

is ldet

= (-3)c2 t +(-5)cu + 2c23

[~

I

;] = 14(5) - 2(3)1 = 1141 = 14.

j3.2 l PROPERTIES OF DETERMINANTS We have seen that. for arbitrary matrices. the computation of determinants by cofactor expansion is quite inefficient. Theorem 3.2, however. provides a simple and efficient method for evaluating the determinant of an upper triangular matrix. Fortunately. the forward pass of the Gaussian elimination algorithm in Section 1.4 transforms any matrix into an upper triangular matrix by a sequence of elementary row operations. If we knew the effect of these elementary row operations on the detenninant of a matrix. we could then evaluate the detem1inant efficiently by using Theorem 3.2. For exan1ple. the following sequence of elementary row operations tr;msforms

A=

I 2 3] [7 9 8 4 5 6

into an upper triangu lar matrix U:

2

3]

-3 -6 9 8 2

-3 0


-7r 1+ r 1- r 1

[I0 0

2 -3

3]

-6

- 5 - 13

-~] = u

-3

Page 4 of 10


212


The three elementary row operations used in this transformat ion are row addition operations (operations that add a multiple of some row to another). Theorem 3.3 shows that this type of elementary row operation leaves the determinant unchanged. Hence the determinant of each matrix in the preceding sequence is the same. so detA =del U = 1(- 3)(- 3) = 9. This calculation is much more efficient than the one in Example 3 of Section 3. 1. The following theorem enables us to use elementary row operations to evaluate determinants:

THEOREM 3.3 Let A be an

11

x

11

matrix.

(a) 1f B is a matrix obtained by interchanging two rows of A, then det B = - detA. (b) If B is a matrix obtained by multiplying each entry of some row of A by a scalar k. then det B = k · det A. (c) If B is a matrix obtained by adding a multiple of some row of A to a different row, then detB = detA. (d) For any n x 11 elementary matrix E. we have det EA = (det E)(det A).

Parts (a), (b), and (c) of 1l1eorem 3.3 describe how the determinant of a matrix changes when an elementary row operation is pe1formed on the matrix. Its proof is found at the end of this section. Note that if A =I, in Theorem 3.3, then (a), (b), and (c) give the value of the determinant of each type of elementary matrix. In pmticular, det E = I if E performs a row addition operation, and det E = - I if E performs a row interchange operation. Suppose that an n x n matrix is transformed into an upper triangular matrix U by a sequence of elementary row operations other than scaling operations. (This can always be done by using steps 1- 4 of the Gaussian e limination algorithm. The elementary row operations that occur are interchange operations in step 2 and row addition operations in step 3.) We saw in Section 2.3 that each of these elementary row operat ions can be implemented by multiplying by an elementary matrix. Thus there is a sequence of elementary matrices E, , £ 2, . . . , Ek such that

By Theorem 3.3(d). we have (det Ek) · · · (det £ 2 )(det EJ)(det A) = det U . Thus (-I)' dct A = det U , where r is the number of row interchange operations that occur in the transformation of A into U. Since U is an upper triangular matrix. its determinant is the product u 11 un · · · of its d iagonal entries, by Theorem 3.2. Hence we have the foll owing important resu lt. which provides an efficient method for evaluating a determinant:

u,.,


Page 5 of 10


3.2 Properties of Determinants

2 13

If an 11 x n matrix A is transformed into an upper triangular matrix U by elementary row operations other than scaling operations, then detA = ( - I )' 11 , 11122 · • · u,., where r is the number of row interchanges perfonned.

Example 1

Use elementary row operations to compute the determinant of

I

3

-3]

0 4 -2 0 4 -7 . -4

4

15

Solution We apply steps 1-4 of the Gaussian elimination algorithm to transform A into an upper triangular matrix U.

A~

I

-3]

3 4 -2 [ : 4 -7 -2 4 - 4 4 15

2r , + r4-. r-l

0 0

n

n -7]

4 4 -2 3 - 3 - 4 12 I 0 0

-7]

0 4 0 4 -2 3 - 3 - 4 4 15

r 1-r3

r,2 ++ r.l

n 0 I

0 -4

-7]

4 3 4 12

-3 -2 I

[ -~ ~ ~ =~] = u

4

3

0 0 4 - 2

4

24

0

0

=

0

I

=

Since U is an upper triangular matrix, we have det U (- 2) · I · 4 · I - 8. During the transformation of A into U, two row interchanges were perfonned. Thus detA Practice Problem 1 ..,..

= (- Ji

· det U

= -8.

Use e lementary row operations to evaluate the determinant of

A=[-~ -~4 --~]6 - 4 In Section 3. 1, we mentioned that the cofactor expansion of an arbitrary 11 x 11 matrix requires approximately e · n! arithmetic operations. By contrast, evaluating the determinant of an 11 x 11 matrix with the use of elementary row operations requires only about ~11 3 arithmetic operations. Thus a computer capable of perfom1ing I billion ari thmetic operat ions per second cou ld calculate the determinant of a 20 x 20 matrix in about 5 millionths of a second by using elementary row operations, compared w ith the more than 209 years it would need to evaluate the dete1111inant by a cofactor expans ion.


Page 6 of 10



FOUR PROPERTIES OF DETERMINANTS Several useful resulls about determinants can be proved from Theorem 3.3. Among these is the desired test for invcrtibility of a matrix.

THEOREM 3.4 Let A and B be square matrices of the same size. The following statements are true: (a) A is inverti ble if and on ly if detA i= 0. (b) det AB = (detA)(det B). (c) detAr = detA. , , , I l (d) l f A IS mvcrublc. then detA- = - - . detA We first prove (a), (b), and (c) for an in vertible n x 11 matrix A . If A is invertible, there arc elementary matrices Et , £ 2, . . . , Ek such that A = Ek · · · £2£1 by the In vertible Matrix Theorem on page 138. Hence, by repeated applications of Theorem 3.3(d). we obtain PROOF

detA = (det£k ) · · · (det£2)(det £,). Since the determinant of an elementary matrix is nonzero, we have det A i= 0. This proves (a) for an invertible matrix. Moreover, for any 11 x 11 matrix B , repeated applications of Theorem 3.3(d) give (det A)(det B) = (det £*) · .. (det £2)(det £ 1)(det B ) = (det £*) · · · (det £2)(det £,B)

= dct (Ek · · · £2£,B) = detAB . This proves (b) when A is invertible. Furthermore, we also have

We leave as an exercise the proof that det ET

= det E for every elementary matrix

E. It follows that

detA r = del (Ei E[ · · · E{)

= (det £i)(det E[) · · · (det £[) = (det£1)(det£2)· .. (det£k)

= (dct Ek) · · · (dct £2)(det Et) = det (Ek · · · £2£1) =detA, proving (c) for an invertible matrix.


Page 7 of 10



215

Now we prove (a), (b), and (c) in the case that A is an 11 x 11 matrix that is not invertible. By Theorem 2.3, there ex ists an invertible matrix P such that PA R, where R is the reduced row echelon form of A. Since the rank of A is not 11 by the Invertib le Matrix Theorem. the 11 x 11 matrix R must contain a row o f zeros. Perform ing the cofactor expansion of R along this row yields det R = 0. Because p - I is invertible, (b) implies that

=

detA

= det (P- 1R ) = (detP- 1)(det R ) = (det p - 1) · 0 = 0.

compl eting the proof of (a). To prove (b). first observe that since A is not invertible, AB is not inve1tible, because otherwise, for C = B(AB )- 1, we obtain AC = !,., which is not poss ible by the Inverti ble Matrix Theorem. Therefore det A8

= 0 = 0 · det 8 = (det A)(det 8),

completing the proof of (b). For the proof of (c). observe, by Theorem 2.2. that AT is not invertible. (Otherv
+;:f!..!.!tfl

For what scalar c is the matrix

A= [

-l -~ ~]

not inverti ble? Solution To answer this question, we compute the determinant of A. The following sequence of row addition operations transforms A into an upper triangular matrix:

- I 0

[-l n

rl + rz-r2

3q +r.\~ r3

=

[~

- I - I c! 2 ] 7

[~

- I

-2r1 +r3~r3 [~

- I - I

0

3

c

~2]

2 2] c+ 0 3c +9

- 1

=

Hence det A I( - 1)(3c + 9) - 3c - 9. Theorem 3.4(a) states that A is not inveltible if and only if its determinant is 0. Thus A is not invertible if and only if 0 = - 3c - 9. that is, if and only if c = - 3.


Page 8 of 10



Practice Problem 2 II>

For what value of c is

~

not invertible?

For those famil iar with partitioned matrices, the following example illustrates how Theorem 3.4(b) can be used:

Example 3

Suppose that a matrix M can be partitioned as

where A is an m x m matrix, C is an n x Show that dctM = (dctA)(dctC ).

Solution

11

matrix , and 0 is the n x m zero matrix.

Using block multiplication, we see that

where 0 ' is the m x

11

zero matrix. Therefore, by Theorem 3.4,

0'] [A0 C ·det

~] =

detM.

As in Example 5 of Section 3. 1, it can be shown that det [

~ ~] =

detA.

A simi lar argument (using cofactor expansion along the first row) yields det [

~ ~'] = det C.

Hence dctM = dct

[~

0'] C

[A ~] = (detC)(detA).

· dct 0

Several important theoretical results follow from Theorem 3.4(c). For example. we can evaluate the determinant of a matrix A by computing the determinant of its transpose instead. Thus we can evaluate the determinant of A by a cofactor expansion along the rows of A 1'. But the rows of AT are the columns of A, so the determinant of A can be evaluated by cofactor expansion along any column of A. as well as any

row.


Page 9 of 10


3.2 Properties of Determ inants

O CAUT/ON

217

Let A and 8 be arbitrary 11 x 11 matrices. Although detAB = (detA)(detB) by Theorem 3.4(b), the corresponding property for matrix addition is 1101 true. Consider, for instance, the matrices and Clearly, detA = det 8 = 0, whereas det (A + B) = det h = I. Therefore det (A + B)

oft detA + detB.

So the determinant of a sum of matrices need not be equal to the sum of their determ inants.

CRAMER'S RULE One of the original motivations for studying determinants was that they provide a method for solving systems of linear equations having an invertible coefficient matrix. The following result was published in 1750 by the Swiss mathematician Gabriel C ramer ( 1704-1752):

THEOREM 3.5 (Cramer' s Rule 2 ) Let A be an inveJ1ible II X II matrix, b be in n"' and M; be the matrix obtained from A by replacing column i of A by b . Then Ax = b has a unique solution u in which the components are given by

detM; detA

u; = - - for

. 1

= 1.2, .... 11.

PROOF Since A is invertible, Ax = b has the unique solution u = A- 1b, as we saw in Section 2.3. For each i, let U; denote the matrix obtained by replacing column i of !, by

u

=

Ill] . [,=, 112

Then AU; =A[e t

= IAe1 =[a t · · · a; - 1 b a i+ l · · · a,] = M;. • The remainder of this section may be omitted without loss of continuity. 2 Cramer's rule was first stated in its most general form in a 1 750 paper by the Swi ss mathematician Gabriel

Cramer (1704- 1752), where it was used to find the equations of curves in the plane passing through given points.


Page 10 of 10



Evaluat ing U; by cofactor expansion along row i produces det U; =II;· det/11-1 =

11;.

Hence. by Theorem 3.4(b). we have detM; =detAU; =(detA)·(det U;)=(detA)·II;. Because detA

tf 0

by Theorem 3.4(a), it follows that Ll;

Example 4

dctM; = detA.

•

Use Cramer's rul e to solve the system of equations Xt Xt Xt

Solution

+ 2x2 + 3x3 = 2

+ +

X] = X2 -

X3

3

= I.

The coefficient matrix of this system is

~ - ~]. 1 Since detA = 6, f\ is invertible by Theorem 3.4(a), and hence Cramer's rule can be used. In the notation of Theorem 3.5, we have

2 0 Therefore the un ique solution of the given system is the vector with components 111 =

detM1 15 5 detM2 -6 detM3 3 I detf\ = 6 = 2' 112 = detA = = - I, and !IJ = detA = 6 = 2·

6

It is readily checked that these values satisfy each of the equations in the given system.


Solve the follow ing system of linear equations using Cramer' s rule : 3x, 2~,

+ 8x2 = 4 + 6x2 = 2

We noted earlie r that evaluating the determinant of an 11 x n matrix with the use of elementary row operations requi res about ~11 3 arithmetic operations. On the other hand, we saw in Section 2.6 that an entire system of 11 linear equations in 11 unknowns can be solved by Gaussian elimination wi th about the same number of operations. Thus Cramer's rule is not an efficient method for solving systems of linear equations; moreover, it can be used on ly in the special case where the coefficient matrix is inve rtible. Nevertheless, C ramer's rule is usefu l for certain theoretical pu rposes. It can be used, for example, to analyze how the solution of Ax = b is influenced by c hanges in b. We conclude this section with a proof of Theorem 3.3.


Page 1 of 10



219

Proof of Theorem 3.3. Let A be an 11 x n matrix with rows a~, a;,. . a;,, respectively. (a) Suppose that 8 is the matrix obtained from A by interchanging rows r and s . where r < s. We begin by establishing the res ult in the case that s = r + l. In this case, a rj = bsj and A rj = 8 sj for each j. Thus each cofactor in the cofactor expansion of 8 along row s is the negative of the corresponding cofactor in the cofactor expansion of A along row r. It follows that det 8 = - det A. Now suppose that s > r + I. Beginning with rows rand r +I, successively interchange a~ with the following row until the rows of A are in the order I

f

I

I

I

I

I

a , , • • • ' ar-1' a r+ l ' , • • , 3 ,P 3 r, 3 .t+ l ' •, •' 311.

A total of s - r interchanges are necessar y to produce this ordering. Now successively interchange a~ with the preceding row unti l the rows are in the order

1l1is process requ ires another s - r - I interchanges of adjacent rows and produces the matrix 8. Thus the preceding paragraph shows that det8 = (-1)1 - r(- 1)'- r - l · dctA = (- 1)21s-r)- l · detA = -detA. (b) Suppose that 8 is a matrix obtained by multiplying each entry of row r of A by a scalar k. Comparing the cofactor expansion of 8 along row r with that of A, it is easy to sec that det 8 = k · detA. We leave the detai ls to the reader. (c) We first show that if C is ann x 11 matrix having two identical rows, then det C = 0. Suppose that rows r and s of C are equal, and let M be obtained from C by interchanging rows rands. Then det M det C by (a). But since rows r and s of C are equal, we also have C = M. Thus det C = det M. Combining the det C. There fore det C 0. two equations involving det M. we obtain det C Now suppose that 8 is obtained from A by adding k times row s of A to row r, where r =fi s. Let C be the 11 x 11 matrix obta ined from A by replac ing a~ = [u 1.112 .... ,u,] by a~ = [v 1.v2, .... v,]. Since A. 8, and C differ on ly in row r , we have A rj = 8 rj = C,7 for every j. Using the cofactor expansion of 8 along row r. we obtain

=-

=-

=

det 8 = (u, + kv 1)(- 1)'+ 1 det 8 , 1 + · · · + (u, + kv,)( - I)'+" det8m

= (u,(-

1)'+ 1 det8rt +··· + u,(- l )r+" det8m) + k

=

[u, (-

(v 1(-l)r+l dct8rt

+ · · · + v,(- 1)'+" dctB,)

1

1)'+ detAr t + · · · + u,(- 1)'+" detAr, ] + k [ v,( - 1)' + 1 det C,t + · · · + v,( - 1)'+" det Cm].

In this equation. the first expression in brackets is the cofactor expansion of A along row r , and the second is the cofactor expansion of C along row r. Thus we have det 8 =detA +k· detC .

However, C is a matrix with two identical rows (namely, rows r and s, which are both equal to a~ ). Since det C "" 0 by the preceding paragraph, it follows that det8 = detA.


Page 2 of 10


220 CHAPTER 3 Determinants (d) Let E be an elementary matrix obtained by interchanging two rows of I, . Then det EA = - detA by (a). Since det E = - I, we have det EA = (det E)(detA). Similar arguments establish (d) for the other two types of elementary • matrices.

EXERCISES In Exercises I- I 0. evaluate the determinant ofeach matrix using a cofactor expansion along rile indicmed column.

I.

[_: ~ -!] 2

3

-2

2.

second column

3.

[ ~ -! - 1

0

-~] I

[~

21.

-2

3

-I

fi rst column 23.

7.

I 3 2] [2 2 3

4.

6.

3 I I third column

[! _: ~] [ ~ ~ -:]

8.

first column

9.

-2

-1

I

25.

first column

27. [ c2

[~ _: ~]

29.

3

[

I

-:]

-l

I

third column

In Exercises I 1- 24, evaluate the dererminam of each matrix using e/eme/1/my row operations.

15.

17.

19.

- :]

14. [-;

-2

0

3

=~] 6

[j [~I 241 ~I]

2

-9 - I

20. [


24.

I -I

4

-2

[-i

-I

2

-2

-3

-~]

3 1.

-! -~]

-i]

-~ -~ ~]

30.

0

0

-I 2

37[_j

~

0

~

c -I 2 l

-I - I l -I -I

I]

c-6 2 4

34.

[

-~

2 -2

- I

l

~

-2

6

c

-I~]

-!]

- 10

_:]

-I

0 - I -I l

"u - I

-( 2

c-

[j -n

32. [

c

0 -l

c

-8

-5

[-l -~ :] -I] [~ ~ -(~

,u 33.

~]

26. [; 28. [

[~

38. [

[j -l ~]

i]

[~ -~J

-I l

[~ ~~

-1

[~

I

~ ~

-2

second column

13.

10

[-~ ~ ~l

I 3 2] [2 2 3

third column

10.

-2

22.

4 -I 2 3 0 I - 2 18 6 -4 For each of tile mmrices in £rerr:ises 25- 38, determine tile value(s) of c for wlliclltlre given matrix is not invertible.

second column

5.

[j =i _-;l _:]

0

0

-:] -I c

_:]

c

- I

0

2

-I 2 l 0

-;]

In Exerci:.·es 39- 58. determine wlrether the state-

~ ments ore true or false.

39. The determinant of a square matrix equals the product of its diagonal entries.

Page 3 of 10



221

40. Performing a row addition operation on a square matrix does not change its determinant.

70. Under what conditions is det ( -A)= - detA ? JuMify )Our :tns\\er.

41. Performing a >ealing operation on a •quare matrix does not change its determinant.

71. Let A and 8 be 11 x 11 matrices ~uch that 8 i' in\'ertible. Prove that det (B - 1AB ) = detA.

42. Performing an interchange operation on a square matrix changes its determinant by a factor or - 1.

72. An 11 x 11 m;urix A is called 11ilpore111 if. for some positive integer k. A* = 0. where 0 is the 11 x 11 £cro matrix. Prove that if A is nilpotent, then det A = 0.

43. For any 11 x 11 matrices A ~md 8, we have del (A+ 8 ) = dctA + detB. 44. For any

11

x 11 matrices A and B. dctA8 = (detA)(detB).

45. If A is any invenible matrix. then detA = 0. 46. For any square matrix A, detA 1 = - detA. 47. The determinant of any square matrix c;m be evaluated by a cofactor expansion along any column.

73. An 11 x 11 matrix Q is called orrhogorra/ if Qr Q = 1,. Prove that if Q is orthogonal. then det Q = ::1: 1. 74. A square matrix A is called skew·.rymmerric if AT = - A. Prove that if A is a skew-symmetric 11 x 11 matrix and 11 is odd. then A 1S not invcnible. What if 11 1s e,·en'! 75. The matrix

48. The determinant of any square matrix equals the product of the diagonal entries of its reduced row echelon form. 49. If detA

:f= 0.

50. The detemtinant of the

11

x

11

identity matrix is l .

51. If A i• any square matrix ;md dct cA = cdetA.

I'

b any scalar. then

is called

a Vwulermo11de mmrix. Show that dctA = (b - a)(c - a)(c - b).

52. Cramer's mle can be used to solve :my syMem o f 11 linear equations in 11 variables. 53. To solve a system of 5 linear equations in 5 ''ariables with Cmmer's rule. the determinants of s ix 5 x 5 matrices must be evaluated. 1 54. If A is an invertible matrix, then det A 1 = - -. detA 55 . If A is a 4 x 4 matrix. then del (- A)= dctA. 56. If A is a 5 x 5 matrix. then del (- A) =
51. For any square matrix A and any po:,iti\'e integer k. det (At )= (detA)k. 58. If an 11 x 11 matrix A is transformed into an upper triangular matrix U by only row interchanges :md 1·ow addition operations, then det A = 11111122 · · · 11,.,. /11

A = [:

then A is an invenible matrix.

a b c

E.tucius 59-66, soll·e each sysrem11sing Cmmer's rule.

76. Use pi'Openies of determinants to show that the equation of the line in 'R. 2 passing through the points (x1. y 1) and (x2. )'l ) c:m be written

\'t ]

n

=0.

)'

77. l.ct 8 = fb t. b2.. .. . bnl be a subset of 'R.n containing 11 distinct vectors, and let B = (b 1 b 2 . . . bn 1. Prove that 8 is linearly independent if and only if det B f. 0.

a;,

and B 78. Let A be an 11 x 11 matrix with rows a'1, a; , .. . . be the 11 x 11 matrix with rows a~ . . .. , a;. a). flow arc the detcrminunts of A and 8 related? Jus tify your answer. 79. Complete the proof of Theorem 3.3(b). 80. Complete the proof of Theorem 3.3(d).

59. 61.

'• + 2.tz = 6 3.tt + 4xz = -3

60.

lit + 4.{2 = -2 7xt + llrz = 5

62.

63 .

6 ·xl + Xz + 3x1 = - 5 4 Lf2 + .lj =

65 .

Lit - .12 + X3 = -5 XI Xj = 2 - x1 + 3x2 + 2xJ =

.lJ

2tt+3t 2 = 7 3.tt + 4.12 = 6 l rt + ltz = -6 6.11 + 5xz = 9

-.11 + 2.\z + Xj = -3 64. ·' 2 + Lfl = - I 4 IJ Xz + 3XJ =

- 2TJ =

66.

11

:f= k detA

69. Pro~e that if A is :m in"enible mmnx. then dctA - I =

I


matrix and b1, denote the (k .j)·cofactor

(a) Pro'e that if P is the matrix obtained from A by replac1ng column k by e1 . then det P = bi.J . (b) Show th:ll for each j , we have

for some

x 11 matnx and k is a scalar.

detA

11

- 2.rt + Jx2 + XJ = -2 I 1rt + X2 - Xj = -,r1 + 2.\2 +xJ = - I

67. Give an example to show that detkA matrix A and scalar k .

68. E\•aluate dct kA if A is an Ju,tify your an>\\er.

81 . Prove that det £T = dct £ for every elementary matrix E. 82. l.ct A be an 11 x of A.

/lim: Apply Cramer's rule to Ax = e1. (c) Deduce that1f B is the" x 11 matrix who;e (i.j )-entry

is b,1 • thenA8 =

(detA)/~ .

This matrix 8 is called the

classict1/ adjoim of A. (d) Sho\\ that if detA

:f= 0.

I

then A- 1 = - - B . dctA

Page 4 of 10


222 CHAPTER 3 Determinants In &ercises 83 -85. IISI' ei1her ft c:alclllmor ll'illt mlllrix capabilities or complller sofMare .welt as MATLAB 10 sol•·e the J>roblem.

(b) U;,e the boxed rc;,uh on page 213 to compute detA. 84. (a) Solve Ax = b by u\ing Cramer's rule. where

A=[~

83. (a) Use elementary row Operdtions other than scaling operations to tran;,form

2.4

-3.0 3.0 6.3

-2 -6

9.6

1.5

5

0.0 A=

[ -4.8

2 3

-~]

4

2 2 - I

0

I

0 -3

-I] -2

24]

-16

and

3

b=

I

[

I~

.

(b) How many determinants of 4 x 4 matrices are evaluated in (a)?

-2 9

85. Compme the classical adjoint (a> defined in Exercise 82) of the matrix A in Exercise 84.

into an upper triangular matrix U.


----------------------------

[~

I. The following sequence of elementary row operations transforms A into tll1 upper triangular matrix U:

[-!

-~]

3 -9 4

Jr1 ~ r2 • r2 4r 1 + r., - r.,

r1• ~r,

-

[~

~ [~

3 0 16

-3] -7 - 18

3 16 0

-3]

-IS = -7

u

- I - I

0

2 ] c+2 3c+6

Because no row interchanges were performed, the determinant of 8 is the product of the diagom~ entries of the previous matrix . which is -(3c + 6) = -3(c + 2). Since a matrix is invenible if and only if its determinant is nonzero. the only value for which 8 is not invenible is c = -2.

3. The coefficient matrix of this system is Smce one row mtcrchange operation was performed. we ha,•e detA = ( 1) 1 det U = (-IX I X 16X-7) = 112.

2. The follo•ving sequence of elementary row operations

A=(;

Since detA = 3(6)- 8(2) = 2. matrix A i;, invenible by Theorem 3.4(a). and hence Cramer' • rule can be used. In the notation of Theorem 3.5. we have

transfonns 8 into an upper triangular matrix:

M1 =

[-l

- I

0

2] ' ++ ('

1 -I 0

4

0

1

-2r1 ' 2r,- - ' 2r_, [

-I

3

M 2 = (;

and

~l

=

8

detMt detA

2 = -1

and

112

detMz -2 dctA = T =-I.

=

REVIEW EXERCISES

~

111 Exercises 1- 11. de/ermine ll'hellter the slmemems ~ an• tnte orfalu. I.

!]

[~

Therefore the unique ;olution of the given 'Y'Icm i;, the vector with components 111

CHAPTER 3

!]-

del[~ ~ J = be- ad.

2. For 11 ~ 2. the (i .j)-cofactor of an 11 x 11 matrix A is the determinant of the (11 - I) x (11 - 1) matrix obt:tined by deleting row j and column i from A . 3. If A is an 11 x 11 m:llrix :md c,1 denotes the (i .j)-cofactor of A. then detll = ll.tCit + a,2c,z +--- + a,.c,. for any i = 1,2..... 11.


4. For all

11

x

11

matrices A :md 8. we have del (A + 8) =

detll + det8 .

5. For all

11

x

11

mat rice~ A and 8, detAB = (dctA){detB ).

6. If 8 is obtained by interchanging two rows of an 11 x

11

matrix A. the n det B = del A. 7. An 11 x is 0.

11

matrix is invcnible if and only if it> determinant

8. For :my square matrix A, del AT = del A. 9. For any invenible matrix A. dctA

1

= - detA.

Page 5 of 10


223

Chapter 3 Review Exercises 10. For any square matrix A and scalar c. dct cA = c del A. JI.

I f A is an upper or lower triangular

11

x n matrix. then

29. Compute the area of the parallelogram in by the vectors

detA = a 11 + an+·· · + a,,.

and

In Exercises 12- 15, compute each indicated cofactor of the matrix

[-~ -~ 12. the ( I. 2)-cofactor 14. the (2. 3)-cofactOr

-!] 13. the (2, 1)-cofactor 15. the (3. I)-cofactor

17. row 3

18. column 2

19. column 1

In Exercises 20- 23. e1•alume the determinam of each matrix by any legitimate method.

20. [;

22.

~]

23.

-i [

[~ =~

3 -2

-6

-I

2

1

2

2 - 1

-i] -2

to transform it into an upper triangular matrix.

In £rerci.~es 25- 28, use a determinalll to find all values of the scalar c for which each matrix is 1101 invertible.

27.

28.

[c+4 -3 c+6

[

- 1 - c- 2 -c -2

-1 3 ] c + 16 -I 3 - 3

26.

c -4 +5] c+1

c-1 2c - 3 c-1

1 -c] 4- c 2


[~

U1

c + 1] 3c+4

+

=

X1

+ 2<3 =

4

Let A be a 3 x 3 matrix such that del A = 5. Evaluate the determinant of each matrix given in Exercises 33- 40. 34. A- 1 35. 2A 33. AT a ,2 - 3az2 0 22 ll)2

(lt2

_:]

(b) Usc your answer to (a) 10 compute detA.

25. [c - 17 20

=

3.0] -2.1

24. (a) Perform a sequence of elementary row operations on

A=

n 3 deter-

In Exercises 31- 32, solve each system of linear equations by Cramer's rule. x 1 - x 1 + 2x, = 7 2
37.

-5.0 21. [ 3.5

[i =: !]

30. Compute the volume of the parallelepiped in mined by the vectors and

In Exercises 16- 19, compllle the detenninam of the matrix in Exercises 12- 15. using a cofactor expansion along each indicated row or column.

16. row 1

n 2 determined

- 2a:n a)2

39.

[ all + 5a3 a 012 + 5032 4a2t 4a22 a31 - 2a21 an - 2an

{132 40. [ all (121 {122 au 012

+ 5033] 4a23 a33- 2an

013

{133] al> {II)

41. A square matrix B is called idemporem if 8 2 = B . What can be said about the delenn inant of an idempotent matrix'? 42. Suppose that an 11 x 11 matrix can be expressed in the form A= PDP - '. where P is an invertible matrix and D is a diagonal matrix. Prove that the detem1inant of A equals the product of the djagonal entries of D. 43. Show that the equation

~'•]

= 0 del[: .: , 0 1 Ill yields the equation of the line through the point (x 1.y1) with slope m.

Page 6 of 10



CHAPTER 3 MATLAB EXERCISES For the following exercises, use MATlAB (or comparable software) or a calculator with matrix capabilities. The MATlAB functions in Tables 0.1, 0.2, 0.3, 0.4, and 0.5 of Appendix O may be useful. In Exercises 1 and 2, a matrix A is given. Use elementary row operations other than scaling to transform A into an upper triangular matrix, and then use the boxed result on page 213 to compute det A.

[-08 -6.5

5.7 - 2. 1 0.2 4.8

I A= .

A~

2

l

[

-2 I

3.5 -2.0 7.9 - 3.1 8.8 10.3

-1.4 -1.4 1.0 0.0 -2.8 -0.4

2 2 4

-2 -2 - 5

3 6

-6

2 -6 3

5 -5 -5

5

2.5 3.2 2.2 - 1.0 5.0 3.7 3 I 6 -4 4 - 3

A-[: -

2 I

3 2 I 0

and

14 -6 5 -I

w

!l

matrix whose rows are x followed by the rows of A in the same order.

-20] -6.5 5.7 - 2. 1 -2.2 3.6

(a) Compute det [:] and det [ ; ] . A + det [w] A . (b) Compute del [v+ A w] and del [v]

(c) Compute del [ 3v -A 2w] and 3det [ Av] - 2det [ w] A .

l

3. Lei 3 3 0

6.7 1.7 - 1.3 3.5 11.4 8.9

v

= (2

2

(d) Usc the results of (b) and (c) to make a conjecture about any function T : R " --. R defined by T(x) = del [ ~]. where C is an

= [4 -4

10

4] .

1] .

For any I x 5 row vector x, let [; ] denote the 5 x 5


(11 -

I) x

11

matrix .

(e) Prove your conjecture in (d). (f) In (a)- (e). we considered the case where row I of a matrix varies and the other rows remain fixed. State and prove a result for the general case, in which row i varies and all of the other rows remain fixed . (g) What happens if rows are replaced by columns in (f)? Investigate this situation. first by experimentation. and then by formulating and proving a conjecture.

Page 7 of 10


4

INTRODUCTION

Solid modeling (three-d imensional geometric modeling) systems have become an indispensable tool for mechanical desig ners. They create a virtua l th ree-dimensional representation of mechanical components for machine design and analysis and also provide tools for visualizing components and computing volumes and surface areas. Eng ineeri ng drawings can be created semiautomatica lly from such a model, and tool paths for machining parts ca n be generated from it. The models are constructed by a variety of techniques. For instance, a polyhedron is constructed by specifying the coordinates of its vert ices and how the vertices are connected to form

.r· .

t he faces of the polyhedron. If the polyhedron is positioned simply relative to the coordinate system, the vertex coordinates can be easy to compute. For example, a cube centered at the origin with an edge length of 2 that has faces parallel to t he coordinate planes has vertex coordinates of(± 1, ± 1, ± 1). (See the figure at the left.) If the polyhedron is not in such a simple position, as in the fig ure on the next page, the construct ion is more complicated. Here again the goal is to constru ct a cube with an edge length of 2 centered at the ori gin having its top and bottom faces parallel to the xy -plane. In t his polyhedron, however, one pair of faces is perpendicular to the line y = x and the third pair is perpendicular to the line y

= - x . In

this case, the new cube can 225


Page 8 of 10


226 4 Introduction

eJ

I

X ••• •

)-

be constructed by rotating the original cube by 45° about the z-axis. For less trivial orientations, multiple rotations might be required, which can be difficult to visualize and specify. A simple alternative is to specify the vertices in t erms of basis vectors (see Section 4.2). Specifically, we can use the standard vectors e1, e2• e3, which are perpendicular


e,

--

to the indicated faces in the second figure. As described in Section 4.4, the coordinates of the vertices relative to these basis vectors are (± 1,±1,±1). Thus the vertices of the cube are ±ea. ± e2 , ±e3 . More generally, if the cube is centered at a point p, its vertices are p ±ea. p ± e2,

P ± eJ.

Page 9 of 10


CHAPTER

4

SUBSPACES AND THEIR PROPERTIES

I

n many applications. it is necessary to study vectors only in a subset of R", which is simpler than working with all vectors in R". For insta nce, suppose that A is an m x n matrix and u and v arc solutions of Ax = 0. Then A(u + v) =Au + Av = 0 + 0 = 0.

so that u + v is a solution of Ax = 0 . Likewise. for any scalars, s u is a solution of Ax = 0. Therefore the solution set of Ax = 0 has the following closure properties: (a) the su m of any pair of vectors in the set lies in the set, and (b) every sca lar multiple of a vector in the set lies in the set. Subsets that have these properties are called subspaces of R". As another example, a plane passing through the origin of R 3 is a subspace of R 3 . In Section 4.1, we define subspaces and give several examples. The techniques learned in Chapter I enable us to find generating sets for particular subspaces. A generating set of the smallest size, called a basis for the subspace, is pm1icularly useful in representing vectors in the subspace. In Section 4.2, we show that any two bases of a subspace have the same number of vectors. This number is called the dimension of the subspace. In Sections 4.4 and 4.5, we investigate different coordinate systems on R" and describe examples in which it is preferable to use a coord inate system ot her than the usual one.

14.1 I SUBSPACE$ When we add two vectors in R" or multiply a vector in R" by a scalar, the resulting vectors arc also in R". ln other words. R" is closed under the operations of vector addition and scalar multiplication. In this section, we study the subsets o f R" that possess this type of closure. Definition A set W of vectors in R" is called a s ubspace of R" if it has the follow ing three prope11ies: I. The zero vector belongs to W. 2. Whenever u and v belong to W, then u + v belongs to W. (ln this case. we say that W is closed under (vector) addition.) 3. Whenever u belongs to W and c is a scalar, then c u belongs to W. (In this case, we say that W is closed under sca lar multiplication.)

227


Page 10 of 10


228

CHAPTER 4 Subspaces and Their Properties

The next two examples describe two special subspaces of R" .

Example 1

The set R" is a subspace of itself because the ze ro vector belongs to R", the sum of any two vectors in R" is also in R", and every scalar multiple of a vector in R" be longs to R".

Example 2

The set W consisting of only the zero vector in R" is a subspace of R" called the zero s ubspace. Clearly, 0 is in W. Moreover, if u and v are vectors in W, then u = 0 and v 0. so u + v 0 + 0 0. Hence u + v is in W. so W is closed under vector addition. Finally, if u is in W and c is a scalar, then cu = cO= 0, so cu is in W. Hence IV is also closed under scalar mu ltiplication.

=

=

=

A subspace of R" other than (0) is called a nonzero subspace. Examples 3 and 4 show how to verify that two nonzero subspaces of R 3 satisfy the three propenies in the definition of a subspace.

•;:f! ..!.!tjl We show that the set W - ( [

::~]

E

R 3 : 6w1

511'2 + 4w3 - 0

I

is a subspace

ofRJ 1. Since 6{0) - 5(0) + 4(0) = 0, the components of 0 satisfy the equation that defines

W. Hence 0

2. Let u

= [~ls in W.

= [::~] IIJ

and v

= [~~]

be in W. Then 611 1 - 5112 + 4113

= 0,

and also

V3

6v1 - 5v2 + 4v.l = 0. Now u

+v =

[::~ : :~] · Since 113 +

6(111 + VJ) - 5(112 + v2) + 4(113 + v3)

VJ

= (6111 -

S112 + 4113) + (6v1 - 5v2 + 4v3)

=0 + 0

= 0, we see that the components of u + v satisfy the equation defining W. Therefore u + v is in W, so IV is closed under vector addition. 3. Let u =

[I )] 112 IIJ

be in W. For any scalar c, we have c u = c

[II)] [CI J ] 112 113

=

c112 .

Cll3

Because 6(c111) - 5(C11z) + 4(cu3)

= c(6111 -

5112 + 4u3)

= c(O) = 0,

the components of c u satisfy the equation defining W. Therefore cu is in W, so W is also c losed under scalar multiplication.


Page 1 of 10


4.1 Subspaces

229

u+ v

IV

Figure 4 .1 The subspace W is a plane through the origin.

Since W is a subset of R 3 that contains the zero vector and is closed under both vector addition and scalar multipl ication, W is a subspace of R 3 (See Figure 4.1.) Geometrically. W is a plane through the origin of R 3 .

Example4

Let w be a nonzero vector in R 3 . Show that the set W of all multiples of w is a subspace of

nJ

Solution First, 0 =Ow is in W . Next, let u and v be vectors in W. Then u = aw and v = bw for some scalars a and b. Since u +v = aw

+ bw =(a+ b)w.

we see that u + v is a multiple of w . Hence u + v is in W , so W is closed under vector addition. Finally, for any scalar c, c u = c(aw) = (ca)w is a multiple of w. Thus c u is in W, so W is also closed under scalar multiplication. Therefore W is a subspace of R 3 . Note that W can be depicted as a line in R 3 through the origin. (See Figure 4 .2.)

)'

X

Figure 4.2 W is the set of all multiples of w.

Example 4 shows that the set of vectors in R 3 Jhat lie along a line through the origin is a subspace of R 3. However, the set of vectors on a line in R 3 that does not


Page 2 of 10


230 CHAPTER 4 Subspaces and Their Properties

pass through the origin is not a subspace, for such a set does not contain the zero vector of R. 3. In the following example, we consider two subsets of R. 2 that are not subspaces ofR.2 :

Example 5

Let V and W be the subsets of R 2 defined by

and

w

=

Wt] e n 2: w12= w,2}. {[wz -

The vectors in V are those that lie in the first quadrant of R.2. and the nonnegative pa.1s of the x- and y-axes. (See Figure 4.3(a).) Clearly, 0 u= [

11 ']

ll2

and v = [v'] are in V. Then u 1 2:. 0, V2

u, + v, 2:. 0 and

112

112

= [~] is in V . Suppose that

2:. 0, v1 2:. 0 , and

v2

2:. 0. Hence

+ vz 2:. 0, so that

is in V. Thus V is closed under vector addition. This conclus ion can also be seen by the parallelogram law. However, V is not closed under scalar multiplication. because u

= [~ Jbelongs to V, but (-

l)u

= [=~Jdoes not. Thus V is not a subspace of n 2 .

v

)'

u + ,,

X

( - l )u

V is closed under vector addition.

\Vis closed under scalar multiplication.

bul 1101 under scalar mull iplication.

bln

1101

under vec10r addition.

(b)

(a)

Figure4.3


Page 3 of 10


4.1 Subspaces

Consider a vector u = [

111

112

J in W. Since

u~ =

231

ui, it follows that uz = ±u1. Hence

the vector u lies along one of the lines y = x or y = - x. (See Figure 4.3(b).) Clearly. 0

= [~]

belongs to W. Moreover, if u

scalar c, c u

= [cu•J cuz

= [:;~]

is in W because (cu 1) 2

is in W. then

ur = u}. So, for any

= c 2 u~ = c 2uf = (cu2)2

Thus W is

closed under scalar multiplication. However. W is not closed under vector addition. For example, the vectors [ : J and [ - : not. Thus W is not a subspace of

J belong to

IV, bu t [:] + [ - :

J=

m

does

R 2•

Our first theorem general izes Example 4.

THEOREM 4.1 The span of a finite nonempty subset of R" is a subspace of R". PROOF

Let S = {w J. w2, . .. , w k ). Since

we see that 0 belongs to the s pan of S. Let u and v belong to the span of S. Then

for some scalars a,. a2 ..... ak and b1. b2, . ... bk. Since u+ v

= (a, w l +

a2 w2 + · · · + at wk) + (b, w l + b2 w2 + · · · + bt wk)

= (a, + b1)w, + (a2 + b 2)w2 + · · · + (ak + bk)w k. it follows that u + v belongs to the span of S. Hence the span of S is closed under vector addition. Fu rthermore, for any scalar c, c u = c(a1w1 + a2w2 + · · · + ak wk)

= (c lal)WJ +

Cc2a2)w2 + · · · + (ckadw k.

so c u belongs to the span of S. Thus the span of S is a lso closed under scalar multiplication. and therefore the span of S is a subspace of R" . •

Example 6

We can use Theorem 4.1 to show that lhe set of vectors of the form

W

=

I[

2a -b

3bl

- a +4b


E

1?} : a and b are scalars

I Page 4 of 10


232 CHAPTER 4 Subspaces and Their Properties is a subspace of 1?.3 . Simply observe that

[2a -3b] [ 2] [-3] =a

b - a + 4b

0 +b

- I

I

,

4

soW =SpanS, where

Therefore W is a subspace of 1?.3, by Theorem 4 .1 .

In Example 6, we found a generating set S for W, thereby showing that W is a subspace. It follows from Theorem 4.1 that the only sets of vectors in R" thar have generating sets are the subspaces of1?.". Note also that a generating set for a subspace V must consist of vectors from V. So in Example 6, even though every vector in the subspace W is a linear combination of the standard vectors in R 3 , the set of standard vectors is not a generating set for W because c 1 is not in W. Practice Problem 1 .,..

Show that

V =

I[35~

] E

R

3

:

s and t arc scalars

1

I

is a subspace of 1?.3 by finding a generating set for V.

SUBSPACES ASSOCIATED W ITH A MATRIX There arc several important subspaccs associated with a matrix. The first one that we consider is the null space, a term we introduced earlier for linear transformations. Defin ition The null spa ce of a matrix A is the solution set of Ax = 0. It is denoted by Null A. For an m x

11

matrix A. the null space of A is the set Null A = {v E 1?.": Av = 0}.

For example, the nu ll space of the matrix

[~

-5 -9

equals the solution set of the homogeneous system of linear equations


+ Jx3 = 0

Xi -

Sx2

2x1 -

9.X2 -

6x3 = 0.

Page 5 of 10


4.1 Subspaces 233 More generally. the solution set of any homogeneous system of linear equations equals the null space of the coefficient matrix of that system. The set W in Example 3 is such a solution set. (In this case, it is the solut ion set of a single equation in 3 variables. 6x 1 - Sx2 + 4x3 = 0.) We saw in Example 3 that W is a subspace of n 3 . Such sets are always subspaces, as the next theorem shows.

THEOREM 4 .2 If A is an m x n matrix, then Null A is a subspace of n" . PROOF Since A is an 111 x n matrix, the vectors in Nu ll A, which are the solutions of Ax = 0 , belong to n". Clearly. 0 is in Null A because AO = 0. Suppose that u and v belong to Null A . Then Au = 0 and Av = 0. Hence, by Theorem 1.3(b). we have

A(u + v) =A u + Av = 0 + 0 = 0. This argument proves that u + v is in Null A , so Nu ll A is closed under vector addition. Moreover, for any scalar c. we have by Theorem 1.3(c) that A(c u) = c(A u) =cO = 0. 1lms c u is in Null A, so Null A is also c losed under scalar multiplication. There• fore Null A is a subspace of n". Another impor1ant subspace associated with a matrix is its column space. Defin ition The column space of a matrix A is the span of its columns. It is denoted by Col A. For example, if

A --

[I

2

-5 -9

then

Recall from the boxed result on page 68 that a vector b is a li near combination of the columns of an 111 x n matrix A if and only if the matrix equation Ax = b is consistent. Hence Col A= (Av: vis inn"}. It follows from Theorem 4. 1 that the column space of an 111 x n matrix is a subspace of n"'. Since the null space of A is a subspace of n", the column space and null space of an m x 11 matrix are contained in different spaces if m ~ 11. Even if m = n , however, these two subspaces are rarely equal.


Page 6 of 10



Example 7

Find a generating set for the column space of the matrix

A=

Is u

= [:]

in Col A? Is v

=

[!]

[i

2

1

4

0

-1]

-8 .

0 2

6

in Col A?

Solution The column space of A is the span of the columns of A. Hence one generating set for Col A is

I[il·[~l ·[~l- [=~J l · To see whether the vector u lies in the column space of A, we must detennine whether A x u is consistent. Since the reduced row echelon form of [A u ] is

=

1 2 0 0

0 - 4 0] 1 3 0 '

[0 0 0

0

I

we see that the system is inconsistent. and hence u is not in Col A. (See Figure 4.4.) On the other hand, the reduced row echelon form of [A v] is

l 0

[0 Thus the system Ax

2 0

0 I

0 0

-4 0.5] 3

1.5 .

0

0

= v is consistent, so v is in Col A. (See Figure 4.4.)

ColA

Figure 4.4 The vector v is in the column space of A, but u is not.


Page 7 of 10


4.1 Subspaces

235

i#ifi,!.!tj:l """ • '''"""'"' ,., ,,,.
;, N'll A' I,

' =[

-~] ;,

•r•oc of ""

m• "'' A ;, E'" mplo 7 . '' " = [ _; ]

N' ll A1

Un like the calculation o f a generating set for the column space of A in Example 7. there is no easy way to obtain a generati ng set for the null space of A directly from the entries of A. Instead, we must solve Ax = 0. Because the reduced row echelon form of A is Solution

[

I 0

2 0

-~] .

0 I

_)

0 0 0

0

the vector form of the general solution of Ax = 0 is

It follows that

So the span of the set of vectors in the vector form of the general solution of Ax = 0 equals Null A . To see if the vector u lies in the null space of A , we must determine whether Au = 0 . An easy calculation con firms this; so u be longs to Null A. On the other hand,

Av=

Since Av

Practice Problem 2

~

¥

[I 2I -1][ ~] [ 0] 2 0

4 0 0 2

- 8 6

-

- IO . 10

2 I

0, we see that v is not in Null A.

Find a generati ng set for the column space and null space of

A= [

I

- 1

2

-3

-1]

4 .

In this book, the subspaces that we consider usually ari se as the span of a given set of vectors or as the solution set of a homogeneous system of linear equations. As Examples 7 and 8 illustrate, when we define a subspace by giving a generating set, there is no work in volved in obtaining a generating set, but we must solve a system


Page 8 of 10


236


of linear equations to check whether a particular vector belongs to the subspace. On the other hand, if V is a subspace that is the solution set of a homogeneous system of linear equations, then we must solve a system of linear equations in order to find a generating set for V. But we can easily check whether a particu lar vector lies in V by verifying that its components satisfy the linear system defining the subspace. Like the column space of a matrix. the row space of a matrix is defined to be the span of its rows. The row space of a matrix A is denoted by Row A. So for the matrix

-I] -8 6

in Example 7, we have

Row A =

S~" IUl U H!ll

Recall that, for any matrix A. the rows of A arc the columns of Ar . Hence Row A Col AT, and so the row space of an m x 11 matrix is a subspace of 1?.". Usually. the three subspaces Null A. Col A, and Row A arc distinct.

=

SUBSPACE$ ASSOCIATED WITH A LINEAR TRANSFORMATION In Section 2.8, we saw that the range of a linear transformation is the span of the columns of its standard matrix. We have just defined the span of the columns of a matrix to be its column space. Thus we can reformulate this statement from Section 2.8 as follows:

The range of a linear transfom1ation is the same as the column space of its standard matrix.

As a consequence of this result, the range of a linear transformation T: 1?." -.. 1?."' is a subspace of 1?."'.

Example 9

Detennine a generating set for the range of the linear transfom1ation T: 1?.4 defined by

Solution

1?.3

The standard matrix ofT is

A= [~0 0~ User: Thomas McVaney

.....

I

-I]

0 -8 . 2 6

Page 9 of 10


4 .1 Subspaces

237

Since the range of T is the same as the column space of A. a generating set for the range ofT is

I[iJ ·[~J ·[iJ ·[=:JI· the set of columns of A.

We also learned in Section 2.8 that the null space of a linear transformation is the solution set of Ax = 0, where A is the standard matrix ofT. We can now restate this result as follows:

The null space of a linear transfomlation is the same as the null space of its standard matrix.

This result implies that the null space of a linear transformation T: 'R!' -+- 'R."' is a subspace of 'R." .

Example 10

Detem1i ne a generating set for the null space of the linear transformation in Example 9. Solution The standard matrix ofT is given in Example 9. Its reduced row echelon form is the matrix

I 2 0I -4] [0 0 0 30 0 0

in Example 8. From the laner example. we see that


For the linear transformation T: R 4 -+- R

3

defined by

find generating sets for the nu ll space and range.

EXERCISES /11 Eurcises I 10. find a genuaring set for each subspace.

2. {[_;;] E 'R.2 : .r

j,

a ;calar}

~:] e n 2 : s is a scalar}

3. {[


Page 10 of 10


238 CHAPTER 4 Subspaces and Their Properties 1

4. { [

~ + r] -3s+l

5. { [;:

e

n 3 : sand 1 are scalars }

26. [ - : ]

~ :] en>: .v and r are scalars }

In Erercises 27- 34, find a generating set for rhe null space of each matrix.

s +3r 6. j [-:·

~ :s]

e n4: r.s. and

t

1[:~~:l .n' ' · '"' ,.,. ~·'·~l 31

8. j [

r ;,:s_+t

-r + 3s + 2I - 2r +s + r

] e

n4 :

r, s, and r are scalars

I

~ ~~

3 2 9. j [ ,.r-4s+3t'] e n4: r. s. and 1 arc scalars ] - r

[-:

28.

[~

arc scalars ]

,. -21

7

27.

29.

30.

31.

+2s

'R' '·' ..

''"~"'"" I

In £ren:ise.1· 11 - 18, determine whether each vecTor belongs to Null A, where

~ -2

-2

-!

-~] .

- 1

2

" [j]

-2

0

[j

-I

-i]

I

-6

I

2

0

-5

I

2

ll

32.

20. [

23. [

" [;j]

~n

21. [

-~]

j]

24. [

-~]


I

0

2

2

I

6

33.

0

-3 -6 3

[j

-~J

-I

- I

" [-l

In Exercises /9- 26, detennine whether each vecTor belongs to Col A. where A is the maTrix used in Exercises 11- 18.

-n

j]

4

-2

[-: [i

2

-I

3

-I -I

-I

-n

-2 2

- I 0

I

-3 -8 5

-2

_:]

2

-I

-3

3

In Exercises 35-42, find generating sers for the range and null space of each linear transformmion. 35. T

([;~])=(XI + ~\'2 -X J(

36. T ([::])

" [-1]

22. [

0

-I [ 2I II -3

37 T

-!]

~]

2

-I

0

" [-j] " [-i] 19. [

3

- I 0 -5

"l['~f~·~l A = [

I 2]

-2

38. T 39. T

([~])

= [~: ~ ~~~] = [ ;:

l

([~~]) = [_2;:~~:~ ~ ~.:~] ([;~]) [XI + ·~ - XJ] =

2xl - X3

X}

40. T

~;;

([;~]) X}

= [

;~ :;~

X1 - X3 X1 + 2x2 + X3

l

Page 1 of 10


4 1 Subspaces

67.

68.

69.

[-~

In £rercises 43--62. detl'nllill<' whether the stareare true fJr falu.

1111'111<

43. If V is a subspace of n• and v is in V. then c v is in V for every scalar c. 45. The subspace {0} is called the null space. 46. The span of a finite nonempty sub;et of n• i> a subspace

ofn•. The null space of an m x n matrix is contained in 'R.". matrix is contai ned in 'R." .

The column space of an m x

11

T he row s pace of an m x

11

mauix is contained in 'R."'.

l11e row space of an {Av : v is in R"}.

111

x

11

~ . 3 vectors

- I I

- I

2

4 -2

matrix A is the set

>pace of A.

52. The null space of every linear transfom1ation is a sub;pacc.

-n

53. The range of a function need not be a subspace.

=

76. Give an example of a matrix for "hich the null 'pace equals the column space. 77. Prove that the inter:.cction of two
v

n• "

:i

={[:~] enL =0} v1

and

of its standard matrix .

IV = {[::] e n2 : v2 = 0}.

56. The null s pace of a li near transformation C
(a) Prove that both V and IV are s ubspnces of 'R.2

; pace of its s tandard matrix.

51. Every nonzero subspace of 'R." contains infinitely m;my VeCtON. 58. A ;ubspace of n• must be closed under vector addition. contains at least two sub;pace~.

= 0.

61. A vector v is in Col A 1f and only 1f Ax = ,. IS consistent. 62. A vector ,. is in Row A if and only if AT x tent.

= v is consis-

(b) Show that V U IV is not a subspace of

80. Prove that 1f V is a s ubspace of n• containing 'ector;. u 1. u 2..... u1 • then V contain' the \pan of {u 1. u 2 ..... U t ). Fo r this reason. the span of {u 1. uz ..... Ut) is called the smallest subspace of 'R." containing the l'ectors u 1. u 2. . . . . Ut. 111 £1·ercises 8 1 -88. show thcu each set is 11ota subspace of the

appmprillle n•.

64. Find a generating set contai ning exactly two vectors for the column s pace of the matrix in Exercise 28.

81. {[::: ] e

65. Find a generating set containing exactly four vectors for the column space of the matrix in Exerci'e 32.

82. { ["

66. Find a gcnemting set containing exactly four vectors for the column space of the m:uri.< in E'erci'e 33.

1

"z]

83. { [

In £ttrl'ists 67- lO.for the column splllt of eaclr matrix. find a g~nullli11g set cofl/aining exacrlv the 1111111bu of• ecrors specified.

nz.

79. Let S be a nonemp1y subset of n•. Prove that S is a ; ubspace of n• if and only if. for all vectors u and v in S and all -.cal= r. the vector u + cv is in S.

63. Find a generating set containing exactly two vectors for the column space of the matrix in Exercise 27.


=

73. Let R be the reduced row echelon form of A. Is Col II Col R? Justify you1· answer. 74. Let R be the reduced ro w echelon form of II. Prove thut Row A = Row R. 75. Give
55. The range of a linear transformation equals the row space

n•

2 vectors

72. Let R be the reduced row echelon fonn of A. Is Null A Null R? Ju,tify your answer.

54. The range of a linear transformation i> a ;ub;pace.

60. A 'ector v is in Null A if and only if A,.

-~ . 3 vectors

space equals the column space.

5 I. For any matrix A. the row space of AT equal; the column

59.

']

-4

5

0 0 - I

[i

-7]

6

-3 -2

71. Octennine the null space. column space. and row space of the m x 11 1ero matrix.

44. E'ery subspace of n• contains 0 .

47. 48. 49. 50.

70.

2 vectors

4

- I

~ ~

n

-3

[~ [-~

239

84.

n

2

: u 1u 2

o}

E 'R.2 : 2u2 + 3u2 2

I

~i,;,] e 'R.

![:::)

=

3

:

= 12}

s and 1 are scalars }

e 'R.2 • ur + ui

~ I}

Page 2 of 10



85.

{[::~] e n

86. {

[::~]

87. {

[::~]

88. {

E

n

3

: u1

3 : Ill

> u2 and " >

~ uz ~

<0}

U) }

Let A and B be two m x 11 matrices. Usc the definition of a subspace to prove that V = {'' E n": Av = Bv} is a subspace of n".

100. Let V and IV be two subspace.~ of n" . Use the defini1ion of a subspace to prove that S ={sEn": s = ,, + w for some ,, in V and win IV}

3

E n : Ill = 112113 }

[::~] e n uu = uj } 3

:

1 2

In Exercises 89- 94, use the definition of a subspace, t1s in Example 3, to prove that each set is a S11bspace of the appropri(l(e n".

89

99.

{[::~] e n

2

is a subspace of n".

Exercises 101- 103, use either a ca/culmor with matrix capa· bilities or comp111er software such as MATL.AB ro solve each problem. /11

101. Let

Ao : u 1 -3112

=0}

9o. {[::;] e n 2 : 5ul + 4u2 = o} 91. {

[::~] en :2u +5u -4u3 = 0 }

92. { [:::]

3

1

u=

2

en3 :-ul+7u2+2u3 =0 }

Jm] •n' ,.,, -.,,+•"='M'"•='] " Jml ,n• "' .-., =uM,.,,_,,, =o] n• --> n"' be a li near transformation. Use the definition of a subspace to prove that the range of T is a subspace of n'".

-I I

[ ']

1.8 -10.3 , 2.3 6.3

2

I

I

0 3

I

- I

3

- 2

-;]

and

v=

[ -6]

-5

1.4 - 1.6 . 1.2 1.8

(a) Is u in the column space of A? (b) Is v in the column space of A?

102. Let A be the matrix in Exercise 10 1, and let

93

95. Let T : n• ~ n"' be a linear transformation. Use the definition of a subspace to prove that the nu ll space of T is a subspace of n".

[-'j

0

u=

[=r:~] 00 47

v=

and

[-~:~]· QO -54

(a) Is u in the null space of A'! (b) Is v in the null space of A?

I03. Let 1\ be the matrix in Exercise I 0 I, and let

96. Let T :

n" --> n"' be a linear transformation. Prove that if V is a subspace of n", then {T(u) En"': u is in V} is a subspace of

97. Let T:

u

nm.

98. Let T: n" --> n"' be a linear transformation. Prove that if W is a subspace of R:", then {u : T(u) is in W} is a subspace of n".

=

[=t~]

and

8.2 2.9

v

=

[=H] 2.9 4.2

(a) Is u in the row space of A? (b) Is v in the row space of A?

SOLUTIONS TO THE PRACTICE PROBLEMS I. The vectors in V can be wrinen in the form

Hence

is a generating se1 for V.


Page 3 of 10


4 2 Basis and Dimension

241

The null space of T is the ;,:tmc as the null ;pace of A. so it is the solution set of Ax = 0. Since the reduced row echelon form of A is consi~ting of the columns of A is a generating set for the column space of A. To find a generating set for the null >pace of A. \\C mu" \Ol\'e the equation Ax= 0. Since the reduced row echelon form of A is

[-5,\J] [-5] 3~~

=

XJ

~

the solutions of Ax x2· X tl [x,X,t

the vector form of the general >Olution is

Xt] [;~ =

[~ ~ -l =

-ll

= 0 can be written as

[-\j- 2.1~] XJ + x~ .II

x,

[-1] [-2] 1

= A.I

1

I + .I~

0 .

0

I

Thus •

Wllfill

lienee

is a generating set for the null space ofT. The range of T is the same as the column space of A. lienee the set is a gener:lling set for the null space of A. 3. The standard matrix ofT i>

A = [~

{[~l [-n. [_:J. [:J l

0 -I

3

- I

of columns of A is a generating :.et for the range ofT.

j4.2 l BASIS AND DIMENSION In the previous section. we saw how to describe ~ub:,pacc> in temls of generating sets. To do so. we write each vector in the subspace as a linear combinauon of the vectors in the generating set. While there are many generating >Cb for a given nonzero subspace. it is best to usc a generating set that contains as few vectors as possible. Such a generating set, which must be linearly independent, is cal led a b<1sis for the subspace. Definition Let V be a nonzero subs pace of linearly independent generating set for V.

n".

A basis (plural, bases) for V is a

For exrunple, the set of standard vectors le t, c2, ... , c, ) in 1?." is both a linearly independent set and a generating set for n". Hence {c t. c2•... , C11 ) is a basis for 1?.". We call this basis the s tandard basis for 1?." and de note it by £. (See Figure 4.5.) I lowever. there are many other possible bases for 1?.". For any angle 0 such that 0° < (} < 360°, the vectors Aeet and Aoe2 obtained by rotating the vectors e 1 and C2 by(} form a basis for n 2. (See Figure 4.6.) There arc also other bases for 1?. 2 in which the vectors are not perpendicular. TilUs there arc infin itely tmmy bases for 1?.2.

In many applications, it is natural and convenient to describe vectors in tenns of a basis other than the standard basis for 1?.".


Page 4 of 10


242 CHAPTER 4 Subspaces and Their Properties y

y

X

X

The standard ba.~is for

n2

The standard ba.~i s for

n3

Figure4.S

Figure 4.6 Rotated standard vectors

We can restate some of our previous results about generating sets and linearly inde pendent sets, using the concept of a basis. For example, recall that the pivot columns of a matrix are those cotTespond ing to the columns containing leading ones in the reduced row echelon form of the matrix. We can restate Theorem 2.4 (which stales that the pivot columns of a matrix arc linearly independent and fonn a generating set for its col umn space), as follows:

The pivot colu mns of a matrix form a basis for its column space.

The following example illustrates this fact:

IJifh.!.!tjl

Find a basis for Col A if 2

A=

[ - II

2

-3


-2

- I I

2 2

1

3 4 -3 2 0 -6 2 0 3

!l Page 5 of 10


4.2 Basis and Dimension

243

Solution In Example I of Section 1.4, we showed that the reduced row echelon form of A is

I 2 0I 00 -1 -5] 0 - 3 [0 0 0 0 0 0 0 0

0

0 0

I

I

2

Since the leading ones of this matrix are in col umns one, three, and four, the pivot columns of A are

As previously mentioned, these vectors form a basis for Col A . Note that it is the pivot columns of A. and not those of the reduced row echelon form of A. that form a basis for Col A.

0 CAUTION

As seen in Example I, the column space of a matrix is usually different from that of its reduced row echelon form. In fact. the column space of a matrix in reduced row echelon form always has a bas is consisting of standard vectors, wh ich is not usually the case for other matrices. We can apply the method in Example I to find a basis for any subspace if we know a finite generating set for the subspace. For if S is a finite generating set for a subspace of 'R", and A is a matrix whose columns are the vectors in S, then the pivot columns of A constitute a basis for Col A, which is the span of S. Note that this basis is also contained in S. For example, if W is the span of

s~ I[~ll ·[jJ ·[~ll ·m·m·ml· then we can find a basis for W by fonning the matrix whose columns are the vectors in S and choosing its pivot columns.

THEOREM 4 .3 (Reduction Th eorem)

V of 1?:'. Then

Let S be a finite generati ng set for a nonzero subspace

S can be reduced to a basis for V by removing vectors from

S. Let V be a subspace of 'R" and S = {u ,. u 2.. . . , Uk} be a generatuk], then CoiA=Span(u 1, u2····· uk}= ing set for V. If A=[u 1 uz V. Since the pivot columns of A form a basis for Col A, the set consisting of the pivot columns of A is a bas is for V. Thi s basis is clearly contained PROOF

~£


•

Page 6 of 10



Example 2

Find a basis for SpanS consisting of vectors in S if

s= Solution

1[il· m. [-~J· [ll· [-~JI·

Let

A_

-

[~ ~ I I

2 2

2 - 1 1 0 I 2

-:] 2 ' 1

the matrix whose columns are the vectors in S . II can be shown that the reduced row echelon f01m of A is

[ ~~~ :

~]

000001' 0 0 0 0

Since the leading ones of this matrix are in columns I. 3. and 5. the corresponding columns of A fonn a basis for S. TI1us the set

I[il hl hJ I is a basis for the span of S , consisting of vectors of S.

If {u t, ll2, ... , uk} is a generating set for R" , then [111 112 ... ud must have a pivot position in each row, by Theorem 1.6. Since no two pivot positions can lie in the sa me column, it follows that this matrix must have at least 11 columns. Thus k ;:::: 11; that is, a generating set for R" nwsr comain at least 11 vectors. Now suppose that {v 1, v 2, . . . , vi} is a linearl y independent subset of R". As noted in Section 1.7. every subset of R" containing more than 11 vectors must be linearly dependent. Hence, in order that {v 1, v2 , ... , vi) be linearly independent, we must have j ~ 11. That is , a linearly independent subset ofR" must comain at most n vectors. Combining the observat ions in the two preceding paragraphs, we see that evet)' basis for R" must contain exactly n vectors. In summary, we have the following resull:

Let S be a finite subset of R". TI1en the following are tme: I. If

S is a generating set for R" , then S contains at least n vectors.

2. If S is linearly independent, then S contains at most 3. If S is a basis for R", then S contains exactly

11

11

vectors.

vectors.

Our next theorem shows that every nonzero subspace of R" has a basis.


Page 7 of 10



245

THEOREM 4 .4 (Extension Theorem) Let S be a linearly independent subset of a nonzero subspace V of R". Then S can be extended to a basis for V by inclusion of additional vectors . .In particular, every nonzero subspace has a basis. PROOF Let S = {u t, u 2, ... , uk} be a linearly independent subset of V. If the span of S is V, then S is a basis for V that contains S. and we are done. Otherwise, there exists a vector v 1 in V that is not in the span of S . Property 3 {u 1, u 2 , . . . , u k. v 1J is linea rl y independent. If the on page 81 implies that S' span of S ' is V, then S' is a basis for V that contains S, and again we are done. Otherwise, there exists a vector vz in V that is not in the span of S'. As before, S" = fu 1, u 2 , ..• , uk> v 1• v2 } is linearly independent. We cominue this process of selecting larger linearly independent subsets of V containing S untll one of them is a generating set for V (and hence a basis for V that contains S ) . Note that this process must stop in at most 11 steps, because every subset of R" containing more than n vectors is linearly dependent by property 4 on page 81. To prove that V actually has a basis. let u be a nonzero vector in V. Applying the Extension Theorem to S = {u}, which is a linearly independent subset of V (property I on page 81), we see that V has a basis. •

=

We have seen that a nonzero subspace of R" has infin itely many bases. Although the vc.ctors in two bases for a nonzero subspace may be different, the next theorem shows that the number of vectors in each basis for a particu lar subspace must be the same.

THEOREM 4.5 Let V be a nonzero subspace of R". Then any two bases for V contain the same number of vectors. Suppose that {u 1, u 2•. .. , uk} and {v 1, v2, ... , vp) are bases for V, and let A= [u t uz ... Uk] and 8 [vt v2 ... vi' ]. Because {u t, U2·· .. , uk} is a generating set for V, there are vectors c; in R_k for i = I, 2, ... ,p such that Ac; v;. Let C [Ct Cz ... Cp ]. Then C is a k x p matrix such that AC 8. Now suppose that Cx = 0 for some vector x in R 1' . Then Bx = ACx = 0. But the columns of 8 are linearly independent, and hence x = 0 by Theorem 1.8. Applying ll1eorern 1.8 to the matrix C, we conclude that the columns of C are linearly independent vectors in R_k. Because a set of more thank vectors from R_k is linearly dependent (propetty 4 on page 81 ), we have that p ::;: k. Reversing the roles of the two bases in the preceding argument shows that k ::;: p also. Therefore k = p; that is, the two bases contain the same number of vectors. • PROOF

=

=

=

=

The Reduction and Extension Theorems give two different characteristics of a basis for a subspace. ll1e Reduct ion Theore m tells us that vectors can be deleted from a generating set to form a basis. ln fact, we learned from Theorem I.7 that a vector can be deleted from a generating set without chang ing the set's span if that vector is a linear combination of the other vectors in the generating set. Furthermore. we saw in item 5 on page 81 that if no vector can be removed from a generating set without changing the set's s pan. then the set must be linearly independent.

A basis is a generating set for a subspace containing the fewest possible vectors.


Page 8 of 10



On the other hand, if we adjoin an additional vector to a basis, then the larger set cannot be contained in a basis, because any two bases for a subspace must contain the same number of vectors. Thus the Extension Theorem impl ies that the larger set cannot be linearly independent.

A basis is a linearly independent subset of a subspace that is as large as possible.

According to Theorem 4.5, the size of every basis for a subspace is the same. This pennits the following definition: Defin ition The number of vectors in a basis for a nonzero subspace V of R" is called the dimension of V and is denoted by dim V. It is convenient to define the dimension of the zero subspace of R" to be 0 .

=

11. In Section 4 .3, Since the standard basis for R" contains 11 vectors, d im R" we discuss the dimensions of several of the types of subs paces that we have previously encountered.

Practice Problem 1 ....

Is

I[-n.[-lll

a basis for R 3 ? Justify your answer.

The next result follows from the preceding two theorems. It contains important information about the size of linearly independent subsets of a subspace.

THEOREM 4.6 Let V be a subspace of R" with dimension k . Then every linearly independent subset of V contains at most k vectors; or equivalently. any finite subset of \f containing more than k vectors is linearly dependent. PROOF Let {v r, v 2.•.. , v 1,) be a linearly independent subset of \f. By the Extens ion Theorem, this set can be extended to a basis {v 1, v2, . .. , v1, . .. • vk} for V. It follows that p ::; k. •

I§Ji,,j,)tji

Find a basis for the subspace

v

~ I [2]'R' ., _,., + ,,, -~.~·I

of R \ and detennine the dimension of V. (Since \f is defined as the set of solutions of a homogeneous system of linear equations. V is, in fact. a subspace.) The vectors in V arc solutions of Xt - Jx2 + Sx3 - 6x4 = 0. a system of I linear equation in 4 variables. To solve this system, we apply the technique described in Section 1.3. Since

Solution


Page 9 of 10



247

the vector form of the general solution of this system is

[~~]- [3x2 - ~~3 X3 X4

-

X3

+ 6x4] . - .t 2

[~]O + [-~]I + . [~]O · .t4

X3

0

X4

0

I

As noted in Section 1.7, the set

containing the vectors in the vector form is both a generating set for V and a linearly independent set. Therefore S is a ba~ is for V. S ince S is a basis for V containing 3 vectors, dim V = 3. Pra ctice Problem 2 •

Find a basis for the column space and null space of

2

-4 -2

-~ -~]. 0

3

To find the dimension of a subspace, we must usually determine a basis for that subspace. In this book, a subspace is almost always defined as either (a) the span of a given set of vectors. or (b) the solution set of a homogeneous system of linear equations. Recall that a basis for a subspace as defined in (a) can be found by the technique of Example I. When a subspace is defined as in (b), we can obtain a basis by solving the system of linear equations. The method was demonstrated in Example 3.

CONFIRMING THAT A SET IS A BASIS FOR A SUBSPACE As we mentioned earlier in this section, a nonzero subspace has many bases. Depending on the application, some bases are more useful than others. For example, we might want to find a basis whose vectors are perpendicular to each other. While the techniques of this section enable us to find a basis for any subspace, we might not produce a basis with the most desirable properties. In Chapters 5 and 6. we describe methods for finding bases with special properties. In the remainder of this section, we show how to determine whether or not a set of vectors forms a basis for a subspace. Consider, for example, the subspace

of 7<.3 and the set


Page 10 of 10


24 8


By using the method of Example 3, we can find a basis for V . Since this basis conta ins two vectors, the dimension of V is 2. We would l ike to show that S, which contains two perpend icular vectors, is also a basis for V. ( In Section 6.2, we discuss a method for transforming any basis into one whose vectors are perpendicular to each other.) By definition, S is a basis for V if Sis both linearly independent and a generating set for V. ln this case, it is clear that S is linearly independent because S contains two vectors, neither of which is a multiple of the other. It remains to show on ly that S is a generating set for V, that is, to show that S is a subset of V such that every vector in V is a linear combination of the vectors inS. Checking that S is a subset of V is straightforward. Because I - l + 2(0) = 0 and -I - I + 2( I) = 0, the vectors in S sat isfy the equation defining V. Hence both

[l]

and

are in V . Unfortunately, checking that eve1y vector in V is a li near combination of the vectors in S is more tedious. This requires us to show that. for every

such that Vt -

v2

+ 2v3 = 0, there arc scalars Ct

and

c2

such that

The following result removes the need for this type of calculation:

THEOREM 4.7 Let V be a k -dimensional subspace of 'R". Suppose that S is a subset of V with exactly k vectors. Then S is a basis for V if either S is linearly independent or S is a generating set for V. PROOF Suppose that S is linearly independent. By the Extension Theorem, there is a basis l3 for V that contains S. Because both l3 and S contain k vectors, l3 = S. Thus S is a basis for V. Now suppose that S is a generat ing set for V . By the Reduction Theorem, some subset C of S is a basis for V. But because V has dimension k. every basis for V must contain exactly k vectors by Theorem 4.5. Hence we must have C = S. so S is a basis for V. •

This theorem gives us three straightforward steps to show that a given set l3 is a basis for a subspace V:

Steps to Show that a Set B is a Basis for a Subspace V of 'R.." I. Show that l3 is contained in V. 2. Show that l3 is line

Page 1 of 10



249

3. Compute the dimension of V, and confitm that the number of vectors in B equals the dimension of V.

In the preceding example. we showed that S is a linearly independent subset of V containing two vectors. Because d im V = 2, all three of these steps are satisfied, and hence S is a basis for V. Therefore we need not veri fy that S is a generating set for V. Two more examples of this technique follow.

IJif! ..!.!tjl

Show that

is a basis for

Clearly, the components of the three vectors in B all satisfy the equati on B is a subset of V . so step I is satisfied. Because the reduced row echelon form of

Solution v1

+ v2 + v4 = 0. Thus

is

it follows that B is linearly independent, so step 2 is satisfied. As in Example 3, we find that

is a basis for V. Hence the dimension of V is 3. But B contains three vectors, so step 3 is satisfied, and hence B is a basis for V. Practice Problem 3 .,..

Show that

l[=ilHll is a basis for the null space of the matrix in Practice Problem I.


Page 2 of 10



IJifi,!.)fJW

Let W be the span of S . w here

s

Jl·

~ I m·[~!] ·[-l]·[

Show that a basis for W is

Solution

Let - 1

3

3

and

- I

- I You shoul d check that the equation Ax

I

_!] . - I

= b is consistent for each b

in 8 . Therefore

8 is a subset of W , so step I is satisfied. We can easily verify that the reduced row echelon form of 8 is

1 0I 00] [0 00 0I ' 0 0

so 8 is linearly independent. Thus step 2 is satisfied. Since the reduced row echelon form of A is

we see that the first 3 columns of A are its pivot columns and hence are a basis for Col A = W . Thus dim W = 3, w hich equals the number of vectors contained in 8, and step 3 is satisfied. Hence 8 is a basis for W.

EXERCISES In Exercises 1- 8. find a basis for (a) the column space tmd (b) the null space of each mmrLt. I.

[_:

3.

[-: -n

-3 3

4 -4

-~]

2.

2

-I

-I 0

4.


[~

[-i

-2 -3

0 - I

3

-3 6

5.

!]

-~]

-4

[-~

-2 2

C'~

l

-4 3 I

7.

0

0 - I

2 -5 -I

-2

-n [-l 6.

-ll

8

- I l

I

-2 3 -I

2

-3 0

5 3

-!]

[! -ll -2

Page 3 of 10



In Exercises 9- 16. a linec1r transformation T is gi••en. (a) Find a basis for The mnge ofT. (b) If The null space ofT is nonzi'm, find a basis for the null space ofT.

([x'] ) [x++ 1

9. T

=

x2

lr1

XJ

XJ

22.

251

{[;~] e R 3:-x.+x2+lr3= 0and

+lr,+ X3] 3x2 + Jx3

lr1 - J.r2 + 4.rJ = 0 }

2x2 + 4x3

23 j[~]•R' . -~+3.,-4,.~0) 24.

ml

" -<>+'-'>h,~O•OO lr1 -

25. Span {

26. Span { 14. T

XIll = [ l 0 x, + lr2 + 3.r3 + 4x• ([ lr + 3x2 + 4.r + x4

1

;:]]

15. T

X)

3

7XI

+ 4X2 +

X)

-lrs

xs

16 T

·

29

([;~]] - [-;:! ;:+ !~:~! :;: ! ;::~] + x

-

4

3x, .r2 lr4 - 3xs x 1 + lr2 + Sx3 + 9x• + 9xs

xs

JO.

In Exercises 17-32. find a basis for each subspace. 17. {[_

2

;]en

2 : s isascalar}

-; ~~;] e nL sand

l[ l l[ 5r -

19.

20.

3sl

-~S

E

5r3s 2r + 6s S _ t 4 7

R

4 :

1

are scalars }

r and s llrC scalars

21.

{ ~:] en :x,-3xd5x =0 } 3

)

E R 4 : r,s . and I are scalars

3r - s +9t

3


-ll [-:l[~l [-~l [=~] }

j]. [-~~]. [_n. [_~]. [~]} l[-!H-i] ,[;j]·[-j] [-l]l ,~. ml.[ll·m.[_!]. [:m s~ :1]· [=ll·[j]. [-ll

)

s,m

l[ l ,,.., l[lH=!l [!HiH=lll 31

2

18. { [

0}

28. Span { [

.r, + lr2 + 3.r3 + 4.rs] = [ 3.rl + X2- X3- 3xs

([X4

=

L:J. [jJ. [~]. [-n}

27. Span { [

5x4

X4

GJ, [!].[-~] }

-2ti-X2+X4

X2 XJ

Jx2 - 5X) -

~

~

In E.urcises 33- 52. de1ennine wheTher The stateare true or false.

//WillS

33. Every nonzero subspace of R" has a unique basis. 34. Every nonzero subspace o f R" has a basis.

35. A basis for a subspace is a generating sel thai is as large as possible. 36. If Sis a linearly independent set and Span S = \1, then S is a basis for V.

Page 4 of 10


25 2


37. Every finite generating set for a subspace contains a basis for the subspace. 38. A basis for a subspace is a linearly independent subset of the subspace that is as large as possible. 39. Every basis for a particular subspace contains the same number of vectors. 40. The columns of any matrix form a basis for its column space. 41. The pivot columns of the reduced row echelon form o f A form a basis for the column space of A. 42. The vectors in the vector form of the general solution of Ax = 0 fmm a basis for the null space of A. 43. If V is a subspace of dimension k. then every generati ng set for V contains exactly k vectors. 44. lf V is a subspace of dimension k, then every generating set for V contai ns at least k vectors.

59. Show that { [

47. The dimension of

n· is

:::'::.:'![_j]. [-lll;,' ,, :,::~~ !t-l]· [-!]· [-i] j;, 63.

M.

49. Every linearly independent subset of a subspace is contained in a basis for the subspace. 50. Every subspace of vectors.

n•

~plO" 1[-;] [-l]· [j]j;, "~ •g~o< why

ing set for

n4

54. Explain why { [

-!]. [-~l [~:]. [_n }

is not

linearly independent.

55. Explain why { [

-~] , [

56. Explain why { [-

oo,;, '" ,,, """

-n }

fdo

.,~.

of

~::: mn!Jl ;, •

bul• '"'
6S. : : :

52. A basis for the range of a linear transformat ion is also a basis for the column space of its standard matrix.

'00<1< '" "'

:::::.:rrfnm ;, '

has a basis composed of standard

51. A basis for the null space of a linear transformation is also a basis for the null space of its swndard matrix.

S3

oo,;, '" "' "'"'=

~ ::::":"1 l1ifll Dll;, '~;,

/1 ,

48. The vectors in the standard basis for n• arc the standard vectors of n•.

is a basis for the subspace in

oo.

45. If S is a linearly independent set of k vectors from a subspace V of dimension k, then S is a basis for V. 46. If V is a subspace of dimension k. then every set containing more than k vectors from V is linearly dependent.

l [-n}

~

mlfl· [-lll;,'.., '"' '"'

::ml

column space of the matrix in Exercise 7.

~.

Show
1Ul· [-lll;, •,.,;, ~l•m• foo '"

space of the matrix in Exercise 8. 67. What is the dimension of Span {v}. where v 'I 0? Justify your answer. 68. What is the dimension of the subspace

m·

1

R" • •• =

0

1·,...;r, ,... '"""'

69. What is the dimension of the subspace

is not a basis ror

nJ

i] ,[-~] }

is not a generating set for

1[I]'

R" " = 0 '"' " =

+·"""'"' ~·-

70. Find the dimension of the subspace

nJ 57. Explain why

{[_:J.[-;J.[-~J.[-m

is not lin-

early independent. 58. Explain why { [ _;] , [

-7]. [_:]}


is not a basis for n 2 •

Justify your answer.

Page 5 of 10


4 .2 Basis and Dimension 71. Let A= {u 1• u2..... Uk} be a basis for a k -dimensional subspace V of'R!'. For any nonzero scalars Ct,c2, . .. • ct. prove that 8 = {ct Ut, c2u 2, . . . ,Ct Utl is a lso a basis for V.

8 1.

72. Let A = {u t, u 2, . .. , Uk} be a basis for a k-dimensional subspace V of 'R!'. Prove that B={ u 1. u, + u2. u J + UJ ..... u ,+ utl is also a basis for V. 73. Let A= {UJ. u2..... Ut } be a basis for a k -dimcnsional subspace V of 'R!'. Prove that {v. 112, UJ .... , uk} is also a basis for V, where v = u , + u2 + · · · + Uk.

82.

l[l]] ~Noll U ~ l[l]] ~Col =l

£=

l

83. Let

V

v

= {

[:~] e n

3:

74. Let A = {u 1• u 2 •. .. , ud be a basis for a k -di me ns ional subspace V of'R!', and let8 = {vt, v2.... ,vk}, where

v, = Prove that

u, + Ut+ J + · · · + Ut

for i = I. 2, ... , k.

S=

B is also a basis for V.

nm

75. Let T: R" -> be a linear transformation and {u t. u 2..... u.,} be a basis for R".

2

4

_i]

-6

-I I I

-3

-2

-6

[

V

-I

-2

3 - I

v, - v2 + "J =

253

-i]

o}

;md

{[-ll[-!J.[!]}

(a) Show that S is linearly independent. (b) Show that

(a) Prove that S = {T(u 1). T(u2), .... T ( u.,)} is a generating set for the range ofT.

(-9v,

(b) G ive an example to show that S need not be a basis for the range ofT. 76. Let T: R" -+ 'R!" be a one-to-one linear tran sformation and V be a subspace of n•. Recall from Exercise 97 of Scct ion4. 1 that W = {T(u): u is in V} is a s ubspace of

+ 6v2) [-

i] n [r] [:J + (7v,

+ (-2VJ + 21'2)

- 5v2) [ -

=

'R."'. (a) Prove that if {u ., ul, .. .. ud is a basis fo r V, {T(u J), T(u 2).... , T(ut )} is a basis for IV.

fo r every vector [::: ] in V. II)

(b) Prove that dim V = dim W. 77. Let V and W be nonzero s ubspaces of R" such that each vector u in R" can be uniquely expressed in the form u = v + w for some v in V and some w in IV. (a) Prove that 0 is the only vector in both V and W. (b) Prove that dim V + dim W =

(c) Determine whether S is a bas is fo r V. J ustify your answer.

84. " ' "

~ lml"'' , , -,, ~o '"'" ~ o] ·"

11.

78. Let V be a subspace o f n•. According to Theorem 4 .4, a linearly independent subset £ = {UJ, U2.... , uml of V is contained in a basis fo r V. S how that if S = {b 1 , b 2, .. .. bt } is any generating set for V. then the pivot bkl columns of the matrix [u , u 2 ... u,. b 1 b2 form a basis for V that contains £.

In Exercises 79-82. use tile procedure described in Etercise 78 to find a basis for the subspace V tilat comains the given Iinearly independenT subseT £ of V .

s~ l[llm [-j]] (a) Show that S is linearly independent. (b) Show that

" " - I , •• - , , .,,

+ (2v 1 -

[ll .

O.Sv2 - O.)VJ)

for every vector

[..'•'] 112

+

(-S., + .. + ,.,,

m

[ '] ["'] -1

~

=

V?

:~

in V.

V)


Page 6 of 10


254


(c) Determine whether S is a ba>is for V. Justify your answer.

87. Show that

In E>.l'l'dsl'.< 85· 88. 11se eithu a calclllator with matrix capabilitit's or romp111er safn.-art! Sllch as MATLAB to soh·e each problmL

1.1] [ - 7.6 2.7] . [ ~-5 2.5] } { [ =7.8 . 9.0 4.0 6.5

85. Let

A =

0.1 0.2 0.7 0.9 [ -0.5 0.5

0.3-J 1.23 1.75

0.5 - 0.5 -0.5

-0.09]

- 1.98 . -2.50

is a basis for the column space of the matriJ< A in Exerci-e 85. 88. Let

A=

(a) Find a basis for the column space of A.

r-·~

(b) Find :1 basis for the null space of A.

0.3 1.2 -0.6

- 0.21 0.63 2.52 -1.26

0.2 0.58 - 0.1 - 0.59 0.6 - 0.06 0.2 1.18

0.4 -0.5 0.6 - 0.2

0.61] -0.81 0.12 . 0.30

86. Show that (a) Compute the rank of A. the dimension of Col A. and the dimension of Row A.

- 57.1 53.8 29.0] [-26.6] 16.0 . -7.0 -9.1 [ 4.9 - 7.0 13.0

(b) Usc the result of (a) to make a conjecture about the relation;.hips among the rank of A, the dimension of Col A, and the di mension of Row A, for an arbitrary matrix A. (c) Test your conjecture, using r;mdorn 4 x 7 and 6 x 3

is a basis for the null space of the matrix A in Exercise 85.

malrices.

SOLUTIONS TO THE PRACTICE PROBLEMS I. Since the given set contains 2 vectors, and 'R3 has dimension 3. this set cannot be a basis for 'R) .

of Ax

= 0. which is

2. The reduced row echelon form of the given matrix A is

[~

-2 0 3] 0

I

2 . The set of vectors in this representation.

0 0 0

![llPll

Hence a basi> for the column space of A is

is a basis for the null space of A. Thus the dimension of Null A is also 2 .

the set consist ing of the pivot columns of A. Thus the dimension of Col A is 2. Solving the homogeneous system of linear equations having the reduced row echelon form o f A as its coefficient matrix. we obttti n the vector form of the general solution

3. Let A be the matrix in Practice Problem I. Since each of the vector~ in the g iven set 8 is a 'olut ion of Ax = 0. 8 is a subset of the null space of A. Moreover, 8 is linearly independent since neither vector in 8 is a mu ltiple of the other. Because Practice Problem I shows that Null A has dimension 2. it follow;. fmm Theorem 4.7 that 8 i ~ a basis fOJ' Null A.

14.3 1THE DIMENSION OF SUBSPACES ASSOCIATED WITH A MATRIX In this section, we investigate the dimension of several important subspaces, including those defined in Section 4.1.


Page 7 of 10


4.3 The Dimension of Subspaces Associated with a Matrix

255

The first example illustrates an importal1! general result.

•;:f!..!.!tjl

For the matrix

A=

_:

[

2

-2

I

2

4

-3

-3

-6

2

in Example I of Section 4.2. we saw that the set of pivot columns of A,

is a basis for Col A. Hence the dimension of Col A is 3.

As mentioned in Example I, the pivot columns of any matrix fom1 a basis for its column space. Hence the dimens ion of the column space of a matrix equals the number of pivot columns in the matrix. However, the number of pivot columns in a matrix is its rank. TI1e dimension of the column space of a matrix equals the rank of the matrix.

The dimensions of the other subspaces associated with a matrix are also determined by the rank of the matrix. For instance, when finding the general solution of a homogeneous system Ax = 0. we have seen that the number of free variables that appear is the nulli ty of A. As in Example 3 of Section 4 .2, each free variable in the vector fonn of the general solution is mu ltiplied by a vector in a basis for the solution set. Hence the dimension of the solution set of Ax= 0 is the nulli ty of A.

The dimension of the null space of a matrix equals the nullity of the matrix.

The preceding boxed res ult can sometimes help to find the dimension of a subspace without actually determining a basis for the subspace. In Example 4 of Section 4.2, for instance, we showed that

is a basis for

Our approach required that we know the dimension of V. so we solved the equation + v2 + v4 = 0 to obtain a basis for V. There is an easier method, however, because

Vt


Page 8 of 10



V =Null [I I 0 1]. Since [ I I 0 I] is in reduced row echelon form, its rank is I and hence its nullity is 4 - I = 3. Thus dim V = 3.

1¥£lrr!•)tfW

Show that

is a basis for the null space of

A~

[

I

-2

I

0

I

0

- :5 - 2 -2 - 1

5 3

'] I

3 . -5 2 - 10

Solution

Because each vector in B is a solution of Ax = 0, the set B is contained in Null A. Moreover, neither vector in B is a multiple of the other, so B is linearly independent. Since the reduced row echelon form of A is

[i ~ -i ~ -~oil , 0 0

0 0

the rank of A is 3 and its nullity is 5 - 3 = 2. Hence Null A has dimension 2, so B is a basis for Null A by Theorem 4.7.

P"cti
Show ""

1[~r

]·[~]J;_,

b"'l' foo

~

'"" 'P'" of oh< m.ori•

[! ~ .l -i -j] We now know how the dimensions of the column space and the nu ll space of a matrix arc related to the rank of the matrix. So it is natural to turn our attention to the other subspace assoc iated with a matrix, its row space. Because the row space of a matrix A equals the column space of AT. we can obtain a basis for Row A as follows: (a) Form the transpose of A, whose columns are the rows of A. (b) Find the pivot columns of AT, which form a basis for the column space of AT. This approach shows us that the dimension of Row A is the rank of A T. In order to express the dimension of Row A in terms of the rank of A, we need another method for finding a basis for Row A.


Page 9 of 10



2 57

It is important to note that, un like the column space of a matrix, the row space is unaffected by elementary row operations. Consider the matrix A and its reduced row echelon form R , given by

A

= [~ ~J

R

and

l

= [~ ~

Notice that Row A but

= Row R = Span { [ : J} ,

= Span { [ ~J} ~ Col R = Span { [ ~ J}.

Col A

It follows from the next theorem that the row space of a matrix and the row space of its reduced row echelon form are always equal.

THEOREM 4.8 The nonzero rows of the reduced row echelon form of a matrix constitute a basis for the row space of the matrix. PROOF Let R be the reduced row echelon form of a matrix A. Since R is in reduced row echelon form, the leading entry of each nonzero row of R is the only nonzero entry in its column. Thus no nonzero row of R is a linear combination of other rows. Hence the nonzero rows of R are linearly independe nt, and clearly they arc also a generating set for Row R. Therefore the nonzero rows of R arc a basis for Row R. To complete the proof. we need show only that Row A= Row R. Because R is obtained from A by elememary row operations, each row of R is a linear combination of the rows of A. Thus Row R is contained in Row A. Since elementary row operations are reversible, each row of A must also be a linear combination of the rows of R. Therefore Row A is also contained in Row R. It follows that Row A= Row R, completing the proof. •

Recall that for the matrix

A= [-!

I

- 2

I

0

I 5 3

0

-2

- 2 - 1

-5

5] I

-3 2 - 10

in Example 2, the reduced row echelon form is

[j l -~ !-~] Hence a basis for Row A is

!lilHlUJI

Thus the dimension of Row A is 3, wh ich is the rank of A.


Page 10 of 10



As Example 3 ill ustrates, Theorem 4.8 yields the following important fact.

The d imension of the row space of a matrix equals its rank.

We have noted that the row and column spaces of a matrix are rarely equal. (For instance, the row space of the matrix A in Example 3 is a subspace of R 5 • whereas its column space is a subspace of R 4 .) Nevertheless, the resu lts of th is section show that their dimensions are always the same. lt follows that dim (Row A)= dim (Col A)= dim (Row AT). and thus we have the following result:

The rank of any matrix equals the rank of its transpose.

We can easily extend the results in this section from matrices to linear transformations. Recall from Section 4. 1 that the null space of a linear transformation T: 'R" -+ 'R"' is equal to that of its standard matrix A , and the range of T is equal to the column space of A. Hence the dimension of the null space of T is the nullity of A. and the dimens ion of the range ofT is the rank of A. It follows that rile sum of rite dimensiom of rite null space and range of T equals rile dimension of rile domain ofT. (See Exercise 71.) Pract ice Problem 2 ....

Find the dimensions of the column space, null space, and row space of the matrix in Practice Problem I. •

SUBSPACES CONTAINED WITHIN SUBSPACES Suppose that both V and W are subspaces of 'R" such that V is contained in W . Because V is contained in W. it is natural to expect that the dimension of V wou ld be less than or equa l to the d imension of W. TI1is expectation is indeed correct, as our next result shows.

THEOREM 4.9 If V and W are subspaces of R" with V contained in W , then dim V Moreover, if V and W also have the same dimension, then V = W.

~

dim W.

PROOF It is easy to verify the theorem if V is the zero subspace. Assume, therefore, that V is a nonzero subspace, and let B be a basis for V. By the Extension Theorem, l3 is contained in a basis for W, so dim V ~ dim IV. Suppose also that both V and W have dimension k. Then B is a linearly independent subset of W that conta ins k vectors. and l3 is a bas is for W by Theorem 4.7. Therefore V =Span B = W. •

Theorem 4.9 enables us to characterize the subspaces of 'R" . For example, this theorem shows that a subspace of 1?..3 must have dimension 0, I, 2, or 3. First, a subspace of dimension 0 must be the zero subspace. Second. a subspace of d imension I


Page 1 of 10



259

must have a basis {u ) consisting of a single nonzero vector, and thus the subspace must consist of all vectors that are multiples of u. As noted in Example 4 in Section 4.1, we can depict such a set as a line through the origin of 1?_3. Th ird, a subspace of dimension 2 must have a basis (u, v} consisting of two vectors, neither of which is a mult iple of the othe r. In this case, the subspace consists of all vectors in 1?.3 having the form a u + bv for some scalars a and b. As in Example 2 of Section 1.6. we can visual ize such a set as a plane through the origin of 1?.3 . Finally, a subspace of 1?.3 having dimension 3 must be 1?.3 itself by Theorem 4.9. (See Figure 4.7.) L

'R 3 is the only IV

3·dimcnsional

.subsp:tce of itself.

-

The 2-dimensional s ubspace wirh ba,o,is {u. v)

The 0-dimensional subspace~

The l-dimensional

subspace with ba.
Figure 4 .7 Subspaces of'R.3

We conclude th is section with a table summarizing some of the facts from Sections 4.1 and 4.3 concerning impo1tant subspaces associated with an m x 11 matrix A. (This table also applies to a linear t ransf01mation T: 'R" --+ 'R"' by taking A to be the standard matrix ofT .) The Dimens ions of the Subspaces Associated with an m x Subspace ColA

Containing Space

11

Matrix A

Dimension rank A

NullA

nullity A = n - rank A

Row A

rank A

You can use the following procedures to obtain bases for these subspaces:

Bases for the Subspaces Associated with a Matrix A ColA : The pivot columns of A form a basis for Col A. NullA: The vectors in the vector form of the solution of Ax= 0 consti tute a basis for Null A. (Sec page 80.) Row A : The nonzero rows of the reduced row echelon form of A constitute a bas is for Row A. (See page 257.)


Page 2 of 10


260 CHAPTER 4 Subspaces and Their Propert ies

EXERCISES In Exercises 1-4, the reduced row echelon form of a matrix A is given. Detennine the dimension of (a) Col A, (b) Null A, (c) Row A. and (d) Null AT. 1.

3.

[~

- 3 0 0 I 0 0

[~

- I

-~] 2 6

0

n

I 0 0

0 0

2.

4

[~

0

- 2

0

5 0

[j

0

-4 2

0

I 0 0 I

-30

0

0

~]

20.

[-l

2{;

-I I

In Exercises 5- 12. a matrix A is given. Determine the dimension of(a) Col A. (b) Null A , (c) Row A. and (d) Null AT.

[-i

10.

[-~

II .

[j

6.

6]

~]

-I -3

7. [;

9.

-4

-8

5. (2

I

-2 3 2 -4 -4

8. [

2 2 0

[~ -~

2

-I 4

-2 6

-I I -3

5

2

-~J

-~]

_l]

12.

u 2

I

0

23 [

-I

-i] -I

- I

-3] -I

0

-I I

21

-l]

[~

I 3 I

-2 -6 -4

24. [

_j

~

_;]

0

-2

2 5 -8

-I

-3

-2

-~]

-2

-I

2

0

- I -3 -I -8 - 2 - I -2 9 2

-I I

j]

5

I -I - I 2 0 -I -3 I 0 -I - I - I 3 - I - I

3 -I -I 4 I

-!] -5 I 3 0

J]

-4 3 -I - 3 0 In Exercises 25-32. use the method described on page 256 to find a basis for Rmv A. -I

25. Exercise 17 28. Exercise 20 31. Exercise 23

-2 4

-I -3 -2] -6 I I - I -3 I -I -8 3

26. Exercise 18 29. Exercise 21 32. Exercise 24

27. Exercise 19 30. Exercise 22

In Exercises 33-40, determine the dimen.~ion of the (a) range and (b) null space of each linear tronsformmion T. Use this information to detennine whether T is one-to-one or omo.

In Exercises 13- 16. a subspace is given. Determine its dimensiou.

In Exercises 17- 24. a matrix A is given. Find a basis for Row A. 17.

[~

18.

[

19. [

I

-:

;J

-1 - 1 0 2 -24]

-~ =~ =~ -~]


Page 3 of 10



~

In Exercises 41 - 60, determine whether tire state-

~ ments are true or false.

63. 13= {

41. If V and W are subspaces of n" having the same dimension. then V = W. 42. If V is a subspace of

n"

having dimension

n"

having dimension 0, then

11.

261

[!l [_:] }· 1

then

V

v=n".

= { [ ; + 1] En 3: sand 1 arc scalars } - 3s +t

43. If V is a subspace of v (0).

=

64. 13 = {

44. The dimension of the null space of a matrix equals the rank of the matrix.

[!l [-~] }·

45. The dimension of the column s pace of a matrix equals the nullity of the matt·ix. 46. The dimension of the row space of a matrix equals the rank of the matrix. 47. The row space of any matrix equals the row space of its reduced row echelon form.

48. The column space of any matrix equals the column space of its reduced row echelon form. 49. The null space of any matrix equals the null space of its reduced row echelon form. 50. The nonzero rows of a matrix fonn a basis for its row space. 51. If the row space of a matrix A has dimension k. then the first k rows of A form a basis for its row spuce. 52. The t'OW space of any matrix equals its column space. 53. The dimension of the row space of any matrix equals the dimension of its column space. 54. T he mnk of any matrix equals the rank of its transpose. 55. The nullity of any matrix equals the nullity of its transpose.

67.

56. For any m x 11 matrix A, the dimension of the null space of A plus the dimension of the column space of A equals m. V

57. For any 111 x 11 matrix A. the dimension of the null space of A plus the dimension of the row space of A e<1ual s 11. 58. If T is a linear transformation. then the dimension of the range of T plus the dimension of the null space of T equals the dimension of the domain of r. 59. lf V is a subspace of W. then the dimension of V is less than or equal to the dimension of W. 60. If IV is a subspace of 1?.!' and V is a subspace o f W having the same dimension as W. then V IV.

=

In Exercises 61 - 68, 11se tile res11lrs ofrilis section to show that 13 is a basis for each Stlbspace V. 61.

2

B= (e1 .e 2). V = {[ ;

62. 13= (e 1,e2). V = { [

~ 3:] E n 2 : s and 1 arc scalars }

s-1'

5

,] 3

en2 : sand 1 are scalars }


=

l[

r -

s+3tl En :

2r -1

-2r + s +

68.

4

-r+ 3s+ 21 1

"= IUlHlUll ~ ~ ~+ 2tl n : = V

l[

3rr _ s Jt 4 -r + 2s

r , s,

E

4

r, s , and I are scalars

]

69. (a) Find bases for the row space and null space of the matrix in Exercise 9. (b) Show that the union of the two bases in (a) is a basis for n•. 70. (a) Find bases for the row space and null space of the matrix in Exercise II. (b) Show that the union of the two bases in (a) is a basis for 3 .

n

Page 4 of 10


262 CHAPTER 4 Subspaces and Their Properties 71. Let T: n" ~ n"' be a linear transformation. Prove the Dimension Theorem: The sum of the dimensions of the null space and range ofT is n.

85. Let

_:

72. Is there a 3 x 3 matrix whose null space and column space are equal? Justify your answer. 73. Prove that, for any m x 11 matrix A and any 11 x p matrix B , the column space of AB is contained in the column space of A. 74. Prove that, for any m x 11 matrix A a nd any 11 x p matrix 8 , the null space of 8 is contained in the null space of A8. 75. Use Exercise 73 to prove that. for any m x and any n x p matrix 8. rankA8 ::; rank A.

11

matrix A

76. Usc Exercise 75 to prove that. for any m x 11 matrix A and
A=

-I

then there is some nonzero vector that belongs to both I' and IV.

86. Let

·-[] Hl Ul Show that B is linearly independent and hence is a basis for a subspace W of n 5 • 87. Let

8 I. Let V beak -dimensional subspace of R." having a basis {UI' U2, .. . , U k}. Define a function T: nk -+ n" by

2.0 1.0

A1=

1n .

+ " "'

r- . r=1n 3.0 1.0

- 2.0 1.0

~o

l[ 1 1 11 1.0 0. I

T ( []] ) = '"" + ' ' "' +

= 0. = 0.

/11 Exercises 86-88. use either a calculator with matrix capo· bi/ities or complller software suclr as MATLAB to solve eaclr problem.

5

80. Prove that if I' and Ware 3-dimensional subspaces ofn 5•

-;]

2 -5 . 7 -4

(b) Prove that if C is a 4 x 4 matrix such that AC then rank C ::; 2.

78. Use Exercise 77 to prove that. for any m x 11 matrix A and any 11 x p matrix 8 , nullity 8 ::; nu llity AB.

n

3 6

-1 2

(a) Find a 4 x 4 matrix 8 with rank 2 s uch that A8

77. Usc Exercise 75to prove that. for any m x 11 matrix A and any 11 x p matrix 8. rank A8 ::; rank 8. Him: The ranks of AB and (A8)7' are equal.

79. Find 2-di mcnsional subspaccs V and W of s uch that the only vector belonging to both I' and IV is 0.

[

I

0 I

2.0 -0.6

1.4

-0.3

and

(a) Prove that T is a linear transformation. (b) Prove that T is one -to-one. (c) Prove that the mnge ofT is V. 82. Let v be a subspace of nn and u be a vector in is not in I'. Define W

= {v +cu : v is in V

n"

that

and c is a scalar}.

(b) Is A 2 a basis for the subspace IV in Exercise 86? Justify your answer.

(a) Prove that IV is a subspace of n". (b) Detem1ine the dime nsion of W . Justify your answer. 83. (a) Show that for any vector u in onl y if u 0.

=

n", u r u =

0 if and

(b) Prove that for any matrix A. if u is in Row A and v is in Null A. then u r v = 0. Him: The row space of A equals the column space of Ar

(c) Show that, for any matrix A, if u belongs to both Row A and Null A, then u 0.

=

84. Show that, for any m x 11 matrix A. the union of a basis for Row A and a basis for Nu ll A is a basis for R." . Him: Use Exercise 83(c).


(a) Is A 1 a basis for the subspace W in Exercise 86? Justify your answer.

(c) Let 8, A1, and A2 be the matrices whose columns are the vectors in /3, A 1 , and A2, respectively. Compute the reduced row echelon fom1s of 8 , A 1• A2 • Br, A[, and

Af.

88. Let 13 be a basis for a subspace W o f n•. and 8 be the matrix whose columns are the vectors in 13. Suppose that A is a set of vectors in n•. and A is the matrix whose columns arc the vectors in A. Usc the results of Exercise 87 to devise a test for A to be a basis for IV. (The test s hould involve the matrices A and 8 .) Then prove that your test is valid.

Page 5 of 10


4.4 Coordinate Systems

263

SOLUTIONS TO THE PRACTICE PROBLEMS I. U:t A denote the given matrix . Clearly,

[j

2 0

3

-2

I

4

7 II

3 0

6

-mlm

belong to Null A . Moreover, neither of these \'ectors is a multiple of the other. and so they are linearly independent. Since Null A ha~ dimenSIOn 2. 11 follows from Theorem .-.7 that

and is a basis for Null A .

[j

2 0

3

-2

I

4 6

7 II

3 0

-il[J=m

2. The reduced row echelon form of the given matrix A i>

I 2 0I 00 - 4]1

0 0 0 0 0 0

[

0 l 0 0

I • 0

Thus both

Pl

Hl

:md

and so its rank is 3. The dimensions of both the column space :md the row space of A equal the rank of A. 'o these dimensions are 3. The dimension of the null space of A is the nullity of A. which is

5 - mnkA = 5 - 3 = 2 .

14.4 1COORDINATE SYSTEMS In previous sections. we have usually expres!.ed vector> in term' of the standard basis for n•. For example.

However. in many applications, it is more natural to express vectors in terms of some basis other than the standard one. For example. consider the ellipse shown in Figure 4.8. With respect to the usual ..\)'·COOrdinate system, it can be shown that the ellipse has the equation

13x 2 - l Oxy + J3y2

= 72.

Observe that the axes of symmetry of the ellipse arc the lines y = x andy = - x, and the lengths of the semimajor and sem iminor axes arc 3 and 2, respectively. Consider a new x 'y ' -coordinate system in which the x'-axis is the line y = x ;md the y'-axis is the line y = - x . Because the ellipse is in standard position with respect to this coordinate system, the equation of this ellipse in the x 'y '-coordinate system is


Page 6 of 10


264 CHAPTER 4 Subspaces and Their Properties x'

Figure 4.8 The ell ipse with equation 13x1

-

1Oxy + 13y1

= 72

But how can we use this simple equation of the ellipse in the x'y'-coordinate system to derive the equation of the same ellipse in the usual xy-coordinate system? To answer this question requires a careful examination of what is meant by a coordinate system. When we say that v

= [!].we mean that v = 8e 1 + 4ez. That is,

the components of v are the coefficients used to represent ,. as a linear combination of the standard vectors. (Sec Figure 4.9(a).) Any basis for a subspace provides a means for establishing a coordinate system on the subspace. For the ellipse in Figure 4.8, the standard basis docs not lead to a simple equation. A better choice is (u ,, uz}. where u 1 x'-ax is and u 2

= [- :]

= [:] is a vector that lies along the

is a vector that lies along the y'-axis. (Note that since

u,

and Uz are not mu ltiples of one another, B = (u,, Uz} is a linearly independent subset of two vectors from 2 . Hence, by Theorem 4.7, (u 1• u 2} is a basis for 2 .) For th is

n

basis, we can wri te the vector v

= [!]

in Figure 4.9(a) as v

n = 6u 1 + (-2)uz. Thus

the coordinates of v are 6 and - 2 with respect to B, as shown in Figure 4.9(b). In this case, the coordinates of v are different from those in the usual coordinate system for n 2.

······~· ···

...... ··i······

······~

.....;...... j •••••

···•······ ······:····" ' "' j······ l 4e2 ... ;······ ...... ..... ·r............ ...... ~

....

·~

(a)

(b)

Figure 4.9 Two coordinate systems for 'R-2


Page 7 of 10



265

In order to identify the vector v with the coefficients 6 and - 2 in the linear combination v = 6u 1 + (-2)u2, it is important that this is the only possible way to represent v in terms of u 1 and Uz. The following theorem assures us that whenever we have a basis for R", we can w rite any vector in n" as a unique linear combination of the basis vectors :

THEOREM 4 .10 Let B = {b t, b 2, ... , bk} be a basis for a subspace V of R" . Any vector v in V can be uniquely represented as a linear combination of the vectors in B; that is, there are unique scalars a,, a2 • . .. , ak such that v = a , b 1 + a2b2 + ·· · + ak bk. Let v be in V . S ince B is a generating set for V, every vector in V is a linear combination of b t. b2... . . bt . Hence there exist scalars a,. a2 •... , ak such that v = a, b t + a2b2 + · ·· + ak b k. Now let c, . c2, . .. ,Ct be scalars such that v = c, b t + c2b 2 + · ·· + Ck bk. To show that the coefficients in the linear combination are unique, we prove thai each c; equ als the correspondi ng a; . We have PROOF

+ a2 b2 + · · · + at bk)- (Ct b l + C2b2 + · ·· + Ck b k) =(a,- Ct)bt + (az - c2)b2 + · ·· + (ak - ck)bt. = (at b t

Because B is linearly independent. this equation implies th at at - Ct = 0, c2 = 0, · · ·, ak - Ck = 0. Thus at = c,, a 2 = c2, · • ·, ak = Ct. which proves that the representation o f v as a linear combination of the vectors in B is un ique .

a2 -

•

The conclusion of Theorem 4 . 10 is of great practical value. A nonzero subspace V of n" contains infi nitely many vectors. But V has a finite basis B. and we can uniquely express every vector in V as a linear combination of the vectors in B. Thus we are able to study the infinitely many vectors in V as linear combinations of the finite number of vectors in B.

COORDINATE VECTORS We have seen that we can uniquely represent each vector in R" as a linear combination of the vectors in any basis for R" . We can use thi s representation to introduce a coordinate system into R" by making the following definition: Definition Let B = (b 1, bz, ... , b11 ) be a basis 1 for R" . For each ' ' in R" , there are unique scalars c , . c2 ..... c,. such that v = Ct b t + c2b2 + · ·· + c,. b11 • The vector

l 1

~:~] c,.

In order for the definition of a coordinate vector to be unambiguous, we must assume that the vectors in B are listed in the spedfic sequence b1, b2, . . . , bn. When working with coordinate vectors, we always

make this assumption. If we wish to emphasize this ordering, we refer to B as an ordered basis.


Page 8 of 10


266


in 1?}' is called the coordinat e vector of v relat ive to 13, or the 13-coordinate vector of v. We denote the 13-coordinate vector of v by (v (a .

+;:f!,!.!tjl

Let

Since 13 is a linearly independent set of 3 vectors in 1?.3 , 13 is a basis for 1?.3 . Calculate u if

(u ]s

=[

_n.

Solution Since the components of [u ]a are the coeffic ients that express u as a line;u· combination of the vectors in 13, we have

=

Let 13 (b 1, b2} be the basis for 1?.2 pictured in Figu re 4. 10. RefetTi ng to that figure, find the coordinate vectors [u]a and [vJa.


Figure 4 .10

l££!,j.jfW

For

v = [ - :]

and

13=

[ [:].[-:] . [~]! ·

determ ine [vJa . Solution To find the 13-coordinate vector of v, we must write v as a linear combination of the vectors in 13. As we learned in Chapter 1, this requires us to find scalars Ct, q , and C3 such that

Ct [:]


+c2 [ - :]

+c3

[i] = [ -!]· Page 9 of 10



26 7

From this equation, we obtain a system o f linear equations with augmented matri x

[l

I - I

2

I

2

Since the reduced row echelon form of this matri x is

-64] '

I 0 0 0 I 0 [0 0 I we see that the desired scalars are

CJ

3

= - 6, cz = 4, C3 = 3. Thus

is the B-coord inate vector of v.

We can easily compute coordinate vectors re lative to the standard bas is£ for 'R". Because we can write every vector

VJ] vz V=

[

~II

in 'R" as v = v 1e 1 + v2ez + · · · + v,e,. we sec that (v(t; = v . It is also easy to compute the B-coord inate vector of any vector in a basis B. For if B = {b 1• b2, . . .. b,} is a basis for 'R". we have

b;

= Ob 1 + Ob2 + · ·· + Ob;- J + lb; + Ob;+J + · · ·+ Ob,,

and so [b;]B = e; . In general, the following result provides a simple method for calcul ating coordinate vectors relative to an arbitrary basis for 'R":

THEOREM 4.11 Let B be a basis for 'R" and 8 be the matrix whose columns are the vectors in B. Then 8 is invertible, and for eve ry vector v in 'R", B[v]B v, or equivalently, [V]B = 8 - 1v.

=

PROOF


Let B = (b 1, b2, . .. , b,) be a basis for 'R" and v be a vector in 'R". If

Page 10 of 10


268


then

= [b r b2 ... b,]

[~~] c,

= B[v]s, where 8 = [b r b2 . . . b,]. Since 6 is a basis. the columns of B are linearly independent. Hence B is invertible by the Invertible Matrix Theorem, and thus • B[v]s =v is equivalenr to [v]s = s - •v. As an altemative to the calculation in Example I, we can compute the vector u by using Theorem 4.11. The res ult is

in agreement with Example l.

Verify that B =

Practice Problem 2 -..

I[l].[:] ,[n I

is a basis for

n 3. Then find u if lu ls =

[

=~l ~

We can also use Theorem 4.1 1 to compute coordinate vectors, as the next example demonstrates.

IJ£iuj.jll

Let

and

as in Example 2, and let B be the matrix whose columns are the vectors in B. Then

Of course. thjs result agrees with that in Example 2.

Practice Problem 3 -..


Find [v] 8 if v = [

-rJ

and 6 is defined as in Practice Problem 1.

Page 1 of 10



269

CHANGING COORDINATES To nbtain an equation of the ellipse in Figure 4.8, we must be able to switch between the x'y' -coordinate system and the xy-coordinate system. This requires that we be able to conve11 coordinate vectors relative to an arbitrary basis B into coordinate vectors relative to the standard basis, or vice versa. In other words, we must know the relationship between [v]B and v for an arbitrary vector v in 'R.". Accord ing to Theorem 4.11, this relationship is [v]B = s - 1v. where 8 is the matrix whose columns are the vectors in B. Although a change of basis could be useful between any two bases for 'R.". the examples we give next arise as rotations of the usual coordinate axes. Such changes of bases arc especially important in Chapter 6. Consider the basis B = (b 1, bz) obtained by rotating the vectors in the standard basis through 45° . We can compute the components of these vectors by using the 45° -rotation matrix, as in Section 1.2: and In order to write the x', y'-equation

as an equation in the usual xy -coord inate system, we must use the relationship between a vector and its B-coordinates. Let [v]B =

[xy''],

and

Since 8 is a rotation matrix. we have [v]B s - 1v 8 7 v, and so

=

=

s- 1 =

Br by Exercise 53 in Section 2.3. Hence

Therefore substituting

x'=

into an equation in the x'y' -coordinate system converts it into one in the xy-coordinate system. So the equation of the given ellipse in the standard coordinate system is


Page 2 of 10



which s impl ifies to 13x 2 - IOxy

Example4

+ 13/

= 72.

Write the equation - J3x 2 + 2.xy + J3y 2 = 12 in terms ofthex'y'-coordinate system, where the x'-axis and they' -axis are obtained by rotating the x -axis andy-axis through the angle 30°. Solution Again, a change of coord inates is required. This time, however, we must change from the xy-coordinate system to the x'y'-coordinate system. Consider the basis B = {b,, b2l obtained by rotating the vectors in the standard basis through 30°. As we did earlier, let

and Since v = B[v]e, we have

Hence the equations relating the two coordinate systems are

y

I , .J3 , = Zx + 2>' ·

These substitutions change the equation - J3x 2 + 21")' + J3y 2 = 12 into 4x'y' = 12; that is. x'y' 3. From this equation. we see that the graph of the given equation is a hyperbola (shown in Figure 4. 11).

=

Figure 4 .11 The hyperbola with equation - ./3x 2


+ 2xy + .j3y2 = 12

Page 3 of 10



271

EXERCISES In Exercises 1-6, the 8-coordinate vector of\' is gi••en. Find v if

1. [v] y

= [~]

2. [v] u = [ -~]

4. [v]8

= [-!]

5. [v]u =

3. [v] y=

m

6. [v]u =

m m

In Exercises 7- 10, the 8-coordinate vector of v is given. Find v if

23. Find the unique representation of u =

[~ J as

a linear

[~ J as

a linear

combi nation o f

24. Find the unique represcnt<•tion o f u = combination of

[-!]

7. [v[ u=

9 (v]G = [

8. [viG = [

~n

10. [vlo = [

11. (a) Prove that 8 =

{[~]. [ -~] }

is a basis for R.2 .

-n

a basis for R.2 .

(b) Fmd [v] 0 1f \ ' = -2 _

13. (a) Prove that 8= { [ for

26. Find the unique representation of u = [ ~] as a linear combi nat ion o f

_;J, [-~] } is [ 3] [-1]

12. (a) Prove that 8 = {[

.

J as a linear

combi nation of

-n

(b) Find [vJ 6 if v =5m - 3 [

.

25. Find the unique representation of u = [ ~

_!]

2

-~l

+4

27. Find the unique representation o f

.

[_:]. [-l]}

-~] _ [

-H

(b) Find [v[ G if v = [ - : ] -4

= [

~]

as a linear

11

= [

~]

as a linear

11

= [

~]

as a linear

combination of

28. Find the unique representution o f

14. (a) Provethat8= { [ -:]·[- : ]. [: ] } isabasisfor

nJ.

11

is a basis

nJ

(b) Find (v] G if v = 3 [

2

combi nation o f

[:l

In Exercises 15- 18, a vector is gi••en. Find its 8-coordinate vector relative to the basis 8 in Exercises 1- 6.

18.

m

29. Find the unique representation of combination of

!11 Exercises 19- 22. a••ecTor is given Find iTs 6-coottlinme vector relative to the basis 8 in Exercises 7- 10.


Page 4 of 10



30. Find the unique representation of u = [

~] as a linear

combination of

(y')2

(r')2

48. The graph o f an equation of the form .:....,- - - ,- = I a· b· is a parabola.

49. The graph of an ell ipse with center at the origin can be (x' )2 (y')z written in the form - - + - - = 2 2 rotation of the coordin~te axes. b

I by a suitable

50. The graph of a hyperbola with center at the origi n can (x')2 (y' )2 be written in the form - bZ = I by a suitable 7 rotation of the coord inate axes. ~ In Exercises 31 - 50, derennine whether The Slate~ mellls are true or false.

31. If S is a generating set for a subspace V of

n".

then every vector in I' can be unique ly represented as a linear combination of the vectors in S.

32. The components of the 8-coordinate vector of u are the coefficients that express u as a linear C·Ombination of the vectors in B.

33. If[; is the standard basis for in 'R.".

n•.

51. Let B = lb , b 2J. where b t =

[il

[~] and bz =

(a) Show that B is a basis for n 2. (b) Determine the matrix A = [ [e dB [e2lo I. (c) What is the relationship between A and B = [b t b 2l?

52. Let B = (b 1• b 2l- where b t =

then [v)c = v for all v

[~] and b2 = [ =~l

(a) Show that l3 is a basis for n 2. (b) Determine the matrix A = [ [e , )6 [e2)6 ).

34. The 8-coordinate vector of a vector in B is a s tandard

(c) What is the relationship between A and 8 = [b 1 b 2 )?

vector.

35. If B is a basis for n" and B is the matrix whose columns are the vectors in 8 , then s - t,, = lv iB for all v in n".

53. Let B=( b ,, b2, bJ}. where b t = [

_:J [=:J b 2=

36. If B = lb 1, bz, ... , b,l is any basis for n", then, for every vector v in n". there are unique scalars such that V = Ct b t + c2 b2 + · · · + c, b, . 37. If the columns of an 11 X II matrix form a basis fOI' n". then the matrix is invenible.

(a) Show that B is a basis for n 3 •

38. If B is a matrix whose columns are the vectors in a basis for n", then, for any vector v in n". the system Bx = v has a unique solution.

(c) What is the relationship between A and 8 = lb , b 2 b3 I?

39. If 8 is a matrix whose columns arc the vectors in a basis 8 for n•. then, for any vector v in n" , [v)a is a solution of 8 x = v.

and bJ

54. Let B= (b1 . b 2. bJ}. where bt = [

[u )g + [v)0 . n•. B is any basis

for n". and c is a scalar, then [cv) 0 = c[v] 0 . 44. Suppose th
[~:] = Ae [~ ]-

45. Suppose that the x' -, y' -axes are obtained by rotating the usual x -. y -axes through the angle 0. 1l1en

[~] = Ao [ ; :] .

is an ellipse.


(x')l

. b2 = [

-il

and bJ =

[~]-

n 3.

(b) Determine the matrix A = [ [e,]B [e2lo [eJIB ). (c) What is the relationship between A and B = [b t b 2 bJJ?

In Exercises 55- 58. an angle 9 is given. Let v [ v] 6

= [~]

and

= [;:]. where l3 is the basis for n 2 obmined by mrarin!l e t

and ez through The angle 9. WriTe equations expressing x' and y' in terms ofx and y . 55. 0 = 30° 56. 0 = 60° 57. 0 = 135° 58. 0 = 330° In Exercises 59- 62. a basis B fo r n 2 is given. If v = [;] and

46. lf Ao is a rotation matrix, then A'/; = A01. 47. The graph of an equation of the form

-n

(a) Show that 8 is a basis for

41. If 8 is any basis for n", then 1016 = 0. 42. If u and v are any vectors in n• and B is any basis for

n". then (u + v)B =

=n

(b) Determine the matrix A = [ [e ,]5 [e2lo [eJIB ).

40. If 8 is a matrix whose columns are the vectors in a basis B for n", then. for any vector v in n", the reduced row echelon form of [8 vJ is 11,. [v iB l·

43. If v is ;my vector in

=[

(y')2

7 + --;r- = I

[v)Ci

= [;,:]. wriTe equarians e.rpre.<.rin8 x ' andy' in terms a[x

and y .

Page 5 of 10



59. {[

61

[

-~l -~]}

H-~J.[-:J}

60.

{[!J. [m

62. {[

-~l [-~Jl

In Exe rcises 63- 66. a basis B for 'R3 is given.

r''lz; =

If v =

[

~] and

[x'] ~:

.

65.

In Exercises 79- 86. an equation of a conic section is given in the x'y' ...cotudintlle system. Determine the equa1irm of the conic section in the u.wal xy-coordinate system if the x'-a.xis and the y'-axis are obrained by rorating the usual x-axis andy-axis through the given angle 0.

79. (x' )

, write equations expressing x'. y', and z' ill tenns

t[~l

[ll [-nl

{[~].[-~]. [=:] }

G4.

66.

l

t[=:J. [= n. [n

{[-~].[~].[:] } = [;, J and

In Exercises 67-70, an angle 0 is given. Let ,.

81.

= [;}where B i.f the basis for n 2 obtained by rmming

e 1 and e2 through The angle 0. Write equations expressing x and y in terms of x' andy'.

67. 0 = 60° 68. 9 = 45° 69.

(J

= 135° 70. 9 = 330°

In Exercises 71-74, a basis B for 'R 2 is given. If v [v]s

= [;:]. write equarions expressing x

= [; Jand

andy in Terms ofx'

andy' . 71.

{[~]. [!]}

73. {[-

~]. [~]}

In Erercises 75- 78, a basis B for

[v]s =

72

.

H-~l[~Jl

74. {[;].

[~]}

n 3 is given. ({ v =

[

~] and

[~:]. wriTe equarions expressing x, y. and z in Terms of

2

+ (y') = I. 9 = 60° )2

(x' )2

(y')2

(x')2

{)")2

22 - y

y -

( ')2 82. .::__ 6 (x')2

= I. 9 = 45

-o

52= I. 9 = 13)

I, 9 = 150°

22 =

I, 9 = 120°

4 (y')2

(x')2 (y')2 84. - - + -22 52 2

85 (x')

32 (x' )2

o

+ ''"')2 _v_ =

83. """j2-

.

[v]s

2

42

80.

ofx. y, and z.

63

273

- (y')

2

42

= I ' 9 = 330°

= I ' 9 = 240°

()•')2

86. -,- + -,- = I , 9 = 300° 32/n Exercises 87- 94, an equation of a conic section is given in the .\')'·Coordinate system. DeTermine the equarion of the conic section in the x 'y '-coordinme system if the x'-axis and they'· axis are obtained by Tvtating the usual x-axis and y-(~ris tluv ugh the given angle 9 .

87. -3x 2 + 14A)'- 3y 2 = 20. 9 = 45° 88. fu· 2 - 2J3xy + 4y 2 = 2 1. 9 = 60° 89. 15x 2 - 2../3xy + 13y 2 = 48. 9 = 150° 90. x 2 - 6xy + y 2 = 12. 9 = 135° 91. 9x 2 + 14../3xy- 5y 2 = 240. 9

= 30° + 33y2 = 720. 0 = 60° + lly 2 = 40. 9 = 150° 2 xy + 2y = 12. 0 = 135°

92. 35x 2 - 2../3xy 93. 17x 2 - 6../3A)'

94. 2\·2 95. Let A= tu•. U2.. .. . u., l be a basis for n· and c 1 , c2 •. ••• c. be nonzero scalars. Rec;~ll from Exe rcise 7 1 o f Section 4.2 thai 8 = {c 1u 1,c2 u2, ... ,c, u.) is a lso a basis for 'R". If v is a vector in 'R" and

{IJ]

x'. y', and z'. [v]A=

7.

[

a.,

compute [ v]s.

96. Let A= {u 1. u2. .... u.,) be a basis for 'R". Recall that Exercise 72 o f Secli on 4.2 shows lhat

8= {u J, UJ + u 2, . . . , u 1 + u.,)


Page 6 of 10


27 4

CHAPTER 4 Subspaces and The ir Propert ies is a lso a basis for n•. If v is a vector in R" and

(\']A =

ll2il O : . [

in 8 (that is. if v = a 1u 1 + 02 112 + · · · + akut for unique scalars a .. a 2 , .•. • ak), then /3 is a ba~is for V . (This is the converse of Theorem 4.10.)

106. Let V = {

[:!]

e n3 :

-2v, + v2 + v3 = 0 } and

a,.

compute (v]s.

97. Let A

=

{ u ~o u 2 • ... , u,} be a basis for R". Recall that Exercise 73 of Section 4.2 shows that

8 = (u 1 + u 2 + · ··+ u,, u 2, u 3, . .. , u,} is a lso a basis for n•. If v is a vector in R" and

["']

(a) Show that Sis linearly independent. (b) Show that

(2v, - 5v2) [-:]

l/2

(\']A=

:

+ (2v,

- 2v2)

G]

•

a,.

compute (v (s. { u ~o u 2 , ... , u, } be a basis for R". Recall that Exercise 74 of Section 4.2 shows that

98. Let A =

for every vector [ where

~J in S.

(c) IsS a basis for V? J ustify your answer. fori = 1,2, ... ,k

v 1 = u 1 + u 1+ 1 + · · · + uk

is a lso a basis for n•. If v is a vector in R" and

[{/'] :

if and only if {(u d s. [u 2Js.... , [uk Is} is linearly independent.

108. Let 8 be a basis for R", (u 1, 112, .. . , uk} be a subset of R", and v be a vector in R". Prove that ' ' is a linear com-

l/2

[\']A=

107. Let B be a basis for R" and (u 1. u 2..... utl be a subset of R". Prove that {u 1 • u 2 ,. ... ut} is linearly independent

•

(Ill

compute [v)s.

99. Let A and 8 be two bases for R". If (v iA = (' 'Is for some nonzero vector v in R " . must A= 8'? Justify your answer.

bi nation of (u ,. 112 . . . . . Ut} if and only if [vlts is a linear combination of ([u dlJ.[U2)6..... [utltsl ·

In Exercises 109- 112. use eillwr a wlcul{IIOrwitfr matrix capa· bilities or compwer sofn••are such as MATLAB to solve each problem.

109. Let

100. Let A and /3 be two bases for n•. lf [v].A = [v)s for every vector v in R", must A = 8'1 Justify your answer.

101. Prove that if Sis linearly dependent, then every vector in the span of S can be written as a Ii ne:tr combination of the vectors in S in more than one way. 102. Let A and 8 be two bases for R ". Express [v].A in terms of (v (6 .

103. (a) Let 8 be a basis for R". Prove that the function T: R" -+ R" defined for all v in R" by T(v) =( v is is a linear transformation. (b) Prove that T is one-to-one and onto.

104. What is the standard matrix of the linear transformation T in Exercise 103? 105. Let V be a subspace of R" and 8 = {u 1 . u 2 , .•• • Ut} be a subset of V , Prove that if every vector v in V can be uniquely represented as a linear combinat ion of the vectors


8

=

l[

25 73 -56 -68 - 118 0] [ 14] [ -6] [- 14] [ -12] -21 ' -66 ' 47 ' ~0 ' 102 23 64 - 50 - )9 - 106 12 42 -29 -39 -62

'"' ·{il (a) Show that 8 is a basis for

ns.

(b) Fi nd [v)s. I I 0. For the basis /3 in Exercise I 09, lind a nonzero vector u in 5 such that u = [u ]s .

n

I l l. For the basis 13 in Exercise 109, find a nonzero vector v in 5 such that (vis= .5v.

n

Page 7 of 10


4.5 Matrix Representations of Linear Operators 11 2. Lc1 Band v be as m Exerci..e 109. and lei

275

(a) Show 1ha1 v is a linear combinalion of u 1 • u2. and

U J.

(b) Show thaJ(v)a is a linear combinalion of (u da. (u 2]6, and (u l (B. (c) Make a conjeclure lhal generali;r.cs lhc results of (a) and ( b).

SOLUTIONS TO THE PRACTICE PROBLEMS I. Based on Figure 4.1 0. we ~ee lhal

u = ( l) bJ t-2bl

and

By Theorem -1.11. we have

V=4b. + 2bl.

h follows lh'u

u

and

(vJu =

= B(u Ja =

[l

[~]. 3. By Theorem 4 .11. we also have

2. Lei B be lhe matlix whose columns are 1he veclors in B. Si nce lhe reduced row echelon form of 8 is /3. the columns of 8 arc linearly independent Thus 8 is a linearly indepcndcn1 >CI of 3 vcclOrs from 7?. 3, and hence il is a basis lor 3.

n

14.5 1MATRIX REPRESENTATIONS OF LINEAR OPERATORS Our knowledge of coordinate systems is useful in the study of linear transformations to A linear transformation where the domain and codomain equal is from Most of the linear transformations that we encou nter called a linear operator on from now on are linear operators. Recall the reflection U of 1?.2 about the x-axis defined by

nn nn.

nn

nn.

u ([:::]) =

[

-:~~]'

given in Example 8 of Section 2.7. (See Figure 2. 13 on page 175.) ln that example. we computed the standard matrix o f U from the standard basis for 1?.2 by using that U(e J) e1 and U(ez) - ez. The resulting matrix is

=

=

n

The reflection of 1?.2 about the x -axis is a special case of a rPjlection of 2 about a line. In general, a reflection about a line C through the origin of 1?.2 is a function T : 1?.2 --. 1?.2 defined as follows: Let v be a vector in 1?.2 with endpoint P . (See Figure 4.12.) Construct a line from P perpendicular to C. and let F denote the point of intersection of this perpendicular line with C. Extend segme nt PF through F to a point P' such that PF = FP'. The vector with endpoint P' is T(v). II can be shown that reflections are linear operators. For such a reflection. we can select nonzero vectors b t and b2 so that b 1 is in the direction of C and b z is in a direction perpendicular to C. It then follows


Page 8 of 10


276 CHAPTER 4 Subspaces and Their Properties \'

p

c.

.. P'

f igure 4 .12 The reflection of the vector v about the line .C through the origin of'R-2

=

=

=

that T (b 1) b 1 and T( b2) - b2. (See Figure 4.13.) Observe that B (b ~o b 2} is a bas is for 'R 2 because B is a linearly independent subset of 'R 2 consisting of 2 vectors. Moreover, we can describe the action of the reflection T on B. In part icular, since and the coord inate vectors of T(b 1) and T (b2) relative to [T(b l))B =

m

and

B are given by

[T(bz)]B

= [_~].

We can use these columns to form a matrix 1 !T(b 1liB [T(b2liB 1=

[o'

0 - I]

that captures the behavior ofT in relation to the basis B. ln the case of the previously described reflection U about the x-axis, B is the standard basis for 1?.2, and the matrix [ [U (bl )lB [U(b2liB] is the standard matrix of U.

c.

.. T(bV

= -bz

f igure 4.13 The images of the basis vectors b 1 and b 2 under T


Page 9 of 10


4.5 Matrix Representations of linear Operators

277

A similar approach works for any linear operator on 1?.!' in wh ich the images of vectors in a particular bas is are given. This motivates the following definition: Definition Let T be a linear operator 2 o n for 1?:'. The matrix

n"

[ [T(b J)]s [T(b2lls ..

and !3

= {b 1, bz, ... , b

11 )

be a basis

[T(b")Js l

is called the matrix representation of T with respect to B, or the !3-matrix ofT. It is denoted by [T]s . Notice that the jth column of the !3-matrix of T is the !3-coordinatc vector of !3. Also, when !3 = £, the !3-matrix ofT is

T ( bj ), the image of the j th vector in

lTle = I [T(b ,) le IT(b2l le ... JT(b") le I= IT(e, ) T(e2) . .. T(e")). which is the standard matrix of T . So this definition extends the notion of a standard matrix to the context of an arbitrary basis for n". For the reflect ion T about line C and the b
=

Calling [T]s a matrix represemation ofT suggests that this matrix describes the action of T in some way. Recall that if A is the standard matrix of a linear operator T on n", then T(v) = Av for all vectors v in n". S ince [v]e = v for every v in n", we sec that [T]e[v]e Av T(v) [T(v)]t:. An analogous result is tme for )TJe: If T is a linear operator on n" and !3 is a basis for n", then the !3-matrix of T is the unique 11 x n matrix such that

=

=

=

[T(v)]s = [T]e[v]e for all vectors v in

Example 1

n". (See Exercise

!00.)

Let T be the linear operator on n 3 defi ned by

Calculate the matrix representation of T with res pect to the basis B = {b ,, b2, b3}, where

2 The definition of matrix representation of

T generalizes to the case of any linear transformation

T: 'R." .... 'R.m. (See Exercises 101 and 102.)


Page 10 of 10



Solution

Applying T to each of the vectors in 6, we obta in

We must now compute the coordinate vectors of these images with respect to 6. Let 8 = !b t b2 b 3). Then

and

from which it follows that the 6 -matrix of T is

As in Section 4.4, it is natural to ask how the matrix representation of T with respect to 6 is related to the standard matrix ofT (which is the matrix representation ofT with respect to the standard basis for 1?!'). The answer is provided by the next theorem.

THEOREM 4.12 Let T be a linear operator on R". 6 a basis for R", 8 the matrix whose columns are the vectors in 6, and A the standard matrix ofT. Then [T]u = s - 1AB , or equivalently, A= 8[T]u8- 1• PROOF Let6={ b t, b2, ... , b,)andB= Ib t b 2 . . . b,[. RecallthatT(u) = Au and [v] u = 8 - 1v for all u and v in R" . Hence

[TJu = [ [T(b t)]u [T(b2))13 ... [T(b,)]u I = [ (A b tlu [A b2J6 1

[Ab, )u ]

1

= [8- (A b t) 8 - (A bz)

8 - 1(A b,))

= [(B - 1A)b t (B - 1A)b2 =

s - 1A!b t

b2

b,J

= s-•As. Thus [T]u =

s - 1AB.

which is equivalent to

• User: Thomas McVaney

Page 1 of 10



279

If two square matrices A and B are such that B = p - 1AP for some invertible matrix P, then A is said to be similar to B. It is easi ly seen that A is similar to B if and only if B is similar to A. (See Exercise 84 of Section 2.4.) Hence we usually describe th is situation by saying that A and B are similar. Theorem 4.12 shows that the B-matrix representation of a linear operator on 1?:' is similar to its standard matrix. Theorem 4.12 not only gives us the relationship between [Tis and the standard matrix of T, it also provides a practical method for computing one of these matrices from the other. 1l1e following ex;unples illustrate these computations:

Example 2

Calculate (TJs by Theorem 4.12 if T and Bare the linear operator and the basis given in Example I:

T

([;~])

and

B=![il[il[:Jl

Solution The standard matrix of T is

Taking B to be the matrix whose columns are the vectors in B. we have

1

[TJu = B - AB = [

-r

Note that our answer agrees with that in Example I.


Example 3

Find the B-matrix representation of the linear operator T on 1?.3, where

Let T be a linear operator on 1?) such that

T

([i]) j], = [

T

([~])

= [

-n,

and

T ([:])

[~]

Find the standard matrix ofT.

Solution


Let A denote the standard matrix ofT,

Page 2 of 10


280 CHAPTER 4 Subspaces and Their Properties and

Observe that 13 = {b 1, b2, b3} is linearly independent and thus is a basis for n 3. Hence we are given the images of the vectors in a basis for n 3. Let 8 [b 1 b2 b 3] and C = [c1 c2 CJ] . Then Ab , = T(b l) = c 1, Ab2 = T (b2) = c2, and Ab3 = T(b3) = CJ . Thus

=

Because the columns of 8 arc linearly independent, B is invertible by the Invertible Matrix Theorem. Thus

A= A(BB- 1 ) = (AB)B -

1

-~

= CB- 1 = [

-.:>

0 1.5

- .5

-I.;]. 1.5

Therefore the standard matrix ofT, and hence T itself. is uniquely determined by the images of the vectors in a basis for n 3 .

Example 3 suggests that a linear operator is uniquely determined by its action on

a basis because we are able to determ ine the standard matrix of the operator in this example solely from this inf01mation. This is indeed the case. as Exercise 98 shows. To conclude this section, we apply Theore m 4.12 to find an explicit formula for the reflection T of n 2 about the line .C, with equation y = ~x. In the earl ier discussion

of this problem, we have seen that nonzero vectors b 1 and- b2 in n 2 must be selected so that b 1 lies on .C and b 2 is perpendicu lar to L. (Sec Figure 4.13.) One selection that works is and

because b 1 lies on .C, which has slope ~. and b2 lies on the line perpendicular to .C, which has slope -2. Let 13 (b t, b 2), B lb 1 b2.l. and A be the standard matrix of T. Recall that

=

=

Then, by Theorem 4.12,

A= B[T] 6 B -

1

=

[:~

.8]

- 6 .

It follows that the reflection of n 2 about the line with equation y

T


([~']) =A [·:I] = [·6 X2 X2 .8

= tx is given by

.8] [Xt] = [.6Xt + .8x2] . .8xJ -

- .6

X2

.6X2

Page 3 of 10




281

Let B be the basis in Practice Problem I , and let U be the linear operator on n 3 such that

[U)s

3 0 0] [0 0 I

=

0 2 0 .

Determine an explicit formu la for U(x ).

EXERCISES In £rercise.1· 1- 10, determine (T]a fiJr each linear operatar T and basis 13.

·~ [[-l]·[-j].[-l]·[J] 10. T

XI]) [XI- X~ ([ =

-'2

+

2r4]

and

+X2 +X3 -3X3 +X4

Xi

X4

•=

+ XJ

21 I - 3x4

X3

[[J[l][j] [l])

In £rercises 11- 18, determine the standcud matrix of tile linear operawr T tuing the given bt1.fis 8 and the matrix represenlllliOn of T with respect to 13.

13={[=:J.[=n.[=:Jl 13

= {

[~l

[-il [-~]}

17. [T]s= [ ~ ~ -:]and -1


I

0

Page 4 of 10


282 CHAPTER 4 Subspaces and Their Propert ies 35. The onl y matrix that is si mii
11

x 11

zero m
36. The o nly matrix that is similar to In is /,.

= [- :

18. [T)s

;

-~] and

37. If T is a linear operator on R" and 8 is a basis for R". then (T]s [ v]s T (v).

=

38. If T is a linear operator on n• and 8 is a basis for n•, then [T] 8 is 1he unique 11 x n ma1rix such that [T]s[v]s = [T(v)]B for all v in n•.

- I

~

~

In Exercises 19- 38. determine whetlter the statemems are true or false.

19. A linear transformat ion T : R " ~ R"' that is one-to-one is called a linear operator o n R". 20. The matrix representation of a linear operator on R " \\~th respect to a basis for 'R!' is an 11 x 11 matrix. 21. If Tis a li near operator on n• and 8 = {b , . b2, ... , b,. I is a basis for R". then column} of (TJB is the 8 -coordinate vec tor of T (bj ). 22. lf T is a linear operator on R" and 8 = {b1. b2. .. .. b, I is a basis for n•. then the matrix representation ofT with respect 10 8 is [T(b ,) T (b2) · ·

T(b,)].

23. If£ is the s tandard basis for R ", then [T k is the standard matlix of

r.

24. If T is a linear operator on R", 8 is a basis for R", B is the matrix whose columns are the vectors in 8. ;md A is the standard matrix of T, then [T)s B- 1A .

=

25. lf T is a linear operator on n•. 8 is a basis for R". 8 is the matrix whose columns are the vcc1ors in 8, and A is the standard matrix ofT, then [T)s BAB - 1 .

=

26. If 8 is a basis for R" and T is the identity operator on R", then (T]s 1, . 27. If T is a reflection of R 2 about a line L , then T(v) - v

=

=

for every vector v on L.

=

28. If T is a reflection of n 2 about a line L, then T (v) 0 for every vector v on L. 29. If T is a reflection of n 2 about a li ne, then there exist~ a basis 8 for R" such that l Tis

l

= [~ ~

30. If T is a reflection of n 2 about a line L. then there ex ists a basis 8 for R" such that ITIB =

[~ _~l

31. If T. 8. and L are as in Exercise 30. then 8 consists of two vectors on 1he line L. 32. Ann x 8 if 8

11

=

matrix A is said to be simi lar to an 11 x PTAP .

33. An 11 x 11 matrix A is similar to an is s imilar to A.

11

x

11

11

matrix

malrix 8 if 8

34. If T is a linear operator on R" and 8 is a basis for 'R!',

then the 6 -matrix of T is si milar to the standard matrix ofT.


39. Let 8 = {b 1. b 21be a b
=

42. Let B = {b , . b2. bJI be a basis for R 3 and T be a linear operator on n 3 such thai T(b 1) b 1 - 2b 2 + 3bJ. T (i11) 6b 2 - b3, a nd T (b3) 5b 1 + 2b2 - 4b J. Determine [T]s.

=

=

=

=

43. Let 8 {b 1, b2, bJ I be a basis for R 3 and T be a linear operator o n n 3 such that T(b ,) = - Sb2 + 4bJ, T(b2) = 2b, - 7bJ. and T(b3) = 3b, + b3. Dctennine [T]lJ. 44. Let B = {b ,, b2, b3} be a basis for R 3 and T be a linear operator on R 3 such that T(b 1) = 2b 1 + Sb 2. T(b 2) = - b , + 3b2, and T(IH) = In - 21)3. Dete rmine [T]s. 45. Let 8 = {b ,. b2. b3. b~} a basis for R 4 and T a linear operator on R 4 s uch thai T (b 1) = b , - b2 + b3 - b~. T(ln ) = 2b2 - b~. T(b3) -3b , + Sb3, and T(b~) 4b2 - b3 + 3b~. Determine [T]s.

=

=

46. Let 8 = {b 1, b2, b3, b4 l be a basis for R 4 and T be a linear operator on R 4 such that T(b 1) = - b 2 + b 4 , T (b2) = b , - 2bJ, T(bJ) = 2b, - 3b4. and T(b4) = - b2 + 2bJ + b 4. Determine [T]s.

In Exercises 47- 54, deiennine (a) [T)s, (b) the standard marrix of T, and (c) an explicit formula for T(x) from the given itiformarion 47. 8

={[:]. [~]}·

48. 8 =

mJ. [~H·

T ([:])

T

=[;].

([~D =

T ([;])

=3[:]

m-2m.

r([~J) = 2GJ-[~J 49. 8 = {[

-~]. [ _:] } • T ( [ -;]) = 3 [ -~]- [ _:].

r([_:J) =2[-;J {[~J.[m . r([;]) r([~]) =3GJ -2m

50. 8=

= -[;]

+4m.

Page 5 of 10



283

58. Find T(b2 - 2b3) for the operator T ;md the basis B in Exercise 42. 59. Find T(2b 1 - b2) for the operator T and the basis B in Exercise 43 . 60. Find T(b 1 + 3b2 - 2b3) for 1he opera10r T and lhe basis B in Exercise 44. 6 1. Find T( - b 1 + 2b2 - 3b3) for1he operator T and lhe basis B in Exercise 45. 62. Find T(b 1 - b3 + 2b• ) for 1he opera1or T and the basis B in Exercise 46. 63. U:1 I be !he idenl ily operalor on 'R.", and lei l3 be any ba~is for 'R.". Delermine !he matrix representation of I with respect to 13. 64. U:t T be the zero operalor on 'R.", and let B be any basis for 'R.". Determine the matrix representation of T with respect to B. In Exetdses 65- 68, find an explicit description of the reflection T of'R.2 abolllthe line with each equation.

65. )' = ~X 66. )' = 2x 67. )' = -2X 68. )'=//IX The orthogonal projection of n 2 on line C through the origin is afimction u: 'R.2 .... 'R.2 defined in the following 111(1/IJier: Let v be a vector in 'R.2 with endpoim P. Construct alinefmm P petpendicul(lr Ill C. and let F denMe the point of intersection of thi.f perpendicular line with C. The vector wirh endpoim F is U(v). (See Figure 4.14). It can be shown that orthogonal projections of 'R. 2 on a line containing 0 arc linear.

'·'

c

p

Figure 4.14 The orthogonal projection of a vector v on a line

through the origin of n

2

69. Find U ([:~~]).where U is the orthogonal projection of 'R.2 on the line with equation )'= X. Him: First, find I u IB· where B = {[:]. [ - : ]} 55. Find T (3b 1 - 2b2) for the operator T and the basis B in Exercise 39. 56. Find T( - b 1 + 4b2) for the operator T and the basis B in Exercise 40. 57. Find T(b 1 - 3b2) for the operator Exercise 41.


T and the basis 13 in

70. Find U (

[::~

J).

where U is the orthogonal projection of

'R.2 on the line with equation y = -

4x.

([:;~]). where U is the orthogonal projection of n 2 on the line with equation)' = - 3x.

71. Find U

Page 6 of 10



72. Find U ( [

;~]). whe re

U is the orthogonal projec tion o f

n 2 on the line with C
In each of Exercises 74- 80. find an explicit formula fo r

Tw (

3

Let IV be a plane thmugh the origin of'R • tmd let v IJe tl vector in n 3 with endpoi111 P. Constmct a line from P perpendicular to IV. and let F dmote the point of illlersection of this perpendicular line with W . Denote the vector with endpoilll F as Uw(,•). Now extend the petpendicular fmm P to F an equal di.w mce to a point P' on the other side of IV, and denote the vecTor with endpoilll P' as Tw(v). In Clwpter 6, it is shown tlwt tire functions Uw and Tw are linear operators on n 3. We call Uw tire orthogonal proj ection of 3 on IV and Tw the reflection of nl abow IV. (See Fig ure 4.15).

n

[:~]} the ref lection of'R

3

abowtlre plane IV defined

by tire given equation.

+ z =0 + 2y - 5z = 0 - 3y + 5~ = 0 + 5y + 7: = 0

+ 3z + 6y - 2z

74. 2.x - y

75. x - 4y

76. x 78. X 80. X

77. x

=o = 0

79. x - 2y - 4z = 0

81. Let IV and 6 be as in Exercise 73, and let Uw be the orthogonal projection of n 3 on W. (a) Find Uw(v) for each vector ' ' in B. (b) Fi nd [Uw )G.

p

(c) Find the standard matrix of Uw. (d) Determine an explicit form ula for Uw

([~:]}

In eaclr of Exercises 82- 88, find an explicit formula f or

F

Uw

([;~]} rlre orthogonal projection of'R

3

o11tlre plane IV

defined by tire given equmion.

82. x Figure 4.15 The orthogonal projection of a vector v on a subspace W of 'R.3 and the reflection of a vector about W

n

73. Let Tw be the reHection of 3 about the plane IV in with equation x + 2y - 3z = 0. and let

n3

84. x 86. x 88. x

83. x - 2y + 5z = 0 85. ,t - 3y - 5z 0 87. x- 5y + 7z = 0

+ y - 2z = 0 + 4y - 3z =o + 6y + 2z = o + 2y - 4z = 0

=

89. Let 6 be a basis for n" and T be a linear operator on 'R" . Prove that T is invertible if and only if [T]G is invertible. 90. Let 6 be a basis for 'R"
Note that the first two vectors in 6 lie in W , and the third vector is perpendicu lar to IV . In general, we can apply a fact from geometry that the vector

91. Let 6 be a basis for 'R" and T be a linear operator o n 'R" . Prove that the dimens ion of the range ofT equals the rank of [Tk 92. Let 6 be a basis for n • and T be a linear operator o n n". Prove that the di mension of the null space ofT equals the nullity of [TJG93. Let 6 be a basis for 'R" and T and U be linear operators on 'R". Prove chat [T + U)!l = (T]!l + [U ]!l. (Sec page 178 for the definition ofT + U .)

whose components are the coefficients of the equation of the plane ax +by + cz =d. is normal (perpendicular) to the plane. (a) Find Tw (v) for each vector v in 6 . (b) Show that

B is

a bas is for

n

3.

(c) Find [Tw ]G. (d) Find the standard matrix of Tw . (e) Determine an explicit fo rmula for T w ( [::]).


94. Let 6 be a basis for n• and T be a linear operator on n". Prove that [cT ]6 c[T] 6 for any scalar c . (See page 178 for the definition of cT .)

=

95. Let T be a li near operator on 'R", and let A and 6 be two bases for n•. Prove that [TJA and LTJG arc s imi lar. 96. Let A and 8 be similar matrices. Find bases A and 6 for n" s uch that IT..lA =A and [TAI!l = 8 . (Thi s proves that s imilar matrices are matrix representations of the s ame linear operator.)

97. Show chat if A is the standard matri x of a reHection of about a line. then detA = - I.

n2

Page 7 of 10



n". lllld let c 1. c 2, . .. • c., be (not necessarily distinct) vectors in n•.

(ii) fsT)~ = s(T]~ (see page 178 for the definitions of T + U and sT); (iii) [T(v)Jc = [TJ~[v) 6 , for any vector v in n•.

98. Let l3 = {bJ. b2.. ". b,} be a basis for

(a) Show that the matrix transfonnation T induced by cs - 1 satisfies T(bJ) = CJ for j = 1,2, ... . n. (b) Prove that the linear transformation in (a) is the unique linear transformation such that T(bJ) = CJ for j = 1.2..... 11. (c) Exte nd these results to an arbitrary linear transformationT :'R"~nm.

99. Let T be a linear opera10r on n" and l3 = {b 1 , b 2. . .. • b, } be a basis for n". Prove that [T] 6 is an upper triangular matrix (see the definition given in Exercise 6 1 of Section 2. 1) if ;md only if T(b1) is a lincm· combi nation of b 1.... , b1 foreveryj. I -:;;j '5;11 . 100. Let T be a linear operator on n" ;md l3 be an ordered basis for n·. Prove lhe following results: (a) For every vector v in

n", [T(v)]B =

[T]B[v]B.

285

(c) Let U : n"' -+ nP. Then

nP be linear. and let V

be a basis for

(d) Let 8 and C be the bases in Exercise 101, and let u: n 3 ..... 2 be a linear transfonnation s uch that

n

-2 -3 Usc (a) to find an explicit formula for U (x).

In Exercises 103- 107. use eirller a calcularor will1marrLr capabiliries or compwer software such as MATI.AB 10 solve each pmblem. 103. Let T and U be the linear operators on

n" defined by

(b) If C is an n x n matrix such that [T(v))B = C[v)B for every vector v in n", then C = [T] 6 .

The following deji11ilion of matrix represemarion of a linear rransformarion is used in Exercises /01 and 102:

n" --. nm

Definition Let T: be a linear transformation, and let 8 = {b 1. b2. .. .. b.,} and C ={c t. c2..... c..,) be bases for and respectively. The matrix

n"

and

nm,

[T(b.,)Jc I

[ [T(b1)lc [T(b2)lc

is called the matrix representation of T with respect to l3 a nd C. It is denoted by [T]~. 101. Let

6= {[:].[-:J.[_:]}

C=WJ.[~]} ·

and

(a) Prove that l3 and C are bases for n 3 and n 2 , respectively. (b) LetT: 3 --. n 2 be the linear transfom1ation de fined by

n

T

([;~]) =[XI + 2x2+-X3]. X) -

X3

X2

'·=

m·~ =Ul·,.= [-il· . . = [=!l· dh.

(a) Compute ITIG- (U 16- and (UT IB· (b) Determine a relationship a mong [T ] 6 . [ U 16, and (UTIB· 104. Let T and U be linear operators on n• ;md l3 be a basis for n". Use the result of Exercise I 03(b) to conjecture a relationship among [T] 6 , [ U ] 6 , and [UT] 6 • and then prove that your conjecture is true. 105. Let l3 and bt . b 2, b3, b" be as defined in Exercise 103.

n"

2XJ

Find [T]~.

n"--. nm

102. Let T: be a linear transformation. and l3 = {b J, b 2, .. . , b.,} and C = {CJ, C2, . ... c.,..} be bases for n• and n'". respectively. Let B and C be the matrices whose columns arc the vectors in l3 and C. respectively. Prove the following results: 1AB. (a) If A is the standard matrix ofT. then [T)~ =

(a) Compute [T]0 , where T is the linear operator on s uch that T(b 1) = ~ . T(b 2) = b 3, T(b 3) = b 4 , and T(b4 ) = b 1• (b) Determine an explicit formula for T (x), where x is an arbitrary vector in n~. 106. Let l3 = {b 1. b 2, b3. b 4 } be as in Exercise 103. and let T be the linear opemtor on n• defined by

c-

n• --. n'" is linear and s [T + U)~ = fTl~ + [U ]~:

(b) If U: (i)


is any scalar. then

Page 8 of 10


286 CHAPTER 4 Subspaces and Their Properties (a) Detennine an explicit formula for an ari>itrary vector in 4 . (b) Compute [T )s and [T 1]u.

n

r - 1(x), where x is

(c) Detennine a relationship between [T]s and [T- 1]s.

107. Let T be an invcniblc linear opcrntor on n• and B be a basis for n•. Usc the result of Exerci>c 106(c) to conjecture a relationship between [T) 8 and [T 1]8 , and then prove that your conjecture is true.

SOLUTIONS TO THE PRACTICE PROBLEMS I. The standard

matn~

of T IS

2. The standard matnx of U 1s 0

A-

[- 1

Let

8 =

~

- I

~l

A=

[l ~]

CHAPTER 4

1

AB = [

~ -3

-I]

5 [-; - I

- 1 . 2

6

He nce

= A

be the matrix who>e columns are the vectors in B. Then the 8 -mutrix repre>entation of T is

iTiu= B

B[U) 6 8~ 1 =

3 -I

,1I] [-2xt + 5.12 - XJ] XJ . [o•llil = --3Xt~ I ++ 6.1z~2 + 2x.l

12] 5 . -6

REVIEW EXERCISES

~

In E.1erc:ises I 25. determine whether tile staremems

~

are Inti' or false.

I. If Ut, uz.... , u, are vccton. in a subspace V of n•. then every linear combination of Ut. u 2..... Ut belongs to v.

2. The 'pan of" finite nonempty 'ubsct of n• is a subspace ofn•. 3. The null 'pace of an m x 11 matrix i' contained in n•. 4 . The COlumn >p:tCC Of an Ill X II matriX is COntained in 'R". 5. The row space of an Ill X II matrix is COntained in n•, 6. The rnngc of every linear tran>fonnation is a subspace. 7. The null space of every linear transformation equals the null space of its >tandard matrix. 8. The runge of every linear transformation equals the row space of its s tandard matrix. 9. Every nonzero subspace of 'R" has a unique bas is. I 0. It is pos.~iblc fo1· different ba>cs for a panicular subspace to contain different numbers of vectors. I I. Every finite gcncruting set for a nonzero subspace contains a basis for the s ub,pacc. 12. T he pivot columns of every matrix form a basis for its column s pace. 13. The vectors in the vector form of the general solution of Ax = 0 con
17. The dimension of the null space of a matrix equals the rank of the matrix. 18. The dimension of the column space of a matrix equals the rank of the matrix. 19. The dimension of lhc ro11 space of a matrix equals the nullity of the matnx. 20. The column space of an) matrix equal> the column space of its reduced row echelon fonn. 21. The null space of any m.11rix equal> the null >pace of its reduced row echelon fonn. 22. If B is a basis for n• and B is the matrix whose columns arc the vectors in B. then 8 tv = {V)B for all v inn·. 23. If T is a linear operator on n•. B i> a basis for 'R". 8 is the matrix whose columns arc the vccton. in B. and A is the >ta11dard matrix ofT, then [TJu = BAB - t. 24. If T is a linear opcrntor on 'R" and B is a basi> for 'R", then (T )s iS the unique II X II matrix MICh that [T)s [v)s = (T(v))s for a ll v in 'R".

25. If T i> a reOcction of 'R 2 about a line, then there e xi,ts a basis B for

n 2 s uch that (T]u =

[:)

_

26. Dctcnni ne whether each phru>c ;, :1 misuse of terminology. If so, explain what is wrong . (a) a ba>is for a matrix

14. No sub.ion greater than"·

(b) the mnk of a ;ub>pacc

15. There is only one subspace of n• having dimension "·

(c) the dimension of a square

16. There is only one s ub>pacc of n• having dimension 0.

(d) the dimension of the zero sub>pacc


~l

matri~

Page 9 of 10


Chapter 4 Rev iew Exercises

287

(e) the di mension of a basis for a subspace (f) the column space of a linear transformation (g) the di mension of a linear transformation (h) the coordi nate vector of a linear opemtor

27. Let V be a s ubspace of 'R!' with dimension k, and let S be a subset of \1. What can be said about m. the number of vectors in S. under the following cond itions? (a) S is linearly independent.

(a) Prove that B is a basis for n J

(b) S is linearly dependent. (c) S is a generating set for V.

(b) Find \' if [ v) B = [

28. Let A be the standard matrix of a linear transformation T: n 5 --. n 1 • If the mngc ofT has d imension 2. determine the di mension of each o f the following subspaces: (a) Col A (d) Null AT

(b) Null A (c) Row A (e) the null space ofT

In Exercises 29 and 30, determine wherlwr rite given set is a subspace of'R4 • Jusrify your answer.

![} ·"· lUl;, 29.

n• •I = "! " = o n· · ... = , ...

... =

column space of the matrix in Exercise 32.

~' '"

=

I

T(b t) = -2b2 + b 3.

= 5b t -

4b 2

T(~)

n 3 and T be the linear

= 4 b t - 3b).

and

+ 2 bJ.

(a) Determine [T ] B. (b) Express T(v) as a linear combination of the vectors in B if v 3b t - b2 - 2bJ.

=

I

39. Detennine (a) [T]B. (b) the standard matrix of T, and (c) an explicit formula for T(x) from the given information about the linear operator Ton

nz.

8= { [-~J. [-~]}.

In Exercises 31 and 32. find bases for (a) rite n11ll space if ir is nonzem, (b) rite column space, and (c) the row space of each

r([-~J)=[!J.

matrix. 2

-2

2 -I

-3

-I - I

4

8

I

= [ -~] .

38. Let B = {b, . b2. b3} be a basis for operator on n 3 such that T(b3)

o

o.••, •• = 0

(c) Find (W]B if W

=~] .

I

j]

In Exercises 33 and 34, a linear Transformation T is given. (a) Find a basis for the range ofT. (b) If tire 111111 space of T i.s nonzero, find a basis fo r rlre n11ll space tifT. 33. T: 7?. 3 -+ 'R4 defined by

r([-;J)=[-:]

and

40. Let T be the linear o perator on 2 defined by

n

=

r([;~J) [~;t_-~~J

n 2 and B be the basis for

and

B=

{[~]. [~] } -

Detennine [T]s. 41. Detennine an explicit description of T(x), using the g iven

basis B and the matrix representation of T with respect to B. [T IB = [

-l ; -i]

and

42. Let T be the linear operator on

B= {[:],

[~]. [ :] }

n 3 such that

r ([~])= [_n· r([-:])=[jJ. and

r(C:D [-nFind an explicit description of T(x). In Exercises 43 and 44. 011 equation of a conic set·riOII is given in

the x'y'-coorriinate system. Determine file equarion of the conic of the linear transformation in Exercise 34.


secrion in the us11al .1)>-coordinme sysrem

if rite x ' -axis and the

Page 10 of 10


288 CHAPTER 4 Subspaces and Their Propert ies y'•a.t is are obwined by rotating the usual X·tuis and y-axis through the gil'en angle B. (x')2 (y')2 43. + = I () = 120°

22 J2

44. -J3(x')2 + 2r'y'

= 12,

+ J3(y'f

= 330°

B

In exercises 45 and 46, {1/1 equation of a conic section is g iven in the xy-coordinate system. Determine the equation of the conic section in the x' y' ·COIJJtlinate sysTem if the x' -axis and the y'axis are obtained by rotming the usual X·{Lt is and y-axis through the g iven angle B.

45. 29x 2 - 42.xy 46.

+ 29y 2 = 200. () + lly 2 = 576.

-39x 2 - 50J3xy

= 3 15°

51. Let B be a basis for n" and T be an invet1ible linear operator on n•. Pro ve that [T- 1] 8 ([T]B)- t.

=

Let V and W be subsets of n " . We defin e the sum of V and W. denoted V + W. as

v + w = lu ill n· : u = v + w fo r some v in V and same w in W) . In Exercises 52- 54, use the preced ing definition.

52. Prove that if V ;md IV arc subspaces of n". then V is also a subspace of n•. 53. Let

() = 210°

47. Find an explicit description of the reflection T ofn 2 about the line with equation y = - ~-r.

V

= {[~]

48. Find an explicit description of the orthogonal projection U of n 2 on the line with equation y = -

and

49. Prove that if lvt , v2, .. . , v, } is a basis for n " and A is an invertible 11 x 11 matrix. then [A vt.Av2, .. .. Av. l is also a basis for n•.

IV = {

&x.

50. Let v and w be subspaces of n". Prove that v u w is a subspace of n" if and only if V is contained in W or W is contained in V .

+W

[~!]

in n 3: in n

Find a basis for V

3

Vt

+ •·2 = Oand

: w1 -

2 v t - V3 = 0 }

2w3 = 0 and w 2 + w3 = 0 } .

+W.

54. Let 5 1 and 5 2 be subsets ofn", and let 5 = 5 1 U 5 2. Prove that if V = Span5t and IV= Span 52. then SpanS= V+IV .

CHAPTER 4 MATLAB EXERCISES For the following exercises, use MATLAB (or comparable software) or a calculator with matrix capabilities. The MATLAB functions in Tables D. I, 0 .2, 0.3, 0.4, and 0.5 of Appendix 0 may be useful.

I . Let

2. Let

A=

["

2.1 - 1.2 2.2

1.3 3.1

0.0 -1.5 4.1 2. 1 1.2 - 4.0

2.2 2.7 1.7

6.5 3.8 2.2

-1.3 2.2 1.4 2. 1 - 1.7 - I. I

-0 2]

4.3 0.2 4.3 . -0.4 2.0

4.7 2.3 1.2 3.2 - 3.1 - 1.0 - 3.3 1.1 2.1 5.5 -4.3 123 A = [ 2. -1.2 1.4 -1.4 -1.2 -1.4 . 1.1 -4.1 5.1 2. 1 3. 1 0. 1 - 2. 1 1.2 - 0.8 3.0 For each of the following parts, determine whether each vector belongs to Null A.

For each of the fo llowing parts, use Theorem 1.5 to determine whether each vector belongs to Col A.

(a)

(c)

[_,] 11.0 - 10.7 0.1 - 5.7 12.9

(a) [

(b)

[' ] 2.0 - 3.8 2.3 4.3 2.2

["] [' ] -2.8 4. 1 2.0 4.0 - 3.7

- 32

(d)

3.0 4.4 8.4 0.4


-58]

- 1.1

H]

(b)

i·~1

(c) [ - 03 3.7

[-~-~] 3.4

-2 6 - 0.9

4 5

(d)

[-~ ~1 - 1.3 - 1.0

3. Let A be the matrix of Exercise 2. (
(b) Use Exercise 78 of Section 4.2 to extend this basis to a basis for

n6 •

Page 1 of 10



(c) Find a basis for Null A.

Let T be a linear operator T on

289

n 5 such that

(d) Find a basis for Row A. 4. Let

~:~ _;:~1

2. 1 1.3 2.2 -1.4 1.3 - 3.7 - 1.2 5.3 4.0 2.7 1.7 4. 1 - 0.7 -3.1 1.0 - 7.2

A =

3.8

1.4 . 6.5

r

5.1

(a) Find a basis for the column space of A consisting of columns of A. (b) Usc Exercise 78 of Section a basis for n6.

4.2 10 extend this basis to

Find the standard matrix o f the linear operator T from the given information.

7. Find the s tandard matrix of the li ne:tr transformation

u: n6 --> n 4 such that

(c) Find a basis for Null A. (d) Find a basis for Row A . 5. Let 2 1

2

33

- 1.3

4. 1

-1~

~4

4~

3 1

43 -32 1.5 ' - 4.2 ' 3.1 · · ' 4.2 1.6 3.8 ~2 -3. 1 0.4

l.l1 r 1r-1. 1r 1

2.2 •

r

0.7 ~I

r

5.3 -4.5

4.5 2.5

5.3 1.3 . [ - 1.4 2.5

1.8 4.1

u

.

-2.4

1 1 -2.3

(a) Show that B is a basis for

n6 •

(b) For each of the following parts. represent the given vector as a linear combination of the vectors in B.

5.1 7.41

(i)

-

r

:~:~

- 5.3 4.21 (ii)

-8.0

- ~:~ 2

r

26.6

- 19.31 6.6 "') 30.2 ( Ill -7.7

7.5 -8.2

r

2.2

-18.9

(c) For each vector in (b). find the corresponding coordinate vector relati vc to B. Exercises 6 and 7 apply the l't!sult.s of Exercise 98 of Section 4.5. 6. Let B = {b t. bz, b 3. b4. bsl be the basis for

B=

n5 g iven by

u

-ll ~l1J {ll ([!]) U l (r-il) m (r:l1J

2.9

2.3

4. 1

1.0

0.0

1.0


1.3 3.0

u

=

=

n" can be described as the column space of a matrix A. Si mply choose a finite generating set for IV and let A be the matrix whose columns are the vectors in this set (in any order). What is less clear is that IV can also be described as the null space of a matrix. The method is described next. Let IV be a subspace of n". Choose a basis for W and extend it to a basis B = {b t . b2.. ... b., } for n". where the first k vectors in B constitute the original basis for IV. Let T be the linear operator on n" (whose existence is guaranteed by Exercise 98 of Section 4.5) defined by

8. It is clear that any subspace IV of

T(b ·) =

10.0 3.0 2.5 2.0 8.0 - 9.0 1.41 . r-1.91 4.0 ' r - 2.31 1.0 . r-3.11 4.0 . r0.71l 5.0 .

[ 4.4 4.0

u ([

= [

I

Prove thai W

{0

bj

1.f.<

'·

J - "

ifj > k.

=Null A, where A is lhe slandard malrix

ofT.

Page 2 of 10


290 CHAPTER 4 Subspaces and Their Properties 9. Let IV be the subspace of n 5 with basis

Hlf!H!ll

matrix whose rows consist of the rows of A followed by the rows of 8 . Prove that Null C = Null An Null 8 = V n IV . (b) Let

Use the method described in Exercise 8 to find a matrix A such that IV = Null A. 10. An advantage of representing subspaccs of 'R." as null spaces of matrices (sec Exercise 8) is that these matrices can be used to describe the intersection of two subspaces, which is a subspace of 'R." (see Exercise 77 of Section 4.1). (a) Let V and IV be subspaces of 'R.", and suppose that A and 8 arc matrices such that V = Null A and IV= Null 8. Notice that A and 8 each consist of 11

columns. (Why?) Let C = [


~

l

that is. C is the

and

w

l[

~s~· -!]'m'm)'

Use (a) and the MATLAB funct ion null(A, 'r') described in Table 0 .2 of Appendix D to fi nd a basis for V (I IV.

Page 3 of 10


5

INTRODUCTION

The control of vibrations in mechanical systems, such as cars, power plants, or bridges, is an important design consideration. The consequences of uncontrolled vibrations range from discomfort to the actual failure of the system with damage to one or more components. The failures can be spectacular, as in the collapse of the Angers Bridge (Maine River, France) in 1850 and the Tacoma Narrows Bridge (Washington State, United States) in 1940. The vibration problem and its solutions can be explained and understood by means of eigenvalues and eigenvectors of the d ifferential equations that model the system (Section 5.5). It is convenient to model a mechanical system as a mass attached to a spring. The mass-spring system has a natural frequency determined by the size of the mass and the stiffness of the spring. This frequency is natural in the sense that if the mass is struck, it will vibrate up and down at this frequency. A final element of the model is a periodic external force, Fo sin wr, applied to the main mass. In a car, this could come from the engine or from regular defects on the highway. In the Angers Bridge collapse, the applied force came from soldiers marching in step. In the Tacoma Narrows Bridge collapse 3, the applied force came from wind-induced vibrations. If the frequency, w, 3 Video clips from this collapse are viewable at http://www.pbs.org/wgbh/novalbridge/tacoma3.html and http'l/www.enm.bris.ac.uk!anm/tacoma/tacoma.html#mpeg.

291


Page 4 of 10


292

5 Introduction

of the external force equals or is close to the natural frequency of the system, a phenomenon ca lled resonance occurs when the variations of the applied force are in step with, and reinforce, the motions of the main mass. As a consequence, those motions can become quite large. In the case of the bridges mentioned above, the motions led to the collapse of the bridges.


The solution to the vibration problem is either to redesign the system to shift its natural frequencies away from that of the applied forces or to try to minimize the response of the system to those forces. A classic approach to minimizing response (the Den Hartog vibration absorber) is to attach an additional mass and spring to the main mass. This new system has two natural frequencies, neither of which is the same as that of the original system. One natural frequency corresponds to a mode of vibration where the masses move in the same direction (the left pair of arrows in the figure). The other natural frequency corresponds to a mode of vibration where the masses move in opposite directions (the right pair of arrows in the figure). If the size of the absorber mass and the stiffness of the new spring are adjusted properly, these two modes of vibration can be combined so that the net displacement of the main mass is zero when the frequency of the applied force, w, equals the natural frequency of the original system. In this case, the absorber mass absorbs all of the vibration from the energy applied to the main mass by the external force.

Page 5 of 10


CHAPTER

5

EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION n many applications. it is important to understand how vectors in R" arc transformed when they are multiplied by a square matrix. In this chapter, we see that in many circumstances a problem can be reformulated so that the original matrix can be replaced by a diagonal matrix, which simplifies the problem. We have already seen an example of this in Section 4.5. where we learned how to describe a reflection in the plane in terms of a diagonal matrix. One impor1ant class of problems in which this approach is also useful involves sequences of matrix-vector products of the form Ap, A2 p , A3 p, .... For example, such a sequence arises in the study of long-term population trends considered in Example 6 of Section 2. 1. The central theme of this chapter is the investigation of matrices that can be replaced by diagonal matrices. the diagona/izab/e matrices. We begin this investigation with the introduction of special scalars and vectors, called eigen values and eigenvectors. respectively. that provide us with the tools necessary to describe diagonalizable matrices.

I

[[!]

EIGENVALUES AND EIGENVECTORS

In Section 4.5, we discussed the reflection T of R 2 about the line with equation = Recall that the vectors

y

tx.

ba=m

and

played an essential role in determining the rule for T. The key to this computation is that T (bJ) is a multiple of ba and T( b2) is a multiple of b2. (Sec Figure 5.1.)

Figure 5.1 Th~ imag~s of th~ basis v~ctors br and b2 ar~ multi pl~s of the vectors.

293


Page 6 of 10


294

CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalizatio n

Nonzero vectors that are mapped to a multiple of themselves play an important role in understanding the behavior of linear operators and square matrices. Defin itions Let T be a linear operator on 1?!'. A nonzero vector v in 1?!' is called an eigenvector ofT if T(v) is a multiple of v: that is. T{v) AV for some scalar A. The scalar1 A is called the eigenvalue of T that corresponds to v.

=

For the reflection T of R 2 about line C and the vectors b 1 and b2, where b 1 is in the direction of C and b2 is in a direction perpendicular to C. we have

T herefore b 1 is an eigenvector of T with corresponding eigenva lue I. and b2 is an eigenvector of T with corresponding eigenvalue - I. No nonzero vectors other than multiples of b 1 and b2 have this property. Since the action of a linear operator is the same as multiplication by its standard matrix, the concepts of eigenvector and eigenvalue can be defined sim ilarly for square matrices. Defin itions Let A be an 11 x n matrix. A nonzero vector v in 'R" is called an eigenvector of A if Av = AV for some scalar 2 A. The scalar A is called the eigen value of A that corresponds to v.

For example, let A=

[:~

-:!l in

Section 4.5, we showed that A is the standard

matrix of the reflection of R 2 about the line

C with the equation y =

!x. Consider

the vectors

It is a simple matter to verify directly that

Therefore neither u 1 nor u 2 is an eigenvector of A . However, b 1 is an eigenvector of A with conesponding eigenvalue I , and b 2 is an eigenvector of A with conesponding eigenvalue - I. Because A is the standard matr ix of the reflection T of R 2 about the line C, we can verify that b 1 is an eigenvector of A without calculating the matrix product Ab 1. Indeed, since b 1 is a vector in the d irection of C, we have Ab 1 = T(b J) = b 1. Similarly. we can also verify that b 2 is an eigenvector of A from the fact that b2 is an e igenvector ofT. In general, if T is a linear operator on 'R", then T is a matrix transformation. Let A be the standard matrix ofT. S ince the equation T(v) = AV can be rewritten as Av = AV. we can determine the eigenvalues and eigenvectors ofT from A. 1

In Jhis book. we are concerned primarily with scalars that are real numbers. Therefore. unless an explicit statement is made to the contrary, the term eigenvalue should be interpreted as m eaning reo/eigenvalue. There are situations, however, where it is useful to allow eigenvalues to be complex numbers. When complex eigenvalues are permitted, the definition of an eigenvector must be changed to allow vectors inC". the set of n·tuples whose components are complex numbers. 2 See footnote 1.


Page 7 of 10


5.1 Eigenvalues and Eigenvectors

295

The eigenvectors and corresponding eigenvalues of a linear operator arc the same as those of its standard matrix. In view of the relationship between a linear operator and its standard matrix. eigenvalues and eigenvectors of linear operators can be studied simultaneously with those of matrices. Example I shows how to verify that a given vector v is an eigenvector of a matrix A.

i#ifh•!,)tji

For

A=[-~! -!l

and

v = [ - :] show that v is an eigenvector of A.

Solution

Because v is nonzero. to verify that v is an eigenvector of A. we need only show that Avis a multiple of'' · S ince

Av=[-~!

-!] [-:]

[ - :]=4 [ - :]=4v,

we see that v is an eigenvector of A with corresponding eigenvalue 4. Practice Problem 1 .,.

Show that and are eigenvectors of

v=

A=[-~!

[-!]

- !l

To what e igenvalues do u and v cnrrespond? An eigenvector v of a matrix A is associated with exactly one eigenvalue. For if )qv = Av = A2V, then AJ = A2 because v "I 0. ln contrast. if vis an eigenvector of A correspond ing to eigenvalue A, then every nonzero multiple of v is also an e igenvector of A corresponding to A. For if c "I 0. then A(cv) = cAv = CAV = A(cv). The process of finding the eigenvectors of an 11 x 11 matrix that correspond to a particular eigenvalue is also straightforward. Note that v is an eigenvector of A corresponding to the eigenvalue A if and only if v is a nonzero vector such that Av=AV Av - AV = 0 Av - AI,v= O (A- Al,)v

= 0.

Thus v is a nonzero solution of the system of linear equations (A - Al,)x = 0.


Page 8 of 10


296

CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization

Let A be ann x n matrix with eigenvalue J... The eigenvectors of A corresponding to J.. are the nonzero solutions of (A - J..l,)x 0.

=

In this context. the set of solutions of (A - J..l,)x = 0 is called the eigenspace of A corresponding to the eigenvalue J... This is just the null space of A - AI,. and hence it is a subspace of 'R". Note that the eigenspace of A corresponding to J.. consists of the zero vector and all the eigenvectors corresponding to J... Similarly, if J.. is an eigenvalue of a linear operator T on 'R" , the set of vectors v in R" such that T(v) = J..v is called the eigenspace ofT corresponding to J... (See Figure 5.2.)

T

0

u

AU

0 IV

Figure 5.2 W is the eigenspace ofT corresponding to eigenvalue .l..

In Section 5.4, we see that under certain conditions the bases for the various eigenspaces of a linear operator on 1?." can be combined to form a basis for 'R". This basis enables us to find a very simple matrix representation of the operator.

Example 2

Show that 3 and - 2 are eigenvalues of the linear operator Ton R 2 defined by

T

([;~]) = [ -3~~~ x2] '

and find bases for the correspond ing e igenspaces.

Solution

The standard matrix of T is

A= [

-~

-2] I

.

To show that 3 is an eigenvalue ofT, we show that 3 is an eigenvalue of A. Thus we must find a nonzero vector u such that Au = 3u . ln other words, we must show that the solution set of the system of equations (A - 3h)x 0, wh ich is the null space of A - 3h, contains nonzero vectors. The reduced row echelon fom1 of A - 3/2 is

=

Because th is is a matrix of nu llity

2 - rank (A - 3/2)

= 2 - I = I,

nonzero solutions exist, and the eigenspace corresponding to the eigenvalue 3 has dimension equal to I. Furthermore. we see that the eigenvectors of A corresponding to the eigenvalue 3 have the form


Page 9 of 10



297

for x2 :j: 0. It follows that

is a basis for the eigenspace of A corresponding to the eigenvalue 3. Notice that by taking x 2 3 in the preceding calculations, we obtain another basis for this eigenspace (one consisting of a vector with integer components), namely,

=

ln a similar fashion, we must show that there is a nonzero vector v such that Av (- 2)v. In this case, we must show that the system of equations (A + 2/z)x 0 has nonzero solutions. Since the reduced row echelon fonn of A + 212 is

=

=

[I -1] 0

0 ,

another matrix of nullity I. the eigenspace corresponding to the e igenvalue -2 also has dimension equa l to I. From the vector form of the general solution of this system, we see that the eigenvectors corresponding to the eigenvalue - 2 have the fonn

for x2 :j: 0. Thus a basis for the eigenspace of A corresponding to the eigenvalue - 2 is

{[:]}. (See Figures 5.3 and 5.4.)

)'

• = 1:] ;r

;r

Av = - 2v

Figure 5.3 A basis vector for the eigenspace of A corresponding

Figure 5.4 A basis vector for the eigenspace of A corresponding

to eigenvalue 3

to eigenvalu e -2

Since 3 and - 2 are eigenva lues of A, they are also eigenvalues ofT. Moreover, the corresponding eigenspaces of T have the same bases as those of A, namely, and


{[!] }.

Page 10 of 10


298 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalizatio n Practice Problem 2 .,.

Show that I is an eigenvalue of the linear operator on 1?.3 defined by

and find a basis for the conesponding eigenspace. The two eigenspaces in Example 2 each have dimension I. Th is need not always be the case. as our next example shows.

IJif!,j,)§i

Show that 3 is an eigenvalue of

3 0 0] [0 I 2 ,

8 =

0

2

I

and find a basis for the corresponding eigenspace. As in Example 2, we must show that the null space of 8 - 3/3 contains nonzero vectors. The reduced row echelon form of 8 - 3/3 is

Solution

0

I

0 0 [0 0

-I]

0 . 0

Hence the vectors in the eigenspace of 8 corresponding to the eigenvalue 3 satisfy x3 = 0, and so the general solution of (8 - 3/3)x = 0 is

x2 -

x,

free

X2 =X3 X3

free.

(Not ice that the variab le x 1, which is not a basic variable in the equation x 2 - x 3 = 0, is a free variable.) Thus the vectors in the eigenspace of 8 corresponding to the eigenvalue 3 have the form

Therefore

is a basis for the e igenspace of 8 corresponding to the e igenvalue 3.

Not all square matrices and linear operators on 1?." have real eigenvalues. (Such matrices and operators have no e igenvectors with components that are real numbers either.) Consider, for example, the linear operator T on 1?.2 that rotates a vector by

90°. If th is operator had a real eigenvalue>., then there would be a nonzero vector v in 1?.2 such that T(v) = J..v. But for any nonzero vector v, the vector T(v) obtained by


Page 1 of 10



299

rotating v through 90° is not a mu ltiple of v. (See Figure 5.5.) Hence v cannot be an eigenvector of T. so T has no real eigenvalues. Note that this argument also shows that the standard matrix of T , which is the 90° -rotation matrix

[0 -1] I

0 '

has no real eigenvalues. y

X

Figure S.S The image of vis not a multiple of v .

EXERCISES In Exe1t·ises 1- 12, a matrix and a vector are given. Show that

the vector is em eigenvector of the matrix and determine the corresponding eigenvalue.

[

~n -~J

l. [ - 10

24

3.

[

[-~ -~]. -~J

[-97 -6

-8 6 -6

7

-5] . -11

8

8.

9.

10.

-2 2

[_! [-~

-7

14 5 -10

[-f]

10] 2 . -7

[=rJ

[-l] -I -n [-n -6 9 16

-~~].

[:]

~~ ~~]. [-~]

12.

6 2

[j

-3

-66]

-13

[_!

6

-5


[-I] -1 I

15] , 15 - 14

In Exercises JJ- 24, a matri,r and " scalar !.. are given. Show that !.. is an eigenvalue of tile mmrix and detumine a basis for its eigenspace.

-14

15. [ 11 -3 17.

-5 7 -3

-3

18.

[-II-7 14] 'A=-4 16. [-II 'A= -1 -30 ~n

-In !..=3

14.

18]. ). = 5 -4

[-; 9 3 7

10'

-n

'A=3

[~ [-3 ~l

- 10] - 4 ' ), = 5

-9

12

19.

-I

-2

~n [=n

5

13. [ 10

-n [-n

[~ -II [-3 6

7.

4. [

-~am

5. [1 9 42 6.

2. [ 1;

11.

20. [

-3 3

-!-5

6

-9 -2

-3

-~l

-2

'A=O

).

=2

Page 2 of 10


300 CHAPTER 5 Eigenvalues, Eigenvectors, an d Diagonalizatio n 1

-4 7

38.

T

39.

T

-4

22.

-2 [-2 -I -I

-4 -;]. A= 0

I

23.

24.

3

([;~]) = [ -2rz:::: ~~: ~~ ]· A= 3 6x3 X3

[-~

-3

[j

3

40.1'

-2

-3

-

12.r1

~

[-2]

6xl ]

14x2 ·

18x1 -

.r']) [-s.r,+ 9xz - 3x3 ] ([xz = -5x, ++ 6xz - 3xJ , XI]) [-2xl- Xz- 3x3] . 30. T ([x2 = - 3x + 4xxz + 9.\) 2r3 X3

- XI

X3

X1

1 -

XJ

32. T ( [;:] ) XJ

2XJ

XZ -

2 -

[n

[-I]I - 3

-2.t,-~z+6x3

I

34.

T ([x']) = [

::~ ~ ~:]. [-~]I 2r, - 3xz - 2x3

xz

49. Every li near operator o n

35. T

([x']) = [ 20x, + 8xz]. Xz -24x, - Bxz

A=

36. T

([::~]) = [ -~~~:~:2 J.

A= 3


are the same

nn al'e the same

has real eigenvalues.

50. O nly square matrices have eigenvalues.

51. Every vector in the eigenspace of a matrix A COITCsponding to an eigenvalue A is an eigenvec1or corresponding to A.

52. The linear operator on n 2 that rotates a vector through < 9 < 180°. has no eigenvectors. the angle 9. where

o•

53. The standard matrix of the li near operator o n n 2 that rotates a vector through the angle 8. where < 9 < 180°. has 110 e igenvalues.

o•

aiOI' T , then ,, is an e igenvector ofT.

6xz]. A= _5

XI]) [ X( - Xz - 3x3] 37. T Xz = -3x,-xz-9XJ, ([X3 x, +xz + 5x3

n"

n•

54. If a nonzero vector ' ' is in the null space of a linear opcr-

-12r,- l3xz 4x, +

e igenvalue A is the column space of A -AI,..

47. The e igenvalues of a linear operator on as those of its standard matrix. 48. Tile e igenspaces of a linear operator on as those of its standard matrix.

9

([x']) = [6x, x,-- 2rz]. A= -2 Xz 6xz

T

42. If Av = AV for some veclor v. then v is an eigenvector of

46. The eigenspace of
In Exercises 33-40, a linear opermor tmd a scalar A are given. Show that A is 011 eigenvalue of the operator and determine a basis for its eigenspace.

33.

41. If Av = ), v for some veclor v. lhen A is an eigenvalue of lhe matrix A .

45. If A is an e igenvalue of a linear operator, then there are infinitely many eigenvectors of the opemtor that correspond to>•.

[-~:~: :~: ~ ~~] , [-~]

= [-~: ~

In Exercises 41- 60. determine wl1ether the statements are true or false.

44. If v is an eigenvector of a matrix. then there is a unique eigenvalue of the matrix that corresponds to v.

3

29. T

31. T ( [ :::] ) =

3

=

[ -1 2r,-1 2t2] . 20x 1 19x2

T

2

43. A scalar A is an e igenvalue of an n x 11 matrix A if and only if the equation (A - H,.)x 0 has a nonzero solution.

2

6x,

Xz

1

2

the matrix A .

3

([x'xz] ) = [S.r, -2rzJ. [I] + xz 27. r([x'] ) = [-~] + 4x, - 1xz 6x2]. [2 ] 28. ([x']) = [1 xz

26. T

x']) [ 2r5x +- 5x2rz-+ 12.t4x3],A= l ([x = -1 -4x, - lr2 + 5x3

~

-7

([x']) _[-3x,-+ x2

lOxz-

X3

~J A= 2

5

-3

1 1

4 -3]3 . A= I

/ 11 Exercises 25-32. a linear operator and a vector are given. Show that the vector is an eigenvector of the operator and determine the corresponding eigenvalue.

25. T

([~.1:2]) = [~;: ~~ = ~;~] . A= - 3 4.q= - 4xz - 3xJ

8 A=2

55. I f v is an eigenvector of a matrix A. then c v is also an eigenvector for any scalar c.

56.

If v is an eigenvector of a matrix A. then e igenvector for any nonzero scalar c .

cv is ;llso an

57. If A and 8 are 11 x n matrices and A is an eigenvalue of both A and 8 , then A is an e igenvalue of A + 8.

58. If A and 8 are 11 x 11 matrices and v is an eigenvector of both A and 8 , the n v is an eigenvectol' of A + 8 .

59. If A and B are 11 x n ma trices and ). is an eigenvalue of both A and 8 , the n A is an e igenvalue of AB.

Page 3 of 10


51 Eigenvalues and Eigenvectors 60. If A and 8 arc 11 x bolh A and 8 , then

11

mat rice; and v is an eigenvector of

v i< an eigenvector of AB.

61. What arc the eigenvalues of the identity operator on 'R."? Ju,o,tify your an\\ver. De\Cribe c
62. What arc the eigenvalues of the zero operator on 'R."? Ju,o,tify )'OUr :m\\,cr. De\Cribe each eigensp
63. Pro'c that if v is an ctgenvector of a matrix A. then for any nontero \Calar c. n• i' aho an eigen,ector of A. 64. Prove that tf v is an eigenvector of a matrix A. then there is a unique M:alar A >uch that A' '= AY. 65. Suppoo.e that 0 i> an eigenvalue of a matrix A. Gi' e another name for the eigenspace of A corresponding to 0. 66. Prove that a square matrix i< invenible if and only if 0 is

301

/u E.rercises 76~2. use eithu a ca/cularor 11·irh matrix cat>abiliriu or computu sofnmrt' such trs MATLAB IQ so/1•e each problem.

76. Let

-1.9 A= [

1.6 1.2 1.6

14.4 -2.7 -8.0 1.6

-8.4 3.2 4.7 3.2

~·]

-1.6 -18.2 . 2.7

Show that

not an eigenvalue.

67. Prove that if A is an eigenvalue of an invenible matrix A, then A ,p 0 and 1/A is an eigenvalue of A 1• 68. Suppose th:ll A is a square matrix in which the s um of the entries of each •·ow equals the same scalar r. Show that r is an eigenvalue of A by finding an eigenvector of A COI'I'esponding to /',

69. Prove that if A is an eigenvalue of a matrix t\ , then A2 is :111 cigenv:1lue of A2. 70. State and prove a generalization of Exercise 69. 71. Determine necessary and sufficient conditions on a vector ' ' ,uch that the •p:m of {Av} equ:tb the ,o,pan of{ ' '}.

72. An

11 x 11 matrix A i> called 11ilporem if. for some positive integer J.. At = 0. where 0 b the 11 x " Lero matrix. Prove that 0 b the only eigenvalue of a nilpotent matrix.

73. Let Vt and ,.2 be eigenvectors of a linear operator T on

n•. and let At and >. 2• respectively, be the corresponding etgenvalues. Prove that tf )q independent.

# A2.then {vt. v2l is linearly

74. Let "t· v2 • and ' 'J be ctgenvectors of a hnear operator T

n•.

on and let At. A~. and ).J. respectively. be the corre>ponding eigenvalue>. Pro'e that if these eigenvalues are di>tinct. then {' 't. ''2· VJ} is linearly independent. Him: Lettmg Ct Vt

+ c~ V2 + ('j\ 'l = 0

at'C eigenvectors of A. What are the eigenvalue> corresponding to each of these eigenvectors'? 77. Are the eigenvalues of A (determined in Exercise 76) also e igenvalues of 3A'? If so. lind an eigenvector corresponding to each eigenvalue.

78. Are Vt, vz, v3 , and v4 in Exercise 76 also eigenvectors of 3A? If so. what eigcnv:due corresponds to each of these eigenvector>? 79. (a) Based on the results of Exercises 76- 78. make a conjecture about the relntion~hip between the eigenvalues of an 11 x 11 matri' 8 and tho_-e of c·B, where ,. is a nonzero scalar. (b) Ba>ed on the re>ult> of Exerci>e> 76- 78. make a conjecture about the relation~hip bet\\een the eigenvectors of an 11 x 11 matnx B and those of cB. where c is a nonzero >Calar. (c) Justify the cOnJeCtures made m (a) and (b).

80. Are vt. ' '2· "J· and "• in Exercise 76 also eigenvectors of AT? lf so. what eigenvalue corresponds to each of these eigenvectors?

81. Are the etgenvalues of A (detemtined in Exercise 76) also

for sc:tlars Ct. c2 • and CJ. apply T to both sides of this equat ion. Titcn multiply both side> of this equation by AJ, and subtract it from the first equation. Now use the result of Exercise 73.

75. Let T be a line:~r operator o n 'R.2 with an eigenspace of dimension 2. Prove that T ).! for some scalar A.

=

eigenvalues of AT? If so. find an eigenvector corresponding to e:tch eigenv:due.

82. Based on the results of Exercises 80 and 81. make a conjecture about any p
SOLUTIONS TO THE PRACTICE PROBLEMS u is an eigenvector of A corresponding to the eigenvalue

I. Since

3. Likewhe. 2 I 2

_!J [-n ~ [-~J ~ -n ~


3[

3u.

2 I

2

Page 4 of 10


302

CHAPTERS Eigenvalues, Eigenvectors, and Diagonalization

;o v i> also an eigenvector of A corre>ponding to the eigenvalue 3.

Therefore the vector fom1 of the general solution of (A - IJ)x = 0 is

2. The standard matrix of T i;

t] [•'~I] ·:~ [X =

A=[-l -: :J· 0

~ -~]

0

[I] ~

•

so

{[nl

and the naw echelon form of A - I 3 is

[~

=Xt

is a basi' for the eigen,pace ofT corre,ponding to eigenvalue I.

0

0

ls.2 1THE CHARACTERISTIC POLYNOMIAL In Section 5. 1, we learned how to find the e igenvalue corresponding to a give n e igenvector and the eigenvectors for a given eigenvalue. But ordinarily. we know neither the e igenvalues nor the eigenvectors o f a matrix. Suppose, for instance, that we want to find the eigenvalues and the eigenspaces of an11 x 11 matrix II. If A. is an e igenvalue of A. there must be a nonzero vector v inn" such that Av = A. v. Thus, as on page 295, vi s a nonzero solution of (A - Al, )x = 0. But in order for the homogeneous system of linear equati ons (A - Al,}x = 0 to have nonzero solut ions, the rank of A - ).!, must be less than 11. Hence, by the Invertible Matrix Theore m. the 11 x 11 matrix A - AI, is not invertible. so its determinant must be 0. Because these steps are all reversibl e, we have the following result: The eigenvalues of a square matrix A arc the values oft that satisfy dct (A - rl. ) = 0. T he equation det (A- rl.) = 0 is called the characteris tic equation of A. and det (A - tl, ) is called the characteristic pol)nOmi al of A. Thuo the eigenvalues of the matrix A arc the (real) roots of the charactcnstic polynomial of A.

lj@,!.itjl

Determine the eigenvalues of

3]

A -- [ 43

6 .

and the n find a basis for each eigenspace. Solution

We begi n by fonning the matrix A - t/2

= [- 43-

I

-3]

6- t

0

The characteristic polynomia l o f A is the determinant of this matrix. which is det (A - r/2) = ( - 4 - t )(6 - t} - ( - 3) · 3

= (-24 -

2t

+ r 2) + 9

= r 2 - 2t - 15 = (1 + 3)(1 - 5).


Page 5 of 10


5.2 The Characteristic Polynomial

303

Therefore the roots of the characterist ic polynom ial are - 3 and 5; so these are the eigenvalues of A. As in Section 5. 1, we solve (A + 3h)x 0 and (A - Sh)x 0 to find bases for the eigenspaces. Since the reduced row echelon fonn of A + 312 is

=

the vector fonn of the general solution of (A

+ 3h)x =

=

0 is

Hence

is a basis for the e igenspace of A correspondi ng to the eigenvalue - 3. In a s im ilar manner, the reduced row echelon form of A - 512, which is

produces the basis

for the eigenspace of A corresponding to the eigenvalue 5. The characteristic polynomial of the 2 x 2 matrix in Example I is t 2 - 2t - IS= (t + 3)(t - 5), a polynom ial of degree 2. In general , the characteristic polynomial of an 11 x n matrix is a polynomial of degree 11.

DCAUTION

Note that the reduced row echelon form of the matrix A in Example I is h. which has the characteristic polynomial (I - 1)2 • Thus the characteristic polynomial of a matrix is not usually equal to the characteristic polynomial of its reduced row echelon form. In general. therefore. the eigenvalues of a matrix and its reduced row echelon form are not the same. Likewise, the eigenvectors of a matrix and its reduced row echelon form are not usually the same. Consequently, there is no way to apply elementary row operations to a matrix in hopes of finding its eigenvalues or e igenvectors. Computing the characteristic polynomial of a 2 x 2 matrix is straightforward, as Example I shows. On the other hand, calculating the characteristic polynomial of a larger matrix by hand can be quite tedious. Although a programmable calculator or computer software can be used to determine the characteristic polynomial of matrices that are not too large, there is no way to fi nd the precise roots of the characteristic polynomial of an arbitrary matrix that is larger than 4 x 4. Hence, for large matrices, a numerical method is usually used to approximate the eigenvalues. Because of the difficulty in computing the characteristic polynomial, in this book we usually usc on ly 2 x 2 and 3 x 3 matrices in our examples and exerc ises. It follows from Theorem 3.2, however. that there arc some matrices for which the eigenvalues can be easi ly determined: The eigenvalues of an upper triangular or lower triangu lar matrix are its diagonal entries.


Page 6 of 10


304 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalizatio n

Example 2

Determine the eigenvalues of the matrix

- 3

- 1

0 0

6 0

-9 7 -5

0

0

0

A_

-

[

-2' ] 3 .

8

Solution Because A is an upper triangular matrix. its eigenvalues arc its diagonal entries, which are - 3, 6, - 5, and 8. Practice Problem 1 ..,.

Detennine the eigenvalues of the matrix

4

0

- 2

- 1

00 0]0

7 -2 0 . [ 8 6 3 9 -5 Recall that the problem of finding eigenvalues and eigenvectors of a linear operator can be replaced by the corresponding problem for its standard matrix. In the context of a linear operator T , the characteristic equation of the standard matrix ofT is called the characteris tic equation of T , and the characteristic polynomial of the standard matrix of T is called the character istic polynomial of T. Thus rhe characteristic polynomial of a linear operator T on R" is a polynomial of degree n IVIIose roots are rile eigenvalues ofT.

Example 3

We have noted on page 298 that the linear operator T on R 2 that rotates a vector by 90° has no real eigenvalues. Equivalently. the 90° -rotation matrix has no real eigenvalues. In fact, the characteristic polynomial ofT, wh ich is also the characteristic polynom ial of the 90° -rotation matrix. is given by

-- tIJ = t2 +1. which has no real roots. This confim1s the observation made in Section 5. I that T, and hence the 90° -rotation matrix, has no real eigenvalues.

THE MULTIPLICITY OF AN EIGENVALUE Consider the matrix

A=

-10 0] [ 0 0

I 2

2 I

.

Using the cofactor expansion a long the first row, we see that

det (A - t/3)

= det

[

- 1- 1 ~

=( - 1 - r)· det [


0 1- t 2

I ~,]

I- 1 2

Page 7 of 10


5.2 The Characteristic Polynomial

= (- 1 -1)[(1 -1)2 -

305

4]

=(- 1 - 1)(12 - 21 - 3)

= -(1 + 1)(1 + 1)(1 = -(1 +

3)

1)2(1 - 3).

Hence the eigenva lues of A are - I and 3. A similar calculation shows that the characteristic polynomial of

8 =

3 0 0] [ 0 0

I 2

2 I

is - (1 + 1)(1- 3)2. Therefore the eigenvalues of 8 are also - I and 3. But. as we explain next, the status of the eigenvalues - I and 3 is different in A and 8. If A is an eigenvalue of an 11 x 11 matrix M. then the largest positive integer k such that (1 - A)k is a factor of the characteristic polynomial of M is called the multiplicity 3 of A. Thus. if

then the eigenvalues of M are 5, which has multiplicity 2; - 6, wh ich has multiplicity 1; 7, which has multiplicity 3; and 8. which has multiplicity 4. Practice Problem 2 .,..

If the characteristic polynomial of a matrix is

determine the eigenvalues of the matrix and their multiplicities. For the preceding matrices A and 8, the eigenvalues - I and 3 have different multiplicities. For A. the multiplicity of - I is 2 and the multiplicity of 3 is 1. whereas for 8 the multiplicity of - I is I and the mult iplicity of 3 is 2. It is instructive to investigate the eigenspaces of A and 8 corresponding to the same eigenvalue. say. 3. Since the reduced row echelon form of A - 3/3 is

I 0 0] [00 01 - 01 ' the vector form of the general solution of (A - 3/J)X = 0 is

[:~] = [-~] = [~] · X3

X3

X3

I

Hence

I[:J l 3 Some

authors use the term algebraic multiplicity. In this case, the dimension of the eigenspace

corresponding to!- is usually called the geometric multiplicity of)•.


Page 8 of 10


306 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization is a basis for the e igenspace of A corresponding to the eigenvalue 3. Therefore this eigenspace has dimension I. On the other hand , in Example 3 of Section 5.1 , we saw that

is a basis for the eigenspace of 8 corresponding to the eigenva lue 3. Therefore this eigenspace has dimension 2. For these matrices A and 8 , the dimension of the eigenspace corresponding to the eigenvalue 3 equals the multiplicity of the eigenvalue. This need not always happen, but there is a connection between the dimens ion of an eigenspace and the multiplicity of the correspondjng eigenvalue. lt is described in our next theorem, whose proof we omit. 4

THEOREM 5.1 Let A be an eigenvalue of a matrix A. The dimension of the eigenspace of A corresponding to A is less than or equal to the multiplicity of A.

Example4

Determine the eigenvalues, their multiplic ities, and a basis for each eigenspace of the linear operator T on n 3 defined by

Solution The standard matrix of T is -I

A = [

~

-~0 -~]. - 1

Hence the characteristic polynomial ofT is det (A - 1/3)

= det [

- 1- I ~

0 - 1 -1

0

- 0I ] - 1- I - 1 ]

- 1- t =(- 1 - 1)3

= -(1 + 1) 3. Therefore the only e igenvalue of T is - I, and its multiplicity is 3. The eigenspace ofT corresponding to - I is the solution set of (A + /3)x 0. Since the reduced row echelon form of A + / 3 is

=

4


For a proof ofTheorem S.l, see (4, page 264].

Page 9 of 10


5.2 The Characte ristic Polynom ial

307

we see that

![!l [i] I is a basis for the eigenspace of T corresponding to - 1. Note that this eigenspace is 2-dimensional and the multiplicity of - 1 is 3, in agreement with Theorem 5.1.

Determine the e igenvalues of


A= [!0 =~0 =~]2 , their mullipl ici ries, and a basis for each eigenspace.

THE EIGENVALUES OF SIM ILAR MATRICES Recall that two matri ces A and 8 are called similar if there ex ists an invert ible matrix P such that 8 = p - 1AP . By Theorem 3.4. we have del (8 - rl,.)

= det (P - 1AP - rP - 1l ,.P ) = dct (P - 1AP- p - 1(tf )P) 11

1

= det (P - (A - tl")P)

= (det P- 1)[det (A -

ll11 ))(det P )

1 = (- ) [del (A detP

tl,.)](det P )

= det (A - tl ,). Thus the cluu·acteristic polynomial of A is the same as that of 8. T herefore the following statements are tme (see Exerc ise 84):

Similar matrices have the same characteristic polynomial and hence have the same eigenvalues and mu ltiplici ties. In addition. the ir e igenspaces corresponding to the same eigenvalue have the same dimension.

In Section 5.3, we investigate matrices that are similar to a diagonal matrix.

COMPLEX EIGENVALUES* We have seen in Example 3 that not all 11 x n matrices or linear operators on 1?/' have real eigenval ues and eigenvectors. The characteristic polynomial of such a matrix • The remainder of this section i s used only in t he description of harmonic motion (an optional topic in

Section 5.5).


Page 10 of 10


308 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalizatio n must have no real roots. However, it is a consequence of the fundamental theorem of algebra that every 11 x 11 matrix has complex eigenvalues. (See Appendix C.) In fact, the fundamental theorem of algebra implies that the characteristic polynomial of every 11 x 11 matrix can be written in the fonn c(1 - )q)(t - >.2) · · · (t - A11 )

for some complex numbers c,At,A2, . .. ,A11 • Thus, if we count each eigenvalue as often as its multipl icity, every 11 x n matrix has exactly 11 complex eigenvalues. However, some or all of these may not be real numbers. There are applications (in such disciplines as phys ics and electrical engineering) where complex eigenvalues provide useful information about real-worl d problems. For the most pan. the mathematical theory is no different for complex numbers than for rea l numbers. In the complex case, however, we must allow complex entries in matrices and vectors. Thus the set of a ll n x I matrices wi th complex entries, denoted by C", replaces the usual set of vectors R", and the set C of complex numbers replaces n as the set of scalars. Example 5 illustrates the calculations requ ired to find eigenvalues and e igenvectors involving complex numbers. However, with the exception of an application discussed in Section 5.5 and designated exercises in Sections 5.2 and 5.3, in th is book we restrict our attention to real eigenvalues and e igenvectors having real components.

Example 5

Detennine the complex eigenvalues and a basis for each eigenspace of

-IO 5J. Solution The characteristic polynomial of A is det (A- 1/2) = det [

-10]

1- 1 _ 5 2

=(I - t )(5 - t)

1

+ 20 =

r ? - 6t

+ 25.

Applying the quadratic formula, we find that the roots of the characteristic polynom ial of A arc I =

6±

J< -6)2_ 4(1)(25) =

6±

2

F-64 2

6 ± 8i

.

= - - = 3 ± 41.

2

Hence the eigenvalues of A are 3 + 4i and 3 - 4i. As with real eigenvalues, we find the eigenvectors in C2 corresponding to 3 + 4i by solv ing (A - (3 + 4i)/2)x 0. The reduced row echelon form of A - (3 + 4i)h is

=

[~

I-2i] 0 .

Thus the vectors in the eigenspace corresponding to 3

[..1'Xt]2 -_ [(- 1+ 2i)X2] _ x X2

-

so a basis for the eigenspace corresponding to 3


2

+ 4i

have the form

[-1+2i]

+ 4i

I

.

is

Page 1 of 10


5.2 The Characteristic Polynomial 309 Similarly. the reduced row echelon form of A - (3- 4i)/2 is

[~

I +2i] 0 .

li enee the vectors in the eigenspace corresponding to 3 - 4i have the fonn

and thus a basis for the eigenspace corresponding to 3 - 4i is


Determine the complex eigenvalues and a basis for each eigenspace of the 90°-rotation matrix A=

[0 -1] I

0 .

When the e ntries of A are real numbers, the characteri st ic polynomial of A has real coeffic ients. Under these condi tions, if some nonreal number is a root of the characteristic polynomial of A, then its complex conjugate can also be shown to be a root. Thus the non real eigenvalues nf a rea/matrix occur iu romplet roujug(lte pairs. Moreover, if v is an eigenvector of A corresponding to a nonreal eigenvalue A. then the complex conjugate of v (the vector whose components arc the complex conjugates of the components of v) can be shown to be an eigenvector of A corresponding to the complex conjugate of A. Note this relationship in Example 5.

EXERCISES In E.urcise.r I 12. o malrir and liS dwractuistic [HJiytUHIIial art' gil't'n. Find t/r(' t'igenmlllt'S of t'aclr matrix and dt'tennint' a basis for eoclr t'lflt'nSIHICI!.

I.

2.

[i [-7

- 8 3] .

-6

3. [ - 10 -15 4.

5.

[-914

(/ - 5)(/ - 6)

a n

8.

[-5 -I

6 2 6

-8

t(t

-7]

9.

+ I) 10.

(/ + 2)(1- 5)

12 .

[! -5 -5 . - 4]

-(1 + 3)(/ - 2)2

-6 2 2

-6] -2 . 6

II .

[!

-(1 +

2)(/ - 4) 2


-(/ + 4)(/ - 2)(/ - 3)

-1 -n

-(t + 3)(1 + 2)(1- I)

-3

2 -I 2

n

"[j

-(/ - 6){/ + 2)1

-6

!].

2 -I 4

[-~

-4]4 .

4 -2

0

-5

-2

3

[-:

-4 2 -4

(I+ 4)(1 + 5)

-5 -3

6. [-2 -3

7.

7 -6- 6 6

3 -2 9 -9

- 4 I -I

-(1 - 3)(1- l)l

-4] 6

0 . (I

3)(1

4)(1 + l)l

9

5

-·]

12 1 6 . (I+ 5)(/ + 2)(1- 1)

- II

Page 2 of 10


310 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization In Exercises 13- 24. find the eigenvalues of each matrix a/UI determine a basis for each eigenspace.

_!]

13.

[~

15.

[~~

17.

19.

2 1.

23.

-4] 11

[-~

~]

5

-3

-8

9

[-~

0 -2

[-~

0 4 0

[-'~

-2

-1

I

0

2 -2

0

0

14.

[- I~ -~J

33. T

16.

[-; -~J

34. T

18.

20.

_;]

22.

-4

[-~

[_~

0 3 0

~~] -2

[-~

7

_n

[_:

24.

-1]

-6

37.r ([;:])=[ ~~:.-=. s~

25. T

([x•]) = [-8x, +

26. T

([x•]) = [ --x, +2x xz 4x - 1x2

Xz

x3

0 -2

0 -3

0 0

I

_!]

0

-2

(t - 5)(t - 7)

13x2

2].

(t

1

+5)(t +3)

1

7

In Exercises 25- 32. a linear operator and ils characteristic polynomial are gil'en. Find the eigenvalues of each operator and determine lt basis for each eigenspace. 6 -x, + x 2 ].

(~:]) = [~: ~ :;~]

([~:]) = [ -IO.~r~ 9xJ 36. T ([:~~]) = [~~~~+-l~~]

n

3

-s

3 0

([·x2r•]) = [-4~1 +x2] -2x . -x2

35· T

- 12

-4

_n

5

In Exercises 33-40,fimlthe eigenvalues of e
2

]

-x. +xz +2x3

-4x~: 6xz

40_ T ( [;:] ) = [

x3

]

-5x 1 + 5x2 +x3

41. Show that

[~ =~J has no real eigenvalues.

42. Show Jhat [ ~

=;J

has no real eigenvalues.

43. Show thallhe linear opemtor T ([x•]) X2

= [-.<.,(J ~·.++3·\5'2X2]

has 110 real eigenvalues. (1 - 6)(t - 2)

44. Show that the linear operator T

([:~:]) = [~: ~ !;:]

has 110 real eigenvalues.

29.T

([.q])= [-3x, +x2 +3xJ, -2x2 +4XJ

45.

-4i ] [I-24iIOi I+ JOi

46.

[I~i

1- Sx2 -7x3] .~'] ) = [-8x x2 6x2 + 3x2 + 1x3 ,

47.

[_;

48.

[~ -~]

x3 3)(1

49.

50.

[~

52.

['~' ~]

xz

-x , +xz + 5x3

x3

-(t

30. T

31. T

+ 2)(t -

([

-(l

]

4)

2

8x, + 8x2 - 9x3

+ +2)(t +9)

([~~]) = [~: :~~: =!~:]. x 3x + 6x 2 - 4x 3

- (1 -

32. T ( - (1

In Exercises 45- 52, use complex 11111nbers to determine the eigenvalues of each linear operator and determine a basis for each eigenspace.

1

3

51] 11

I+2i

[~

4 0

0 51.

[:

-6- i] 3i

I

0

't]

I

-4] 1- i

-i

0 0

-irJ

0

0

-i

1)(1 - 2) 2

[:x;~~]) = [~;: : ~:~: =~:]. 8x1 + 8xz - 9x3 + 1)2 (t -

3)


~

~

In Exercises 53- 72. determi11e wl1ether the staremems are trite or false.

53. If two matrices have the same characteristic polynomial. 1he11 they have Jhe same eigenveclors.

Page 3 of 10


5 .2 The Characteristic Polynomial 54. If two matrices have the same charJctcristic polynomial. then they have the same eigenvalues. chamcteri~tic polynomial of an polynomial of degree 11 .

55. The

11

x

11

matrix is a

57. The cigcnvecton. of a matrix are equal to those of its reduced row echelon fonn. 58. An 11 x

59. Every II

11

matrix hru.

X II

11

dJotinct eigenvalues.

matrix has an eigenvector in

and -9. respectively. Write :til the po>
56. The eigenvalues of a matrix arc equal to those of its reduced row echelon fonn.

n·.

60. E•ery >quare matrix ha' :t complex cigenv:alue. 61. The characteriotic polynomial of an 11 x 11 matrix can be written c(1 - At )(I - A2) ···(I - A.) for some real numbers c. A1• Al ..... An. 62. The characteristic polynomial of an 11 x 11 matrix can be written c(l- At)(l - .l.2) · · · (1- An) for some complex numbers c. AJ. Az ..... An.

63. If (1 - 4) 2 divides the char:tcteristic polyno mial of A. then 4 is un c igcnvnluc of II with multiplicity 2. 64. The mult iplici ty of an e igenvalue equals the di mension of the corresponding cigen,pace.

65. If A is an eigenvalue of multiplicity I for a matrix A, then the dimension of the eigenspace of A corresponding to A is I.

(b) dim IVz

=3 =I

(c) dim IVt = 2 78. Suppo'e that A i' a 5 x 5 matnx with no nonreal eigenvalues and exactly three real eigenvalues. 4. 6. and 7. Let IVt. \Vz. and IVJ be the eigenspaces corresponding to 4. 6. and 7. rcspcctl\'ely. Write all the possablc characteristic polynomials of A that a.re con,iotcnt with the following information: (a) dim \Vz

=3

(b) d im IV1 = 2 (c) dim IVt

= I and dim IV2 -

79. Show that if A is an upper triangul :1r or a lower ttiangular matrix. then A is an eigenvalue CJf A with mult iplicity k if and only if A appcttrs exactly k times o n the diagonal of A.

80. S how that the rotation matrix A 11 h:ts no real eigenvalues if 0 < 0 < 180 . 81. (a) Determine n ba
A= [

67. The nonreal eigenvalues of a real matrix occur in complex conjugate pairs. 68. If A t~ an 11 x 11 matrix. then the sum of the multiplicities of the cigcn,•aluc> of A equal> 11.

70. The only eigenvalue of 1. is I .

=

11

matnx A are the solutions of

x 11 matrix. and suppose that. for a particular scalar c. the reduced row echelon fonn of A - cl. is 1•. What can be said about c?

74. If f(t) is the chamctcristic polynom ial of a square matrix A. whnt isf(O)? 75. S uppose that the characteristic polynomial of a n mnlrix A is 11

1/ - l

11

x n

+ · ·· + liJI + ao.

Deteamine the chamcteriotic polynomial of -A.

76. What is the coefficient oft" in the characteristic polynomial of an

11

x

11

c

"" 0.

ba~is

for each ctgenspace of

A=[~ ~]. (b) Determine a ba.
11

+ 0 11

(d) Establish a relationship between the cigem·cctors of any ..quare mmrix 8 and tho-.e of c·8 for any scalar

82. (a) Detennine a

71. A square zero matrix hru. no eigenvalues.

72. The eagenvalues of an 11 x dct(A - tl.) 0.

(c) Determine a ba~is for each cigcm,pace of 5A.

(e) E.
69. The scalar I h an cagenvaluc of 1• .

11

-~ ~].

(b) Determine a basis for each eigenspace of -3A.

conjug:ue pairs.

0 11 1

2

(d) d im \Vz = 2 and dim IVJ = 2

66. The nonreal eigenvalue> of a matrix occur in complex

73. Let A be an

311

matrix?

77. Sup)lOSC that A i' a 4 x 4 matrix with no nonreal eigenvalues and exactly two real eigenvalues. 5 and -9. Lei IVt and \Vz be the eigcnspaces of A corresponding to 5


(c) Determine a basis for c:tch eigenspace of A - 6/2. (d) Establish a relationship between the eigenvectors of any 11 x 11 matrix /J :md th<>
(b) Establish a relationship between the characteristic polynomial of :my 'quare m:ttrix 8 and that of 8 r. (c) What does (b) imply about the relationship between the eigenvalues of a squt~re matrix 8 and those of 8 r?

(d) Is there a relationship between the eigenvectors of a square matrix 8 and those of 8 r?

Page 4 of 10


312 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization

=

84. Let A and 8 be 11 x 11 matrices such that 8 p - 'AP. and let ).. be an etgenvalue of A (and hence of 8). Prove the following results:

88. Compme the characteristic polynomial of

[l

(a) A vector ,, in n• is in the eigenspace of A correspondmg to).. tf and only tf p-•v ism the eigenspace of 8 corresponding to A.

0 0 0 0

-17]

I 0

-20

18

-19 .

0

(b ) If (v t. ' 2· ...• vtl is a basis foe the eigenspace of A corre;ponding to).., then (P •,•,. P 1v1 , ... ,P- 1vt} 1s a basts for the etgenspace of 8 corresponding to J..

89. Use the result of Exerctsc 88 to hnd a 4 x 4 matrix whose

(c) TI1e eigcn,pace' of A and 8 that correspond to the same eigenvalue have the same dimension.

(a) Compute the characteristic polynomials of A and A r.

85. Let A be a symmetric 2 x 2 matrix. Prove that A has real eigenvalues.

86. (a) The characteristic polynomial of A = [ ~

!J

has the

form t + rt + s for some scalars rands. Determine r :md .
characteristic polynomial is ,• - lit 1 + 23r 1 + 7t - 5. 90. Let A be a random 4 x 4 matnx. (b) Formulate a conjecture about the characteristic polynonlials of 8 and where 8 is an arbitrary 11 x 11 matrix. Test your conjecture using an arbitrary 5 x 5 ma11ix.

or.

(c) Prove that your conjecture in (b) i;, valid.

91. Let

/11 £um:ius 87 91. rtse either a calculawr ll'irh marrix capa· biliries or com(JI/11!1' sofrll'are srtch as MATLAB ro solve each problem.

-3.5] - 4.0 .

A-- [6.5 7.0

=

(b) Show that 112 + rA + s/2 0. the 2 x 2 zero matrix. (A simi lar result is true for any square matrix. It is called the Cayley-Hamilroll rheorem.)

(a) Find the eigenvalues of A and an eigenvector corresponding to each eigenvalue. (b) Show that A b invertible. and then find the e igenvalues of A 1 and an eigcn vector corresponding to each eigenvalue. (c) Use the results of (a) and (b) to fomlUiate a conjecture about the relationship between the eigenvectors and eigenvalues of an invertible 11 x 11 matrix and those of its inver:.e.

87. Compute the ch:mtcteristic polynomial of

[i : ll

(d) Test your conjecture in (c) on the invertible matrix

(Thb matrix is called the 3 x 3 Hilbert marrix. Computallons wtth I hi bert matrices arc subject to significant roundoff error:..)

[-5!

·~] .

-2 8 2

- -I

(e) Prove that your conJecture in (c) is valid.

SOLUTIONS TO THE PRACTICE PROBLEMS I. Because the given matrix is a diagonal matrix. its eigenvalue' are its diagonal emrie,, which are 4, -I, -2, and 3. 2. The eigenvalues of the matrix arc the roots of its charactcri•tic polynomi:tl, which i< -(r - 3)(1 + 5) 2(t - 8) 4. Thus the eigenvalues of the matrix are 3, -5, and 8. The multiplicity of an eigenvalue A is the number of factors of r -A that appear in the characteristic polynomial. Hence 3 is an eigenvalue of multiplicity I. -5 is an eigenvalue of multiplicity 2. and 8 is an eigenvalue of multiplicity 4. 3. Form the matrix

det 8 = (-I )3+1 b 31 · dct811

0

- 1] -5

2 -I

+ (- I )3+2/>32 · dctBn

+ (- I )3+ 3lm · det Bn =0+0+(- 1)6(2-r)·det [

' - r

4

- 1 ]

-3-r

=(2-t)l(l-t)(-3-t)+41

= (2 -

- I -3- r


To evaluate the dctem1inant of 8. we use cofactor expan'ion along the third row. TI1cn

+ 2t - 3) + 4 1 = + 2r + 1) = -(r -2)(t + 1)2 • I )l(t 2

(2 - r)(t 2

Page 5 of 10


5.3 Diagonalization of Matrices e igenvalue 2 is

Hence the eigenvalues of A arc -I. which has multipl icity 2, and 2, which has multiplicity I. Because the reduced row echelon fonn of A + /3 is

[~

{[-:]}·

OJ ~ ~ '

- .5

4. The chamcteristic polynomial of A is dct (A - 1/2)

we see that the vector form o f the general solution of (A + / J)X = 0 is

Taking

x2

3 13

= det [

-1

-I ] =1 2 +I= (1 +i)(l-i).

1

- I

Hence A has eigenvalues -i and i. Since the reduced row eche lon form of A + i/2 is

= 2, we obtain the basis

a basis for the eigenspace con·esponding to the eigenvalue -i is

{[~] }

Furthermore, the reduced row echelon form of A - i/2 is for the eigenspace of A corresponding to the eigenvalue -I. Also,

I [0

-i]0 '

so a basis for the eigenspace corresponding to the eigenvalue i is

{[:]} .

is the reduced row echelon form of A - 213. Therefore a basis for the eigenspace of A con-esponding to the

I 5.3 1DIAGONALIZATION OF MATRICES In Example 6 of Section 2.1, we considered a metropolitan area in wh ich the current populations (i n thousands) of the city and suburbs arc given by City Suburb>

[500] 700 = p,

and the population shifts between the c ity and suburbs are described by the following matrix: From C ity Suburbs To

City

Suburbs

.85 [ . 15

.03] =A .97

We saw in this example that the populations of the city and suburbs after 111 years are given by the matrix-vector product A"' p . In this section, we di scuss a technique for computing A"' p . Note that when 111 is a large positive integer, a direct computation of A"' p involves considerable work. Thi s calculation would be quite easy, however, if A were a diagonal matrix such as


Page 6 of 10


3 14 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization

For in thi s case, the powers of D are diagonal mat rices and hence can be easily determined by the method described in Section 2. 1. ln fact,

0 J = [(.82)"'

1111

0

OJ

I .

Although A"# D , it can be checked that A= pop-t, where

P= [

- I I

This relationship enables us to compute the powers of A in tenus of those of D. For example.

and

In a si mi lar manner, it can be s hown that

A"'= PD"'P - 1

=

~ [ I + 5(.82)"' 6

5 - 5(.82)"'

I - (.82)"'] 5 + (.82)"' .

Hence

= ~ [' + 5(.82) p 6 s - 5(.82)

111

A"'

111

I - (.82)'"] [500] 5 + (.82)'" 700

= ~ [1200 + 1800(.82)"'] 6 6000 - 1800(.82)"'

= [ 200 + 300(.82)"' ] . 1000 - 300(.82)"' Because lim (.82)"' = 0, we see that the limit of A"' p is m-,.oo

200] [ 1000 . Hence after many years, the population of the metropolitan area will consist of about

200 thousand city dwellers and I million submban ites.


Page 7 of 10


5.3 Diagonalization of Matrices

31 5

In the previous computation, note that calculating PD"' p - l requires only 2 matrix multiplications instead of them - I mult iplications needed to compute A 111 directly. This s implification of the calcu lat ion is possible because A can be written in the form PDP - 1 for some diagonal matrix D and some invertible matrix P . De finition An 11 x 11 matrix A is called d iagonalizabl e if A onal n x 11 matrix D and some invertible 11 x 11 matrix P.

= PDP - 1 for some diag-

Because the equation A= pop - l can be written as p - 1AP = D , we see that a diagonalizable matrix is one that is similar to a diagonal matri x. It follows that if A= pop - L for some diagonal matrix D , then the eigenvalues of A are the diagonal entries of D. The matrix

A=

[.85 .03] .15

.97

in Example 6 of Section 2. 1 is diagonalizable because A= pop - l for the matrices

p = [- 1 I 5']

and

Notice that the eigenvalues of A are .82 and I. Every diagonal matrix is diagonalizable. (See Exercise 79.) However, not every matrix is diagonalizable, as the following example s hows:

MS:fi,!.!tjl

Show that the matri x

A= [~ ~] is not diagonalizable.

=

Solution Suppose. to the contrary. that A PDP- 1 • where P is an invertible 2 x 2 matrix and D is a diagonal 2 x 2 matrix. Because A is upper triangular, the only eigenvalue of A is 0, wh ich has multipl icity 2. Thus the d iagonal matrix D must be D = 0. Hence A= pop - L = pop - • = 0, a contradiction.

The follow ing theorem tells us when a matrix A is d iagonal izable and how to find an invertible matrix P and a diagonal matrix D such that A= PDP - 1:

THEOREM 5.2 An 11 x 11 matrix A is diagonalizable if and only if there is a basis for 'R" consist ing of eigenvectors of A. Fluthennore, A pop- L, where D is a diagonal matrix and P is an invertible matrix. if and only if the columns of P arc a basis for 'R" consisting of eigenvectors of A and the diagonal entries of D are the eigenvalues corresponding to the respective columns of P.

=


Page 8 of 10


316 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization PROOF Suppose first that A is diagonalizable. Then A = PDP - 1 for some diagonal matrix D and invertible matrix P. Let AJ,A2, · . . ,A., denote the diagona l entries of D. Since P is invertible, its columns must be linearly independent and hence form a basis for 1?!' . Rewriting A PDP - 1 as AP PD , we have that the jth columns of these matrices are equal; that is.

=

=

Therefore each colu mn of P is an eigenvector of A, and the diagonal entries of D are the corresponding eigenvalues of A. Conversely, suppose that (p 1, P2, ... , p.,} is a basis for 1?." consisting of eigenvectors of A and that Aj is the eigenvalue of A corresponding to Pi. Let P denote the matrix whose col umns are p 1, pz, ... , p ., and D denote the diagonal matrix whose diagonal entries arc AJ. A2 .. . .. A., . Then the jth colunm of AP equals Apj, and the jth column of PD is P (Aj ej) = Aj(Pej) = Aj Pj· But A pj Aj Pj for every j because Pi is an eigenvector of A correspondi ng to the eigenvalue Aj. Thus AP = PD . Because the columns of P are linearly independent, P is invertible by the Invertible Matrix Theorem. So multiplying the preceding equation on the right by p - t gives A = PDP - 1, proving that A is • diagonali zable.

=

Theorem 5.2 shows us how to diagonalize a matrix such as

.03] .97

[·8~

A=

,))

in the popu lation example discussed earlie r. The characteristic polynomial of A is

85 det (A - 1/z) = det [ · .!

= (.85 2

=1

-

= (1 -

.03 ]

I

.97 -

5

I

1)(.97 - I) - .03(. 15)

1.821

+ .82

.82)(1 - 1),

so the e igenvalues of A are .82 and I. Since the red uced row echelon form of A- .82/z is

[~ ~]'

we sec that

is a basis for the eigenspace of A corresponding to .82. Likewise,

is the reduced row eche lon form of A -


h. and

thus

Page 9 of 10



317

is a basis for the eigenspace of A correspondi ng to I. The set

obtained by combining Bt and 82 is linearly independent, si nce neither of its vectors is a multiple of the other. Therefore 8 is a basis for n 2 , and it consists of eigenvectors of A. So Theorem 5.2 guarantees that A is diagonalizable. Notice that the columns of the matrix

in the diagonalization of A are the vectors in this basis, and the diagonal matrix

has as its diagonal entries the eigenvalues of A corresponding to the respective columns of P. The matrices P and D s uch that PDP- 1 =A are not unique. For example. taking and also gives pop -t =A, because these matrices satisfy the hypotheses of Theorem 5.2. Note, however, that although the matrix D in Theorem 5.2 is not un ique, any two such matrices differ only in the order in which the eigenvalues of A are listed along the diagonal of D. When applying Theorem 5.2 to show that a matrix is d iagonalizable, we normally use the foll owing result: If the bases for distinct eigenspaces are combined, then the resulting set is linearly independent. This result is a consequence of Theorem 5.3, and is proved in [4, page 267].

THEOREM 5.3 A set of eigenvectors of a square matrix that correspond to distinct eigenvalues is linearly independent. PROOF Let A be ann x n matrix with eigenvectors v 1, Vz, ... , v, having corresponding distinct eigenvalues At. A2, .... Am. We give a proof by contradiction. Assume that this set of eigenvectors is linearly dependent. Since eigenvectors are nonzero. Theorem 1.9 shows that there is a smallest index k (2 ::; k ::; m) such that vk is a linear combination of Vt. vz, ... , v k-t. say, ( 1)

for some scalars Ct ,Cz, ... , Ck- t· BecauseAv; = both sides of equation ( 1) by A, we obtain


A; V;

for each i . when we multiply

Page 10 of 10


318 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalizatio n

that is, (2)

Now multiply both s ides of equation ( I) by Ak and subtract the result from equation (2) to obtain

By our choice of k , the set {v 1, Vz, . . . , Vk -

tl

is linearly independent, and thus

But the scalars A; - Ak are nonzero because A1, Az, ... , A, are distinct; so

= Cz = · ·· = Ck - 1 = 0. ( I ) implies that vk = 0, which contradicts CJ

Thus equation the defin ition of an eigenvector. Therefore the set of eigenvectors {v 1, Vz, ... , v111 ) is linearly independent. • It follows from Theorem 5.3 that an 11 x 11 matrix having must have 11 linearly independent eigenvectors.

Every

11

x

11

matrix having

11

11

distinct eigenvalues

distinct eigenvalues is diagonalizable.

The technique used to produce the invertible matrix P and the d iagonal matrix D on page 317 works for any diagonalizable matrix.

Algorithm for Matrix Oiagonalization Let A be a diagonalizable n x 11 matrix. Combining bases for each eigenspace of A forms a basis l3 for R" consis ting of eigenvectors of A. Therefore, if P is the matrix whose columns are the vectors in l3 and D is a diagonal matrix whose diagonal entries are eigenvalues of A corresponding to the respective columns of P, then A= PDP - 1•

+;:f!,!.!tfM

Show that the matrix

is diagonalizable , and find an invertible matrix P and a diagonal matrix D such that A= PDP - 1•

Solution

On pages 304-306 in Section 5.2, we computed the char acteristic polynomial -(1 + 1) 2(1 - 3) of A , and we also found that


Page 1 of 10


5.3 Diagonalization of Mat rices

is a basis for lhe e igenspace of A correspondi ng

10

319

eigenva lue 3. Simil arly,

is a basis for the eigenspace correspond ing to - I. The combi ned sel

is therefore linearly independent by the comment on page 3 17. It follows from the Algorithm for Matri x Diagonalizalion !hat A is diagonalizable and A PDP - 1, where

=

0 - I

and

0 Practice Problem 1 .,..

~] .

- I

The c haracteristic polynomial of

A =

[

-4 ~

-65 OJ0 3 2

is - (1 + I)(t - 2)2 . Show that A is diagonalizable by finding an invertible matrix P and a diagonal matrix D such that A PDP - 1• ~

=

WHEN IS A MATRIX DIAGONALIZABLE? As we saw in Example 1, nol all square matrices are diagonalizable. Theorem 5.2 tells us that an n x n matrix is diagonalizablc when there arc n linearly independent eigenvectors of A . For thi s to occur, two differe nl conditions musl be satisfied; they are given next. 5

Test for a Diagonalizable Matrix Whose Characteristic Polynomial Is Known Ann x n matrix A is d iagonalizable if and only if both of the following cond ilions arc true 6: I. The total number of eigenvalues of A, when each e igenvalue is counted as often as its muhipl ici ty, is equal to 11.

2. For each eigenvalue A of A , lhe dimension of the corresponding eigenspace, which is n - rank (A - AI,), is equal to the multiplicity of A.

Note that, by T heorem 5. 1, the eigenspace corresponding to an eigenvalue of muhi plic ily I musl have dimension I. Hence condition (2) need be checked only

s For a proof of thisresult. see [4, page 2681. 6 It follows

from the fundamental theorem of algebra that the first condition is always satisfied if complex

eigenvalues are allowed.


Page 2 of 10



for eigenvalues of multiplicity greater than I. Therefore, to check if the matrix A in Practice Problem 3 of Section 5.2 is diagonalizable, it is necessary to check condition (2) only for the eigenva lue - I (the one w ith multiplici ty greater than I ) .

Example 3

Determine w hether each of the following matrices is diagonal izable (using real eigenval ues):

A=

[-~ -~] 2 0 0 -I

c = [-6 ;

-3 2 3

B=

-:]

M

[-~

= [-3 3

-3 -4 3 2

-4 8 -8

-5

-~]

-~]

The respective characteristic polynomials of these matrices are - (r

+

l )(r 2

+ 4),

- (r

+

l)(r

+ 4)2 ,

- (1

+

! )(1

+ 4) 2 ,

and

- (r

+

l)(r 2

-

4).

Solution The only eigenvalue of A is - I, w hich has mu ltiplici ty I . Because A is a 3 x 3 matrix, the sum of the multiplicities of the eigenvalues of A must be 3 in order to be diagonalizab le. Thus A is not d iagonalizable because it has too few eigenvalues. The eigenvalues of B are - I. which has multiplicity I. and - 4, wh ich has multiplicity 2. 1l1us B has 3 eigenvalues if we count each eigenvalue as often as its multiplicity. Therefore B is diagonalizable if and only if the dimension of each eigenspace equals the multiplicity of the correspondi ng eigenvalue. As noted, it is onl y the eigenvalues with multipl icities greater than I that must be checked. So 8 is diagonalizable if and only if the dimension of the eigenspace correspond ing to the eigenval ue - 4 is 2. Because the reduced row echelon fonn of 8 - ( - 4)/3 is

I I 2] [ 0 0

0 0 ' 0 0

a matrix of rank I. this eigenspace has dimension 3 - rank (8 - ( - 4)1J) = 3 - I = 2. Since this equals the mult ipl icity of the e igenva lue - 4, B is diagonalizable. The eigenvalues of C arc also - I , which has multiplicity I , and -4, which has multiplicity 2. So we see that C is diagonalizable if and only if the d imension of its eigens pace corresponding to eigenvalue -4 is 2. But the reduced row echelon fonn of C - (- 4)/3 is

I 0

0 1

[0 0

a matrix of rank 2. So the dimension of the eigenspace cotTespondi ng to eigenvalue - 4 is 3- rank (C - ( -4)/3) = 3 - 2 = 1,


Page 3 of 10


321

5.3 Diagonalization of Mat rices

which is less than the multiplici ty of the eigenvalue - 4. Therefore C is not diagonalizable. Finally, the charac teristic polynomial of M is - (1 + l )(t 2 - 4)

= - (1 +

1)(1 +2)(1-2).

We see that M has 3 distinct e igenvalues (- 1, - 2, and 2), and so it is diagonalizable by the boxed result on page 318.


Detem1i ne whether each of the g iven matrices is diagonalizable. If so, express it in the form PDP- 1, w here P is an inve11ible matri x and D is a d iagonal matri x.

A= [~

2 0 - I

5

i]

- 1

2

-~]

-4

The characteristic polynomials of A and B are -(1 - 2)(1 2 + 3) and -(1 - 2)(1 respectively.

+ 1)2, <1111

In S ection 5.4, we consider what it means for a linear operator to be diagonalizable.

EXERCISES In Exercises 1- 12, a matrix A and its characteristic polynomial are given. Find. if possible. an invenib/e matrix Panda diagonal matrix D such that A = PDP - 1• Othenvise, explain wl•y A is 1101 diagmwlizable. I.

[-~

~]

[=~

2.

(t - 4)(1 - 5)

t

+3

[-~

4.

3.

2

~]

[-~ ~ =~] -8

-(1

-2

+ 5)(t 3 3

[!

6.

2)(t - 3)

-~] -5

-5 7.

-3

9.

[!

8.

_:~]

4

- 1

I I.

[-j (t

0

5 0

0 0 4

_!]

0

- I

+ 1)3(1 -

4)

-1

[

6

-~


!]

I

-2

-9] - 14

14.

[-~

;J

15.

[-~ -~J

16.

[_:

17.

[-~

18.

[-~

-n

2 -3 0

[~

0

13. [16 25

1) 2

-l]

~

19.

-I

~]

[~

20.

2

-~]

0

3

0

-1

4 2]

-
- (t - 5)(1 - 3)2

=!]

+ 5)(1 + 4)(1 + 2) 5

10.

~

-2

+ 3)(t + 4

12.

=

[=~ -~ -~] - (t

-~]

II

2

- (t

-(t - 1)(t 2 + 2)

-2

[=~

0 -10 2 -5 0 7 -5 0 -5 (t + 3)(t - 2?

-5 5

In Exercises 13- 20, a matrix A is given. Find, if possible. an invertible mturix P tmd a diagonal matrix D such that A PDP- 1• Othenvise, explain why A is not diagona/izable.

t(t + I)

5.

[-8

0 3 0

-I]

-I

5

In Exercises 21-28, use complex •wmbers to find an invertible matrix Panda diagonal matrix D such that A= PDP - 1• 21.

[~ -~]

23. [

I- i 1

22. [ _;

-4 ]

1- i

51] II

24 . [' - IOi 24i

-4i ] I+ ! Oi

Page 4 of 10


322 CHAPTER S Eigenvalues, Eigenvectors, and Diagonalization

25. 27.

- I

3

G

i

0

[2'

I

0

-n

26.

[~

-l+i I-_2i]

i

~;]

28. [2'

-I

0

0

3iI lriJ

-6-i

48. If. for each eigenvalue i.. of a matrix A. the dimension of the eigcn;pace of A corresponding 10 i. equals the multiplicity of A. then A is diagonalizable.

49. A 3 x 3 matrix has eigenvalues -4. 2. and 5. Is the m:llrix diagonaliLable? Justify your answer.

50. A 4 x 4 matrix h:L< eigenvalues -3, -I, 2, and 5. Is the m:urix diagonalizable? Justify your answer.

~

/11 Exercises 29-48, determine wlletller tile stale·

~ mellls are true or false. 29. Every 11 x n matrix is diagonali1able. 30. An 11 x 11 matrix A is diagonahzable tf and only if there i> a basis for n.• consisting of eigenvectors of A. 31. If P is an invenible 11 x 11 matnx and D " a diagonal 11 x 11 matrix such that A= PDP - 1 • then the columns of P form a basis for

n•

consisting of eigenvectors of A.

32. If P is an invenible matrix and D is a diagonal matrix such that A PDP 1 , the n the eigenvalues of A are the

=

diag01ml entries of D .

33. If A is a diagonalizable matrix, then there e xbts a unique diagonal matrix D such that A = PDP

1•

51. A 4 x 4 matrix has e igenvalues -3, -I. and 2. The e igenvalue - 1 ha; multiplicity 2. (a) Under what conditions is the matrix diagonalizablc? Justify your an>\\er. (b) Under" hat conditions is it not diagonalitable? J u.,tify your ans\1 er.

52. A 5 x 5 matrix has eigenvalues -4. w hich has multiplicity 3. and 6. which has multiplicity 2. The eigenspace corre,ponding to the e igenvalue 6 has dimcmion 2. (a) Under what conditions is the matrix diagonulizable? Justify your answer. (b) Under whm conditions is it not diagonalizable? Justify your answer.

11

distinct eigenvectors. then it is

53. A 5 x 5 matrix has eigenvalues - 3. which has multiplicity 4. and 7. which has multiplicity I.

x

11

matrix has 11 distinct eigen-

(a) Under what conditions is the marrix diagon.1li.wble? J usufy your answer.

16. If Bt. Bz ..... Bt are ba-.es for distinct eigenspaces of a

(b) Under what conditions is it not dmgonahzable'/ Jusufy your answer.

34. lf an 11 x

11 matrix has diagonali7ablc.

35. E'cry diagonalizablc

11

value.. matrix A. then 8 1 U pendent.

Bz

U · · · U Bt i> linearly inde-

37. If the sum of the multiplic ities of the eigenvalues of an 11

x

11

matrix A equals

11,

then A is diagonulizablc.

38. If, for each eigenvalue A of A. the multiplicity of A equals the dimension of the corresponding e igenspace. then A is diagonalizable. 39. If A is a diagonalizablc 6 x 6 matrix havi ng two distinct eigenvalues with multiplicitic.~ 2 and 4. then the corre>ponding eigenspaces of A must be 2-dimensional and 4-dm>ensional. 40. If A is an eigenvalue of A, then the dimension of the ctgenspacc corresponding to A equals the mnk of A- AI•. 4 I. A dlagonal11 x n matrix has 11 dtslmcl eigenvalues. 42. A diagonal matrix is dtagonahzable.

43. The standard vectors are e igenvectors of a diagonal matrix. 44. Let A und P be 11 x 11 matrices. If the columns of I' form u sci o f 11 linearly independent e igenvectors of A. then PAP - 1 i> a diagonal matrix. 45. If S is a set of distinct eigenvectoN of :1 matrix. then S i~ linearly independent. 46. If S is a set of eigenvectors of a matrix A that correspond to distinct eigenvalues of A. then S is linearly independent.

47. If the characteristic polynomial of 3 matrix A factors into a product of linear factors. then A is diagonalizable.


54. Let A be a 4 x 4 matrix with exactly the eigenvalue; 2 and 7. and corre,ponding eigenspaces IV1 and IV2 • For each o f the p:trb shown. e ither write the characteristic polynomial of A. or st:Hc why there is insuffic ient information to determine the churacteristic polynomiul. (a) dim IVt = 3. (b) dim IV2 = 2. (c) A i; diagonalizable and dim IV2

= 2.

55. Let A be a 5 x 5 matrix with exactly the eigen,aluc; 4. 5. and 8. and corresponding eigenspaces IV 1• W 2• and IV,.

For e:tch of the gi\'en parts. either write the characteristic polynomial of A . or state why there is in~uffictent infonnation to detennine the characteristic polynomial.

=

=

(a) dim IV1 2 and dim IV3 2. (b) 1\ is dtagona lizable and dim IV2

= 2.

(c) II is d iugonalizable. dim IV1 = I, and dim IV2 = 2. 56. Let A =

[I -2] I

_

2

and B

= [2 I

(a) Show th:ll AB and BA ha,•e the same cigcnv:t lues. (b) Is AB diagonalizablc? Justify your answer. (c) Is BA diagonalizablc? Justify your answer.

n·

In Exen:iu:. 57 -62, {Ill II X II IIWirix A. a /xJsis far ('OIISistillg of tilll'lll'tctors ofA. a11d the corrl'SpDIItfillg ei~em·aluts a!'l' gil't'll. CalculOit! A1 for an arbi1rary positil't! illlt!!IU 4.

Page 5 of 10



323

74. Find a 2 x 2 matrix having eigenvalues 7 and -4. with corresponding eigen\·ectors

[-~]and [

-!l

75. Find a 3 x 3 matnx having eigenvalues 3. -2. and I. with corresponding

eigemecto~ [

-n [-:J -n and [

76. Find a 3 x 3 matrix having eigen\alue> 3. 2. and 2. with corresponding

eigenvccto~

[:J

[Hand

[:J

77. Give an example of diagonalitable 11 x 11 mat rice> A and B such that A + 8 is not diagonalizable.

/11 £urcise.r 63 72. a mmri..1 tmd its dwracteristic polynomial are gi•·en. Detennilll! all ••tdues of the scalar c for which eaclr matri..l is lim tliagollolizah/r.

63.

65.

[-~ ~ =i]

64.

72.

3

c

~]

+ 3)(t + 2)

[-~

[-: ~ =:]

~ ~]

3

2

66.

I

-(t - c)(t - 2)(t - 4)

~]

[!0 -!0

68.

('

[=~ ~ =~]

70.

[~ =~

-~ -~~]

(1- c)(t

+ 3)(t + l )(t- 2)

0

7

0

4

2

13

0

3

[~ =~

[~ ~ -~J

-Ct - c)(t + 2)(t- 3)

0

+ 2)(t -

corresponding eigenvectors [ :] and

11

x

11

matrix having a single eigenvalue

c. Show that if A is diagonalizablc, then A= cl,.. (b) Use (a) to explain why

[~ ~J is not diagonalizable.

81. If A is a diagonalizablc matrix. p1'0VC that AT is diagon:tl· izable. 82. If A is an invcniblc matrix that is diagon:•lizable. prove that A -I is diagonali:t.:1blc.

83. If A is a di:•gonali1.:1blc matrix. prove that A2 is diagonal· izable. 84. If A is a di:•gonalitable matrix. prove that At i• diagonal· izable for any positive integer k. 85. Suppose that A and 8 are >imi lar matrices such that B P 1AP for some invenible matrix P.

=

(a) Show that A is diagonnlizable if and only if B is diagonaltzable. (b) How are the eigen,·nlues of A related to the eigenvalues of 8? Just1fy your an:.\\er (c) How are the eigenvectors of A related to the eigenvectors of 8? Ju>tify your :an:.wcr.

root.

13 l)(t - 3) 73. Find a 2 x 2 mmrix having eigenvalues - 3 and 5, with (t- c)(t

80. (a) Let A be an

86. A matrix 8 is called a cub~ root of a matrix A if 8 3 =A. Prove that every diagonalizable matrix has a cube

[0~ =:~ ~ =:i] 10

-n

-(t - c)(1 2 - 8)

-Ct- c)(t + 2)(t +I)

71.

-I~

-(1 - c)(t

-(t - c)(t - 3)(t- 4)

69.

- I

-(1- c)(l - 2)(1 - 3)

-(t - c)(t 2 + 7)

67.

[

-7

78. Give an example of tliagonalitablc" x 11 mat rice> A and B such that AB is not diagonalitable. 79. Show that every diagon:1l 11 x 11 matrix is diagonalizable.

l

[~

&7. Prove that if a nilpotent matrix is diagonalizable. then it must be the zero matrix. /lim: Use Exercise 72 of Section 5.1. 88. Let A be a diagonalizable 11 x 11 matrix . Prove that if the characteristic polynomial of A is /(t) = a,t• + tt., _ 1t" - 1 + · · · + a 1t +au. thcn/(A) = 0. where /(A)= a., A" + a., 1A" - 1 + ·· · + at A + ao/11 • (This result is called the Cayll'y·llamiltoll theorem. 7 ) Him: If A = PDP 1 • show that /(A)= Pf(D)P 1.

7 The Cay1ey-Hamthon

theorem ftrst appeared in 1858. Anhur Cayley (1821-1895) was an English mathematician who contributed greatly to the development of both algebra and geometry. He was one of the first to study matrices. and th1s work contributed to the development of quantum mechanks. The Irish mathematician William Rowan Hamilton (1805-1865) is perhaps best known for hi! use of algebta in optics. His 1833 paper ftrst gave a formal structure to ordered pairs of real numbers that yielded the system of complex numbers and led to hts later development of

quorernions.


Page 6 of 10


324 CHAPTER 5 Eigenvalues, Eigenvectors, an d Diagona lizatio n

89. The t ra ce of a square matrix is the sum of its diagonal entries. (a) Prove that if A is a d iagonalizable matrix, then the trace of A equals the sum of the eigenvalues of A. Him: For all 11 x 11 matrices A and 8. show that the trace of A8 equals the trace of 8A. (b) Let A be a diagonalizable 11 x 11 matrix with c haracteristic polynomial ( -l )"(t - )q)(t - /, 2) • · · (t - ).,). Prove that the coefficie nt of t• - 1 in th is polynomial is (- 1)" - 1 times the trace of A.

90.

9 1. [

92.

rj

-I~

6

24

-2

5

18

0

- 6 - 25 -8

-36

2 3

4

18

-2

:

~

~

-5

18

I

2

5

-10

r-:

-29 32 3

24 -171 -3 17 -II

94. Let 1.00 0.16 [ 0.00

4.0 0 .0 - 0.5

~]

and

(b) What are the eigenvalues of A when c

= 8. 1?

(c) lf c = 8.0, what appears to happen to the vector A111 u as m increases?

For each of the matrices in £urcise.r 90- 93. ji11d, if po.r.sible. a11 i11vertible matrix P a11d a diago11al matrix D such that A = PDP - 1• /fllo such matrices exist. explai11 why 110t.

-7 -4

7

-2

-3 -5

0

(a) If c = 8.1, what appears to happen to the vector A111 u as m increases?

!11 Exercises 90- 94. 11se either a calc11lator with matrix capabilities or compwer software such as MATLA8 to solve each problem.

-5 -6

-5

A=

(c) For A as in (b), what is the constant term of the characterist ic polynomial of A?

-4 - 1

93.

13 - II

(d) What are the eigenvalues of A when c

-3] -4 I 3

12 141 --2014

9

= 8.0?

(c) lf c = 7.9, what appears to happen to the vector A111 u as m increases?

(f) What are the eigenvalues of A when c

= 7 .9?

(g) Let 8 be an 11 x 11 mattix havi ng 11 distinct eigenvalues, all of which have absolute value less than I. Let u be any vector in R!' . Based on your answers to (a) through (f), make a conjecture about the behavior of 8"' u as 111 increases. Then prove that your conjecture is valid.

SOLUTIONS TO THE PRACTICE PROBLEMS I. T he given matrix A has two eigenvalues: -I, which has multiplicity I, and 2. wh ich has multiplicity 2. For the eigenvalue 2, we see that the reduced row echelon form of A - 213 is

form of A+ l3. which is

we see that

Therefore the vector form of the general solution of (A - 21))x = 0 is

so

is a basis for the eigenspace of A corresponding to the eigenvalue 2 . Li kewise, from the reduced row eche lon


is a basis fo r the eigenspace of A corresponding to the eigenvalue -I . Take

the matrix whose columns are the vectors in the eigenspace bases. l11e correspondi ng diagonal matrix D is

0 2

~].

0 -I

Page 7 of 10


5.4 Diagonalization o f Linea r Operators

325

the matrix whose do:ogonal entries are the eigenvalues th:u correspond 10 the re;pecth•e columns of P. Then A = PDP 1•

Checking the eigenvalue - I (the eigenvalue with multi plicity greater than I). we see that the reduced row echelon form of 8 - (-I )/J = 8 + l.l is

2. The ch:oracteri,tic pol) nomial of A 'hows that the only eigenvalue of A is 2. and that its multiplicity is I. Because A is a 3 x 3 matrix. the ;urn of the multiplicities of the eigenvalues of A mu>l be 3 for A to be diagonalizablc. lienee A h not diagonalitable. The matrix 8 has two eigenvalues: 2. which has multiplicity I: and - I. which has multiplicity 2. So the sum of the muhiphcoloe; of the c•gcnvaluc; of 8 •s 3. Thus there arc enough eigenvalue> for 8 10 be diagonalizable.

Since the rank of lhos matrix os 2. the d•men;•on of the eigenspace corresponding to e•genvalue -I •• 3 - 2 I. which is le~ than the multiplicity of the eigenvalue. Therefore 8 is not diagonali£able.

=

is.4•1DIAGONALIZATION OF LINEAR OPERATORS In Section 5.3, we defined a diagonalizable matrix and saw that an 11 x 11 matnx ts diagonalizable if and only if there is a basis for n" consisting of eigenvectors of the matri x (Theorem 5.2). We now define a linear operator on n" to be d iagonalizable if there is a basis for n" cons isting of eigenvectors of the operator. Since the eigenvalues and eigenvectors of a linear operator arc the same as those of its standard matrix, the procedure for fi ndi ng a basis of eigenvectors for a linear operator is the same as that for a matrix. Moreover. a basis of eigenvectors of the operator or its standard matrix is also a basis of eigenvectors of the other. Therefore

a li11ear operator is diago11ali:ab/e if a11d 011/y if irs sta11dard matrix is diago11ali:able. So the algorithm for matrix diagonalization on page 3 18 can be used to obtain a basis of eigenvectors for a diagonalizable linear operator. and the test for a d iagomtlizable matrix on page 319 can be used to identify a linear operator that is diagonalizable.

Example 1

Find, if possible, a basis for on n 3 defined by

Solution

n 3 consisting of eigenvectors of the linear operator T

The standard matrix of T is

A= [

-~

9

-7 3

~].

- I

Since T is d iagonali zable if and only if A is, we must determine th e eigenvalues and cigcnspaces of A. The characteristic polynomial of A is - (1 + I ) 2(r - 2). Thus the eigenval ues o f A are - I. which has multi plicity 2, and 2, which has multiplici ty I. The reduced row echelon fonn of A + / 3 is I

I

0 0 [ 0 0 • This section can be omined without loss of continuity.


Page 8 of 10


326 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization so the eigenspace corresponding to the eigenvalue - I has

as a basis. Since - I is the only eigenvalue of multipli city greater than I and the dimension of its eigenspace equals its multiplicity. A (and hence T) is diagonalizable. To obtain a basis of e igenvectors, we need to exam ine the reduced row echelon form of A - 213. which is

It follows that

is a basis for the e igenspace of A correspond ing to the eigenvalue 2. Then

is a basis for 1?.3 consisting of eigenvectors of A. This set is also a basis for 1?.3 consisting of eigenvectors ofT.

Example 2

Find, if possible, a basis for 1?.3 consisting of eigenvectors of the linear operator T on 1?.3 defined by

T

Xt]) [-xt +x2+ l.r3] ([X3 = 0 X2

Xt - X2

.

Solution The standard matrix ofT is - I

A=

[

~

-~ ~] .

To detcnnine if A is diagonalizable, we must find the characteristic polynomial of A, which is - 12(1 + 2). Thus A has the eigenvalues 0, which has multiplicity 2, and - 2, which has multiplicity l. According to the test for a diagonalizable matdx, A is diagona lizable if and only if the eigenspace correspond ing to the eigenvalue of multiplic ity 2 has dimension 2. Therefore we must examine the eigenspace corresponding to the eigenvalue 0. Because the reduced row echelon fonn of A - 0/3 = A is


Page 9 of 10


5.4 Diagonalization of li near Operators

327

we see that the rank o f A - 0/3 is 2. Thu s the e igenspace corresponding to the eigenvalue 0 has dimension I, and so A, and hence T , is not diagonalizable.

Practice Problem 1

~

The c haracteristic polynomi al of the linear ope rator

T

x']) [ x, + +x3] ([x3 = - x, +2x, + 3x3 2xz

X?

2xz

is - (1 - 2) 3 . Detennine if this linear operator is diagonalizable . If so, li nd a basis for R 3 consisting of eigenvectors of T.
Determine if the linear operator

is diagonalizable. If so, find a basis for R 2 consisting of e igenvectors ofT.

Let T be a linear operator on R" for which there is a basis B = (v 1. v2. .. .. v,.) cons isting o f e igenvectors of T. Then for each i, we have T (v;) A; V;, where).; is the eigenvalue corresponding to v;. Therefore IT (v;)] s = A; e; for each i, and so

=

[Tis= [ [ T (vJ)]e [ T (vz)] e ... [T (v,.)]s ] = [A JeJ Azez . . . >..,. e,.] is a diagonal matrix. The converse of this result is also true. which explains the use of the term diagona/izable for such a linear operator.

A linear operator T on R" is diagonalizable if and only if there is a basis B for R" such that [Ti s, the B-matri x ofT , is a diagonal matrix. Such a basis B must cons ist of eigenvectors ofT .

Recall from Theore m 4 . 12 that the B-matrix ofT is given by [Tis= s- 1AB , where 8 is the matrix whose columns are the vectors in B and A is the standard manix of T . Thus. if we take

in Example I, we have

[T ]s= B - 1AB =

- J [

0

0

0 0OJ ,

- 1

0 2

which is the diagonal matrix whose di agonal entries are the eigenvalues corresponding to the respective columns of B.


Page 10 of 10


328


A REFLECTION OPERATOR We conclude this section with an example of a diagonalizable linear operator of geometric interest, the reflection of 1?) about a 2-dimensional subspace. Although previously defined in the Exercises for Section 4.5, we repeat the definition here. Let W be a 2-dimensional subspace of n 3. that is, a plane containing the origin, and consider the mapping Tw : n 3 -+ 'R} defined as follows: For a vector u in n 3 with endpoint P (sec Figure 5.6). drop a perpendicular from P to W, and extend this perpendicular an equal distance to the point P ' on the other side of W. Then Tw( u ) is the vector with endpoint P' .

IV

Figure 5.6 The reflection of 'R. 3 about a subspace W

It will be shown in Chapter 6 that Tw is linear. (See Exercise 84 in Section 6.3.) Assuming that Tw is linear. we show that reflections arc diagonalizable. Choose any two linearly independent vectors b 1 and b2 in W, and choose a th ird nonzero vector b3 perpendicular to W. as shown in Figure 5.6. Since b3 is not a linear combination of b 1 and b2, the set l3 = {b ~o b2, b3} is linearly independent and hence is a basis for nJ. Furthermore, Tw(b t) b , and Tw(b2) b2 because a vector in W coincides with its reflection. In addition, Tw(b3) = - b3 because Tw reflects b3 through W an equal distance to the other s ide. (See Figure 5.6.) It follows that b , and b2 are eigenvectors of Tw with corresponding eigenvalue I, and b3 is an eigenvector of Tw with corresponding eigenvalue - I. Therefore Tw is diagonalizable. In fact. its columns are

=

[Tw(b ,)]s = So

Example 3

[~],

[Tw(b2)ls =

=

[!] ,

[Tw]B=[~

0 I

0

and

~l

~]

.

(4)

- I

Find an explicit formu la for the reflection operator Tw of where


[Tw(b3)]B = [ _

n3

about the plane W,

Page 1 of 10


5.4 Diagonalization of linear Operators

329

Solution As in Chapter 4, we can obtain a basis for W by solving the equation 0: that defines it, namely. Xt - X2 + X3

=

Therefore

is a basis for IV. Recall from analytic geometry that the vector [

~] is perpendicular (normal) to

the plane with the equation ax+ by+ cz =d. So setli ng b 3 = [ - :]. we obtain a nonzero vector perpendicu lar to W. Adjoin b3 to B 1 to obtain a basis

for

n 3 consisting of eigenvectors of Tw. Then [Tw ] 8

is as in Equation (4). By Theorem 4. I 2, the standard matrix of A is given by

A= BJTw JsB-

1

=

[

~ ~I -~] 2

2

3

3

2

2

- 3

Tw

3

3 . I

3

([;~]) =A[;~]

is an explicit formula for Tw.


Find an explicit formula for the reflection operator of n 3 about the plane with equation + 3z 0. <1111

x - 2y

=

EXERCISES In l:.xercises 1-8. a linear operator Ton n 3 and a basis 8 for n 3 are gi•·en. Compwe [Till- and determine whether B is a basis

for n 3 con.ri.rting of eigenllec/OrS ofT.


Page 2 of 10


330 CHAPTER 5 Eigenvalues, Eigenvectors, an d Diagonalizatio n In Exercises 9- 20. a linear operator T on R " and its clwracterisric polynomial are given. Find, if possible, a basis for R" consisTing of eigem•ecrors ofT. If no such basis exists. explain why.

9. T

([:~]) = [~;: =~;~].

(r

+

r ([::~]) = [10r:,-:_ ~: ] ,

+ 3)(t

([::~]) = [ -~~~+-5::2 ].

(t

+ 3)(1- I)

= [

5

7

12. T

13.

+5

_;~,+~r~xJ

10. T ([::])

11.

2 t

T

- 2)

([;~]) = [ 7x~!x~r2 x - 7x, + 3x3

] •

+ x2

3

-(t

14. T

2

2 )2

+ 5)(1 -

2 )(1 - 3)

([x'] ) [ -3x,+ ] X2

=

4Xl

X]

X2

-(t

•

x'] ) [-x,-xz] ([ = 16. T ([x'] ) = [3x + 3x22x,]

15. T

+ 3)(t -

1)2

X3

X2

-X2

X]

XI+ X2

•

-1(1

+ 1) 2

1

X2

X2

•

-t(t - J)(t - 3)

4x 1 -

x3

6

17. T

([;~]) = [ x~~r~x~ ;x~XJ] , X3

-(t

18. T

+ 3)(1 -

4X3

4)(t - 6)

([x'] ) = [ -x X2

XJ

20. T

1 ] '

-X2

X1 -

2x2

-

([;~]) [~;~=~;~] , .l,l

X4

=

2<3

5XJ

-(1

+ 1) 3

XJ

(r

+ 2)(r - W

+ 3X4

In Exercises 21- 28, a linear operator Ton n• is given. Find, if possible. a basis B for 'R" such thar [T ] B is a diagonalmarrix. If no such basis exists, explain why. 2 1. T 22. T


([X' xz ])= [X 3.x,I-x2 - J ([x']) = [-4x, -x, ++3x26x2J x2 ,\.2

Page 3 of 10


5.4 Diagonalization of linear Operators

23. T

([:~~]) = [~~~~_+3~~2 J

24. T

([:~ ]) = [ t';;;•~ 193~:2 ]

25. T

= [4x -~ 5x2] 1

X3

27. T

28. T

42. Let W be a two-dimensional subspace of n 3 . If Tw: n 3 - n 3 is the reflection ofn3 about W.then each nonzero vector that is perpendicular to IV is an cigcnvcc· tor of Tw corresponding to the eigenvalue 0. 43. If T is a diagonalizable linear operator having 0 as an eigenvalue of multiplicity m, then the dimension o f the null space of T equals m. 44. If T is a linear operator on 1?.", then the sum of the multiplicities of the eigenvalues ofT equals n.

([;~])

26 . T ( [;:])

2

45. If T is a diagonalizable linear operator on n", then the sum of the multiplicities of the eigenvalues ofT equals 11. 46. If T is a linear operator on n• having 11 di stinc t eigenvalues. then T is diagonalizable.

-X3

([;~]) = [ - xt ~~;2

47. If 8 1 • 8 2, .•• , Bk are bases for distinct eigenspaces of a linear operator T. then Bt U Bz U · · · U 8k is a linearly independent set.

- x3]

([;i]) = [3x~·~23~:X3]

~

~

48. If 8,. 82 ..... Bt arc bases for all the distinct eigenspaces o f a linear operator T, then 8 1 U Bz U · · · U Bt is a ba~is for 1?." consisting o f eigenvectors ofT.

In Exercises 29-48, derennine wheTher rite srareme/lis are True or false.

29. If a linear operator on n" is diagonalizable. then its standard matrix is a diagonal matrix. 30. For every linear operator on n•. there is a bttsis B for n• such that [T]G is a diagonal matrix. 31. A linear operator on n• is diagonalizablc if and only if its standard matrix is diagonalizablc. 32. If T is a diagonalizable linear operator on n•. there is a unique basis B such that [T] 6 is a diagonal matrix. 33. lf T is a diagonalizable linear operator on n•. there is a unique d iagonal matrix D such that [T]B D . 34. Let W be a 2-dimcnsional subspace of n 3 . The reflection of n 3 about W is one-to-one. 35. Let W be a 2-dimcnsional subspace of nJ The reflection of n 3 about w is onto. 36. If T is a linear operator on n" and B is a basis for n • such that [T] 6 is a diagonal matrix , then 8 consists of eigenvectors of T. 37. The characteristic polynomial of a linear operator T on 1?." is a polynomial of degree 11. 38. If the characteristic polynomial o f a linear operator T on 1?." factors into a product o f linear factors, then T is diagonalizable.

=

39. If the characteristic polynomial of a linear operator T on n" does not factor into a product of linear factors. then T is not diagonalizablc. 40. If, for each eigenvalue A of a linear operator T on n•, the dimension of the eigcnspac·c ofT corresponding to A ec1uals the multiplicity of A, then T is diagonalizable. 3 • If 41. Let IV be a two-dimensional subspace of Tw : n 3 - n 3 is the reflection of n 3 about W . then each

n

nonzero vector in W is an eigenvector of Tw corresponding to the eigenvalue - I.


331

In Exerci.
49. T ( [x x2 ] )

1

= [ -5Xt12r+ +10x3 cx2 - Sx3] - sx, -

.1.1

3x3

- (1 - c)(r - 2)(1 - 7) 50. T

([Xt]) X2 = [Xt +2x2 C.
-X3]

XJ 6x1 - X2 + 6x3 -(t - c)(t - 3)(1 - 4)

51. T

([;~]) = [ -8x -x, -~.~2 -x3 ] + X2 - 5X3 X3

1

2 X2] 52. T ([XI]) X2 = [-4xt+ - 4X2 XJ CXJ -(t - c)(1 + 4) 2 - (1 - c)(r + 4)

53. T

([x

54_ T

([;~]) = [ -4x~x~ 2x

1

~2 ] )

= [ 2r, -

1

cx 3x2 + 2x3] ~3 -3x1 - x 3 -(t- c)(r + 3)(1 + I) 2

x3 - (1 - c)(t

SS. T (

4x 1

+ 4xz -

]

2r3

+ 4)(1 + 2)

[;~]) _

[ -Sx,

X3

-9xt

::2

+ + 3x3 ] + 13x2 + 5x3

-(1- c)(t 2 + 2)

Page 4 of 10


332 CHAPTER 5 Eigenvalues, Eigenvectors, an d Diagona lizatio n

56. T

([:~:]) = [wx:·~ 2.r3] 6xz + XJ

72. Exercise 64

l1·3

75. Let [u . v. wf be a basis for

-(t - c)(t - 6)(t - 7)

operator on

7

57.

T (

[:~]) = [ ~1 0~~x:~:2]

-(t - c)(t

58. T

73. Exercise 65

74. Exercise 66

T(au

+ 3)(t + 2)

In Exercises 59- 64, the equation of a plane W through the origin of n 3 is given. Determine an explicit formula for the reflection Tw ofn3 about IV.

=

59. x +y +z o 6 1. X+ 2y- Z 0 63. x + 8y - 5z = 0

60. 2x

=

+ y + z =0

62. x +z = o 64. 3x - 4y + 5z

-4]

-4

3 -1

7

=0

-I

-1] 2

-2

Exercises 67- 74 use the definition of the orthogonal projection Uw ofn 3 on a 2-dimensimwl subspace IV, which i.v gil•en in the exercises of Section 4.5. 67. Let IV be a 2-dimensional subspace of

n

(a) Prove that there exists a basis 8

=

[~ ~

(b) Prove that [Tw ]e

=

n 3• for n 3 s uch

G ~ ~].

that

where 8 is the

basis in (a).

= !
is the

(d) Usc (c) and Exercise 59 to find an expli cit fo•mula for the orthogonal projection Uw, where IV is the plane with equation x + y + z 0.

=

In Exercises 68-74, find an explicit fomwla for the orthogonal projection Uw of n 3 on the plane IV in each specified exerci.ve. Use Exercise 67(c) and the ba.vis obtained in the .vpec· ijied exercise to obu1in your answer. 68. Exercise 60 70. Exercise 62

69. Exercise 61 71. Exercise 63


let T be the linear

= a u + b'' -

cw

(b) Is T d iagonalizable? J ustify your answer. 77. Let T be a linear operator on n• ru1d 8 be a basis for n" such that [T]e is a diagonal matrix. Prove that 8 must consist of eigenvectors of T . 78. If T and U are diagonalizable linear operators on n", must T + U be a diagonal izable linear operator on n"? Justify your answer. 79. If T is a diagonal izable linear operator on n". must cT be a diagonalizable linear operator on 'R." for any scalar c? Justify your answer. 80. If T and U are diagonalizable linear operators o n 'R.", must TV be a diagonalizable linear operator on n"? Justify your answer. 81. Let T be a linear operator on n". and suppose that v 1 • v2..... Vt arc eigenvectors of T con espond· ing to distinct nonzero e igenvalues. Prove that the set (T(v, ), T(v2), ... , T(vtH is linearly independent.

- 1

(c) Prove that [Uw ] G basis in (a).

T(a u + b' ' + c w)

n 3 • and

for all scalars a, b, and c.

-2

-2

2 -1 [ -2

66

76. Let (u . v. wf be a basis for operator on n 3 defined by

(a) Find the eigenvalues of T and determine a basis for each eigenspace.

In Exercises 65 and 66. the standard matrix of a reflection ofn 3 about a 2-dimensional subspace IV is giwn. Find an equation for IV.

[Uwle

+ bv + c w) = a u + b v

(b) Is T diagonalizable? J ustify your answer.

+ 5)

-8

be the linear

(a) Find the eigenvalues of T and determine a basis for each eigenspace.

3

65. -1 [ -81 I 9 -4 -4

n 3 • and let T

for all scalars a, b, and c.

([~:]) = [x, :·~-~ ~~r3] x -2.r, - 3x3

-(t - c)(t 2

n 3 defined by

82. Let T and U be diagona lizable linear operators on n". Prove that if there is a basis for n" cons isting o f eigenvectors of both T and U. then TU = UT. 83. Let T and U be linear operators on 'R.". If T 2 = U (where T 2 = 7T), then T is called a square root of U. S how that if U is diagonalizablc and has only nonnegative eigenvalues, then U has a sqm•re root. 84. Let T be a li near operator on n• and 6 1 • 62 ..... 8t be bases for al l the d istinct e igenspaces ofT. Prove that T is diagonalizable if and only if 6 1 U 8 2 U · · · U 6k is a generating set for n".

In Exercises 85 and 86. use either a calculator with matrix capabilities or computer software such as MATLAB to find a basis for n 5 consi.tting of eigenvectors of each linear operaTor T; or explain wl1y no such basis exist.f.

Page 5 of 10


5.4 Diago naliza tion of linear Operators 85. T is the linear operator on R 5 defi ned by

333

86. T is the linear o perator on R 5 de fined by

Xtl

[- I Lt 1 - 9x2 + 13x3 + 18x4 - 9xsl 6.r1 + Sx2 - 6x3 - 8x4 + 4xs T x3 = 6x1 + 3x2 - 4x3 - 6x4 + 3xs . - 2rt + 2<3 + 3x• - 2rs [ x4 xs 14x t + 12r2 - 14xJ - 20X4 + 9xs x1 ]

x2

x2

T

x3

[x,

[ -2r, - 4x2- 9x3 - 5x 16x5 ] x + 4x2 + 6x3 + 5x• + 12rs 4

=

xs

-

1

4x 1 + IOx2 + 20x3 + 14x4 + 37x5 • 3x, + 2r2 + 3x3 + 2x• + 4xs -4xt - 6x2 - 12<.1 - 8.r.t- 2 1xs

SOLUTIONS TO THE PRACTICE PROBLEMS I. Since the characteristic polynomial ofT is -(1 - 2) 3 . the only e igenvalue of T is 2. which has mullipl ic ity 3. The standard matrix of T is

A= [

is a basis for the eigenspace of A corresponding to - I . Combi ning these e igenspace bases, we obtain the SCI

I] 2 0 . ~ 2 2 3

-I

and the reduced row echelon form of A - 2 /J is

I

- 2

0

0 0

[0

-I]

0 . 0

a matrix of rank I . Since 3- I < 3, the dimens ion of the eigenspace corresponding to e igenvalue 2 is le.~s than its mulliplicity. Hence the second condition in the test for a diagonalizable matrix is not true for A. and A (and thus T) is not diagonalizablc.

which is a basis fo r R 2 cons isting of eigenvectors o f A and T.

3. We must construct a basis B fo r R 3 cons isting of two vectors from W and one vector perpendic ul ar 0. we obtain the equation x - 2y + 3z

=

10

W. Solvi ng

z] [2 ] [-3] ~ ~ +z ~ . [.r]~ [2y-3 =

=y

Thus

2. The sttmdard matrix of 1' is

A=

-7 [ 3

-I 4OJ.

;md its characteristic polynomial is (1 + I)(1 + 2). So A has two eigenvalues (-I and -2), each o f mullipl ic ity I . Thus A, and hence T, is diagonalizable. To find a basis for R 2 consisting of eigenvectors of T. we find bases for each of the eigcnspaces of A. Since the reduced row echelon form of A + 212 is

we sec that

n-m

is a basis for the eigenspace of A con-esponding to -2. A lso, the reduced row echelon form of A + h is

is

;I

basis for W . In addition. the vector

is nonnal to the phme W. Combining these three vectors. we obtain the desired basis

for R 3 . For the 0 11hogonal project ion operator Uw on W, the vectors in B are eigenvectors corresponding to the e igenvalues I. I, and 0, respectively. Therefore

[Uw]B= Hence

n-m


I 0 OJ [00 0I 00 .

Lening 8 denote the matrix whose columns are the vectors in B. we see by Theorem 4.12 that the standard matrix

Page 6 of 10


334

CHAPTER S Eigenvalues, Eigenvectors, and Diagonalization A of Uw is

A = B[Uw!eB

So the formula for Uw is I--

_I [ 132

14

( [;_~: ] ) = A [~121 ] = 1

1

U"

- 3

1

14

[ll11 2rz3xJ ] . 2x1 ++IQ12 + 6.rJ - hi + 61 2 + 513

Is.s• l APPLICATIONS OF EIGENVALUES In this section. we discuss four applications involving eigenvalues.

MARKOV CHAINS Markov chains have been used to analyze situations as diverse a~ land use in Toronto. Canada [3]. economic development in ew Zealand 161. and the game of Monopoly [I I and [2[. This concept is named after the Russian mathematician Andrei Markov (1856- 1922). who developed the fundamentals of the theory at the beginning of the twentieth century. A M arkov chain is a process that consists of a fin ite number of states and known probabil it ies fJij, where fJij represents the probabi lit y of movi ng from s tate j to state i. Note that this probability depends only on the present state j and the future state i. T he movement of population between the c it y and suburbs described in Exam ple 6 of Section 2. 1 is an example of a Markov chain with two states (l iving in the city and liv ing in the suburbs), where Pii represents the probability of moving from one location to another during the coming year. Other possible examples include pol itical affiliation (Democrat, Republ ican, or Independent). where Pii rcprc~cnts the probability of a son having affiliation i if his father has affiliation j; cholesterol level {high. normal. and low), where Pij represents the probability of moving from one level to another in a specific amount of time; or market share of competing products. where p,1 represents the probability that a consumer switches brand~ in a ccnain amount of time. Consider a Markov chain with 11 states, where the probability of moving from state j to state i during a cenain period of time i!. ,,,, for I S i .j S 11. The 11 x 11 matrix A with (i .))·entry equal to p,1 is called the tr ansit ion matrix of this Markov chain. It is a stochastic matrix. that is. a matrix wi th nonnegmivc entries whose column sums arc all I. A Markov chain often has the propeny that 11 is possible to move from any state to any other over several periods. In such :1 case. the transition matrix of the Markov chain is called r egul ar. It can be shown that the transition matrix of a Markov chain is regular if and only if some power of it contains no zero entries. Thus. if .5 A= 0 .4 .7 . [ .5 .6 0

0 .3]

then A is regular because

A2 =

.40 .35

[ .25

.18 . 15] .58 .28 .24 .57

has no zero entries. On the other hand,

.5 0 8 =

0

I

[ .5 0

.3] .7

0

• This section can be omined without loss of continuity.


Page 7 of 10


5.5 Applications of Eigenvalues

335

is not a regular transition matrix because. for every positive integer k, B* contains at least one zero entry, for example, the (I, 2)-entry. Suppose that A is the transition matrix of a Markov chain with n states. lf p is a vector in 'R" whose components represent the probabilities of being in each state of the Markov chain at a particular time, then the components of p arc nonnegative numbers whose sum equals I . Such a vector is called a p rob ability vector. In this case. the vector A"' p is a probability vector for eve1y pos itive integer m. and the components of A"' p give the probabil iti es of being in each state after m periods. As we saw in Section 5.3. the behavior of the vectors A"' p is often of interest in the study of a Markov chain. When A is a regular transition matrix , the behavior of these vectors can be easily described. A proof of the following theorem can be found in [4, page 300]:

THEOREM 5.4 If A is a regular

11

x n transition matrix and p is a probability vector in 'R", then

(a) I is an eigenvalue of A; (b) there is a unique probability vector v of A that is also an eigenvector corresponding to eigenvalue I; (c) the vectors A"' p approach v for m = I, 2, 3, ....

A probability vector v such that Av = v is called a stea dy-state vector . Such a vector is a probabi lity vector that is also an eigenvector of A correspond ing to eigenvalue I. Theorem 5.4 asserts that a regular Markov chain has a unique steadystate vector, and moreover, the vectors A"' p approach v form I , 2, 3, . . . , no mauer what the original probabil ity vector p is. To illustrate these ideas, consider the following example:

=

Example 1

Suppose that Amy jogs or rides her bicycle every day for exercise. If she jogs today, then tomoiTOW she will flip a fair coin and jog if it lands heads and ride her bicycle if it lands tails. If she rides her bicycle one day, then she wi ll always jog the next day. This situation can be modeled by a Markov chain with two states (jog and ride) and the transition matrix today ride tomorrow

jog ride

~]=A.

For example, the (I, I )-entry of A is .5 because if Amy jogs today, there is a .5 probability that she will jog tomorrow. Suppose that Amy decides to jog on Monday. By taking p = [

~] as the original

probability vector, we obtain .5

Ap

= [ .5

~] [~] = [:~]

and


Page 8 of 10


336 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization Therefore on Tuesday the probabilities that Amy will jog or ride her bicycle are both .5, and on Wednesday the probability that she will jog is .75 and the probabi li ty that she will r ide her bicycle is .25. Since A 2 =

[:~;

:;] has no zero entries, A

is a regular transition matrix. Thus A has a unique steady-state vector v, and the vectors A"' p (m = I , 2, 3, ...) approach ,. by Theorem 5.4. The steady-state vector is a solution of Av = v. that is, (A - /2)v = 0. Since the reduced row echelon form of A - his

[I -2] 0

0 .

we see that the solutions of (A - /z)v = 0 have the form

In order that v = [

~~]

be a probabi lity vector, we must have vr

2vz

+

= ~·

Thus the unique steady-state vector for A is"=


I. Hence

vz = 3vz = vz

j ogs ~ of the time and rides her bicyc le

+ vz =

[!J

Hence, over the long nm, Amy

J of the time.

A car survey has found that 80% of those who were driving a car five years ago are now driv ing a car. 10% are now driving a minivan. and 10% are now driving a sport uti lity vehicle. Of those who were driving a min ivan five years ago, 20% are now driving a ca r. 70% are now driving a miniva n. and 10% are now driving a sport util ity vehicle . Fi nally, of those who were driving a sport utility vehicle five years ago, 10% are now driving a car, 30% are now drivi ng a mini van, and 60% are now drivi ng a sport utility vehicle. (a) Determi ne the transition matrix for this Markov chain. (b) Suppose that 70% of those questioned were driving cars five years ago, 20% were driving minivans. and 10% were driving sport utility vehicles. Estimate the percentage of these persons driving each type of vehicle now. (c) Under the cond itions in (b), estimate the percentage of these persons who wi ll be drivi ng each type of vehicle five years from now. (d) Detem1ine the percentage of these persons driving each type of vehicle in the long run, assu ming that the present trend conti nues indefinitely. ~

GOOGLE SEARCHES In December 2003. the most popular search engi ne for '.Veb searches was Google. which performed about 35% of the month's Web searches. Although Google does not reveal the precise details of how its search engine prioritizes the websites it lists for a user' s search, one o f the reasons for Google's s uccess is its PageRank"'' algorithm,


Page 9 of 10



337

which was created by its developers, Larry Page and Sergey Brin, wh ile they were graduate students at Stanford Univers ity. Intuitively, the PageRankTM algorithm ranks Web pages in the following way: Consider a dedicated Web stufer who is viewing a page on the Web. The surfer moves to a new page by e ither randomly choosing a link from the cu rrent page (this happens 85% of the time) or moves to a new page by choosing a page at random from all of the other pages on the Web (th is happens 15% of the time). We will see that the process of moving from one page to another produces a Markov chain, where the pages are the states, and the steady-state vector g ives the proportions of times pages are visited. These proportions give the ranks of the pages. It follows that a page with many links to and from pages of high rank will itself be ranked high. In response to a user' s search, Google dctcnnines which wcbsitcs arc relevant and then lists them in order of rank. To investigate this process formal ly. we introduce some notation. Let 11 denote the number of pages that are considered by a Google search. (In November 2004, 11 was about 8.058 billion.) Let A be the n x 11 transition matrix associated with the Markov chain, where each of the 11 pages is a state and aij is the probabi lity of moving from page j to page i for I :5. i.j :5. 11. To determ ine the entries a;j of A, we make two assumptions: I. If the current page has links to other pages, then a certain portion p (usually abou t 0 .85) of the time, the surfer moves from the current page to the next page by choosing one of these links at random. The complementary portion of time, I - p, the surfer randomly se lects one of the pages on the Web. 2. lf the current page does not link to other pages. then the stufer randomly selects one of the pages on the Web. To aid in the calculation of A. we begin with a matrix that describes the page links. Let C be the 11 x 11 matrix defined by .. _ { CI ) -

1 if there is a link from page j to page i 0 otherwise.

For any j, let Sj denote the number of pages to which j links: that is, Sj equals the sum of the entries of column j of C. We determ ine a;j , the probabi lity that the reader moves from page j to page i, as follows. If page j has no links at all (that is. Sj = 0) then the probabil ity that the surfer picks page i random ly from the Web is simply l / 11. Now suppose that page} has links to other pages: that is. Sj ,P 0. Then to move to page i from page j, there are two cases:

Case 1. The surfer chooses a link on ]>age j. and this link takes the surfer to page i. The probability that the surfer chooses a link on page j is p, and the probabi lity that th is link connects to page i is I / sj cij J.~j . So the probabil ity of this case is p(cij J.~j ) .

=

Case 2. The sur fer decides to choose a page on the Web randomly, and the page chosen is page i. The probabil ity that the surfer randomly chooses a page on the Web is 1 - p , and the probabil ity that page i is the chosen page is l j 11 . So the probabil ity of this case is (I - p)j n. Thus. if Sj ,P 0, the probability that the surfer moves to page i from page j is pcij + (l - p). Sj


II

Page 10 of 10


338


So we have

+~

pC;j G;j

=

Sj

II

jI

=0.

if Sj

II

For a simple method of obtaining A from C , we introduce the m 2 . . . m"). where

11

x

11

matrix

i\1 = [m t

I

-;.Cj

if Sj

#

0

J

lllj

=

{]

if

It follows that A

= pM + -1 -- IpV , where W

s, = 0.

is the

11

x

11

matrix whose entries are

II

all equal to I. (See Exercise 34.) Observe that a;i > 0 for all i and j. (For n 8.058 billion and p 0.85. the smallest possible value for aij is equal to (I - p)/ n <::: 0.186 x J0- 10 .) Also observe that the sum of the entries in each column of A is I, and hence A is a regular transition matrix. So, by Theorem 5.4, there exists a unique steady-state vector v of A. Over time. the components of v describe the distribution of sutfers visiting the various pages. The components of v are used to rank the Web pages listed by Coogle.

=

Example 2

=

Suppose that a search engine considers only I 0 Web pages. which are linked in the following manner:

Page

links to pages 5 and 10

2

1 and8

3

1, 4, 5, 6, and 7

4

I, 3, 5, and 10

5

2, 7, 8, and 10

6

no links

7

2and4

8

1, 3, 4, and 7

9

1 and3

10

9

The transition matrix for the Markov chain associated with this search engine, with p 0 .85 as the probabil ity that the algorithm follows an outgoi ng link from one page to the next, is:

=


Page 1 of 10


5.5 Applicat ions of Eigenvalues

0.0150 0.0150 0.0 150 0.0 150 0.4400 0.0150 0.0150 0.0 150 0.0 150 0.4400

0.4400 0.0 150 0.0 150 0.0 150 0.0 150 0.0150 0.0150 0.4400 0.0 150 0.0150

0. 1850 0.0150 0.0150 0.1850 0. 1850 0. 1850 0. 1850 0.0150 0.0 150 0.0150

0.2275 0.0150 0.2275 0.0150 0.2275 0.0150 0.0150 0.0150 0.0150 0.2275

0.0150 0.2275 0.0150 0.0150 0.0150 0.0150 0.2275 0.2275 0.0150 0.2275

0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0. 1000 0.1000

0.0150 0.4400 0.0150 0.4400 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150

0.2275 0.0150 0.2275 0.2275 0.0150 0.0150 0.2275 0.0150 0.0150 0.0150

0.4400 0.0150 0.4400 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150 0.0150

339

0.0150 0.0150 0.0 150 0.0 150 0.0150 0.0150 0.0150 0.0 150 0.8650 0.0150

The s teady-state vector for this Markov chain is approx imately 0. 1583 0.0774 0. 1072 0.0860 0. 1218 0.0363 0.0785 0.0769 0. 1282 0. 1295 The components o f this vector give the rankings for the pages. 1-!ere, page I has the highest rank of . 1583, page 10 has the second-highest rank of .1295, page 9 has the third-highest rank of .1282, etc. Notice that. even though page 3 has the greatest number of links (5 links to other pages and 3 links from other pages), its ranking is lower than that of page 9. which has only 3 links (2 links to other pages. and I link from another page). This fact illustrates that Google · s method of ranki ng the pages ta kes into account not just the number of links to and from a page, but also the rankings of the pages that are linked to each page.

SYSTEMS OF DIFFERENTIAL EQUATIONS The decay of radioactive material and the unrestricted growth of bacteria and other organ isms are examples of processes in which the quanti ty of a substance changes at every insta nt in proportion to the a mou nt present. If y f(l) re presents the amount of such a substance present at time t, and k represents the constant of proportionality, then thi s type of growth is described by th e differential equation f '(r) J..f(r), or

=

=

y' =ky.

(5)

In calcu lus, it is s hown that the general solution of equation (5) is

y = aekt. where a is an arbitrary constant. That is, if we substi tute aek' for y (and its derivative akek' for y') in equation (5), we obtain an idenlity. To find the value of a in the general solution. we need an i11itial conditio11. For instance, we need to know how much of the subslance is presenl at a partic ular ti me, say, r substance are present initially. then y(O) = 3. Therefore

= 0.

If 3 unils of lhe

3 = y(O) = aek


Page 2 of 10


340 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization Now suppose that we have a system of three d ifferential equations:

y; = 3y, y~ = 4y2

Y3 = Sy3 This system is just as easy to solve as equation (5). because each of the three equations can be solved independent ly. Its general solution is Yt

= ae3'

Yl

= be4' = ces'.

Y3

Moreover, if there are initial conditions y 1(0) = 10, yz(O) = 12, and )'3(0) = 15, then the particular solution is g iven by

= !Oe 3' 4 Y2= 12e ' Y3 = l 5e 5' .

Yt

As for systems of linear equations, the preceding system of differential equations can be represented by the matrix equation

J [y'] y; [3 0 O [y;l y) =

0

4

0

Y2 .

0 0 5

Y3

Letting

y'

=

and

D

=

[~

we can represent the system as the matrix equat ion

y' = Dy w ith initial condition

y(O)

= [i~]

.

More generally, the system of linear differential equations

y; = 011)' 1 + 012Y2 + · ·· + o,,y, Y~ = 02tYt

+ a22Y2 + ··· + 02trYn

ca n be w ri tten as y' =Ay,

(6)

where A is an n x n matrix. For example, such a system could describe the numbers of animals of three species that are dependent upon one another, so that the growth rate of each species depends on the present number of animals of all three s pecies.


Page 3 of 10



341

This type of system arises in the context of predator-prey models, where y 1 and )'2 might represent the numbers of rabbits and foxes in an ecosystem (see Exercise 67) or the numbers of food fish and sharks. To obtain a solution of equation (6). we make an appropriate change of variable. Define z =p - ly (or equ ivalently, y = Pz), where P is an invertible matrix. It is not hard to prove that y' = P'L. (Sec Exercise 66.) Therefore substituting Pz for y and Pz' for y' in equat ion (6) yields Pz' =APz,

or

z'

= P- 1APz.

Thus, if there is an invertible matrix P such that p - 1AP is a diagonal matrix D, then we obtain the system z' = Dz. which is of the same simple fonn as the one just solved. Moreover, the solution of equation (6) can be obtained easi ly from z because y=Pz. If A is diagonalizable, we can choose P to be a matrix whose columns form a basis for 'R." consisting of eigenvectors of A. Of course, the diagonal entries of D are the eigenvalues of A. This method for solv ing a system of differential equations with a diagonalizable coefficient matrix is sununarized as follows:

Solution of y'

= Ay When A Is Oiagonalizable

l. Find the eigenvalues of A and a basis for each eigenspace.

2. Let P be a matrix whose columns consist of basis vectors for each eigenspace of A, and let D be the diagonal matrix whose d iagonal entries are the eigenvalues of A corresponding to the respective columns of P. 3. Solve the system z' Dz. 4. The solution of the original system is y Pz.

=

ijifi ..!.!tjl

=

Consider the system

y; = 4yt + )'~ = 3y t

.Y:l

+ 2)'2·

The matrix form of this system is y' = Ay, where y

= [Y•J Y2

and

Using the techniques of Section 5.3, we sec that A is diagonalizablc because it has distinct eigenvalues I and 5. Moreover, and

{[:]}

are bases for the corresponding eigenspaces of A. Hence we take

P= [


- I 3

:]

and

Page 4 of 10


342


Now solve the system z'

= Dz, which is

z; =

Zt

z~ = 5Z2 ·

The solution of this system is

z

aet J = [ beSt .

Thus the general solution of the origi nal system is

y

= Pz = [ -

I]

I

1

3

or

[aet] best

=

[- aet + beSt] 3aet + best ·

y , = - aet + beSt Y2 = 3aet + beSt.

Note that it is not necessary to compute P -I. If, add itionally, we are given the initial conditions y 1(0) 120 and Y2(0) 40, then we can find the particular solution of the system. To do so. we must solve the system of linem· equations

=

120 = y 1(0) = - ae 0 40 = Y2(0)

=

3ae 0

=

+ beS =-a+ b + be 51°> = 3a + b

for a and b. S ince the solution of this system is a = - 20 and b = 100, the pm1icu lar solution of the original system of differential equations is

Yt = 20et Y2 = - 60et


+ IOOeSt + I OOeSt .

Consider the foll owing system of differentia l eq uations:

y; = - 5y, y~

=

4y2

8y, -7y2

(a) Find the general solution of the system. (b) Find the particular solution of the system that satisfies the initial conditions y, (0) I and )'2(0) 4. ~

=

=

It should be noted that, with on ly a slight modification , the procedure we have presented for solving y' Ay can also be used to solve a nonhomogeneous system y' = Ay + b , where b =F 0. However, th is procedure for solving y' = Ay cannot be used if A is not diagonalizable. In s uch a case. a si milar technique can be developed from the Jordan canonical form of A . (See [4, pages 515-5 16].) Under some circumstances, the system of differential equations (6) can be used to solve a higher-order different ial equation. We illu strate this technique by solving the third-order di fferential equation

=

111 )'


-

6y" + Sy' + 12y

= 0.

Page 5 of 10



343

= y, Y2 = y', and Y3 = y", we obtain the system y; = Y2

By making the substitut ions y 1

Y~ = Y:i = - 12y,- Sy2

Y3

+ 6y3.

The matrix form of this system is

y;] [ [ y~

=

Y:i

0 0 - 12

The characteristic equation of the 3 x 3 matrix

~

A= [

- 12 is -(13 - 6t2 + 5t + 12) = 0. which resembles the original differential equation y"'- 6y" + 5y' + 12y = 0. This s imi larity is no accident. (See Exercise 68.) Since A has distinct eigenvalues - 1. 3, and 4. it must be diagonalizable. Using the method previously described, we can solve for y 1, Y2· and Y3· Of course, in this case we are interested only in Yt = y . The general solution of the given third-order equation is

y

= ae- + be3' + ce4' . 1

HARMONIC MOTION Many real-world problems involve a differential equation that can be solved by the preceding method. Consider, for instance, a body of weight w that is suspended from a spring. (See Figure 5.7.) Suppose that the body is moved from its resting position and set in motion. Let y(l) denote the distance of the body from its resting point at time r, where positive distances are measured downward. If k is the spring constant, g is the acceleration due to gravity (32 feet per second per second), and -by'(t) is the damping force, 8 then the motion of the body satisfies the differential equation

~y"(t) + by'(t) + ky(t) g

= 0.

0 y

Figure 5.7 A weight suspended from a spring • This section requires knowledge of complex numbers. {See Appendix C.)

aThe dampi119 force is a frictional force that reflects the viscosity of the medium in which the motion occurs. II is proportional to the velocity, but acts in the opposite direction.


Page 6 of 10



=

For example, if the body weighs 8 pounds, the spring constant is k 2 . 125 pou nds per foot, and the damping force constant is b = 0 .75, then the previous differential equation reduces to the form

y" + 3y' By making the substitut ions Yt

+ 8.5y = 0.

= y and yz = y', we obta in the system y; = y~ = - 8.5y• -

Y2

3n

or. in matrix form.

The characteristic polynomial of the preceding matrix is 1

2

+ 3/ + 8.5,

which has the non real roots - 1.5 + 2.5i and - 1.5 - 2.5i. The genera l solution of the differential equation can be shown to be

y

= ae<-l.S+2.Silt + be<-l.S-2.Sill.

Us ing Eu ler's formula (see Append ix C), we obtain

y

= ae-l.St (cos 2.51 + i s in 2.51) + be-1.5 (cos 2.51 1

i sin 2.51),

which can be written as

y

= ce-1.5

1

cos2.5t

+ de-l.Sr sin 2.51 .

The constants c and d can be determined from in itial cond itions, such as the init ial displacement of the body from its resting position, y{O), and its initial velocity. y'(O). From th is solution, it can be shown that the body oscillates with ampl itudes that decrease to zero.

DIFFERENCE EQUATIONS To introduce difference equations. we begin with a counting problem. This problem is typical of the type that occurs in the study of combinatorial analysis, which has gained considerable attention in recent years because of its applications to computer science and operations research. Suppose that we have a large number of blocks of three colors: yellow. red. and green. Each yellow block fills one space, and each red or green block fills two spaces. How many different ways are there of arranging the blocks in a line so that they fill n spaces? Denote the answer to this question by r 11 , and let Y , R, and G represent a yellow, a red, and a green block, respectively. For conven ience, we let ro = I. The following 0, 1. 2. and 3: table lists the possible arrangements in the cases n

=


Page 7 of 10



34 5

y 2

YY,R,G

3

YYY, l'R, \'tj, RY, GY

Now suppose that we have to fill n spaces. Consider the three possible cases: C a se I. The last block is yellow. In this case, the yellow block fills the last space. ;md we can till the first in r,_ 1 ways.

11 -

I spaces

Case 2. The last block is red. In th is case, the red block fills the last two spaces, and we can fi ll the first n - 2 spaces in 1'11 -2 ways. C ase 3. T h e last block is gr een . This is similar to case 2, so the total number of ways to fill the first n - 2 spaces is ru-2 ·

Putting these cases together, we have (7)

Notice that this equation agrees wi th the table for n

= 2 and n = 3:

rz

= r1 + 2ro = I + 2 · I = 3

I'J

= r2 + 2rt = 3 + 2 · I = 5.

and

With this formula, we can easily determine that the number of arrangements for 11 is r4

= r3 + 2rz = S + 2 · 3 =

=4

I I.

Equation (7) is an example of a differ ence equation, or r ecu rren ce r elation. Difference equations are analogous to differential equations. except that the independent variable is treated as discrete in a difference equation and as continuous in a differential equation. But how do we find a formula expressing r 11 as a function of 11 ? One way is to rewrite equation (7) as a matrix equation. First, write

r,

= r,

r, + 1 = r,

+ 2r, _ 1.

(The second equation is formed by replac ing can now be written in the matrix form

[ ~'11/'1+1 t ] =


11

by

11

+I

in equation (7) .) The system

[Q2 JI ] [1'1r,1-l] '

Page 8 of 10


346 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization or s, = As, _ 1, where Sn =

r, ] [ Yn + l

A=[~

and

Furthermore, from the solutions for

11

=

0 and

11

:] .

= I. we have

Thus

s,

= As,_ 1 = A2s,_z =···=A" so= A"[: ].

To compute s, . we must compute powers of a matrix , a problem we considered in Section 5.3 for the case in which the matrix is diagona li zable. Us ing the methods developed earlier in this chapter, we find matrices

such that A= PDP -

- I

~]

p

= [-

1•

Then, as in Section 5.3, we have A"= PD" p - t. Thus

and

1I 2' ]

or

r, ] [ I '] [(-1)" [ ~'n + l = - 1 2 0

D= [ 0

OJ [~!

2"

3

1 I [ (- 1)" +2"+ ] (- 1)"+1 + 2"+2 .

=3 Therefore

r,

=

(-1)" +2"+1

3

As a c heck, observe that this formula gives ro = I, rt = I, r2 = 3, and r 3 = 5, which agree with the values obtained previously. It is now easy with a calculator to compute r, for larger values of 11. For example, I'JO is 683, rw is 669,051 , and rn is al most 3 bi llion! In general , a kth -order h omogeneous linear difference equation (or recurrence relation) is an equation o f the fom1 (K)

where the a;'s are scalars, 11 and k are positive integers such that 11 > k, and ak f. 0. Equation (7) enables us to compute successive va lues of r, if we know two consecutive val ues. In the block problem, we found that ro = I and rt = I by enumerati ng the possibilities for filling 0 spaces and I space. Such a set of consecutive values of r 11 is called a set of initial conditions. More generally, in equation (K) , we need to know k consecutive va lues of r, to have a set of initia l conditi ons. Thus the number of consecutive values required equals the order of the difference equation. As in th e preceding exa mple, we can represent the kth-order equation (8) by a matrix equation of the fom1 S11 = As,_J, where A is a k x k matrix and S11 is a vector


Page 9 of 10



347

in n*. (See Exercise 82.) It can be shown (see Exercise 83) that if A has k distinct eigenvalues, )< 1, .i..2, .. . , Ak. then the general solut ion of equation (8) has the form

(9) The b; 's are detennined by the initial conditions, which are given by the components of the vector so. Fu rthennore, the A; 's, which are the distinct e igenva lues of A, can also be obtained as solutions of the equation ( 10) (See Exercise 84.) Equations (9) and ( 10) offer us an altemative method for finding r11 without computing either the characteristic polynomia l or the eigenvectors of A. We ill ustrate this method with another example. It is known that rabbits reproduce at a very rapid rate. For the sake of simplicity, we assume that a pair of rabbits does not produce any offspring during the first month of their lives, but that they produce exactly one pa ir (male and female) each month thereafter. Suppose that, initially, we have one male- female pair of newborn rabbits and that no rabbits die. How many pairs of rabbits w ill there be after 11 months? Let r11 denote the number of pairs of rabbits after n months. Let's try to answer th is question for n 0, I , 2 , and 3. After zero months, we have on ly the original pair. Similarly. after I month. we still have only the original pair. So r, = ro = I. After 2 months, we have the original pair and their offspring; that is, rz 2. After 3 months, we have all the previous pairs and . in addition. the new pair of offspring of the original 3. In general , after 11 months. we have the pairs pair; that is, r3 rz + r 1 2 + I we had last month and the offspring of those pairs that are over I month old. Thus

=

=

=

=

=

( II )

a second-order difference equation. The numbers generated by equation (II) are I, I , 2. 3. 5, 8. 13. 21. 34, .. . . Each number is the sum of the preced ing two numbers. A sequence w ith this property is called a Fibonacci sequence. It occurs in a variety of settings. including the number of spirals of various plants . Now we use equations (9) and ( 10) to obtain a formula for r11 • By equations ( I I) and (10), we have

which has solutions ( I± ~)/2. Thus, by equation (9), there are scalars b 1 and bz such that

To find b 1 and b2. we use the initial conditions

I =

ro =

(J)b,

+

( I) bz

This system has the solution


Page 10 of 10


3 48 CHAPTER S Eigenvalues, Eigenvectors, and Diagonalization

Thus, in general,

This complicated formula is a surprise, because r, is a positive integer for every value of 11. To find the fiftieth Fibonacci number. we compute rso. which is over 20 billion! Our final example. which can also be solved by differential equations. involves an application to heat loss.

Example4

The water in a hot tub loses heat to the surrounding air so that the difference between the temperature of the water and the temperature of the surrounding air is reduced by 5% each minute. The temperature of the water is 120°F now. and the temperature of the surrounding air is a constant 70°F. Let r, denote the tempcmture difference at the end of 11 minutes. Then

r, = .95r,_ 1 for each 11

and

ro = 120 - 70 = so•F.

=

=

By equations (9) and ( I 0), r, b'A." and >.. = 0.95, and hence r, b(.95)". Furthermore. 50= ro = b(.95)0 = b, and thus r, = 50(.95)". For example, at the end of 10 minutes, r, 50(.95) 10 ~ 30°F, and so the water temperature is approximately 70 + 30 = I00°F.

=

In Exercises 85-87. we show how to find a formula for the solution of the first-order nonhomogeneous equation '• = ar, _ 1 +c. where a and c are scalars. This equation occurs frequently in financial applications, such as annuities. (See Exercise 88.) Practice Problem 3 ""

In the back room of a bookstore there arc a large number of copies of three books: a no\'el by Nabokov. a novel by Updike, and a calculus book. The novels are each I inch thick. and the calculus book is 2 inches thick. Find r,. the number of ways of arranging these copies in a stack 11 inches high. ~

EXERCISES ~

~

In £.urdsu 1- 12. detemrim! whethu the statemellls are true or Jalli'.

7. Every regular transition matrix has a unique probability vector that is an eigenvector corresponding to eigenvalue I.

I. Tile row •un~; of the tr:tnsition matrix of a Markov chain are :111 l.

8. The general solution of y'

2. If the transition matrix of a Markov chain contains zero emries. then it is not regular.

9. If p - I AP is a diagon:1l matrix D. then the change of variable z Py transform> the matrix equation y' Ay into

3. If A is the tr:msition matrix of a Markov chain and p is any probability vector, then Ap is a probability vector. 4. If A is the tr:msition 1mttrix of a Markov chain and p is any probability vector. then as m approaches infinity, the vectors A"' p approach a probability vector. 5. If A is the transition matrix of a regular Markov chain, then as m approaches mhnity. the vectors A"' p approach the same probability vector for every probability vector p. 6. Every regular tranSitiOn matrix has I as an eigenvalue.


z'

=ky is y = ke

= = Dz.

1

•

=

10. A differential equat ion a.ly"' + a2y" + cqy' + ao.v

=

0, where lll,ll2,tli.tiO are scalars. can be written as a system of linear difTerential equmions.

II. If A =PDP - '. where P i~ an inveltible matrix and D is a diagonal matrix. then the solution of y' = Ay 1s p - •z. where z is a solution of z' Oz.

=

12. In a Fibonacci sequence. each term after the fir.t two is the sum of the two preceding teml\.

Page 1 of 10



In EurdSI!S 13- 20, dt'termill<' 1\'llether ettcll tmnsitionmatri.x is regular.

~]

13. [0.25 0.75 15. ["5 .5

0 0

0 17.

['

-~

[~ .5] .5

16.

r-~

.7 .3

.9

0 .5

.2

0

0 .8

.5

0

18.

:]

20.

.9 0

next time !hey buy baking powder. IO'l \\ill buy bmnd B. and 20'k "ill buy bmnd C. Of 1ho,e "ho bought brand B last. IW will buy bmnd A the next time they buy baking powder. 60~ will buy brand B. and 3W will buy brand C. Of those" ho bought bmnd C last. IO'k will buy each of brand~ A and B the next time they buy b:•~ing po" der. and 80'l: will buy bmnd C.

.6

[.2 .8 .70 -~] 0

.3

.9

['

0 .5

.I

0 .5

.9 0

-~

0

1]

.8

In Exercises 21 -28. a IY!8ular transition matrix is given. Determine its j'/Ctldy-stllt~ \'ector. .I

.3] .7

[.5

.6

.2] .I

.3

.7

21. [-9

23.

25.

.2 .3

r-~ .2

27.

[: .4

0

.I

0 .4 .6

. I] .9

0 .5

0 .2

.2 .3

.8

22.

[·6.4

.I] .9

24.

["6.3

.9 0

.4

0 .4

.8

26.

0

0

~]

.9

28.

.I

["6.3 .6

[: .-1 0

0 .5 0

.5

·~]

2] 0 0

.2 0 .8

:]

.9 0

29. When Ali>on goe' to her fa\'Orite ice cream store. she bu)> etther a root beer float or a chocolate sundae. Lf she bought a float on her last visit. !here is a .25 probability that 'he "ill hoy a Hom on her nex1 vi,il . If she bought a >undae on her laM 'i>it. there i> a .5 probability that she will buy a float on her next vi~it. (a) Assuming thai this information describes a Markov cham. wnle a 1rans1tion matnx for this situation.

(b) Lf Alison bought a sundae on her next-to-last visit. what i> the probabi lity that she will buy a float on her

nex t vi~il? (c) Over the long run. on what proponion of Alison's 11ips docs she buy a Mmdae?

30. S uppose that the probability that the child of a college-educmed P are college-educated. what (approxtm;ne) proponion of rhe populalion will be college-educated in one, 1wo. and three generations?


(c) Without any knowledge of the present proponion of college-educaled parents. determine the eventual proponion of college-educated people.

31. A supertn:ll"ket >ells three brands of baking powder. Of those who bought bmnd A last. 70'-t will buy brand A the

-~]

.5 .5 0

.I

0

[8.2 0 -~] 0

19.

..7] 3

14.

349

(a) Assuming that the infom1ation describes a Markov chain. write a tr:.msition matrix for this situation.

(b) If a customer la>t bought brand B. what is the probability that his or her next purchase of baking powder is brand B? (c) If a customer la>l bought brand A. what is the probability that hi> or her >ccond future purchase of baking powder is brand C? (d) Over the long run. what proportion of the s upermar· ket'• bak ing powder >ales arc :1ccounted for by e<~ch brand? 32. Suppose that a panicular region with a constant population is divided into three areas : the city. suburbs. and country. The probability thai a pen.on living in rhe city moves to the •uburb' (in one year) i• . I 0. and the probability that someone moves to the country i' .50. The probability is .20 !hat a pen.on living in the suburbs moves to the city and is . I0 for a move to the country. The probability is .20 that a pe"on li,·ing in the country mo' e' to the city and is .20 for a move to the >Uburb<. Suppose initially that 50'k of the people hve in the city. 3W live in lhe suburbs. and 20'l live in the country. (a) Dctenninc the tmnsition matri~ for the three states.

(b) Determine the percentage of people living in each area after I. 2. and 3 years. (c) Use either a calculator with matrix capabilities or computer .oftware >Uch as MATLAB to find the per· cent:~ge of people living in e:1ch are:1 after 5 and 8 years. (d) Determine the evenlllal percentage• of people in each area.

33. Prove that lhe sum of the e ntries of each column of the matrix A in the subsection on Google searches is equal to I. 34. Verify that the matrix A in the s ubsection on Googlc searches satisfies the equation

A = pM

1- p

+ - - W, II

where M and IV arc a> defined in the subsection.

/11 Exercise 35. use either a mlculator with mmrix CII[Jabilities or computer softl•·all' such as MATI..AB.

Page 2 of 10


350 CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization 35. In [5]. Gabriel and Neumann fou nd that a Markov chain could be used to describe the occurrence of rainfall in Tel Aviv during the rainy seasons from 1923-24 to 19-19- 50. A day "a> clas~ified as ner if at least 0.1 mm of precipitation fell at a certain location in Tel Aviv during the period from 8 AM one day to 8 AM the ne~t day; otherwise. the day wa; cla;,;,ified a~ tlry. The data for November follow:

(c) What are the probabilities that an object in the third state will next move to each of the other s t:ues? (d) U'e your an\Weh 10 (a). (b), and (c) 10 predict the steady-,tate vector for A (e) Verify your predictton tn (d).

38. Gi\e an example of a 3 x 3 regular tr:tn\ition matrix A such that A. A 2• and A 1 all contain zero entries. 39. Let A be an 11 x 11 Mocha.,tic matrix. and let u be the vector in 'R." with all components equal to I. (a) Compute AT u . (b) What does (a) imply about the eigenvalues of AT? (c) PrO\e that det (A - 1.) = 0.

(a) As>uming that the preceding information describes a Markov chain. write a transition matrix for this situation.

(d) What docs (c) imply about the eigenvalues of A? 40. Usc ideas from Exercise 37 to con,truct two regular 3 x 3 'tocha,tic matrices having

(b) If a November day wa' dry. what is the probabi lity that the following day will be dry?

UJ

(c) If a Tuesday in November was dry, what is the probability that the fo llowing T hursday will be dry? (d) If a Wednesday in November was wet. what is the probability that the fo llowing Saturday witt be dry? (e) Over the long run. what is the probability that a November day in Tel Aviv is wet? 36. A company leases rental cars at three Chicago offices (located at Midway Airport. O'Hare Field. and the Loop). It' record' show that 60'k of the car;, rented at Midway are returned there and 20'k are returned to each of the other location;. Also. So<:t of the cars rented at O'Hare are returned there and IO'k arc returned to each of the other locations. Finally. 70'k of the car> rented at the Loop are returned there. I<>'< are rewrned to Midway. and 20'k are returned to O'llare. (a) rusummg that the prccedmg mformatton describes a Markov chain. wnte a transition matrix for this situation. (b) If a car is rented at O' llare. what is the probability that it will be returned to the Loop? (c) If a car ts rented at M idway. \\hat is the probability that it witt be returned to the Loop a fter its second rental? (d) Over the long run. if all o f the cars are returned, what proportion of the company's fleet will be located at e"ch office?

as their steady-state vector. 4 1. Prove that if A is a stochastic matrix and p is a probability vector. then Ap i> a probability vectot·. 42. Let A be the 2 x 2 stocha.>tic matrix [

.I .8 .I

43. Let A be an

state will next move to each of the other states?


11

x

11 ~tocha,tic

matrix.

(a) Let v be any vector m R" and k be an index such that 1•:,1 !: l••t for each 1. Pro'·e that the ab.>olute value of every component of AT v is less than or equal to I••• I· (b) Use (a) to ;how that tf ' ' tS an eigenvector of AT that corresponds to an eigem•alue A. then IAI·I••tl!: 1•'4 . (c) Deduce that if A is an eigenvalue of A. then IAI !: I.

44. Prove that if A and 8 arc >tochastic matrices. then AB is a stochastic matrix. /11

Exen:ises 45- 52. ji11d rite geneml .wlurio11 of each sysrem of

diflereurial Nfllariam.

45.

y; = 3y, + 2)'2 = 3,vt - 2y2

Yi

=

26y, + _ 48)):22

48

Y2 =- Yt

y; = 49.

(a) What arc the probabilities that an object in the first ; tate will next move to each of the other states?

(b) What are the probabilities rhar an object in the second

.

(b) Determine a ba>is for each eigenspace of A.

.3] .3 . .4

b

(c) Under what condition> is A diagonaltzable?

47. )'; .90 .05 [ .05

I-b]

(a) Detennine the eigenvalue' of A.

37. S uppose that the tnmsition mmrix of a Markov chain is

A=

°

1- ( I

51.

>'2

.vi

2y, = 3y, = -3y,

+ 2n + 3y, -

)'J

= -5y , + 6yz = -15y , + 14y2 y; = .l'• + 2y2 - n Yi = .Vt + Yl .vi = 4y , - 4y 2 + Sn

_ y;

50.

Yi

Yl = -3y, + n + .n .vi= S.n - 2n- 4.n .v.l = -lOy, + 2.12 + 4.n

Page 3 of 10



y; = 12y t -

!Oy2 - lOy; 8yz - lOYJ Syz - 3y; In Exercises 53-60. find the particular solution of each system ofdifferemial equations thlll satisfies the given initial conditions.

= !Oyt Yi = Sy t -

52. y;

53.

)'; = )' t Y2 = 4y t

+ )'2 + yz

with )'t(O) = 15, )'2(0)

y; =

54.

>'2 = -

y; = Y2 =

8yt -4yt

with Yt(O) 56.

pounds per foot and the damping force constant is 0.5 . 65. Find the general solution of the differential equation that describes the harmonic motion of a 10-pound weight allached to a spring. where the s pring constant is 1.25 pounds per foot and the damping force constant is 0.625.

67. Let Yt denote the number of rabbits in a certain area at time t and y 2 denote the number of foxes in this area at time 1. Suppose that at time 0 there are 900 rabbits and 300 foxes in this area and assume that tl1e system of differential e<1uations

+ 2)'2 + 2y2 )'2(0)

=1

y ; = - 5yt - 8yz Y2 = 4yt + 7y2 with Yt(O)

y; = 2)'1 Yl = )'t

= 1, n(O) = -3

y; = 6yt -

57.

y;

y; = -3yt + 59.

y; = - 7y• +

68. Show that the characteristic polynomial of

+ 3n

y~ = 13)'t - 20y2 - 8}'3 with Yt(O)

(c) Approximately what is the eventual ratio of foxes to rabbits in this area? Does this number depend o n the initial numbers of rabbits and foxes in the area?

= 1, y3(0) = 2

2)'2

9>'2

= - 4, Y2(0) = -

5 . .v;(O)

[_~

=3

)'; =

60.

3y2

(b) Approximate ly how many of each species will be present at times 1 = I. 2. and 3? For each of these times. compute the ra1io of foxes to rabbits.

y; =

=

-

{a) Find the partic ular solution of this system.

= 0, )'2(0) = 2. )'3(0) = 1

)'t + 2y3 = 2yt + 3.vz - 2y; )'~ = 3y3 with Yt(O) -1 , n(O)

- 4)'2

expresses the rate at which the number of animals of each species changes.

5yz - 7)'3 = )'t )'3 y) = 3)' t - 3}'2 - 4y;

>'l

with .l't (0)

58.

=

=

= 7, Y2(0) = 5 = 2,

64. Find the general solution of the differential equation that describes the harmonic motion of a 4-pound weight auached to a spring. where the spring constant is 1.5

66. Let z be ;m 11 x I column vector of differentiable functions. and let P be any 11 x 11 matrix. Prove that if y P7~ then y' P7! .

+ 2y2 y. + 5)'2

2y t

with Yt (0)

55.

= -10

351

5)'t - 2)'2 - 2y; = 18yt - 7yz - 6y3 y~ = -6yt + 2yz + )'3 with Yt(O) = 4, yz(O) = 5. )'3(0)

y;

=8

61. Convert the second-order differential equation

y" - 2y' - 3y

=0

into a system of differe ntial equations,
y"' - 2y" - 8y'

-1 3 -at 2

_!]

-bt -c.

69. Show that if )q, ),2 . and ),3 are distinct roots o r the polynomialt 3 + at 2 + bt + c, then y aeJ·t' + beJ·2' + cel.JI is the general solution of y'" + ay" +by' + cy 0. Him: Express the differential equation as a system of differential equat ions y' = Ay. and show that

=

=

=0

into a system of differential equations, and then find its general solution. 63. Convert the third-order differential equation

y"' - 2y" - y' + 2y = 0 into a system of differential equations. and then find the particular solution s uch that y(O) "" 2, y' (O) "" -3, and y"(O) = 5.


is

0

-b

is a basis for

n 3 consisting of eigenvectors of A.

In Erercises 70-78, use either of the two methods developed in this section to find a fonnula for rn. Then use your result to find r 6. 70. r,

= 2rn- t : ro = 5

= -3rn- t: ro = 8 = '"-t + 2r11-2: ro = 7 and rt = 2 r, = 3rn-l + 4r,.-2; ro = I and rt = I

7 1. ,.,

72. r, 73.

Page 4 of 10


352 CHAPTER S Eigenvalues, Eigenvectors, and Diagonalization 74. r. 15. r. 76. r.

77. r. 78. r.

= 3r•-• -

2rn-!: ro = I and TJ = 3 = -r._ 1 + 6r._z: ro = 8 and r 1 = I = - Sr._ 1 - .tr.-z: ro = 3 and r1 = 15 = r•-1 + 2rn-2: ro = 9 and r 1 = 0

=

(c) Dcri,·c equation t9l by comparing the last componenls of the veclor equation in (b).

=

=

=

2r.,_ 1 + r•- 2- 2r.,_3; ro 3, r, I, and rz 3 79. Suppose 1ha1 we have a large number of blocks. The blocks are o f five colors: red, yellow, green. orange, and blue. Each of I he red and yellow blocks weighs one ounce: and each of the green. orange. and blue blocks weighs two ounces. Let r. be lhe number of ways I he blocks can be arranged in a stack that weighs 11 ounces.

84. Prove 1h:u a scalar A is an eigenvalue of lhe malrix A in Exerci~e 82 if and only if A is a solution of equa1ion 1101. Him: Lei

(a) Detennine ro. r., rz, and r .1 by li>ting the possibilities. and pro,·e each of the foUowing results:

(b) Write the difference equarion in,•olvmg r•.

(i) If A IS an eigenvalue of A. then w l i> a corre>pondmg eigcnvecror. and hence A is a solulion of equa1ion r 101.

{c) Use (b) 10 find a fonnula for r •.

(d) Use your answer 10 (c) to checl. your an>wers in (a).

w,

80. Suppo'e 1hat a bank pay' inlere>l on >:JVing' of 8% compounded annually. Use the appropriare difference equation to delermine how much money would be in a savings uccounl aflcr 11 years if inilially !here was $1000. What is lhc v:1lue of the accounl aflcr 5 yca•·s. 10 ye:1rs, <~nd 15 years?

(ii) IfA is a >olution of equation (IUl, lhen i> an eigcnvecror of A. and A is the corresponding eigenvalue.

ft1 Exercises 85- 87. we examine rlre nonhomogeneous firM· order dif!Prc'tlce equation which is of the form

l'n = ar, _J +c.

81. Wrile lhe third-order difference equ:lliun

r. = 4r._, - 2r.

whet~

2 +Sr. __,

a a11d cart' ronswms. For the purpose oftheu eurcises.

" '<' let

= As. _ 1• as we d1d m th1s section.

m matri~ notation. s.

and

82. Justify the matrix fonn of equation Sl gh•en in this >cction: s. = As._ 1 • where

s.,

=[

:~

l

11

A=

[l ak

0

0

= ro + nc. For rhi> exercise, we '1..~smne 1ha1 '' ¥ I.

(b) U;,c (u) 10 derive lhc solution r. 87.

I

~l

85. Prove lhal s., =A" so for any positive integer 11. 86. For !his exercise, we assume rhat a = I. 1 (a) J>rovc 1ha1 A" = [ c ~] for any posilive inlcgcr 11 .

l n+J. - 1

and

A=[!

0 0

0

0

0

at-1

Ot -2

az

(a) Prove that I and a are eigenvalues of A.

1J

(b) Prove thai there exist eigenvectors v 1 and v2 of A corresponding to I and a. respecti,·ely. and 'cal= 1 1 and 12 such that

83. Consider a klh-order difference equation in the fonn of equation ( ~ l with a sel of k inilial condilions. and let lhe matrix fonn of the equalion be s., =As, _ ,, Suppose. furlhcrmorc. thai A has k distincl eigenvalues A•· A2 •.... Ak, and that v 1, v2.... , Vt are correspondi ng c igenveclors. (a) Prove thai there exis1 scalars t 1.12.... . I t such lhat

(b) Prove that for any positive inlegcr 11.

(c) Usc (b) to prove that there exisl scalars b 1 and b2 such lhal r, b 1 + a"b2 • where

=

-c b,= - a- 1

and

!J,

.

c

= ro + a- I

88. An invcs1or opened a savings accoum on March I with an inillal depo>~t of $5000. Each year thereafrer. he added $2000 10 the account on March I. If the account cam> 1ntcres1 al the rare of 6'k per )Car. find a formula

for the value of this account after

11

years. Hmt: Use

Exercise 87(c).


Page 5 of 10



In Exercises 89 9/, liSt' l'itlru ft c:alclllaror ll'itlr matrix capabilities or comp11tU so/Mart' s11ch as MATUB to sol1•t' t'aclr J>roblem. 89. Soh·e the system of differential equations

y; =

+ 4. h·2 +

3.2)'t

+ 3.7y,

7.7_\'J

.v.i = -1.8n y~

=

- 1.8n - 4.4y3 0.1yz + 2.9yJ

1.7yt

subject to the initial conditions YJ(O) - 2. y 4 (0) - 3.

)'I (0)

-

for each purpose is :1> follows: 10'!. 20%. 25%. O'K, O'l:, 5%, 15'1:. IO'l. IO'l. and 5~. respectively. What percentage of land will be used for each purpose two decades later? IS

regular.

(c) After many decades. "hat percentage of land will be used for each purpose'/

l.&y,

+ 0.4y, =

(a) Suppose that at some time the percentage of land usc

(b) Show that the transiuon matrix

Yi = -0.3.rt + 1.2n + o.2y1 + o.5y,

353

I. )'2(0)

=

-4.

91. A search cngmc consoders only 10 Web pages. which arc linked in the folio\\ ing manner:

90. In (3], Bourne examined the changes in land use in Toronto. Canada. during the years 1952- 1962. Land was classified in ten ways:

2.6,8,and9 4 and7

I. low-dem,ity residential 3. office 5. automobi le commercial 7. warehousing 9. transpo o1ation

2. high-density residential 4. gener~l commercial 6. parking 8. industry 10. vac:mt

1,4,S.8.and 10

no links 2 and 10

sand 9

6

I, s. 6. •nd 9

T he following transition matrix s hows the changes in land usc fro m 1952 10 1962:

4 and 8

Sand 10 5

1952 6

.00

.08 o~

m

. II .07 70 06 .00 01 00 OS

.t4 .08 12 .39 .04 00 .01 .09

.o2 .11 OJ .II J8 21 .01 .08

U~oo(' IR

3

tl

C
J 4 S 6 7

s

9 10

w I() 04 04 22 03 02 00 08

.02 4t OS 04 00 04 .00 00 00

00 07 .41 OS 01 28 14 ()() 00

02 01 09

.JO 09 27 OS 0' 01

.a.t

01

Ol<

oo

9

.Ot

.01 02 .02 .03 03 .08 .18 61 .00 .Ol

.0 1 18 . 14 .04 10 .39 .OJ .03 .08 .00

no links

10

10

.251 .08 .03 .03 OS .t5 .U 13 .00 .06

j

Assume that the trend in land use changes from 1952 to 1962 continues indetimtcly.

(a) Apply the PageRank algorithm to find the transition matrix for the Markov cham :bsociated with this search engine. based on p 0.85. the portion of the time a surfer mo' es from the current page to the ne
=

(b) Find the steady-state \ector for the Marko\• chain obtained in (a). and use 11 to rank the pages.

SOLUTIONS TO THE PRACTICE PROBLEMS I. (a) The Markov chain ha> three states. which correspond to the type of vehicle driven- car. van. or spon utility vehicle (suv). The transotion matrix for this Markov chain is

Hence the probability that someone in the survey is now driving each type of vehicle is given by

Ap f-Ive Year' Ago ~..:ar v~ul ' u''

Nn\1

~::~, [:~ :~ :~] = A. 'liV

.I

.I

.6

(b) The probability vector that gives the probability of driving c:teh type of vehicle five years ago is

p

= [ .70] .20 . .10


.8

.2

.I

.7 .I

= [ .I

.1] .3 [.70] .20 . 10 .6

= [.61] .24

.

.15

Therefore 6 1% of those surveyed arc now driving cars, 24% arc now do·iving minivans. and 15% arc now driving >j)Ort ulility vehic les. (c) The probability that five years from now someone in the survey will be driving each type of vehicle is g iven by

A(A p)

.8

.2

= [ .I

.7 .I

.I

. 1] [.61] .3 .24 .6

. 15

= [.551] .274 . .175

Page 6 of 10


35 4

CHAPTER 5 Eigenvalues, Eigenvectors, and Diagonalization So we estimate that in five years 55.1 % of those surveyed will drive cars, 27.4% minivans, and 17.5% sport utility vehicles.

(d) N01e that A is a regular transitio n matrix , so, by Theorem 5.4, A has a steady-state vector v. This vector is a solution of Av v. that is. of the equation (A - i))v = 0 . Since the reduced row echelon form of A is 0 -2.25 ] I -~75 .

=

[~

0

we sec that the solutions of (A - /3)v form VI = 2.25V) 112 = 1.75VJ v3 free.

ln order that v

2.2Sv3

+

have the

= [Y•J Y2

be a probability vector. we

= I. Hence

=

where

A= [

and

-4]

-5

7 .

8

- I I

=

6e-' -

5e''

= -6e - ' + 10e3' .

3. As in the block example d iscussed earlier, r 0 I because there is one "empty" stack. Furthermore, rt 2 because each of the two novels is I inch high. Now suppose that we have to pile a stack 11 inches high. There arc three cases to consider:

Ca~~ 2. The b ottom hook is the n ovd by Updike. This is simi larto case I. so there arc r, _ 1 ways of stacking the rest of the books.

C ase 3. The b ottom hook is a calculus hook . Since this book is 2 inches thick, there are rn -2 ways of stacking the rest of the books. Combining these three cases, we have the second-order di ffercnce equation rn

= 2r,• - 1 + l"n -2·

We use equation ( 10) to obtain

>. 2 = 2.1. + I,

a nd

D=

z; = - z,

z2 =

which has solutions I ± ./2. Thus, by equation (9 ), there are scalars b 1 and b 2 such that

- I [ 0

Then the c hange of variable y = Pz tran.
Jz2 .

To find b 1 a nd b 2, we use the initial conditions

=

=

I r(O) ( l )bt + (l)b2 2 = r(l) =(I + ,f'i)bt +(I - hlb2. This system has the so lution

Hence bt

Therefore the general solution of the g iven system of differential equations is y

the

and

are bases fo r the eigenspaces of A. Take

P = [

It desired particular solution is

C ase 1. The b ottom hook is the n o,·el hy Naboko ,·. Since this novel is I inch thick, there are r,. _ 1 ways of s tacking the rest of the books.

1.75v3 + "3 = 5V3 I \13 .2.

The c haracteristic polynomial of A is (1 + I )(1 - 3), so A has eigenvalues - I and 3. Since each eigenvalue of A has mull iplicity I, A is diagonali:wble. In the usual manner, we find that

{[- :]}

= Yt(O) = -ae0 - be310> = - a - b = Y2(0) = ae 0 + 2be 3<0> = a + 2b . is easily checked that a = -6 <~nd b = 5. so I

4

=

2. (a) The matrix form of the given system of differential

y

( b) In order to satisfy the initial conditions y 1(0) = I and y2(0) = 4. the constants a and b mu st satisfy the system of linear equations

)'2

=

= Ay.

31

-

)'I

So \It = .45. v2 = .35 , and v3 = .2. Thus we expect that, in the long run, 45% of those surveyed will drive cars. 35% mini vans. and 20% sport uti lity vehicles. ct}ual ions is y'

1

"' = -aebe = ae _, + 21Je31 .

~2

=

= [ :~]

must have v 1 + v2 + v3

=0

that is .

=

J2 ;;' 2..;2

and

b2

=

J2 - J J2 . 2

Thus. in general.

= Pz = [ -II


Page 7 of 10



CHAPTER 5

REVIEW EXERCISES

~ In E.lt'rcisn I 17. dl!tl!nninl' ll'ill'tilu tht' sraremmrs ~ are tme or falst'. I. A scalar >. IS an eigenvalue of an only if det (A- H 4 ) 0.

=

2. If>.

cigcmector of 1he m:llrix th:ll corrc~pond' 10

>..

4. The cigcn,p:1cc of an 11 x n matrix A corresponding 10 cigem'lllue >. is the null >pace of A- AI•.

5. The eigenvalues of a linear operator on

x

11

malrix has

n"

9. Every diagonali zable values.

11

25.

[-~

has real eigenvalues.

distinct eigenvalues.

x

11

malrix has

11

27.

29. T

x

malrix and D is a diagonal 11 x 11 m:urix 'uch thai A= p -' DP. then 1he columns of P forrn a basis for n• con,i>ting of eigenvectors of A.

n•

i' diagonalinble if and only if i1s slandard ma1rix i> diagonalizable.

17. lfO is an eigenvalue of a matrix A, then A is not invertible.

[~:]) = [3x 3:rl+ ~:1

] .\J

3.r,] 2.\J

[_~ ~]

[~

0 c

31.

32.

0

- (t - c)(t - 2)(r - 3)

[~

33.

-~] 10

- I

-10 12

[-!

3-1.

-~]

c

-4

2

- (t - t·)(t

-(t - <")(r - 2)(1 + 2)

sion of the eigenspace corresponding to >. i> the nullity of

A- H •.

j]

In E.rercise.r 31 - 34. o matrix and it.s clrmnctuistic polynomial are gi1•en. Determine ttl/ wt/111!.< of tlrl' scalar c for ll'lriclr l!aclr matrix is trot diogmwli:.ttbll!.

15. If>. is an eigenvalue of an 11 x 11 matrh A. then the dimen-

16. A linear opemtor on

0 2 -3

= [;.~,--~:~]

XJ

11

14. If P is an Invertible 11 x 11 matrix and D is a dia2onal 11 x n m31rix such that A = PDP 1. then the eigen,~alues of A are the diagonal entric• of D .

[ -2 -4 4

26.

XJ

12. An

11

-~J

([:~:]) = [ -3:rl +~::3.t2 -

30. T (

11 x 11 matrix A is diagonalizablc if and only if there i> a b:l'i' for n" COil>i>l ing of eigenveCIO" of A.

- I

= [~:~ 1+_2.;:2 ]

T ([:::])

disli ncl eigen-

II. The muhiplicily of an eigenvalue need not equal the dimension of the corre,pondi ng cigenspllcc.

24. [- 1

l11 E.rurisu 27- ·30. a lint'ar Ol>t'rmor T on n• is gi•·m. Find. ifpouiblt!, a basis for n• cmuisting of eigrmet·wrr ofT. Ifno such basis exists, t'xploin wiry.

are the same

11

_!]

-I

28. T ([:::])

n•

10. I f two 11 x 11 matrices have the same c haracteristic polynomial. then lhcy have the sumc c igenvcclors.

13. If P i> un 1nvcniblc

0 0

n· are the same

as 1hose of ils Sl3ndard matrix.

6. The eigenspaccs of a linear operator on as those ol i1s s landurd mutrix. 11

[-~ ~]

a unique

3. If v is an eigen,eetor of a matrix. then there is a unique eigcm':lluc of 1he matrix that corresponds to v.

7. Every linear opemtor on

23.

x rr matrh A if and

11

i~ an eigenvalue of a matrix. then there is

8. Every

35 5

+

I

0

I )(1 - 2)

~]

-(t - c)(t -

2) 2

111 EXt'rcises 15 and 16. a matri.\ A rs giwm. Fmd At for an arbitrary positil'l' illl<'/lt'r k.

35.

[~

=!]

36. [

-:~ -~]

1

18. Show lhnl [

-~

37. Lei T be 1hc linear oper:Hor on R. defined by _;] has no real eigenvalues.

5 19. [ - 2 21.

[-~

~]

20. [ II

0 - I

22.

0

[-:

- 1

=~J


-

-

Find a basis l3 such 1h:11 ITie is a diagonalmalrix.

0

In Eurdus 23 26, a matri.1 A is xil·en. Find, if 1>0ssible,

.12

38. Find a 3 x 3 mmrix having e igenvalues - I, 2, and 3 with corresponding CigenvCCIOrS [ -

0 - 1

on im•ertible matrix P and a diagonal maTrix D such That A = PDP 1. If no such matrices e.1ist. explain ll'lr)'.

' ']) _[-4. ,-_3.,·'l 3"J] · ([:r3 6.x, + 6.rz + 5x3 2

T

111 Exercises 19 -22, dett!l'llline tire eigem•ttlues of each matrix wul a basis for each eigempaa.

39.

~ovc [~ th:ll

a

0

:l [-n n Wid [ -

~] is nol diagonaliLable for m1y

scalars a and b.

Page 8 of 10


3 5 6 CHAPTER S Eigenvalues, Eigenvectors, and Diagonalization 40. Suppo>e that A i; an

11 x 11 matrix having two distinct eigenvalues. AI and A!. where At has multiplicity I. State and prove a nece;sary and ;ufficienl condition for A to be diagonali7-able.

-II. Prove that 1. A i> mveruble tf and only if I is nor an eigenvalue of A .

-12. Two

matrice' A and 8 are called simultant!ously diagot~ali:oblt! if there exists an invenible matrix P such that both P 1AP and P 1 BP are diagonal matrices. Ptove x

11

11

that if A and 8 are 'imultaneously diagonalitable. then A8 8A.

=

43. Let T be a linear operator on 'R.". 6 be a basis for 'R.", and A be the standard matrix ofT. Prove that [T] 6 and A have the same charocterisuc polynomial.

-14. Let T be a linear operator on n•. A subspace II' of n• is called T -itwariaflf if T(w)" in IV for each win IV. Prove that if V is an eigenspace of T. then I' is T -invariant.

CHAPTER 5 MATLAB EXERCISES For the following exercises, use MA TLAB (or comparable software) or a calculator with matrix capabilities. The MATLAB functions in Tables 0.1, 0.2, 0.3, 0.4, and 0.5 ofAppendix 0 may be useful. I. For each of the following matrices A. find. if possible. an invenible matrix P and a diagonal matrix 0 such that A PDP - 1• If no such matrices cx i'l. explain why.

=

(a)

[-~

(d)

I

2 3

-2 6 4

[ [

-7

6

II

-7

(c)

14

-2 -2 8

-2

-~

(b)

5

18

-3 lO

4. A search engine consider;, only lO Web pages. which me linked in the following manner:

-l]

Page

LinkJ to p•g•s

2,S,7,and 10

II

I, 8.and9 7 and 10

3

Sand6

10

t.4. 7. and 9

6

=~~23

- 12

4, 7,and8

-2 7

1.l. S. and6

3-1

10

[=:

I 2

nol•nks

I 2

-I 0 I - I 0 2. Let A be an 11 x 11 matrix. and let I ~ i
A=

[-~1~ ~!~ ~~~ 107 -2 15

72 - 146

42 - 85

-30 -2 64

-~]

4 .

28

2

- 57

-5

9

2andS

10

landl

(a) Apply the PageRank algonthm to lind the transition matrix for the Marko' chain associated with this search engine, based on p 0.85. the pon1on of the time a surfer moves from the current page to the next page by randomly choming a linl. when one exists.

=

(b) Find the :.teady-Matc vector for the Markov chain obtained in (a). and use it 10 rank the pages.

5. For each of the following linc:tr operators T. find a basis for the domain of T consisti ng of eigenvectors of T. or explain why no such basis ex ists.

3. Find the matrix with eigenvectors

[il [ll Ul lll [1]

and corresponding eigenvalue> I. - I. 2. 3. and 0.


,r'] ) [ 12xt +xz + 13.\J + 12.r4 + 14xs 10.11 + ll.z + 6.1J + I Ln + 1xs .Q

(b)T

x, x,

([

.Is

=

l

.ll - 4x, - 3x, - 13.1t - l1z- llxJ- 14.r,- ll1s J1 1

l11

-'2

+ 2.tz- J13 + 51•

Page 9 of 10


357

Chapter 5 MATLAB Exercises 6. Let

(a) Find the rule for T. (b) Determine whether T is diagonalizable. If T ·~ diagonahlable. find a basis for 'R4 of etgenvector. ofT.

8. Let and let (u) Show that

6 = ( v~o "2· "J• v~). 6 is a basi< for 'R4 •

A=

(b) Find the rule for the un ique linear operato r T on such that T(' 'l)

= 2v1.

T(vl)

= J v2.

T (\'1)

= - \'\.

'R4

=

7. Let 6 be the basis gi,-en in Exercise 6. and let T be the linear operator on n• such that

= " 2·

T(v4)

= 2v4 .

T(v!)

=2v 1.


T(vJ)

= - v3,

0.5

0. 1

0.2 0.2 0. 1 0.3 0 .2

and

0.4 0.2 0. 1 0. 1 0.2

0.3 0. 1 0 .2 0. 1 0.3

'] 0.3 0.4 0. 1 0

and

pc

[!l

(a) Show that A is a regular transition matrix.

and

T (\'4) V3 - " •· (c) Determine whether T is diagonalizable. If T is diagonaliLable. find a ba'i' for 'R4 of e1gen\'ectors of T.

T(v 1)

['' 0 .2 0 .1

(b) Find the

Meady-~tatc

vector v for A.

(c) Compute Ap . A 1op. and A lOOp. (d ) Compare the last vector. A 100p, with 25v. E'platn. 9. Find the solution of the finite difference equation

'• - 3rn

1

+ 5rn-2 -

w ith initial conditions ro

,. = 3.

15rn- J - 4r,_.

+ 12r

11

5

= 5. rt = l. f! = 3. fJ = I. and

Page 10 of 10




Page 1 of 10


6

INTRODUCTION

Identity verification is increasingly important in our modern and highly mobile society. Applications of identity verification range from national secu rity to banking. Most of us no longer conduct transactions in person at a bank where the tellers know our names. Today it is not unusual for a person to obtain money at an Automated Teller Machine {ATM) far from where the money was deposited. Increasingly, biometric authentication is used to verify the identity of a person. Biometric authent ication is the automatic identification or identity verification of living humans based on behavioral and physiological characteristics. In contrast to common identification methods such as ID cards and Personal Identification Numbers (PINs), biometric identifiers cannot be lost or stolen and are more difficult to forge. The most commonly used biometric methods are: Fingerprints Hand Geometry Iris Recognition • Voice Recognition • Facial Recognition The earliest facia l recognition technique is the eigenfaces method, which is based on principal component analysis (see Section 6.8).

359


Page 2 of 10


360 6 Introduction Each facial image is a matrix of pixels in which every pixel is represented by a numerical value corresponding to its intensity. In the eigenfaces method, each image is converted initiall y to a single long vector (see the preceding figure on page 359). Principal component analysis. based on the eigenvalues and eigenvectors of the covariance matrix, yields a new and relatively small set of vectors (the eigenfaces) that captures most of the variation in the original set of vectors (images).


The image in the upper left corner of the preceding figure corresponds to the first principal component of an original image; the ghost-like images in the rest of the figure correspond to its other principal components. The original image can be expressed as a weighted sum of vectors (eigenfaces) from the principal component analysis. These weights provide the identity of the person: Different images of the same person have approximately the same weights, and the image of another person will have significantly different weights.

Page 3 of 10


CHAPTER

6

ORTHOGONALITY

ntil now. we have focused our attention on two operations with vectors, namely, addition and scalar multiplication. In this chapter, we consider such geometric concepts as length and perpendicularity of vectors. By combining the geometry of vectors with matrices and linear transformations, we obtain powerful techniques for solving a wide variety of problems. For example. we apply these new tools to such areas as least-squares approximation, the graphing of conic sections, computer graphics. and statistical analyses. The key to most of these solutions is the construction of a basis of perpendicular eigenvectors for a given matrix or linear transformation. To do this, we show how to convert any basis for a subspace of R" into one in wh ich all of the vectors are perpendicular to each other. Once this is done, we determ ine conditions that guarantee that there is a basis for 'R" consisting of perpendicular eigenvectors of a matrix or a linear transformation. Surprisingly. for a matrix, a necessary and sufficient condition that such a basis exists is that the matrix be symmetric.

U

[I!J

THE GEOMETRY OF VECTORS

In this section, we introduce the concepts of length and perpendicularity of vectors in 'R". Many familiar geometric propetties seen in earlier courses extend to this more genera l space. In particular, the Pythagorean theorem, which relates the squared lengths of sides of a right triangle, also holds in 'R". To show that many of these results hold in 'R" , we define and develop the notion of dot product. The dot product is fundamental in the sense that, from it, we can define length and perpendicularity. Perhaps the most basic concept of geometry is length. In Figure 6 .1 (a). an application of the Pythagorean theorem suggests that we define the length of the vector u

ur ui.

to be + This defin ition easi ly extends to any vector v in 1?." by defining its norm (length), denoted by llvll. by

II vii

= jv~ + vf + ·· · + v,~

A vector whose norm is I is called a unit vector. Using the definition of vector norm, we can now define the distance between two vectors u and v in R" as ll u - vii . (See Figure 6.1 (b).)

361


Page 4 of 10


362 CHAPTER 6 Orthogonality

u - y,.

······

,

... ···

....· ... .u

\\!1·~.- vii

X

(a) The length o f a vector ll in 'R.2

(b) The d istance between vectors

u and v in n_n

Figure 6.1

Example 1

Find Hu ll . ll v ll, and the distance between u and v if

II =

Solution

[i]

and

By defini tion,

and the d istance between u and v is

nu- v ii = J(J - 2)2 + (2 Practice Problem 1 .,..

( - 3))2

+ (3 -

0)2 =

J35.

Let

=[

u

-i]

and

(a) Compute Hu ll and ll vll. (b) Determine the distance between u and v. I 1 . and M v arc Ulllt vectors. (c) Show that both

w"

Just as we used the Pythagorean theore m in R 2 to motivate the definition of the norm of a vector, we use this theorem again to exam ine what it means for two vectors u and v in R 2 to be perpendicular. According to the Pythagorean theorem (sec Figure 6.2), we see that u and v are perpendicular if and only if

2

Uv - u ll = Uu ll

2+ Uvll 2

ud = /If+ uf + Vf + vf "f- 211tVt + u? +vi -2u2v2 + ui = llf + ui + Vf + vf - 2u 1v1 - 2uzv2 = 0 (lit - UJ)2


+ (112 -

Page 5 of 10


6.1 The Geometry of Vectors

····..

363

··...Uv - ull

·····.. ... ··. u

Figure 6.2 The Pythagorean theorem

The expression u 1v 1 + uzv2 in the last equation is called the dot product of u and v, and is denoted by u · v. So u and v arc perpendicular if and only if their dot product equals zero. Using this observation. we define the dot product of vectors u and v in R" by ll•\' =Iii

III+ 112112 + ... + """"'

=

We say that u and v arc orthogonal (perpendicular) if U·" 0. Notice that, in R" , the dot product of two vectors is a scalar, and the dot product of 0 with every vector is zero. Hence 0 is orthogonal to every vector in R". Also, as noted, the property of being orthogonal in R 2 and R 3 is equivalent to the usual geometric definition of perpendicularity.

'%fh.!.il•

Let

and

Detem1ine which pairs of these vectors are orthogonaL Solution

We need only check which pairs have dot products equa l to zero. u.v

= (2)( I) + ( -

li •W= \'•W

I )(4)+ (3)( - 2)

=- 8

(2)(- 8)+ (- 1)(3)+ (3) (2) = - 13

= ( J)(- 8) + (4)(3) + ( - 2)(2) = 0

We see that v and w are the on ly orthogonal vectors.

Practice Problem 2

~

Determine which pairs of the vectors

U=[=i]. \' =[-il

and

are orthogonal.


Page 6 of 10


364 CHAPTER 6 Orthogonality It is useful to observe that the dot product of u and v can also be represented as the matrix product u r v.

r

u v=[u, 112 ··· u,]

v'] =u,v, + uzvz+ · ·· + u,v,= u · v [v, vz :

Notice that we have treated the I x 1 matrix ur v as a scalar by writing it as 112112 + · · · + 11 11 v, instead of lui v, + 112112 + · · · + u, v,]. One useful consequence of ide ntifying a dot product as a matrix product is that it enables us to " move" a mat rix from one side of a dot producl to the other. More precisely, if A is an m x n matrix , u is in R", and ' ' is in R"', then

"' v, +

This follows because

Just as there are arithmetic properties of vector addition and scalar mu ltiplication, there are arithmetic properties for the dot product and norm .

THEOREM 6.1 For all vectors u , v, and w in R" and every scalar c. (a) (b)

U•U U•U

= llull 2 . = 0 if and only if u

= 0.

(c) U•V = V•ll. (d) u• (v + w)= u ·v + u• w.

(e) (v + w)• U =V •U + W•ll. (f) (cu) . v = c(u. v) = u. (cv). (g)

llcull

= lclllull.

We prove parts (d) and (g) and leave the rest as exercises. (d) Us ing matrix properties, we have

PROOF

u . (v + w) = ur( v + w) = ur v + ur w

= U•V + U•W. (g)

By (a) and (t). we have

Ucull2

= (c u) • (cu )

By taking the square root of both sides and using

llcull


=

lclllull.

JCi = lei, we obtain •

Page 7 of 10



365

Because o f Theore m 6. 1(f), there is no amb iguity in wri ting cu. v for any of the three expressions in (f). Note that, by Theorem 6.1 (g), any nonzero vector v can be normalized , that is.

I

transformed into a un it vector by mu ltiplying it by the sca lar - . For if u ll v ll then

1- 11

lln ll =

ll v ll

1

v ii=

=-1

llv ll

v,

1

lll v ll = ll vll = 1. ll v ll ll vll

This theorem allows us to treat expressions with dot products and norms just as we would algebraic expressions. For example, compare the simi larity of the algebraic result (2x + 3.d = 4x 2 + 12xy + 9y 2 with

The proof o f the preced ing equality re lies heavily on TI1eorem 6.1 : 112u + 3vll 2 = (2u = = = = =

+ 3v) • (2u + 3v) (2u) . (2u + 3v) + (3 v) . (2u + 3v) (2u) • (2u) + (2u) • (3v) + (3v) • (2u) + (3v) • (3v) 4(u ·ll) + 6(u • \') + 6(v ·ll) + 9(v • v) 2 by 4llu lf + 6(u· v) + 6(u • v) + 911 v ll 2 2 411 u ll + 12(u • v) + 911 v ll

by (a) by (e) by (d) by (f) (a) and (c)

As noted earlier, we can write the last expression as 4llu ll 2 + 12u. v + 911 v ll 2• From now on, we will om it these steps w hen computing with dot products and norms .

DCAUTION

Expressions such as u 2 and uv are nor defined. It is easy to extend (d) and (e) of Theorem 6. 1 to linear combinations, namely.

and

As an application o f these arithmetic properties. we show that the Pythagorean theorem holds in 'R".

THEOREM 6.2 (Pythagorean Theorem in n,n )

Let u and v be vectors in 'R" . Then u and v

are orthogonal if and only if


Page 8 of 10


366

CHAPTER 6 Orthogonality

PROOF Applying the arithmetic of dot products and norms to the vectors u and v, we have

Because u and v ru·e 01thogonal if and only if u· v diately.

= 0, the result follows imme•

ORTHOGONAL PROJECTION OF A VECTOR ON A LINE Suppose we want to find the distance from a point P to the line C given in Figure 6.3. lt is cleru· that if we can determine the vector w. then the desired distance is given by lin - wU. The vector w is cal led the or thogonal p rojection of u on C. To find w in terms of u and C. let v be any nonzero vector along C. ru1d let z u - w. Then w = cv for some scalar c. Notice that z and v are orthogonal; that is,

=

0 = Z• V = (U - W) • v = ( U - CV) • V = U • V - CV• V = U•V - Cll v ll 2 .

So c

U• V U •V = -, v. Therefore the distance from P to C is given by , and thus w = -ll vll 2 ll v ll -

Uu- wU= lu- ~vii· llvll 2

Figure 6.3 The vector w is the orthogonal projection of u on C..

Example 3

Find the distance from the point (4, I) to the line whose equation is y

Solution

Following our preceding derivation, we let and

Then the desired distance is


= 4x.

U • V

ll v ll 2 v

=

9[2]

S

1 ·

I[~] - ~ [nI = ~ I [-~J I = ~ -15.

Page 9 of 10



Practice Problem 3 Ill>

367

Find the orthogonal projection of u on the line through the origin with direction v, where u and v arc as in Practice Problem 2. ~ We rctlu·n to a more general case in Section 6.4. where we apply our results to solve an interesting statistical problem.

AN APPLICATION OF THE DOT PRODUCT TO GEOMETRY* Recall that a rhombus is a parallelogram wi th all sides of equal length. We use Theorem 6. 1 to prove the following result from geometry:

The diagonals of a parallelogram are orthogonal if and only lelogram is a rhombm.

if the paral-

The diagonals of the rhombus are u + v and u - ' '· (See Figure 6.4.) Applying the arithmetic of dot products and norms, we obtain

From this result. we see that the diagonals are orthogonal if and only if the preceding dot product is zero. l11is occurs if and on ly if llull 2 llvll 2 , that is, the sides have equal lengths.

=

Figure 6.4 The diagonals of a rhombus

THE CAUCHY- SCHWARZ AND TRIANGLE INEQUALITIES Recall that, in every triangle, the length of any side is less than the sum of the lengths of the other two sides. This simple result may be stated in the language of nonns of vectors. Refe1Ting to Figure 6 .5, we see that this s tatement is a consequence of the triangle inequaliry in 2 :

n

v

figure 6.5 The triangle inequality

' The remainder ofthis se,tion may be omitted wittlout loss of (Ontinuity. However, ttle Caudly- S,tlwarz and the triangle i nequalities are frequently used in latercourses.


Page 10 of 10


368


What we show is th at this inequality holds not just for vectors in R 2 • but also for vectors in R" . We begin with the Cauchy-Sclnvarz inequality.

THEOREM 6.3 (Cauchy-Schwarz lnequal ity1 )

For any vectors u and v in R ", we have

PROOF If u = 0 or v = 0, the result is immediate. So assu me that neither u nor v is zero. and let I I W = -U Z= V. and II vii !l u ll Then w.w = z,z = I, and hence

0 ::; ll w ± z ll 1 = (w± z) . (w ± z) = w.w ± 2(w.z) + z,z = 2 ± 2(w.z). It follows that ± w • z ::; I, and so Iw • z I ::; !. Therefore

• The case where eq uality is achieved is exami ned in the exercises. At the end of this section, we see an interesting application of the Cauchy-Schwarz inequality.

Example4

Verify the Cauchy- Schwarz inequality for the vectors

U =

Solution

[-!]

We have u • v = - 12 , !l u ll =

and

../29, and II v ii = J30. So

lu· vl2 = 144 ::; 870 = (29)(30) = ll u ll 1 · ll v ll 1. Taking squ are roots con firms the Cauchy- Schwarz inequality for these vectors.

Example 5 contains another consequence of the Cauchy- Schwarz inequality.

Example 5

1


The Cauchy- Schwarz inequality was developed independently by the French mathematician AugustinLouis Cauchy (1789- 1857), the German Amandus Schwarz (1843- 192 1), and the Russian Viktor Yakovlevich Bunyakovsky (1804- 1899). The result first appeared in Cauchy's 1821 text for an analysis course at the Ecole Polytechnique in Paris. It was later proved for functions by Bunyakovsky in 1859 and by Schwarz in 1884.

Page 1 of 10



Solution

By applying the Cauchy-Schwarz inequal ity to u = [

369

:~] and v = [~] ,

we obtain the desired inequality.

Our next result is the prom ised generalization toR" of the triangle inequa lity.

THEOREM 6.4 (Triangle Inequality)

PROOF

For any vectors u and v in R" . we have

Applying the Cauchy-Schwarz inequality. we obtain

Taking square roots of both sides yields the triangle inequality.

•

The case where equality is achieved is examined in the exercises.

Example 6

Verify the triangle inequality for the vectors u and v in Example 4.

Solution

ll u ll =

Since u + v = [

=~], it follows that ll u + v ii = J35. Recalli ng that

../29 and II vii = J30, the triangle inequality follows from the observation that

ll u + v ii=

J35 < J36 =

6 <5+ 5=

J25 + J25 < ./29 + J3ci =

ll u ll + ll vll.

COMPUTING AVERAGE CLASS SIZE A private school for the training of computer programmers advertised that its average class s ize is 20. After receiving complaints of false adve1t ising, an investigator for the Office of Consumer Affairs obtained a list of the 60 students enrolled in the school. He polled each student and learned the student's class size. He added these numbers and divided the total by 60. The result was 27.6, a figure significantly higher than the adve1tised number 20. As a result. he initiated a complaint against the school. However, the complaint was withdrawn by his supervisor after she did some work of her own. Using the same enrollment list, the supervisor polled all 60 students. She found that the students were divi ded among three classes. The first class had 25 students, the


Page 2 of 10



second class had 3 students, and the thi rd class had 32 students. Notice that the sum of these three enrollments is 60. She then divided 60 by 3 to obtain a class average of 20, confi rming the adve11ised class average. To see why there is a difference between the results of these two computations, we apply linear a lgebra to a more genera l situation. Suppose that we have a total of m students who arc d ivided into 11 classes consisting of v1, "2· •.•• v, students, respectively. Us ing this notation, we see that the average o f the chL~s s izes is given by _ I v = -

n

(vi +

112

m

+ · · · + v,) =

- . 11

This is the method used by the supervisor. Now consider the method used by the investigator. A student in the ith class responds that his or her class s ize is v; . Because there are v; students who give v[ that is contributed by the i th class. this response, the poll yields a sum of v;v; Because this is done for each class, the total su m of the responses is

=

Since m students are polled, th is sum is d ivided by m to obta in the investigator's " average." say v•. given by • 1 ? v = -(Vj' Ill

To see the relationship between

?

,

+ "2 + · · · +

v,;-) .

v and v•, we define the vectors u and v in R" -

and

V-

by

[~~]

..

v, Then U•U =11,

v. v = v~ + v} + · · · + v,;.

and

U•V =Jn,

Hence v

Ill

= -

II

U• V = --, li•U

v•

and

I 2 2 2 V•V = -(v 1 + v2 + · · · + v,) = --. li •V Ill

By the Cauchy-Schwarz inequal ity,

and so. dividing both sides of this inequality by (u. v)(u · u). we have

ii=

~ < ~ =v• li•U -

U •V

Consequently, we always have that il ~ v•. It can be shown that il"" v• if and only if all of the class sizes are equal. (See Exercise 124.)


Page 3 of 10



371

EXERCISES In Exercises 1- 8, rwo vecror., u and v are given. Compure rile norms of rile vecrors and rire disrance d berween rhem.

In Etercises 17- 24, rwo orrhofimrai vecrors u and v are given. Compure rile quamiries llull 2• llvll 2, and Jlu + v11 2. Use your resulrs ro illusrrare rhe Pyrlragorean rileorem.

m

17. u = [ -;J and v =

18. u =

m

and v = [-

19. u = [;]and v =

21. u

22. u

= [-

23.

and v = [

and v = [

u=

24. u =

i]

[i]

and v = [

-~]

[-In

andv=

[-!]

:]

and v = [

-~]

In Etercises 25- 32. rwo vecrors u and v are given. Compure rire quamiries JluJI, llvll. and Jlu +vii. Use your resulrs ro illusrrare rile rriangie inequaiiry.

25. u = 26. u = 27.

m m

29. u = [

30. u = [

31. u =

32.

and v = [

=~J

and v = [

_;J

u = [~Jandv=[-n

28. u = [


[~]

m _;J = nJ ~

20. u =

In Exercises 9- 16, rwo vecrors al'l! given. Compwe lire dol producr of rile vecrors. and detennine wilerher lire vecrors are orrlwgmral.

~]

-~]

m

-n -n [l] and v = [ :]

and v =

[-n

U= [

and v =

andv =

[~]

-~] and v ~~] = [

Page 4 of 10


372


In Exercises 33-40. two vectors u and v are given. Compllte the qualllities nu ll . Uvll. and u • v. Use your results w illustrate the Cauchy-Sclnrar~ inequality.

35. u

n"

-;J m = m = [!]

= [~]

= [-~J

65. The norm of a sum of vectors is the sum of the norms of the vectors.

[~]

66. The squared norm of a sum of 011hogonal vectors is the s um of the squared no•ms o f the vectors.

63. The norm of a vector equals the dot product of the vector with itself.

and v =

64. The norm of a multiple of a vector is the same multiple

and v

and v

36. u = [ - ! ]and v =

37. u =

[-!]

38. u = [ :] and ' ' = [

39. u =

of the norm of the vector.

[_~]

andv=

67. The orthogonal projection of a vector on a line is a vector that lies along the line. 68. The norm of a vector is always a nonnegative real number. 69. If the norm of ,. equals 0. then v = 0.

-!]

70. If u • v = 0. then u = 0 or v = 0.

[n [=:] -n j]

7 1. For all vectors u and v in

andv=

40. u = [

73. The distance between vectors u and ,. in (cu) • v

In Exercises 41- 48. a ••ector u and a line Lin n are given Compllle the orthogonal projection w of u on l. and use it to compute the distance d fmm tlw endpni111 of u ltJ C. 0

43. u = [!] and y = - x

45. u =

47. u =

[~] andy =

3x

m

42. u =

[~] anti y = 2.x

44. u =

[!]

is Uu - vii .

46. u = [

-;J

and y= x

= u • (cv).

75. For all vectors u. v. and w in

n".

U• (\' + w) = li•V + U•W. 76. If A is an 11 x 11 matrix and u and v are vectors in thenAu • v = u. Av.

n".

77. For every vector v in n", Uvll = II -vii . 78. If u and v are onhogonal vectors in n". the n

and y = -2.x

48. u = mand y=-4x

For Erercises 49- 54, suppose thlll u, v, and w are vectors inn" such that !l u ll = 2. nvll = 3. llwll = 5, U•V = - I. U•W =I. andv ·w = -4.

Uu +v ii= Uu ll + Uvll · 79. If w is the orthogonal projection of u on a line through the origin o f 2 , then u - w is orthogonal to every vector on the line.

n

the origin of to u.

n 2• then w is the vector on the line c losest

8 1. Prove (a) of Theorem 6. 1. 82. Prove (b) of Theorem 6. 1.

49. Compute (u + v) . w.

50. Compute 114wll .

83. Prove (c) of Theorem 6. 1.

51. Compute llu + vU2 •

52. Compute (u + w) • v.

84. Prove (e) of Theorem 6. 1.

II'' - 4 wll 2 .

54. Compute 112u + 3vU2 .

For Exercises 55- 60. suppose that u. v, and w are vectors inn" .mch 1/101 U oil = 14, U • V = 7, U o\1' = - 20, V • V = 21, v • w = -5, and w . w = 30. 55. Compute 11' '11

n•

80. If w is the orthogonal projection of u on a line through

and )' = -3x

53. Compute

[u • vi = nu ll • nvll .

74. For all vectors u and v inn" and every scalar c.

and v = [

[~] andy =

n",

72. For all vectors u and v in 'R.". u • v = v • u.

2

4 1. u =

is a vector in

n".

62. The dot product o f two vectors in

33. u = [ 34. u

61. Vectors must be of the same size for their dot product to be defined.

1

56. Compute n3u ll .

57. Compute v • u.

58. Compute w • (u + v).

59. Compute 112u - v U2.

60. Compute [[ v + 3wll .

In Exercises 61 - 80, determine whether the sratemellls are true or false.


85. Prove (f) of Theore m 6.1. 86. Prove that if u is onhogonal to both v and w. then u is orthogonal to every linear combination of v and w . 87. Let {v, w } be a basi s for a subspace W of n", and define \ ' • \V

z = w - - - v. V•V

Prove that {v. z} is a basis for II' consisting of orthogonal vectors.

88. Prove that the Cauchy- Schw
Page 5 of 10


6.1 The Geometry of Vectors 89. Prove that the triangle inequality is an equality if and only if u is a nonnegative multiple of v or v is a nonnegative multiple of u .

[i]

99. u = [ -nand v =

90. Use the lriangle inequalily 10 prove lhal I ll vii- ll w lll:::; ll v - w ll for all vectors ' ' and w in R!'.

100.

u = [~]andv=[-~]

91. Prove (u + v) . w = u • w + v • w for all veclors u , v. and w in R".

101.

u = [-~]andv =[-~]

92. Let z be a vector in R". Let W = (u e R": U• Z = 0}. Prove that W is a s ubspace of 7?.".

102. u = [ - :] and v =

[-~]

and v =

104. u =

[-!]

and v =

105. u =

[-~]

and v =

W = (u e 7?.": u •Z = 0 for all z inS} .

Prove that W is a s ubspace of R". 94. Let W denote the set of all vectm·s 1hat lie along 1he line with equalion y = 2x. Find a vee lor z in R 2 such lhat W = (u e 7?.2 : u • z = 0}. Juslify your answer. 95. Prove the parallelogram law for veclors in R":

106. u = 96. Prove that if u and v are orthogonal nonzero vectors in R". then they are linearly independent. 97 2 Let A be any m x 11 matrix.

n]

103. u =

93. Let S be a subset of 7?." and

[ nand V=

373

[l]

[-~] [-~]

[l]

Ler u a11d v be vectors in 7?.3 . Deji11e u x v to be the vecTor

[::~:~ =::~:~]. IIJ V2 -

which is called rite cross product of u and v.

111V J

(a) Prove 1ha1 ATA and A have the same null space. Hi11t: Let v be a vector in 7?." such that ATAv = 0. Observe lhal ATAV • V =A v· Av = 0.

For Exercises 107- 120. use tile preceding definition of the cross producT.

(b) Usc (a) to prove that r.mk ATA = rank A.

107. For every vector u in 7?.3. prove lhat u x u = 0. 108. Prove that u x v = -(v x u) for all vectors u and '' in

nJ.

!lull

u

109. For every vector u in 7?. 3. prove that u x 0 = 0 x u = 0. 110. For al l vec1ors u and v in 7?.3, prove thai u and v are parallel if and only if u x v = 0 . I I I. For all vec1ors u and'' in 7?.3 and all scalars c, prove 1ha1 c(u x v) = c u x v = u x c v.

Figure6.6 98.3 Let u and v be nonzero vectors in n 2 or n 3, and lei () be the angle between u and v. Then u, v. and v - u delermine a triangle. (Sec Figure 6.6.) The relationship between the lengths of the s ides of this 1riangle and () is called the law of cosi11es. It states lhat

112. For all vcclors u . v. and w in 7?. 3. prove that ll X (V +

w)

= II X V +

II X \\'.

113. For all vectors u , v, and win 7?. 3, prove that ( u + v)

X \1'

=

U X W

+VX

\1'.

114. For all veclors u and ,. in 7?. 3. prove thai u x v is orthog· onal to bolh 11 and v. Use I he law or cosines and Theorem 6. I 10 derive I he formula U•V = ll u llll vJJ cosO.

/11 Exercises 99-/06, use the formula ill Exercise 98to detennine the a11gle betwee11 the vectors u a11d v. 2

3

115. For all vectors

11,

v, and win 7?. 3, pmve 1hat

(u x v) . w

= u. (v x w).

116. For all veclors u . v, and w in 7?. 3, prove thai u x (v x w) = (u • w)v- (u • v)w.

This exercise is used in Section 6. 7 (on page 439). This exercise is used in Section 6.9 (on page 471).


Page 6 of 10


3 74


11 7. For all vectON u.

1',

and w in

n 3' prove that

!11 Exercise 125, 11se either a calculator" ith matrix capabilities or com1'111er software .wch a.r MATL.AB to sol\•e the problem.

(u x v) x w = ( w. u)v - (w. v)u. 118. For all vectON u and v in Uu x " '

2

125. In every triangle. the length of any side is less than the sum of the lengths of the other two sides. When this observation is gcncraliu:d ton·. we obtain the triallgl~ inequality (Theorem 6.4 ). "hich "•Jte'

n J_ prove that 2

2

2

= 1Ju ll 1vll - (u • v) .

119. For all vectors u and v in 'R. 3 • prove that u X ' 'II= t u D' 'll smll. where ll 1s the angle between u and ''· Him: u~c Excrci;cs 98 and 118. 120. For all \'CCtors u. v. and w in R.3 • prove the Jacobi idmtity: (U X v) X W

+ (\' X w ) X U + ( W X U)

X \"

=0

£.1erdw< 121 124 rt'fl'r wtht' IIJ>(IIit"atitm regartli11g the two

meth{)(ls of comp11ti11g tl\'eragl' class si:e given i11 this sectio11. !11 Eurctses 121 123, tlalll a~ give11 for smdems enrolled i11 a thru-uction semilltll" co11rse. Comp11te the average determilletl!Jy the .wpe111isor and the average '' • determi11eti/Jy the illl'e.nigator.

[u + v for any vectors u nnd ,. in

122. Section I contains 15 students. and each of sections 2 and 3 cont:lin' JO st udent•. 123. Each of the three section> contains 22 students. 124. Use Exerci>e 88 to prove that the two averaging methods for determining clas> >ize are equal if and only if all of the class ,izes are equal.

l ull+

"2

=

II,.

n•. Let

"~ [il ·-[=~]

v

12 1. Section I cont:lin> 8 student,, section 2 contains 12 stu dents, and sectio n 3 contains 6 students.

~

v, =

[

2.01 ] 4.01 6.01 .

and

8.01

3.0 11] 6.0 9.0 1 . [ 12.0 1

(a) Verify the triangle inequality for u and ''· (b) Verify the tri:mglc inc4uality for u and v 1 .

(c) Verify the triangle inequality for u tmd v2. (d) From what you have observed in (b) and (c). make a conjecture about when ec1uality occurs in the triangle inequality. (e) lmerpret your conjecture in {tl) geometrically in 2•

n

SOLUTIONS TO THE PRACTICE PROBLEMS I. (a) We have Uu ll = J1 2 + ( 2)2 + 22 = 3 and Dvll =

J62

+ 22 + Jl = 7.

(b) We have u - vt =

2. Taking dot products. we obtain u. \' = (-2)(1) + (-5){-1)+ (3)(2) = 9

I[=nI

U•W V•W

= (-2)(-3)+(-5){ 1)+ (3)(2) = 7 ={ 1)(-3) + (- 1){ 1) + (2){2) = 0.

So u and w arc orthogonal. but u and v are not orthogonal. and v and w arc not orthogonal.

= J(-5)2 + (-4)2 + (-1)2 = J4i. {c) We have

3. Let w be the required orthogonal projection. Then

2)(1)+( 5)( 1) + (3)(2) [ - :] (2

+ (- I )2 + 22

2

and

ln!,·H: [m ~ [!] =/~+~+~= !. 49

49

49

~

= [

-n.

16.2 1ORTHOGONAL VECTORS It is easy to extend the property of orthogonality to any set of vectors. We say that a s ubset of is an orthogonal set if every pair of distinct vectors in the set is orthogonal. The subset is called an orthonormal set if il is an onhogonal set consisling entirely of unit vectors.

nn


Page 7 of 10


6.2 Orthogonal Vectors 375 For example, the set

is an orthogonal set because the dot product of every pair of distinct vectors in S is equal to zero. Also, the standard basis for 1?." is an oJthogonal set that is also an orthonorma l set. Note that any set consisting of just one vector is an orthogona l set. Practice Problem 1 ...

Determine whether each of the sets

and

is an oJthogonal set. Our first result asserts that m most circumstances orthogonal sets are linearly independent.

THEOREM 6 .5 Any oJthogonal set of nonzero vectors is linearly independent. Let (v 1, v2, ... , vk) be an orthogonal subset of 1?." consisting of k nonzero vectors, and let c ,. c2, .... c* be scalars such that

PROOF

Then, for any

V;,

0=

we have

O·V;

= C; (v; • V;) = c;llvd l2 . But llv; 11 2 # 0 because v; are linearly independent.

# 0, and hence c; = 0. We conclude that v,. v2, ... , v* •

An oJthogonal set that is also a basis for a subspace of 1?." is called an orthogonal basis for the subspace. Likew ise, a basis that is also an orthonormal set is called an orthonormal basis. For example, the standard basis for 1?." is an orthonormal basis for 1?.". Replacing a vector in an oJthogonal set by a scalar multiple of the vector results in a new set that is also an orthogona l set. If the scalar is nonzero and the orthogonal set consists of nonzero vectors, then the new set is linearly independent and is a generating set for the same subspace as the ori gin al set. So mu ltiplying vectors in an orthogonal basis by nonzero scalars produces a new oJthogonal basis for the same


Page 8 of 10



subspace. (See Exercise 53.) For example, consider the orthogonal set {v 1, vz, v3}, where and

Th is set is an orthogonal basis for the subspace W =Span (v 1, vz, v3). In order to el iminate the fractional components in v3. we may replace it with the vector 4v3 to obtain another orthogonal set {v 1, ' 'z,4v3) , which is also an orthogonal bas is for W . ln particular, if we normalize each vector in an orthogonal basis. we obtain an orthonormal basis for the same subspace. So

is an orthonormal bas is for W. If S = (v 1• v2. .... vk} is an orthogonal basis for a subspace V of 1?..", we can adapt the proof of Theorem 6.5 to obtain a simple method of representing any vector in V as a linear combination of the vectors in S . Th is method uses dot products, in contrast to the tedious task of solving systems of linear equations. Consider any vector u in V , and suppose that

To obtain c;, we observe that

=c;(V; •V;)

and hence

U •V;

c· - - -2

' - llv;ll

.

Su mmarizing, we have the following result:

Representation of a Vector in Terms of an Orthogonal or Orthonormal Basis Let {v 1, vz, ... , Vk) be an orthogona l basis for a subspace V of 1?..", and let u be a vector in V . Then

If, in add ition, the orthogonal basis is an orthonom1al basis for V, then


Page 9 of 10


6 .2 Orthogonal Vectors

1$if!,j,)tji

Recall that S

377

{v1. v2, v3 }. where and

is an orthogonal subset of 1?.3 . Since the vectors in S are nonzero, Theorem 6.5 tells us that S is linearly independent. Therefore, by Theorem 4.7, Sis a basis for 1?.3 So S is an orthogonal basis for R 3 . Let u =

[~]· We now use the method previously described to obta in the coeffi-

cients that represent u as a linear combination of the vectors of S. Suppose that

Then li •VJ 10 C t = -2- = -.

UvJij

14

and

The reader should verify that


Let

(a) Verify that S is an orthogonal basis for 1?.3 . (b) Let u

= [ -~l

Use the preceding boxed formula to obtain the coefficients that

represent u as a linear combination of the vectors of S.

Since we have seen the advantages of using orthogonal bases for subspaces, two natural questions arise: 1. Docs every subspace of R" have an onhogonal basis?

2. If a subspace of R" has an otthogonal basis, how can it be found? The next theorem not onl y provides a positive answer to the first question, but also gives a met hod for converti ng any linearly independent set into an Ot1hogonal set with the same s pan. This method is cal led the Gram - Schmidt (ortlwgollalizatioll) process. 4 We have the foll owi ng important consequence: Every subspace of R" has an orthogonal and hence an orthonormal basis.

4

A modification of thi~ proc

Page 10 of 10



Figure 6.7 The vector w is the orthogonal projection of u 2 on the line C through v 1.

The Gram-schmidt process is an extension of the procedure in Section 6.1 for finding the orthogonal projection of a vector on a line. Its purpose is to replace a basis {UJ, Uz, .... uk} for a subspace W of R " with an 01thogonal basis {VJ, Vz, . ... vk} for W. For this purpose, we can choose Vt = u 1 and compute Vz using the method outl ined in Section 6.1 . Let w = liz · v~ v 1, the orthogonal projection of Uz on the line Uvt ll C through v 1, and let v 2 = liz - w. (See Figure 6.7.) Then ' ' • and v2 are nonzero orthogonal vectors such that Span {vJ. v2} Span (ll 1, u2lFigure 6.8 visually suggests the next step in the process, in which the vector ll3 is replaced by VJ . See if you can re late this figure to the proof of the next theore m.

=

·····

v, Figure 6.8 The construction of v3 by the Gram -Schmidt process

THEOREM 6.6 (The Gram- Schmidt Process 5 )

of R". Define Vt

Let (u 1, uz, ... , uk} be a bas is for a subspace W

= UJ,

Uz • VJ vz = uz - - - v 1,

uv. n2

U) • Vt

V3 = UJ - ~ V t -

U) •Vz Uvzll 2 Vz.

s The Gram- Schmidt process first appeared in an 1833 paper by the Danish mathematician Jorgen P.

Gram (1850- 1916).A later paper by the German mathematician Erhard Schmidt (1876- 1959) contained a detailed proof of the result.


Page 1 of 10


6.2 Orthogonal Vectors 379

Then (v 1• v2, .... v; ) is an orthogonal set of nonzero vectors such that Span (v, . v2.... , v;) =Span (u , , u 2.... , u; ) for each i. So {v 1, v2, .... vk} is an otthogonal basis for W.

s:

Fori= 1,2, .. .. k, letS;= {u t, U2..... u;) and = (v 1, v2.... , v;}. Each S; consists of vectors in a bas is for W , so each S; is a linearly independent is an orthogonal set of nonzero vectors such that set. We first prove that Spans; = SpanS; for each i. Note that v , = u , :/= 0 because u 1 is in S , . which is linearly independent. = (vd is an orthogona l set of nonzero vectors such that Spans;= Therefore SpanS1• For some i =2,3, .... k, suppose that s;_, is an orthogonal set of nonzero vectors such that Spans;_,= SpanS;- 1 • S ince PROOF

s;

s;

V;

=

U; • Vt

Ul • V2

U; • Vl- 1

ll vlll 2

ll v211- -

llv;-dl-

U; - - -V I - - - . , V? - · ·· - - - - . , Vi -I ·

v; is a linear combination of vectors contained in the span of S;. Furthermore, v; ¥= 0 because, otherwise, u; would be contained in the span of S ;_ 1, which is not the case because S; is linearly independent. Next, observe that, for any j < i ,

U 1 • Vj

- --v ·· II vi 11 2 1

= U; • Vj

-

U; • V;-1

V· -

1

·· · -

- - - v· 1 . v· ll v; - t ll 2 , _ 1

U; • Vj

=0.

s;

It follows that is an orthogonal set of nonzero vectors contained in the span of S;. But is linearly independent by Theorem 6.5. and so is a basis for the span of S; by Theorem 4 .7. Thus SpanS;= Span S; . In particular, when i = k, we see that Spans; is a linearly independent set of orthogonal vectors such that SpanS~ = Span Sk = W . That is, is an 011hogonal basis for W. •

s;

s;

s;

Example 2

Let W be the span of S = (u 1, u z, u3}, where

and

are linearly independent vectors in 1?.4 . Apply the Gram-Schmidt process to S to obtain an orthogonal basis S' for W.


Page 2 of 10



Solution

Let

and

v

3

=

11

3

_

~

11

v _ 3 ' v2 V ? ll v dl2 I ll v211 2 -

~

= [:2] - 4 [ :](-2 I) [ l I

I

~] = ~ [- :] ·

1 0

4

I - 1

Then S' = {v , , vz. v3) is an orthogonal basis for W .


Let

w

~ S~·

I[ ~ll '[j]I' =1] '[

Apply the Gram-Schmidt process to obtain an otthogonal basis for W .

Example 3

Find an orthononnal basis for the subspace W in Example 2 , and write

as a linear combination of the vectors in this basis.

Solution

As noted previously, normalizing v 1, v2. and

VJ

produces the orthonormal

basis

{w, , w,, w,}

~ I i [1]';, [-1] 'i [~il l

for W. To represent u as a linear combination of w 1• wz. and w 3 . we usc the boxed formula on page 376. Since

13

li • W t


= z•

and

U •W3

= 2'

Page 3 of 10


6 .2 Orthogonal Vectors

381

we see that

13

(-3)

= 2 w' + J2 Practice Problem 4 .,.

1

w2 + 2w3.

Let

[_: -~

A= - 1 -3 I

5

~] 3 - 4

(a) Use the result of Practice Problem 3 to find an orthonormal basis B for the column space of A. (b) Let

U=[- 10~] • which is in the column space of A. Write u as a linear combination of the vectors ~ in the basis B.

THE QR FACTORIZATION OF A MATRIX* Although we have developed methods for solving systems of linear equations and finding eigenvalues, many of these approaches suffer from roundoff errors when applied to larger matrices. It can be shown that if the matrix under consideration can be factored as a product of matrices with desirable properties, then these methods can be modified to produce more reliable results. ln what follows, we discuss one of these factorizations. We illustrate the process with a 4 x 3 matrix A having linearly independent columns a 1, a 2, a 3. First, apply the Gram-Schmidt process to these vectors to obtain orthogonal vectors v,, v2, and v3. Then normalize these vectors to obtain the orthonormal set (w1, w2, w3). Note that a 1 is in Span {a t}= Span {wt), a2isin Span(a,, a 2)=Span(w,,w2). and a 3 is in Span (a, , a2, a 3) =Span {w,, w2, w3). So we may write



Page 4 of 10


382 CHAPTER 6 Orthogonality for some scalars r11 , r12 , r22, r13 , rn, r33 . Define the 4 x 3 malrix Q Note that we have

a, = Q

[r

11 ]

~

,

[rp]

a2 = Q r~2

,

and

a3 =

Q

= [wJ

w2 w3].

[~~~] . 1)3

If we lei and

r3 =

[~~~] , r33

then we have

where R is lhe upper triangular matrix YJJ

R=

0

[ 0

Notice that, by the boxed result on page 376. we have r;j = aj . w; for all i and j. This resull can be ex lended 10 any m x 11 malrix with linearly independenl columns. that is. with rank 11.

The QR Factorization of a Matrix Lei A be an

matrix with linearly independent columns. There exists an R"', and an 11 x 11 upper triangular matrix R such that A QR. Furthermore, R can be chosen to have positive diagonal enlri es.6 111

x

11

m x n matrix Q whose columns fonn an orthonormal set in

=

In general, suppose that A is an m x 11 matrix with linearly independent columns. Any factorization A = QR , where Q is an 111 x n matrix whose columns form an orthonormal set in R"' and R is an n x 11 upper triangular matrix, is called a QR factorization of A. It can be shown (see Exercise 66) that for any QR factorization of A, the columns of Q form an orthonormal basis for ColA.

Example4

Find a QR factorization of the matrix

Solution

6


We begin by letting

See Exercises 59-61.

Page 5 of 10


6.2 Orthogonal Vectors

383

To find vectors w 1, w 2. and w 3 w ith the desired prope11ies, we use the resu lts of Examples 2 and 3. Let

1] 2 [_: . I

and

w3 =

- 1

So

As noted, we can quickly compute the entries of R :

So

The reader can verify that A = QR.


Find the QR factorization of the matrix A in Practice Problem 4 .

AN APPLICATION OF QR FACTORIZATION TO SYSTEMS OF LI NEAR EQUATIONS

=

Suppose we are given the consistent system of linear equations Ax b, where A is an QR be a QR factorization of A. Using the result that Q r Q = I, (shown in Section 6.5), we obtain the equivalent systems

m x n matrix with linearly independent columns. Let A

=

Ax= b

QRx

=b

QrQR x =Q 7 b J, Rx = Q7 b Rx

= Q7 b .

Notice that , because the coefficient matrix R in the last system is upper triangular, this system is easy to solve.


Page 6 of 10



IJifi,!.)fJW

Solve t he system

+ 2r2 + X3 = I + X2 + X3 = 3 x1 + 2 x3 = 6 X t + X2 + X3 = 3. XJ Xt

Solution

T he coefficient matrix is

i]

A=[! :

Us ing the results of Example 4, we have A= QR , where

Q=

[l -~ j]

So an equivalent system is Rx

2

h

0

~] . I

2

= Qr b. or 2r2 +

2r t +

¥

~x3 =

.fixz - :l,}-x3 =

-¥ I

2·

Solving the third equation. we obtai n X3 = I. Substituti ng this into the second equation and solving for x2, we obtain

=

or xz - 2. Finall y, substituting the values for x3 and x2 into the first equation and solving for Xt gives

2r t + 2(-2)+

5

13

2 = 2'

or x 1 = 4. So the solution of the given system is

[-n. Practice Problem 6 ..,..

Usc the preceding method and the solution of Practice Problem 5 to solve the system Xt -

X2

+

X3

=

- x1 + 3x2 + 6x3 =

- xt - 3x2 + 3x3 XJ


+ 5xz -

=

6 13

10

4x3 = - 15.

Page 7 of 10


6.2 Orthogonal Vectors

385

EXERCISES

26. 28. 30. 32.

v SpanS

In £rercises 17- 24, an orrlwgonal ser Sand a veeror in are r:iven. Use dot products ( 1101 systems of linear equations) to reprrtsent v as a linear combination of tile vectors in S. 17.

33. Exercise 25, b

= [ -~]

35. Exercise 27. b

=

37 """'" "

- I I] -20 [ -13

b= [

Exercise Exercise Exercise Exercise

I0 12 14 16

34. Exercise 26, b

= [ -~]

36. Exercise 28. b

=

[ -19 13] -2

-l]

S = WJ.[-~]}andv = [!]

18. S

=I[-:].[:]}and ''=[-~J


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Elementary Linear Algebra: A Matrix Approach, 2nd Edition, page: EP-1 No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.


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Elementary Linear Algebra 2nd Edition

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