CHAPTER 7 LEAST SQUARES SOLUTIONS TO LINEAR SYSTEMS
7. LEAST SQUARES SOLUTIONS TO LINEAR SYSTEMS Objectives The major objectives of this chapter are to study least squares solution to the linear system problem Ax = b. Here are the highlights of the chapter.
Some applications giving rise to least squares problems (Sections 7.2 and 7.5). A result on the existence and uniqueness of the solution of least squares problems (Theorem 7.3.1 of Section 7.3). Computational methods for least squares problem: The normal equations (Section 7.8.1) and the QR factorization (Section 7.8.2) methods for the full-rank overdetermined problem; the QR factorization with column pivoting algorithm for the rank-de cient overdetermined problem (Section 7.8.3); and the normal equations (Section 7.9.1) and the QR factorization methods (Section 7.9.2) for the minimum-norm solution of the full-rank underdetermined problem.
Sensitivity analyses of least squares and generalized inverse problems (Section 7.7). Iterative Re nement procedure for least-squares problems (Section 7.10).
Background Material Needed for this Chapter The following background material and the tools developed in previous chapters will be needed for the smooth reading of this chapter. 1. The Cholesky factorization algorithm for solving a symmetric positive de nite system (Section 6.4.7). 2. The QR factorization using the Householder and Givens methods (Algorithms 5.4.2 and 5.5.2), The QR factorization of a non-square matrix. 3. Orthogonal projections (Section 5.6). 4. The QR factorization with column pivoting (Section 5.7). 5. The Iterative re nement algorithm from Chapter 6 (Algorithm 6.9.1). 6. Perturbation analysis of linear systems (Section 6.6). 388
7.1 Introduction In Chapter 6 we discussed several methods for solving the linear system
Ax = b; where A was assumed to be square and nonsingular. However, in several practical situations, such as in statistical applications, geometric modeling, signal processing etc., one needs to solve a system where the matrix A is non-square and/or singular. In such cases, solutions may not exist at all; in cases where there are solutions, there may be in nitely many. For example, when A is m n and m > n, we have an overdetermined system (that is, the number of equations is greater than the number of unknowns), and an overdetermined system typically has no solution. In contrast, an underdetermined system (m < n) typically has an in nite number of solutions. In these cases, the best one can hope for is to nd a vector x which will make Ax as close as possible to the vector b. In other words, we seek a vector x such that r(x) = kAx ; bk is minimized. When the Euclidean norm k k2 is used, this solution is referred to as a least squares solution to the system Ax = b. The term \least squares solution" is justi ed, because it is a solution that minimizes the Euclidean norm of the residual vector and, by de nition, the square of the Euclidean norm of a vector is just the sum of squares of the components of the vector. The problem of nding least squares solutions to the linear system Ax = b is known as the linear least squares problem (LSP). The linear least squares problem is formally de ned as follows:
Statement of the Least Squares Problem Given a real m n matrix A of rank k min(m; n), and a real vector b, nd a real n-vector x such that the function r(x) = kAx ; bk2 is minimized. If the least squares problem has more than one solution, the one having the minimum Euclidean norm is called the minimum length solution or the minimum norm solution. This chapter is devoted to the study of such problems. The organization of the chapter is as follows. In Section 7.2 we show how a very simple business application leads to an overdetermined least squares problem. In this section we simply formulate the problem as a least squares problem and later in Section 7.5 we present a solution of the problem using normal equations. 389
In Section 7.3 we prove a theorem on the existence and uniqueness of solution of an overdetermined least squares problem. In Section 7.7 we analyze the sensitivity of the least squares problems due to perturbations in data. We prove only a simple result here and state other results without proofs. Section 7.8 deals with computational methods for both full rank and rank-de cient overdetermined problems. We discuss the normal equations methods and the QR factorization methods using Householder transformations, modi ed Gram-Schmidt and Classical Gram-Schmidt orthogonalizations. Underdetermined least squares problems are considered in Section 7.9. We again discuss here the normal equations and the QR methods for an underdetermined problem. In Section 7.10 an iterative improvement procedure for re ning approximate solutions is presented. In Section 7.11 we describe an ecient way of computing the variance-covariance matrix of a least squares solution, which is (AT A);1.
7.2 A Simple Application Leading to an Overdetermined System Suppose that the number of units bi of a product sold by a company in the district i of a town depends upon the population ai1 (in thousands) of the district and the per capita income ai2 (in thousands of dollars). The table below (taken from Neter, Wasserman and Kunter (1983)) compiled by the company shows the sales in ve districts, as well as the corresponding population and per capita income. District Sales Population Per Capita Income
i
bi
ai
ai
1
2
1 162 274 2450 2 120 180 3254 3 223 375 3802 4 131 205 2838 5 67 86 2347 Suppose the company wants to use the above table to predict future sales and believes (from past experience) that the following relationship between bi, ai1 , and ai2 is appropriate:
bi = x + ai x + ai x : 1
1
2
2
3
If the data in the table has satis ed the above relation, we have 162 = x1 + 274x2 + 2450x3 390
120 223 131 67
x x x x
+ 180x2 + 3254x3 1 + 375x2 + 3802x3 1 + 205x2 + 2838x3 1 + 86x2 + 2347x3
= = = =
1
or
Ax = b;
where
01 BB 1 BB A=B BB 1 B@ 1
1
0
1
274 2450 162 C B 0x 1 BB 120 CCC 180 3254 C 1 CC B C B C 375 3802 C CC ; b = BBB 223 CCC ; x = B@ x2 CA : B@ 131 CA 205 2838 C x3 A 1 86 2347 67 The above is an overdetermined system of ve equations in three unknowns.
7.3 Existence and Uniqueness As in the case of solving a linear system, questions naturally arise: 1. Does a least squares solution to Ax = b always exist? 2. If the solution exists, is it unique? 3. How do we obtain such solutions? The following theorem answers the questions of existence and uniqueness. Assume that the system Ax = b is overdetermined, or square, that is, A is of order m n where m n. An overdetermined system Ax = b can be represented graphically as: =
m>n x
A
b
The underdetermined case will be discussed later in this chapter. 391
Theorem 7.3.1 (Least Squares Existence and Uniqueness Theorem)
There always exists a solution to the linear least squares problem. This solution is unique i A has full rank, that is, rank(A) = n. If A is rank de cient, then the least squares problem has in nitely many solutions. We present a proof here in the full-rank case, that is, in the case when A has full rank. The rank-de cient case will be treated later in this chapter and in the chapter on the Singular Value Decomposition (SVD) (Chapter 10). First, we observe the following
Lemma 7.3.1 x is a least squares solution to an overdetermined system Ax = b if
and only if x satis es
AT Ax = AT b:
(7.3.1)
Proof. We denote the residual r = b ; Ax by r(x) to emphasize that given A and b, r is a function of x. Let y be an n-vector. Then r(y ) = b ; Ay = r(x) + Ax ; Ay = r(x) + A(x ; y ). So, kr(y)k = kr(x)k + 2(x ; y)T AT r(x) + kA(x ; y)k : 2 2
2 2
First assume that x satis es
2 2
AT Ax = AT b;
that is, AT r(x) = 0. Then from the above, we have
kr(y)k = kr(x)k + kA(x ; y)k kr(x)k ; 2 2
2 2
2 2
2
implying that x is a least squares solution. Next assume that AT r(x) 6= 0. Set AT r(x) = z 6= 0. De ne now a vector y such that
y = x + z: Then
r(y) = r(x) + A(x ; y ) = r(x) ; Az kr(y)k = kr(x)k + kAzk ; 2AT r(x)zT = kr(x)k + kAz k ; 2kz k < kr(x)k ; 2 2
2 2
2
2 2
2 2
2
2 2
392
2 2
2 2
kzk22 if Az 6= 0. This implies that x is not a least for any > 0 if Az = 0, and for 0 < < k2Az k22 squares solution. Proof. (of Theorem 7.3.1) Since A has full rank, AT A is symmetric and positive de nite and is thus, in particular, a nonsingular matrix. The theorem, in the full rank case, is now proved from the fact that the linear system AT Ax = AT b has a unique solution i AT A is nonsingular.
7.4 Geometric Interpretation of the Least Squares Problem Let A be an m n matrix with m > n. Then A is a linear mapping of Rn ! Rm . R(A) is a subspace of Rm . Every vector u 2 R(A) can be written as u = Ax for some x 2 Rn . Let b 2 Rm . Because k k2 is a Euclidean norm, kb ; Axk2 is the distance between the end points of b and Ax. It is clear that this distance is minimal if and only if b ; Ax is perpendicular to R(A) (see Fig. 7.1). In that case, kb ; Axk2 is the distance from the end point of b to the \plane" R(A).
b
b-Ax R(A) Ax
Figure 7.1 From this interpretation, it is easy to understand that a solution of the least squares problem to the linear system Ax = b always exists. This is because one can project b to the \plane" R(A) to obtain a vector u 2 R(A) and there is x 2 Rn such that u = Ax. This x is a solution. Because b ; Ax is perpendicular to R(A) and every vector in R(A) is a linear combination of column vectors of A, then b ; Ax is orthogonal to every column of A. That is,
AT (b ; Ax) = 0 393
or
AT Ax = AT b:
7.5 Normal Equations and Polynomial Fitting De nition 7.5.1 The system of equations AT Ax = AT b is called the normal equations. A well-known example of how normal equations arise in practical applications is the tting of a polynomial to a set of experimental data. The engineers and scientists gather data from experiments. A meaningful representation of the collected data is needed to make meaningful decisions for the future. Let (x1; y1); (x2; y2); : : :; (xn; yn) be a set of paired observations. Suppose that the mth (m n) degree polynomial y(x) = a0 + a1x + a2 x2 + + am xm (7.5.2) is the \best- t" for this set of data. One strategy for \best- t" is to minimize the sum of the squares of the residuals
E= We must then have Now,
n X i=1
(yi ; a0 ; a1 xi ; a2 x2i ; ; am xmi )2:
(7.5.3)
E = 0; i = 1; : : :; m: ai n E = ;2 X (yi ; a ; a xi ; a xi ; ; am xmi ) a i n E = ;2 X xi (yi ; a ; a xi ; a xi ; ; am xmi ) a i 0
1
.. .
0
1
2
2
=1
0
1
2
2
=1
n E = ;2 X xmi (yi ; a ; a xi ; ; am xmi ) am i 0
=1
394
1
(7.5.4)
Setting these equations to zero, we have
X
X
X
X
a n + a xi + a xi + + am xmi = yi X X X m X a xi + a xi + + am xi = x i yi 0
1
2
0
P
a
0
1
X
xmi + a
1
X
2
2
+1
xmi + + am +1
X
.. .
X
xi m = 2
(7.5.5)
xmi yi
(where denotes the summation from i = 1 to n). X Setting xki = Sk ; k = 0; 1; ; 2m and denoting the entries of the right hand side, respectively, by b0; b1 ; ; bm , the system of equations can be written as:
0S BB S BB . B@ ..
0 1
S S
1 2
Sm Sm
+1
Sm 1 0 a 1 0 b 1 C BB a CC BB b CC Sm C C BBB .. CCC = BBB .. CCC : .. C . C A@ . A @ . A S m am bm +1
0
0
1
1
(7.5.6)
2
(Note that S0 = n.) This is a system of (m + 1) equations in (m + 1) unknowns a0; a1; ; am . This is really a system of normal equations. To see this, de ne 0 1 x xm 1 1 BB 1 x1 C xm2 C 2 B C V =B B@ ... ... . . . ... CCA : 1 xn xmn Then the above system becomes equal to
V T V a = b;
(7.5.7)
(7.5.8)
where a = (a0; a1; : : :; am )T , and b = (b0; b1; : : :; bm)T . This is a system of normal equations; furthermore, if the xi 's are all distinct, then the matrix V has full rank. The matrix V is known as the Vandermonde matrix. From our discussion in the previous section, we see that a is the least squares solution to the system (7.5.8).
Example 7.5.1 Suppose that an electrical engineer has gathered the following experimental data consisting of the measurement of the current in an electric wire for various voltages.
x = voltage 0 2 5 7 9 13 24 y = current 0 6 7:9 8:5 12 21:5 35 395
We would like to derive the normal equations for the above data corresponding to the best t of the data to (a) straight line (b) a quadratic, and would like to see a comparison of the predicted results with the actual result when v = 5.
Case 1. Straight line Fit
01 BB 1 BB BB 1 B V =B BB 1 BB 1 BB @1
0 0 1 01 C BB 6 CC 2C CC BB CC C BB 7:9 CC 5C CC B CC 7 C; y = B BB 8:5 CC : CC BB 12:0 CC 9C CC BB CC 13 A @ 21:5 A 1 24 35:0
The Normal Equations are:
V TV a = V Ty = b ! ! 7 60 0:0906 a = 10 : 60 904 1:3385 3
The solution of these equations are:
a = 0:6831 a = 1:4353: The value of a + a x at x = 5 is 0:6831 + 1:4345 5 = 7:8596. 0 1
0
1
Case 2. Quadratic Fit
01 BB 1 BB BB 1 B V =B BB 1 BB 1 BB @1
The normal equations are
0 7 BB @ 60
0 2 5 7 9 13 1 24
0 1 C 4 C CC 25 C CC 49 C CC : 81 C CC 169 C A 576
V TV a = V Ty = b
1
0
1
60 904 0:0091 CC B C 4B 904 17226 A a = 10 @ 0:1338 C A: 904 17226 369940 2:5404 396
The solution of these normal equations is:
0a Ba a=B @
0 1
a
1 0 0:8977 1 CC = BB 1:3695 CC : A @ A 0:0027
2
The value of a0 + a1x + a2 x2 at x = 5 is 7.8127.
Note: The use of higher degree polynomial may not necessarily give the best result. The matrix of the normal equations in this case is ill-conditioned:
Cond2(V T V ) = 2:3260 105:
Indeed, it is well-known that the Vandermonde matrices become progressively illconditioned as the order of matrices increases.
7.6 Pseudoinverse and the Least Squares Problem Denote (AT A);1AT = Ay .
De nition 7.6.1 The matrix
Ay = (AT A); AT 1
is called the pseudoinverse or the Moore-Penrose generalized inverse of A. From (7.3.1), it therefore follows that the unique least-squares solution is x = Ay b.
Least Squares Solution Using the Pseudoinverse The unique least squares solution x to the full-rank least-squares problem Ax = b is given by x = (AT A);1AT b = Ay b: Clearly, the above de nition of the pseudoinverse generalizes the ordinary de nition of the inverse of a square matrix A. Note that when A is square and invertible,
Ay = (AT A); AT = A; (AT ); AT = A; : 1
1
397
1
1
We shall discuss this important concept in some more detail in Chapter 10. An excellent reference on the subject is the book Generalized Inverses by C. R. Rao and S. K. Mitra (1971). Having de ned the generalized inverse of a rectangular matrix, we now de ne the condition number of such a matrix as Cond(A) = kAk kAyk.
De nition 7.6.2 If an m n matrix A has full rank, then Cond(A) = kAk kAyk. Note: If not explicitly stated, all the norms used in the rest of this chapter are 2-norms, and Cond(A) is the condition number with respect to 2-norm. That is, Cond(A) = kAk kAyk . 2
2
Example 7.6.1 01 21 B 2 3 CC ; rank(A) = 2: A=B @ A 4 5
Thus, A has full rank.
;1:2857 ;0:5714 0:8577 ! y T ; T A = (A A) A = 1 0:5000 ;0:5000 Cond (A) = kAk kAky = 7:6656 2:0487 = 15:7047 : 1
2
2
2
Variance-Covariance Matrix In certain applications, the matrix AT A is called the information matrix, since it measures the information contained in the experiment, and the matrix (AT A);1 is known as the variancecovariance matrix. An algorithm for computing the variance-covariance matrix with-
out explicitly computing the inverse is given in Section 7.11.
7.7 Sensitivity of the Least Squares Problem In this section we study the sensitivity of a least squares solution to perturbations in data, that is, we investigate how a least squares solution changes with respect to small changes in the data. This study is important in understanding the di erent behaviors of di erent methods for solving the least squares problem that will be discussed in the next section. We consider two cases: perturbation in the vector b and perturbation in the matrix A. The results in this section are norm-wise perturbation results. For component-wise perturbation results, see Bjorck (1992), and Chandrasekaran and Ipsen (1994). 398
Case 1: Perturbation in the vector b Here we assume that the vector b has been perturbed to ^b = b + b, but A has remained unchanged.
Theorem 7.7.1 (Least-Squares Right Perturbation Theorem) Let x and x^, respectively, be the unique least-squares solutions to the original and the perturbed problems. Then if kbRk 6= 0, kx^ ; xk Cond(A) kbRk :
kxk
Here
kbRk
Cond(A) = kAk kAyk;
and bR and bR are, respectively, the projections of the vectors b and b onto R(A).
Proof. Since x and x^ are the unique least squares solutions to the original and the perturbed problems, we have
x = Ayb x^ = Ay(b + b): Thus,
x^ ; x = Ay b + Ayb ; Ay b = Ay b: Let bN denote the projection of b onto the orthogonal complement of R(A). That is,
(7.7.1)
b = bR + bN : Since bN lies in the orthogonal complement of R(A) = N (AT ), we have AT (bN ) = 0. So
x^ ; x = Ay b = Ay(bR + bN ) = Ay (bR) + Ay (bN ) = Ay bR + (AT A); AT bN = Ay bR : 1
Again, since x is the unique least squares solution, we have
Ax = bR from which (again taking norms on both sides) we get
kxk kkbARkk : 399
(7.7.2)
Combining (7.7.1) and (7.7.2), we have the theorem.
Interpretation of Theorem 7.7.1 Theorem 7.7.1 tells us that if only the vector b is perturbed, then, as in the case of linear systems, Cond(A) = kAk kAyk serves as the condition number in the sensitivity analysis of the unique least squares solution. If this number is large, then even with a small relative error in the projection of b onto R(A), we might have a drastic change in the least squares solution. On the other hand, if this number is small and the relative error in the projection of b onto R(A) is also small, then the least squares solution will not change much. Note that it is the smallness of the relative error in the projection of b onto R(A), namely kbR k , that plays the role kbRk here, not merely the smallness of kbR k.
Example 7.7.1 An Insensitive Problem 01 21 011 B CC BB CC A=B 0 1 ; b = @ A @ 1 A ; b = 10;
4
011 BB CC @1A
1 0 1 1 PA = Projection of A onto R(A) 0 0:9333 0:3333 0:1667 1 B CC =B 0 : 3333 0 : 3333 ; 0 : 3333 @ A 0:1667 ;0:3333 0:8333
(See Example 5.6.2 of Chapter 5.)
0 1:3333 1 B 0:3333 CC bR = PA b = B A @
0:6667 0 0:13333 1 B C bR = PA b = 10;3 B @ 0:03333 CA 0:06667 Cond(A) = 2:4495: Since kkbb Rkk = 10;4, a small number, and A is well-conditioned, from Theorem 7.7.1 above, we R expect that the least squares solution will not be perturbed much. The following computations 400
show this is indeed the case.
!
:6667 x = Ay b = :3333 ! :6667 y x^ = A (b + b) = :3334 kx^ ; xk = 2:4495 10; kxk 4
Example 7.7.2 A Sensitive Problem 0 1 B ; A=B @ 10
1
1 C 0 C A; 10;4
4
0
011 B :1 CC ; b = 10; B @ A 3
x=
! 1 1
;
x^ =
! 1:5005
0:5005
0
0 2 B ; b=B @ 10
4
10;4
1 CC A
Cond(A) = O(104) :
:
Using results of Example 5.6.3 from Chapter 5, we have 0 2 1 001 B 10;4 CC ; b = BB 0 CC bR = PAb = B N @ A @ A ; 4 10 0 Since the product of the condition number and kkbb Rkk is 7.0888, we should expect a substantial R change in the solution. Indeed this is true. kx^ ; xk = 0:5000 (Large!)
kxk
Case 2. Perturbation in the matrix A The analysis here is much more complicated than the previous case. We will state the result here (without proof) and the major consequences of the result. Let the perturbation E of the matrix be small enough so that rank(A) = rank(A + E ): 401
Let x and x^ denote the unique least squares solutions, respectively, to the original and to the perturbed problem. Let EA and EN denote the projections of E onto R(A) and onto the orthogonal complement of R(A), respectively. Then if bR 6= 0, we have the following theorem due to Stewart (1969).
Theorem 7.7.2 (Least-Squares Left Perturbation Theorem) Let x and x^ be
the unique least-squares solutions to Ax = b and (A + E )^x = b, and let rank(A + E ) be the same as rank(A). Then
kx^ ; xk 2Cond(A) kEAk + 4(Cond(A)) kEN k kbN k + O kEN k kxk kAk kAk kbRk kAk
2
2
Interpretation of Theorem 7.7.2 Theorem 7.7.2 tells us that in the case where only the matrix A is perturbed, the
sensitivity of the unique least squares solution, in general, depends upon the squares of the condition number of A. However, if kEN k or kbN k is zero or small, then the sensitivity will depend only on the Cond(A). Note that the residual r = b ; Ax is zero if bN = 0.
Two Examples with Di erent Sensititivities We now present two examples with the same matrix A, but with di erent b, to illustrate the di erent sensitivities of the least squares problem in di erent cases. In the rst example, (Cond(A))2 serves as the condition number of the problem; in the second example, Cond(A) serves as the condition number.
Example 7.7.3 Sensitivity Depending Upon the Square of the Condition Number 0 1 B A=B @ 0:0001 0
1
0 1
1 1 CC BB CC 0 A; b = @1A 0:0001 1
0 1:0001 1 0 ;0:0001 1 B C B C kbN k bR = PA b = B @ 0:0001 CA ; bN = B@ 0:9999 CA ; kbRk = 1: 0:0001
0:9999
402
(Using PA from Example 5.6.3 of Chapter 5.) Let
0 0 ;0:0001 1 B 0 0:9999 CC E = 10; B @ A 4
0 0:9999
0 1 1 1 B C A+E = B @ 0:0001 0:0001 CA ; EN = 10;
4
4
0:0002
0
0 0 10; 1 BB C @ 0 0:9999 CA 0 0:9999
kEN k = kE k = 9:999 10; : kAk kAk Though the product of kkEANkk and kkbbN kk is rather small, since (Cond(A)) = 2 10 is large, we R 5
2
8
should expect a drastic departure of the computed solution from the true solution. This is indeed true, as the following computations show: ! ;4:999 ! 0:5 3 x^ = 10 ; x= 5 0:5
kx^ ; xk = 9:999 10 (Large!) : kxk 3
Note that
kEN k kbN k (Cond(A)) = 9:999 10; 2 10 = 1:9998 10 : kAk kbRk Example 7.7.4 Sensitivity Depending Upon the Condition Number 2
Let
5
8
4
0 0 ;0:0001 1 B 0 0:9999 CC (same as in Example 7.7.3) E = 10; B @ A 4
0 0:9999
0 1 B 0:0001 A=B @ 0
1
0
1
1 2 C B C 0 C A ; b = B@ 0:0001 CA : 0:0001 0:0001
0 1 001 B CC. Note that P = BB 10; In this case, bR = b and bN = B A @ @0A
4
Example 5.6.3 of Chapter 5.)
10;4
0
403
1
10;4 10;4 C. (See 0:5000 ;0:5000 C A ;0:5000 0:5000
Thus, according to Theorem 7.7.2, the square of Cond(A) does not have any e ect; the least squares solution is a ected only by Cond(A). We verify this as follows:
! 1 y x=A b= 1 ! 1:4999 x^ == 0:5000 Cond(A) = 1:4142 10 0 1 1 1 B 10; 0 CC EA = 10; B @ A 4
4
4
10;4
0
kEAk = 10; kAk kx^ ; xk = 0:5000 kxk 4
:
Residual Sensitivity. We have just seen that the sensitivities of the least squares solutions due
to perturbations in the matrix A are di erent for di erent solutions; however, the following theorem shows that the residual sensitivity always depends upon the condition number of the matrix A. We state the result in somewhat simpli ed and crude form. See Golub and Van Loan, MC (1983, pp. 141{144) for a precise statement and proof of a result on residual sensitivity.
Theorem 7.7.3 (Least-Squares Residual Sensitivity Theorem) Let r and r^ denote the residuals, respectively, for the original and the perturbed least squares problems; that is,
r = b ; Ax r^ = b ; (A + E )(^x): Then
kr^ ; rk kEN k + 2Cond(A) kEN k + O kEN k : kbk kAk kAk kAk 2
Interpretation of Theorem 7.7.3 404
The above result tells us that the sensitivity of the residual depends at most on the condition number of A. Example 7.7.5 0 1 B 10; A=B @
1
0 1
1 1 C B C 4 0 C A ; b = B@ 1 CA 0 10;4 1 0 0 ;0:0001 1 B C E = 10;4 B @ 0 0:9999 CA 0 0:9999 ! 0:5 ;4:9999 ! 3 x= ; x^ = 10 0:5 5 0 ;0:0001 1 B 0:9999 CC r = b ; Ax = B @ A 0:9999 0 ;1 1 B C r^ = b ; (A + E )^x = B @ 0:9998 CA 0:9999 0 0 ;10;4 1 kr^ ; rk = 0:5; E = P E = 10;4 B B@ 0 0:9999 CCA N N A kbk 0 0:9999 Cond(A) = 1:4142 104 and
Cond(A) kkEANkk = 1:4142:
The inequality in Theorem 7.7.3 is now easily veri ed.
Sensitivity of the Pseudoinverse. The following result, due to Wedin (1973), shows that it is Cond(A) again that plays a role in the sensitivity analysis of the pseudoinverses of a matrix.
405
Theorem 7.7.4 (Pseudoinverse Sensitivity Theorem) Let A be m n, where m n. Let Ay and A~y be, respectively, the pseudoinverse A and of A~ = A + E .
Then, provided that rank(A) = rank(A~), we have
y y
A~ ; A p kE k
A~y
2Cond(A) kAk :
Example 7.7.6 01 21 0 0:0010 0:0020 1 B2 3C B C A=B @ C A ; E = 10; A = B@ 0:0020 0:0030 CA 4
0:0040 0:0050 ;1:2857 ;0:5714 0:8571 ! y A = 1 0:5000 ;0:5000 0 1:001 2:002 1 B C A + E = A~ = B @ 2:002 3:003 CA 4:004 5:005 ;1:2844 ;0:5709 0:8563 ! y ~ A = 0:9990 0:499995 ;0:4995
y y
A~ ; A
y
= 10;4 kE k : kAk
A~ 4 5
Note that
Cond(A) = 15:7047:
7.8 Computational Methods for Overdetermined Least Squares Problems 7.8.1 The Normal Equations Method One of the most widely used approaches (especially in statistics) for computing the least squares solution is the normal equations method. This method is based upon the solution of the system of normal equations AT Ax = AT b: 406
We assume that A is m n (m > n) and that it has full rank. Since in this case AT A is symmetric and positive de nite it admits the Cholesky decomposition
AT A = HH T : Therefore, the normal equations approach for solving the least squares problem can be stated as follows:
Algorithm 7.8.1 Least Squares Solution Using Normal Equations Given an m n (m > n) matrix A with rank(A) = n, the following algorithm computes the least squares solution x from normal equations using Cholesky factorization. (1) Form c = AT b. (2) Find the Cholesky factorization of AT A = HH T . (3) Solve the triangular systems in the sequence
Hy = c H T x = y:
Flop-Count. The above method for solving the full rank least squares problem requires about
mn + n op: mn for computing AT A and AT b, n for computing the Cholesky factorization 2 6 2 6 of AT A, and n op to solve two triangular systems. Thus, the method is quite ecient. 2
3
2
3
2
Example 7.8.1 Normal Equation Solution of the Case-Study Problem 01 BB 1 BB A=B BB 1 B@ 1
1 0 5 1120 B (1) Form AT A = B @ 1120 297522 14691 3466402 0 703 1 B C (2) Form AT b = B @ 182230 CA 2164253
274 2450 1 C 180 3254 C CC 375 3802 C CC ; 205 2838 C A 86 2347 1 14691 C 3466402 C A 44608873
407
0 112 1 BB 120 CC BB CC b=B BB 223 CCC B@ 131 CA 67
(3) Solve the normal equations: T AT Ax = A 0 7b:0325 1 B C x = B @ 0:5044 CA : 0:0070
Note that the system is very ill conditioned: Cond(AT A) = 3:0891 108: To see how the computed least squares solution agrees with the data of the table, we compute
0x Bx ( 1 274 2450 ) B @
1 2
x
1 CC = 162:4043 A
3
(True value = 162)
0x B ( 1 180 3254 ) B @x
1 2
x
1 CC A = 120:6153
3
(True value = 120)
0x B ( 1 375 3802 ) B @x
1 2
x
1 CC A = 222:8193
3
(True value = 223)
0x Bx ( 1 205 2838 ) B @
1 2
x
1 CC = 130:3140 A
3
(True value = 131)
0x B ( 1 81 2347 ) B @x
1 2
x
1 CC A = 66:847
3
(True value = 67): Suppose that the company would like to predict, using the above results, the sales in a district with the population 220,000 and per capita income of $2,500. Then the best prediction using the 408
given model is
0 7:0325 1 B C ( 1 220 2500 ) B @ 0:5044 CA = 135:5005 0:0070
Note: In this example, the residuals are small in spite of the fact that the data matrix A is ill-conditioned.
Example 7.8.2 01 21 031 B CC B C A=B @ 2 3 A ; b = B@ 5 CA
3 4 9 rank(A) = 2; rank(A; b) = 3. Thus the system Ax = b does not have a solution. We therefore calculate the least squares solution.
!
40 (1) c = AT b = 57 (2) Cholesky factorization of AT A: 3:7417 0 H= 5:3452 0:6547 (3) Solutions of the triangular systems:
y = x =
!
!
10:6904 ; ;0:2182 ! 3:3333 : ;0:3333
!
3:3333 The unique least squares solution is x = . ;0:3333 Note that ! ; 1 : 8333 ; 0 : 3333 1 : 1667 Ay = (AT A);1AT = 1:3333 0:3333 ;0:6667 and ! 3:3333 y A b= : ;0:3333
Numerical diculties with the normal equations method 409
The normal equations method, though easy to understand and implement, may give rise to numerical diculties in certain cases. First, we might lose some signi cant digits during the explicit formation of AT A and the computed matrix AT A may be far from positive de nite; computationally, it may even be singular. Indeed, it has been shown by Stewart (IMC, pp. 225{226) that, unless Cond(A) is less than 10 2 , where it is assumed that AT A has been computed exactly and then rounded to t signi cant digits, then AT A may fail to be positive de nite or even may not be nonsingular. The following simple example illustrates that fact. t
Example 7.8.3 0 1 B 10; A=B @
Consider
1
1 C 4 0 C A ; 4 0 10 with t = 9. The columns of A are linearly independent. Now, compute ! 1 + 10;8 1 T A A= 1 1 + 10;8 Since t = 9, we will get ! 1 1 T A A= ; 1 1 which is singular. Note that Cond(A) = 1:4142 104 > 10 2 = 104. Second, the normal equations approach may, in certain cases, introduce more errors than those which are inherent in the problem. This is seen as follows. From the perturbation analysis done in Chapter 6, we easily see that, if x^ is the computed least squares solution obtained by the normal equations method, then kx^ ; xk Cond(AT A) t
kxk
= (Cond(A))2:
(Exercise):
Thus, the accuracy of the least squares solution using normal equations will depend upon the square of the condition number of the matrix A. However, we have just seen in the section on perturbation analysis of the least squares problem that in certain cases such as when the residual is zero, the sensitivity of the problem depends only on the condition number of A (see Theorems 7.7.1 and 7.7.2). Thus, in these cases, the normal equations method will introduce more errors
in the solution than what is warranted by the data. 410
A Special Note on the Normal Equations Method In spite of the drawbacks mentioned above about the normal equations method, we must stress that the method is still regarded as a useful tool for solving the least squares problem, at least in the case where the matrix A is well conditioned. In fact, it is routinely used in many practical applications and seems to be quite popular with practicing engineers and statisticians. Note that in the example above, if we would use an extended precision in our computations, the computed matrix AT A could be obtained as a symmetric positive de nite matrix and the normal equations method would yield an accurate answer, despite ill-conditioning; as the following computations show: ! 1 : 00000001 1 AT A = 1 1:00000001 ! 2:00000001 T C = A b= 2:00000001 ! 1:000000005 0 T H = the Cholesky factor of A A = : 0:999999995 0:00014142135581 Solution of Hy = c: ! 2 y= : 0:00014142135651 Solution of H T x = y : ! 1! 0:999999999500183 x= : 1:0000000499817 1 (The exact solution.)
7.8.2 The QR Factorization Methods for the Full-Rank Problem ! R
Let QT A = R = 1 be the QR decomposition of the matrix A. Then, since the length of a 0 vector is preserved by an orthogonal matrix multiplication, we have
kAx ; bk = QT Ax ; QT b = kR x ; ck + kdk ; 2
2 2
!
2
1
2 2
2 2
c where QT b = . Thus, kAx ; bk will be minimized if x is chosen so that d 2 2
R x ; c = 0: 1
411
The corresponding residual then is given by
krk = kdk 2
2
This observation immediately suggests the following QR approach for solving the least squares problem:
Least Squares Using QR (1) Decompose Amn = Qmm Rmn. (2)
Form QTnm bm1 =
!
c . d
(3) Solve the upper triangular system
R x=c 1
where
R
R=
1
0
!
:
Use of Householder Matrices Once A is decomposed into QR and the vector QT b is formed, the solution x is obtained by back substitution. Suppose that the Householder method is used to compute the QR decomposition of A. Then, following the notations of Chapter 5, we see that the product QT b can be formed as: For k = 1; 2; : : :; n b = Hk b, where Hk ; k = 1; : : :; n are the Householder matrices such that
QT = HnHn; H H : 1
2
1
Thus, the matrix Q does not have to be formed explicitly. The idea of solving the least squares problem based on the QR factorization of A using the Householder transformation was rst suggested by Golub (1965). 412
Algorithm 7.8.2 The Householder-Golub method for the full-rank least squares problem Given an m n matrix A (m n) with rank(A) = n and an m-vector b, the following algorithm computes the least squares solution x using Householder matrices H1 through Hn. 1. Apply the Householder method to A. Obtain R1 and the Householder matrices H1; H2; : : :; Hn. 2. Form HnHn;1
!
c H H b = , where c is an n-vector. d 2
1
3. Solve R1x = c.
Flop-count. Since the cost of the algorithm is dominated by the cost of the QR decomposition
of A, the overall op-count for the full-rank least squares solution using Householder's method is
n m ; n3 (Exercise): 2
Thus, the method is about twice as expensive normal equations method. Note that the n2m asnthe 3 normal equations method requires about 2 + 6 op.
Round-o Error and Stability. The method is stable. It has been shown in Lawson and
Hanson (SLP, p. 90) that the computed solution x^ is such that it minimizes
k(A + E )^x ; (b + b)k
2
where E and b are small. Speci cally,
kE kF cn kAkF + O( ) and kbk c kbk + O( ); 2
2
2
2
where c (6m ; 3n + 41)n and is machine precision. That is, the computed solution is the exact least squares solution of a nearby problem.
Example 7.8.4 01 21 B 2 3 CC ; A=B @ A 3 4
413
031 B 5 CC b=B @ A 9
(1) A = QR:
0 ;0:2673 B ;0:5345 Q = B @ ;0:8018 0 ;3:7417 BB 0 R = B BB @ 0
(2)
QT b
1
0:8724 0:4082 C 0:2182 ;0:8165 C A ;0:4364 0:4082 ;5:3452 1 C R! 0:6547 C CC = 1 : CA 0 0
!
c = d 0 ;10:6904 1 B C = B @ ;0:2182 CA : 0:8165
(3) Solution of the system R1x = c: ;3:7417 ;5:3452 ! ;10:6904 ! x= 0 0:6547 ;0:2182 The least squares solution is ! 3:3332 x= : ;0:3333 Norm of the residual = krk2 = kdk = 0:8115
Example 7.8.5 0 1 B A=B @ 0:0001 (1) A = QR
0
1
1 C 0 C A; 0:0001
0 2 1 B C b=B @ 0:0001 CA 0:0001
0 ;1 1 0:0001 ;0:0001 B ;0:0001 ;0:7071 0:7071 CC Q = B @ A 0:7071
0
0:7071
0 ;1 ;1 1 BB CC R ! R = @ 0 0:0001 A = 1
0 (See Example 5.4.3 from Chapter 5.)
0
414
0
(2)
0 ;2 1 BB CC c ! T Q b = @ 0:0001 A = d
0
(3) Solution of the system: R1 x = c
;1 ! 0 0:0001 x x ! 1
;1
x x
1 2
!
1 = 1 2 = 1:
The unique least squares solution is
1
;2 ! = 0:0001
. Note that
0:0005 5 ;5 Ay = 103 0:0005 ;5 5 ! 1 Ay b = : 1
!
Norm of the residual = krk2 = kdk = 0.
Use of Givens Rotations We can, of course, use the Givens rotations to decompose A into QR and then use this decomposition to solve the least squares problem. However, as we have seen before, the use of Givens rotations will be more expensive than the use of Householder matrices. Recall that computations of Givens rotations require evaluations of square roots; however, there are \square-root free" Givens rotations, introduced by M. Gentleman (1973), which can be used to solve the least squares problem. The square-root free Givens rotations are also known as fast Givens rotations. For details, see Golub and Van Loan MC 1984 (pp. 156-157).
Use of Gram-Schmidt and Modi ed Gram-Schmidt Algorithms Another technique, known as the Gram-Schmidt process, can be used to decompose A into QR. A properly implemented Gram-Schmidt process, when used to solve the linear least squares problem, is just slightly more expensive than the Householder method, but seems to be as numerically as e ective as the Householder method for solving the least squares problem. We rst state these algorithms for QR factorization and then show how the modi ed Gram-Schmidt can be used to solve the least squares problem. 415
Algorithm 7.8.3 Classical Gram-Schmidt (CGS) for QR Factorization Given A = (a ; a ; : : :; an) of order m n with rank(A) = n, the following algorithm computes a 1
2
matrix Qmn = (q1; : : :; qn) with orthonormal columns and an upper triangular matrix R = (rij )nn such that A = Qmn Rnn. For k = 1; 2; : : :; n do For i = 1; 2; : : :; k ; 1 do
rik qiT ak
qk ak ;
kX ;1 i=1
rikqi
rkk = kqk k qk rqk :
2
kk
The algorithm, as outlined above, is known to have serious numerical diculties. During the
computations of the qk's, cancellation can take place and, as a result, the computed qk 's can be far from orthogonal. (See later in this section for details.)
The algorithm, however, can be modi ed to have better numerical properties. The following algorithm, known as the modi ed Gram-Schmidt algorithm, computes the QR factorization of A in which, at the kth step, the kth column of Q and the kth row of R are computed (note that the Gram-Schmidt algorithm computes the kth columns of Q and R at the kth step).
Algorithm 7.8.4 Modi ed Gram-Schmidt (MGS) for QR Factorization Set qk = ak ; k = 1; 2; : : :; n. For k = 1; 2; : : :; n do rkk = kqkk2
qk rqk kk For j = k + 1; : : :; n do rkj qkT qj qj qj ; rkjqk :
The above is the row-oriented modi ed Gram-Schmidt method. The column-oriented version can similarly be developed (Exercise #17). The two versions are numerically equivalent.
Remarks: The modi ed Gram-Schmidt algorithm is not as satisfactory as the Householder or
Givens method for computing the QR factorization of A. It can be shown (Bjorck (1992)) that the computed Q by MGS, denoted by Q~ , satis es Q~ T Q~ = I + E , where kE k Cond(A), whereas the Householder method produces Q~ T Q~ = I + E where kE k . (For more details, see the discussions in the next section.) 416
Flop-count. The op-count for the MGS is mn , compared to mn ; n needed by the Householder method. (Note that MGS works with the full length column vector at each step, whereas the Householder method deals with successively shorter columns.) Although in the 2 2 case the CGS and MGS algorithms produce the same results, we use a 2 2 example to illustrate here how the computational arrangements di er with the same matrix. 2
2
All computations are performed with 4-digit arithmetic.
Example 7.8.6 0 1 B A=B @ 0:0001 0
k=1:
1
1 C 0 C A; 0:0001
0:0001
Gram-Schmidt
0 1 1 B 0:0001 CC ; q =a =B @ A 1
0 2 1 B C b=B @ 0:0001 CA
1
0
r = kq k = 1 11
1 2
0 1 1 B 0:0001 CC q = rq = B @ A 1
1
k=2:
11
0
001 B CC q = a ; r q = 10; B @ ;1 A
r = 1; 12
2
2
12 1
0 0 1 B C q =B @ ;0:7071 CA
1
1
2
0:7071 q1T q2 = ;7:0711 10;5 Form Q and R:
0 1 1 0 B C Q = (q ; q ) = B @ 0:0001 ;0:7071 CA 1
R=
r
11
2
! r 12
0 r22
0 0:7071 ! 1 1 = 0 1:414 10;4
417
3
3
Form c:
2 c = QT b = 0
!
!
2 The least squares solution is x = . 0
Modi ed Gram-Schmidt q =a ; 1
q =a
1
2
2
k=1: r = kq k = 1 11
1 2
0 1 1 B 0:0001 CC q =B @ A 1
0
0 0 1 B CC q =q ;r q =B ; 0 : 0001 @ A
r = qT q ; 12
1
2
2
2
12 1
0:0001
k=2: r = kq k = 1:4142 10;
0 0 1 B CC q = rq = B ; 0 : 7071 @ A
22
2
2
Form Q and R:
2
22
0 1 1 0 B C Q=B @ 0:0001 ;0:7071 CA ; 0
4
0:7071
!
0:7071
1 1 R= 0 1:4142 10;4
!
2 The least squares solution is x = . (Note that for the same problem, the Householder0! 1 QR method produced x = , which is correct (Example 7.8.5).) 1 418
Modi ed Gram-Schmidt versus Classical Gram-Schmidt Algorithms Mathematically, the CGS and MGS algorithms are equivalent. However, as remarked earlier, their numerical properties are di erent. For example, consider the computation of q2 by the CGS method, given q1 with kq1 k2 = 1. We have
q a ;r q ; 2
2
12 1
where r12 = q1T a2. Then it can be shown (Bjorck (1992)) that
k (q ) ; q k < (1:06)(2m + 3) ka k : 2
2
2 2
Since q1T q2 = 0, it follows that
jqT (q )j < (1:06)(2m + 3) ka k : 1
2
2 2
This shows that in CGS two computed vectors, q and q , can be far from being orthogonal. On the other hand, it can be shown Bjorck (1992) that in MGS the loss of orthogonality 1
2
depends upon the condition number of the matrix A. Speci cally, it has been shown that
2 (A) ;
I ; Q^ T Q^
2 1 ;c1c Cond Cond 2 2 (A)
assuming that c2Cond2(A) < 1, where c1 and c2 are small constants. Since in MGS the loss of orthogonality depends upon the condition number, one could use column pivoting to maintain orthogonality as much as possible. Thus, as far as nding the QR factorization of A is concerned, neither algorithm can be recommended over the Householder or the Givens method. With CGS the orthogonality of Q can be completely lost; with MGS,
the orthogonality of Q may not be acceptable when A is ill-conditioned. Note that in the above 2 2 example, the computed Q^ (in four-digit arithmetic) is such that ! 1 ; 0 : 0001 T Q^ Q^ = : ;0:0001 1
On the other hand, for the same problem using the Householder method, Q^ T Q^ = I (in four-digit arithmetic). To compare the computations with the di erent Gram-Schmidt processes, we also computed QR factorizations of the 5 5 Hilbert matrix using the Gram-Schmidt, the modi ed Gram-Schmidt, and the Householder methods (which is known to be very stable) and tabulated the results of kI ; Q^ T Q^ k in extended precision. The results are displayed in the following table. 419
TABLE 7.1 Comparison of QR Factorization with Di erent Methods Method kI ; Q^T Q^k Gram-Schmidt 1:178648780488241 10; 7
Modi ed Gram-Schmidt 4:504305729523455 10;12 Householder 4:841449989971538 10;16
Remark: The table clearly shows the superiority of the Householder method over both the Gram-Schmidt and modi ed Gram-Schmidt methods; of the latter two methods, the MGS is clearly preferred over the CGS. Use of the MGS method in the least squares solution We have just seen that the MGS is superior to the CGS for QR factorization, but it is still not fully satisfactory. However, as far as the least squares problem is concerned, it is a di erent story. There seems to be a growing tendency to use MGS in solutions of least squares problems even over the Householder method, especially for solving large and sparse problems. Bjorck and Paige (1992) have recently shown that the MGS is numerically equivalent to Householder
QR factorization applied to A augmented with a square matrix of zero elements on top. However, while using MGS in the solution of a least squares problem, special care has to be
taken. Since Q may not be computationally orthogonal, if c = QT b is computed using Q obtained from MGS, and this c is used to solve Rx = c, then the computed least squares solution may not be accurate. The MGS should be applied to the augmented matrix (A; b) to compute the factorization:
!
R z (A; b) = (Q ; qn ) : 0 1
From this it follows that
+1
x Ax ; b = (A; b) ;1
! !
R z x = (Q ; qn ) 0 ;1 = Q (Rx ; z ) ; qn : 1
!
+1
1
+1
Chris Paige is a Australian-Canadian numerical linear algebraist, who rejuvenated the use of the Lanczos algorithm in matrix computations by detailed study of the break-down of the algorithm. He is a professor of computer science at McGill University, Montreal, Canada.
420
If qn+1 is orthogonal to Q1, then kAx ; bk2 will be a minimum when Rx = z . Thus, the least squares solution can be obtained by solving
Rx = z; and the residual r will be given by r = qn+1 . Details can be found in Bjorck (1990). The above discussion leads to the following least squares algorithm.
Algorithm 7.8.5 Least Squares Solution by MGS 1. Apply MGS to Amn to obtain Q = (q1 ; : : :; qn ) and R. 2. For k = 1; : : :; n do k = qkT b
b b ; kqk 3. Solve Rx = ( ; : : :; n)T . 1
Example 7.8.7 Consider solving the least squares problem using the MGS with 0 1 1 0 2 1 1 B :0001 0 CC ; b = BB :0001 CC : A=B @ A @ A 0 :0001 :0001 ! 1 The exact solution is x = . 1 0 1 1 0 ! 1 1 B C Q=B @ :0001 ;:7071 CA ; R = 0 :0001 : 0 :7071 ! 2 T If we now form c = Q b and solve Rx = c, we obtain x = . On the other hand, if we obtain 0 x using !the algorithm above, we get (1 ; 2) = (2; 0:0001), and the solution of Rx = (1 ; 2)T is 1 x . 1
Round-o property and op-count. It can be shown (see Bjorck and Paige (1992)) that the MGS for the QR ! factorization method is numerically equivalent to the Householder method 0 0
applied to
A b
; that is,
HnHn; H H 1
2
1
!
!
R c = : A b 0 c 0 0
1 2
421
From this equivalence, it follows that the MGS method is backward stable for the least squares problem. The method is slightly more expensive than the Householder method. It
requires about mn op, compared to the mn ; n op needed by the Householder method. 2
3
2
3
7.8.3 The QR Factorization Method for the Rank-De cient Case In this section we consider the rank-de cient overdetermined problem. As stated in Theorem 7.3.1, there are in nitely many solutions in this case. There are instances where, the rank-de ciency is actually desirable, because it provides a rich family of solutions which might be used for optimizing some other aspects of the original problem. In case the m n matrix A, m n, has rank r < n, the R matrix in a QR factorization of A is singular. However, we have seen that the use of the QR factorization with column pivoting can theoretically reveal the rank of A. Recall that Householder's method with column pivoting yields
QT AP
R
=
R
11
12
0
!
;
0 where P is a permutation matrix, R11 is an r r nonsingular upper triangular matrix and R12 is r (n ; r). This factorization can obviously be used to solve the rank-de cient least squares problem as follows. Let
PTx = y AP = A~ and
QT b =
Then
kAx ; bk
2 2
!
c : d
= QT APP T x ; QT b 22 (2-norm remains invariant under orthogonal matrix multiplication)
~ ; QT b
=
QT Ay 2 2
R R ! y ! ! c
=
;
0 0 y d
= (R y + R y ; c) + d 11
12
1 2
11 1
12 2
2 2 2
422
2 2
2
2
Thus, kAx ; bk22 will be minimized if y is chosen so that
R y = c;R y : 11 1
12 2
Moreover, the norm of the residual in this case will be given by:
krk = kb ; Axk = kck : 2
2
2
This observation suggests the following QR factorization algorithm for rank-de cient least squares solutions.
Algorithm 7.8.6: Least Squares Solutions for the Rank-De cient Problem Using QR Step 1: Decompose Amn P = QmnRnm using column pivoting. ! c T Step 2: Form Q b = . d
Step 3: Choose an arbitrary vector y . 2
Step 4: Solve the r r nonsingular upper triangular system: R y = c;R y : 11 1
Step 5: Recover x from
12 2
x = Py:
Minimum Norm Solution Since y2 is arbitrary, the above method will yield in nitely many least squares solutions, as Theorem 7.3.1 states. The solution obtained by setting y2 = 0 is called the basic solution. In case R12 = 0, the basic solution is the minimum norm least squares solution, i.e., among the in nite number of least squares solutions, it has the minimum norm. In case R12 6= 0, the basic solution can not be minimum norm solution (Exercise 20). In such a case the complete QR factorization with column pivoting can be used. Recall from Chapter 5 that the complete QR factorization of an m n matrix A is given by ! T 0 T Q AW = : 0 0 If pivoting is used, then ! T 0 T AP = Q W ; 0 0 423
where P is a permutation matrix. The minimum norm solution will then be given by
x = PW
T; c 1
0
!
; PW
T; c 1
!
0
where c is a vector consists of the rst r elements of QT b. (Exercise #21(a))
Remarks: 1. A note on the use of column pivoting: We have shown that the column pivoting is useful for the rank-de cient least squares problem. However, even in the full rank case, the use of column pivoting is suggested (see Golub and Van Loan MC 1984). 2. For the rank-de cient least squares problem, the most reliable approach is the use of singular value decomposition (see Section 7.8.4 and Chapter 10).
Round-o property. It can be shown (Lawson and Hanson, SLP p. 95) that for the minimum length least squares problem, the computed vector x^ is close to the exact solution of a perturbed problem. That is, there exist a matrix E and vectors x^ and b such that x^ is the minimum length least squares solution of (A + E )^x b + b, where
kE kF (6m + 6n ; 6k ; 3s + 84)s kAkF + O( ) 2
and kbk (6m ; 3k + 40)k kbk + O(2 ), where k is the rank of A and s = min(m; n). Moreover,
kx ; x^k (6n ; 6k + 43)s kxk + O( ): 2
Note: In the above result we have assumed that R in R =
R
R R
!
is zero. But in 0 22 practical computations it will not be identically zero. In that case, if R^ 22 is the computed version of R22, then we have 22
11
12
kE kF
R^
F + (6m + 6n ; 6k ; 3s + 84)s kAkF + O( ): 2
22
Example 7.8.8 A Consistent Rank-De cient Problem 031 01 01 B 2 0 CC ; b = BB 6 CC ; rank(A) = 1: A=B @ A @ A 0
0 0
424
!
1 0 Step 1: AP = QR; P = . 0 1
0 ;:4472 :8940 0 1 B ;:8944 ;:4472 0 CC Q = B @ A 0
0
0 ;2:2361 0 1 B 0 C: R = B 0C @ A 0
0 ;6:7082 1 Step 2: QT b = BB@ 0 CCA ; c = (;6:7082).
1
0
0 Step 3: Choose y2 = 0.
Step 4: Solve R y = c ; R y : 11 1
12 2
;6:7082 = 3: y = Rc = ; 2:2361 1
11
The minimum norm least squares solution is
!
3 x=y= : 0
Example 7.8.9 An Inconsistent Rank-De cient-Problem 01 01 B 2 0 CC ; A=B @ A 0 0
011 B 2 CC : b=B @ A 0
0 ;0:4472 ;0:8944 0 1 0 ;2:2361 0 1 B C ; R = BB 0 C: ,Q=B ;0:8944 ;0:4472 0 C Step 1: PA = QR; P = 0C @ A @ A 0 1 0 0 1 0 0 ! c Step 2: QT b = ; c = ;2:2361. ! 1 0
d
Step 3: Choose y = 0. 2
!
1 Step 4: The minimum norm least squares solution is x = y = . 0 (Note that R11y1 = c1 ; R12y2 gives y1 = 1) 425
7.8.4 Least Squares Solution Using the SVD Acting on the remark in the previous section, for the sake of curious readers and those who know how to compute the Singular Value Decomposition (SVD) of a matrix (for instance, using some software package such as MATLAB, LAPACK or LINPACK), we just state the following results which show how the SVD can be used to solve a least squares problem. A full treatment will be given in Chapter 10. Let A = U V T be the SVD of A, where A is m n with m n and = diag(1; : : :; n). Let
0 b0 B ... b0 = U T b = B @
1
b0m
1 CC : A
Least Squares Solutions Using the SVD A family of the least-squares solutions is given by
x = V y; where
0y B ... y=B @
1
yn
1 C C A;
8 b0 > < i; i 6= 0 yi = > i : arbitrary, i = 0:
In the rank-de cient case, an expression for the minimum norm solution is
x=
r uT b0 X i i=1
i vi;
where r = rank(A), and ui and vi are the columns of U and V , respectively. (Note that rank(A) = the number of nonzero singular values of A.)
426
Example 7.8.10
0 1 B 0:0001 A=B @
021 B 0 CC ; b0 = U T b = B @ A 0
1
0
1
1 2 C B C 0 C A ; b = B@ 0:0001 CA 0 0:0001 0:0001 A = U V T gives 01 0 ;0:0001 1 B C U =B @ 0 ;0:7071 0:7071 CA 0 0:7071 0:7071 0 1:4142 0 1 B C =B @ 0 0:0001 CA 0 0 ! 0:7071 ;0:7071 V= 0:7071 0:7071
y y= y
1 2
!
!
1:4142 = ; 0
!
1 x = Vy = : 1
7.9 Underdetermined Systems Let A be m n and m < n. Then the system
Ax = b has more equations than unknowns. Such a system is called an underdetermined system. An underdetermined system can be illustrated graphically as follows:
m
=
A
b x
Underdetermined systems arise in a variety of practical applications. Unfortunately, underdetermined systems are not widely discussed in the literature. An excellent source is the survey paper by Cline and Plemmons (1976). An underdetermined system has either no solution or an in nite number of solutions. This can be seen from the following theorem: 427
Theorem 7.9.1 Let Ax = b be an underdetermined system. Then every solution x of the system can be represented as
x = xR + xN ; where xR is any solution; that is,
AxR = b; and xN is in the null space of A, that is,
AxN = 0:
Note. A full-rank underdetermined system has an in nite number of solutions. In
this case, the complete solution set is given by
An Expression for the Solution Set of the Full-Rank Underdetermined System x = AT (AAT ); b + (I ; AT (AAT ); A)y , where y is arbitrary. 1
1
In the following, we describe two approaches for computing the minimum norm solution assuming that A has full rank, i.e., rank(A) = m.
428
7.9.1 The Minimum Norm Solution of the Full-rank Underdetermined Problem Using Normal Equations Theorem 7.9.2 Let A be m n (m < n) and have full rank. Then the unique minimum norm solution x to the underdetermined system
Ax = b is given by
(7.9.1)
x = AT (AAT ); b: 1
(7.9.2)
Proof. Since A has full rank, AAT is nonsingular and clearly x is the unique solution of the system (7.9.1) given by (7.9.2). To see that x is indeed the minimum norm solution, let's assume that y is another solution and de ne z = y ; x: Then
kyk = kx + zk 2 2
2 2
= kxk22 + kz k22 + 2(xT z ) = kxk22 + kz k22 + 2(bT (AAT );1A(y ; x)) = kxk22 + kz k22 (since A(y ; x) = Ay ; Ax = b ; b = 0). This shows that
kyk kxk : 2 2
2
Thus, x is the minimum norm solution. The above discussion suggests immediately the following algorithm.
Algorithm 7.9.1 Minimum Norm Solution for the Full-rank Undetermined Problem Using Normal Equations Step 1: Solve
AAT y = b: 429
Step 2: Form the minimum norm solution x: x = AT y:
De nition 7.9.1 The equations
AAT y = b
are called the normal equations for the underdetermined system (7.9.1).
Example 7.9.1 !
1 2 3 A= ; 2 3 4
!
6 b= : 9
Step 1: AAT = y =
Step 2:
14 20 20 29 ;1 ! : 1
!
011 B CC x = AT y = B @1A: 1
Remarks: The same diculties with the normal equations, as pointed out in the overdetermined case, may arise in the underdetermined case as well. For example, consider the explicit formation of AAT , with t-digits, when ! 1 0 A= : 1 0 If is such that 10;t < < 10; 2 , then t
1 + 2 1 T AA = 1 1 + 2 is singular.
430
!
7.9.2 The QR Approach for the Full-Rank Underdetermined Problem A QR factorization of A can, of course, be used to solve an underdetermined system and, in particular, to compute the minimum norm solution. Since A in this case has more rows than columns, we decompose AT into QR instead of A. Thus, we have ! R QT AT = : 0 (Note that AT is n m; n m.) So, the system Ax = b becomes (RT ; 0T )QT x = b: Denote y = QT x. Then we have
(RT ; 0T )y = b:
Partitioning y conformably, we get
!
yR =b yN
(RT ; 0T ) or
RT yR = b:
The unique minimum norm least squares solution is obtained by setting
yN = 0: So, the minimum norm solution is given by
x = Qy = (Q ; Q )
yR
!
0 The above discussion suggests the following algorithm. 1
2
= Q1yR :
Algorithm 7.9.2 Minimum Norm Solution for the Full-rank Underdetermined Problem Using QR Step 1: Find the QR factorization of AT : ! R QT AT =
Step 2: Partition Q = (Q ; Q ). Step 3: Solve 1
0
2
RT yR = b:
Step 4: Form x = Q yR. 1
431
:
A note on implementation. Note that if we use the Householder method to compute the
QR factorization of A, the product Q1yR can be computed from the factored form of Q as the product of Householder matrices.
Flop-count. Using Householder orthogonalization, m n ; m ops will be required to imple3
2
3
ment the above algorithm.
Round-o property. It has been shown (Lawson and Hanson SLP (pp. 93)) that the computed vector x^ is close to the exact minimum length least squares solution of a perturbed problem.
That is, there exist a matrix E and a vector x^ such that x^ is the minimum length solution of (A + E )^x b where
kE kF (6n ; 3m + 41)m kAkF + O( ): 2
Example 7.9.2 !
Step 2:
1 2 3 A= ; 2 3 4
6 b= 9
0 ;:2673 B ;:5345 Q =B @ ;:8018 0 ;3:7417 B ;5:3452 R=B @
:8729 1 C :2182 C A ;:4364
1
Step 3: Step 4: The minimum norm solution:
0
1
0 C: :6547 C A 0
;1:6036 ! yR = : 0:6547
011 B 1 CC : x = Q yR = B @ A 1
1
432
!
:
7.10 Iterative Re nement It is natural to wonder if a computed least squares solution x can be improved cheaply in an iterative manner, as was done in the case of a linear system. A natural analog of the iterative re nement procedure for the linear system problem described in a section of Chapter 6 will be the following. The scheme was proposed by Golub (1965).
Algorithm 7.10.1 Linear System Analog Iterative Re nement For Least Squares Solution Let x(1) be an approximate solution of the least squares problem. Then For k = 1; 2; : : :, do until convergence occurs: (1) r(k) = b ; Ax(k) (compute the residual).
(2) Compute c(k) such that Ac(k) ; r(k) 2 is minimum. (3) Correct the solution:
xk
( +1)
= x(k) + c(k):
An analysis of the above re nement procedure by Golub and Wilkinson (1966) reveals that the method is satisfactory only when the residual vector r = b ; Ax is suciently small. A successful procedure used widely in practice now follows. The method is based upon an interesting observation made by Golub that the least squares solution x and the corresponding residual vector r satisfy the linear system
I A AT 0
Note that the above system is equivalent to
! r! x
=
b
0
!
:
Ax + r = b and
AT r = 0;
which means that x is the solution of the normal equation
AT Ax = AT b: Thus one can apply the iterative re nement procedure described in Chapter 6 to the above augmented linear system, yielding the following scheme, which is due to Bjorck (1967a; 1968). Ake Bjorck is a Swedish numerical linear algebraist, well known for his outstanding contributions to solutions of least squares problems. He is a professor of mathematics at Linkoping University in Sweden.
433
Algorithm 7.10.2 Iterative Re nement for Least Squares Solutions Set r(0) = 0; x(0) = 0. For k = 1; 2; : : : do
rk (1) Compute k r
( ) 1
!
( ) 2
b
!
I A = ; T 0 A 0
in double precision.)
! rk ! ( )
xk
( )
! ck !
. (Using accumulation of inner products
!
rk = k . ck r ! rk ! ck ! rk (3) Update the solution and the residual: = k + k . xk x c I A (2) Solve the system T A 0
( ) 1
( ) 1
( ) 2
( ) 2
( +1)
( )
( ) 1
( +1)
( )
( ) 2
Remark: Note that for satisfactory performance of the algorithm, step 1 must be performed
using double precision; which is in contrast with the iterative re nement procedure for the linear system problem given in Section 6.6, where accumulation of inner products is not necessary.
! I A
Implementation of Step 2
Since the matrix T is of order m + n, the above scheme would be quite expensive A 0 when m is large. ! Fortunately, one can implement the scheme rather cheaply. Observe that if R 1 QT A = is the QR decomposition of A, then the system 0
I A AT 0
can be transformed into
QT c
+
! c!
R
c
1
!
1 2
r = r
1 2
c = QT r ;
0 2 (RT1 ; 0)QT c1 = r2 :
1
!
1
This shows that the above augmented system can be solved by solving two triangular systems and two matrix-vector multiplications as follows:
r0 ! (1) . r0 (2) Solve for c0 : RT c0 = r : Form QT r1 =
1 2
2
1
2
2
(3) Solve for c2: R1c2 = r10 ; c02:
!
c0 (4) Form c = Q 0 . r 2
1
2
434
Flop-count. With the above formulation each iteration will require only 4mn ; n op, 2
assuming that the Householder method has been used and that Q has not been formed explicitly. Note that for the matrix-vector multiplications in steps 1 and 3, Q does not need to be formed explicitly; these products can be obtained if Q is only known in implicit form.
Round-o error. It can be shown (Bjorck (1992)) that the solution x s obtained at the ( )
iteration step s, satis es
x s ; x
kx s; ; xk < c Cond(A); s = 2; 3; : : :; and ( )
(
1)
2
2
c is an error constant.
An Interpretation of the Result and Remarks The above result tells us that the iterative re nement procedure is quite satisfactory. It is even more satisfactory for least squares problems with large residuals. Note that for these problems (Cond(A))2 serves as the condition number. On the other hand, the above result shows error at an iterative re nement step depends upon the condition number of A. The procedure \may give
solutions to full single precision accuracy even when the initial solution may have no correct signi cant gures" (Bjorck (1992)). For a well-conditioned matrix, the convergence may occur even in one iteration. Bjorck and Golub (1967) have shown that with a 8 8 ill-conditioned Hilbert matrix, three digits accuracy per step both for the solution and the residual can be obtained.
Example 7.10.1 01 21 B CC A=B @2 3A; 3 4
r = 0; (0)
031 B CC b=B @5A
9 k=0
x = 0; (0)
031 BB 5 CC ! ! BB CC r b = =B BB 9 CCC : r 0 B@ 0 CA (0) 1 (0) 2
0
435
(2) Solve the system:
I A AT 0
! c ! (0) 1
c
(0) 2
r = r
(0) 1
!
(0) 2
0 0:3333 1 BB ;0:6667 CC ! BB CC c B = B 0:3333 C CC : c BB @ 3:3333 CA ;0:3333 (0) 1 (0) 2
(3) Update the solution and the residual:
0 0:3333 1 BB ;0:6667 CC ! ! ! BB CC r r c B = + = B 0:3333 C CC : x x c BB @ 3:3333 CA ;0:3333 (1)
(0)
(0) 1
(1)
(0)
(0) 2
The computations of c(0) and c(0) are shown below. 1 2 ! 0 r1(0) = b; r2(0) = . 0 0 ;10:6904 1 ! 0 r1 CC. Thus, Tb = B B = Q ; 0 : 2182 @ A r20 0:8165 ;10:6904 ! 0 r1 = 0:2182 0 r2 = (0:8165):
!
0 . 2 = 0 ! 3:3333 (0) c2 = . ;0:3333 0 0:3333 1 B ;0:6667 CC. c(0) =B 1 @ A 3:3333 Note that
c0
!
3:3333 x = ;0:3333 is the same least squares solution as obtained by the QR and normal equation methods. (1)
436
7.11 Computing the Variance-Covariance Matrix (AT A);1 In statistical and image processing applications very often one needs to compute the variancecovariance matrix X = (AT A);1 : In order to see how this matrix arises in statistical analysis, consider the classical Linear Regression Model: b = Ax + ; with the properties 1. E () = 0; and 2. cov() = E (T ) = 2I . Here b is the response vector, A is the design matrix, and is the error term. The parameters x and 2 are unknown. Regression analysis is concerned with prediction of one or more (response) dependent variables from the values of a group of predictor (independent) variables. Assuming that A has full rank, then the least squares estimate of x in the above model is, clearly, given by
x^ = (AT A); AT b: 1
Also,
^ = b ; ^b = (I ; A(AT A);1AT )b
(note that AT ^ = 0 and ^b^ = 0). A common estimate of 2 is: 2 = kb ; Ax^k2 , where A is m n. m;n For details, see Johnson and Wichern (1988), and Bjorck (1992). Since vital information might be lost in computing AT A explicitly, one wonders if it is possible to compute X from the QR factorization of A. We show below how this can be done. Consider the QR factorization of A: R~ ! T Q A= : 0 Then AT A = (R~ )T R~ 2
and, thus,
X = (AT A); = (R~ ); (R~ )T : 1
437
1
One can then easily compute (R~ );1, since R~ is an upper triangular matrix. Note that since X is symmetric, only one-half of its entries need to be computed, and X can overwrite R. However, computing (AT A);1 this way will require explicit multiplication of (R~ );1(R~ );T . This explicit multiplication can easily be avoided if the computations are reorganized. Thus, if x1 through xn are the successive columns of X , then from X = (R~ );1(R~);T ; we have
R~(x ; : : :; xn) = (R~ );T : is just r1 en, the last column, xn , is easily computed by solving the 1
Since the last column of R~ ;T upper triangular system
nn
~ n = 1 en: Rx r
(7.11.1)
nn
By symmetry we also get the last row of X . Suppose now that we have already determined xij = xji; j = n; : : :; k + 1; i j . We now determine xik ; i k. For xkk we have n X xkk rkk + rkjxjk = r1 ; kk
j =k+1
from which we obtain
0
1
n X rkj xkj A : xkk = r1 @ r1 ; kk kk j =k+1
(7.11.2)
(Note that xkj = xjk ; j = k + 1; : : :; n have already been computed.) For i = k ; 1; : : :; 1, we have
0
1
k n X X xik = ; r1 @ rij xjk + rij xkj A : (7.11.3) ii
j =i+1
j =k+1
Thus, (7.11.1){(7.11.3) determine all the entries of X .
Algorithm 7.11.1 Computing the Variance-Covariance Matrix X = (AT A); 1. Compute the last column xn by solving the system (7.11.1). 2. Compute xkk from (7.11.2), for k = n ; 1; n ; 2; :::; 2; 1: 3. Compute xik from (7.11.3), i = k ; 1; :::; 1; k = 2; 3; :::; n. 438
1
Flop-count. The computation of X = (AT A); using the above formulas requires about n3
3
1
ops.
Example 7.11.1 01 5 71 B 2 3 4 CC : A=B @ A 1 1 1
A = QR gives
0 ;0:4082 B ;0:8165 Q = B @ ;0:4082 0 ;2:4495 B R = B @ 0 0
0 4 1 B CC The last column x = B @ ;24 A (using (7.11.1))
0:9045 ;0:3015 ;0:3015 ;4:8990 3:3166 0
1
0:1231 C ;0:4924 C A 0:8616 ;6:5320 1 C 4:8242 C A: ;0:2462
3
16:5 x = 35 (using (7.11.2)) 3 x = ;6 5 (using (7.11.3)) . x = 1:5 0 1:5 ;6 4 1 B ;6 35 ;24 CC. Then X = (AT A); = B @ A 4 ;24 16:5 22
12 11
1
7.12 Review and Summary, and Table of Comparisons 1. Existence and Uniqueness. The least squares solution to the problem Ax = b always exists. In the overdetermined case, it is unique i A has full rank (Theorem 7.3.1).
2. Overdetermined Problems. We discussed two methods: the normal equations method and
the QR factorization method. The normal equations method is simple to implement, but there are numerical diculties with this method in certain cases. The QR factorization methods are more expensive than the normal equations method, but are more reliable numerically. If one insists on using normal equations, the use of extended precision is 439
recommended. One can use Householder or Givens transformations or the modi ed Gram-Schmidt method to achieve QR factorization of A needed for solution of the least squares problem. The Householder QR factorization is the most ecient and numerically viable; the modi ed Gram-Schmidt is only slightly more expensive than the Householder method and is not as reliable as the Householder or the Givens method numerically for QR factorization. However, with some reorganization it can be made competitive with the Householder method to compute the least squares solution. Indeed, popularity of using the MGS method in solving the least-squares problem is growing. Some numerical experiments even suggest the superiority of this method over other QR methods. See Bjorck (1992). Rice (1966) rst noted the superiority of this algorithm. For the rank-de cient overdetermined problem, there are in nitely many solutions. Here we have discussed only the QR factorization method with column pivoting. It is reasonably accurate, but the most numerically viable approach to deal with rank de ciency is the SVD approach to be discussed in Chapter 10.
3. Underdetermined Problems. For the underdetermined problem there are in nitely many solutions in the full-rank case. We have discussed the normal equations method and the QR factorization method for the minimum-norm solution. The normal equations here are:
AAT y = b; whereas those of the overdetermined problem are:
AT Ax = AT b: The numerical diculties with the normal equations are the same as those of the overdetermined problem. The round-o error analyses for the underdetermined and the rank-de cient overdetermined problems are somehow complicated, because solutions in these cases are not unique unless imposing the requirement of having the minimum-norm solution. The backward round-o analyses
of the QR factorization methods using Householder transformations in these cases show that the computed minimum-norm solution in each case is nearby the minimumnorm solution of a perturbed problem. This is in contrast with the results obtained by the Householder-QR factorization method for the full rank overdetermined problem, where it is shown that the computed solution is the exact solution of a perturbed problem.
4. Perturbation Analysis: The results of perturbation analyses are di erent for di erent cases of the perturbations in the data.
440
If only b is perturbed, then Cond(A) = kAk kAy k serves as the condition number for the unique least squares solution (Theorem 7.7.1). If A is perturbed, then the sensitivity of the unique least squares solution, in general, depends upon the square of the condition number (Theorem 7.7.2). In certain cases, such as when the residual is zero, the sensitivity depends only on the condition number of A.
5. Iterative Re nement: As in the case of the linear system problem, it is possible to improve
the accuracy of a computed least squares solution in an iterative fashion. An algorithm which is a natural analog to the one for the linear system (section 6.9) is satisfactory only when the residual vector r = b ; Ax is suciently small. A widely used algorithm due to Bjorck is presented in section 7.10. This algorithm requires solution of an augmented system of order m + n (where A is m n). It is shown how to solve the system in a rather inexpensive way using QR factorization of A. The solution obtained by this iterative re nement algorithm is quite satisfactory.
6. Computing the Covariance Matrix: An ecient computation of the matrix (AT A); is 1
important in statistical and image processing applications. In section 7.11, we show how to compute this matrix using QR factorization of A and without nding the inverse explicitly.
TABLE 7.2
COMPARISONS FOR DIFFERENT LEAST SQUARES METHODS
441
Problem
Method
Overdetermined Full-Rank
Normal Equations
Flop-Count mn2 + n3 2 6
Numerical Properties
(1) Diculties with formation of AT A (2) Produces more errors in the solution than there are warranted by data, in certain cases 3 n Overdetermined Householder-QR mn2 ; 3 Stable: The computed solution is the Full-Rank exact solution of a nearby problem Overdetermined MGS-QR mn2 Almost as stable as the Full-Rank Householder-QR 3 Overdetermined Householder-QR 2mnr ; r2(m + n) + 2r3 , Mildly stable: The computed minimumRank-De cient with column pivoting where r = rank(A) norm solution is close to the minimum-norm solution of a perturbed problem 3 m Underdetermined Normal m2 n + 6 Same diculties as in the case of the Full-rank Equations overdetermined problem 3 Underdetermined Householder-QR m2 n ; m3 Same as the rank-de cient Full-rank overdetermined problem
442
7.13 Suggestions for Further Reading The techniques of least squares solutions are covered in any numerical linear algebra text, and in some numerical analysis texts. The emphasis in most books is on the overdetermined problems. For a thorough treatment of the subject we refer the readers to the book by Golub and Van Loan (MC). The books by Stewart (IMC), by Watkins (FMC), and by Gil, Murray and Philip Numerical Linear Algebra and Optimization also contain detailed discussions on perturbation analyses. The books MC and FMC contain a very thorough treatment of the perturbation analysis. A complete book devoted to the subject is the (almost classical) book by C. L. Lawson and R. J. Hanson (SLP). There is a nice recent monograph by Bjorck (1992) that contains a thorough discussion on least squares problems. These two books are a \must" for the readers interested in further study on the subject. The SLP by Lawson and Hanson, in particular, gives the proofs of the round-o error analyses of the various algorithms described in the present book. Any book on regression analysis in statistics will contain applications of least squares problem in statistics. We have, in particular, used the book Applied Linear Regression Models, by John Neter, William Wasserman and Michael Kunter, Richard D. Irwin, Inc. Illinois (1983). A classical survey paper of Golub (1969) contains an excellent exposition of the numerical linear algebra techniques for the least squares and the singular value decomposition problems arising in statistics and elsewhere. A paper by Stewart (1987) is also interesting to read. The readers are highly recommended to read these papers along with other papers in the area by Golub, Bjorck, Stewart, etc. representing the most fundamental contributions in this area. See, for details, the references given in the book on these papers and the list of references appearing in the recent monograph of Bjorck (1992). For more on underdetermined problems, see the paper by Cline and Plemmons (1976).
443
Exercises on Chapter 7
PROBLEMS ON SECTION 7.3 1. Let A be m n; m n. Prove that the matrix AT A is positive de nite i A has rank n. 2. Show that the residual vector r = b ; Ax is orthogonal to all vectors in R(A). 3. Prove that x is a least squares solution i
Ax = bR and b ; Ax = bN ; where bR and bN are, respectively, the range-space and column-space components of the vector b. 4. Prove Theorem 7.3.1, both in the full-rank and rank-de cient cases, based on QR factorization of A.
PROBLEMS ON SECTIONS 7.5, 7.6, AND 7.7 5. Let A have linearly independent columns. Then prove that (Cond2 (A))2 = Cond2 (AT A). 6. Using least squares, t a straight line and a quadratic to
x 0 1 3 5 7 9 12 y 10 12 18 15 20 25 36 Compare your results. Compute the condition number of the associated Vandermonde matrix in each case. 7. Find the generalized inverse and the condition number of each of the following matrices. 0 1 1 1 ! 0:0001 1 B C A= ; A=B 0:0001 0 C @ A 1 1 0 0:0001 01 2 31 ! 7 6:990 B 3 4 5 CC : A= ; A=B @ A 4 4 0 7 8
444
8. Let A and A~ have full rank. Let x and x~ be, respectively, the unique least squares solutions to the problems Ax = b and
A~x~ = b;
where A~ = A + E . Then prove that
kx~ ; xk Cond(A) kE k 1 + kbk kx~k kAk kx~k k 1 + kE k : +(Cond(A)) kkE Ak kAk 2
(Hint: Apply the perturbation analysis of the linear systems with normal equations.) 9. Verify the inequality of problem 8 with the following data:
01 21 B CC A=B @3 4A;
031 B CC b=B @7A
5 6
E = 10; A 4
11
:
10. Verify the inequality of Theorem 7.7.4 in each of the following cases. 01 21 B3 4C (a) A = B @ C A ; E = 10;4A 05 1 6 1 1 B ;4 C ; E = 10;4A (b) A = B 0 C @ 10 A 0 10;4 01 11 B0 1C (c) A = B @ C A ; E = 10;3A: 0 0 11. Work out a proof of Theorem 7.7.2.
445
PROBLEMS ON SECTIONS 7.7 AND 7.8 12. Develop an algorithm to solve the normal equations whose system matrix is positive semide nite and singular, based on complete pivoting of the system matrix. Apply your algorithm to the normal equations with ! 1 1 A= 1 1 and ! 2 b= : 2
01 11 B 2 3 CC ; 13. Let A = B @ A (a)
(b) (c) (d) (e)
001 B 5 CC. b=B @ A
0 1 1 Find the unique least squares solution x using (i) x = Ay b, (ii) the normal equations method, (iii) the Householder and the Givens QR factorization methods, (iv) the CGS and MGS methods. Find Cond(A). Show that for this problem, the sensitivity of the least squares problem, when only A is perturbed, depends upon Cond(A). Let A = A = 10;4A and let x~ = A~y b where A~ = A + E . Find kx~k;xkxk and verify the inequality of Theorem 7.7.2 for this problem. Find r and r^ and verify the inequality of Theorem 7.7.3.
14. Construct your own example where the sensitivity of the least squares problem will depend upon the square of the condition number of the matrix. (Show all your work.) 15. Consider the following well-known ill-conditioned matrix 01 1 11 BB 0 0 CC CC ; jj 1 A=B BB CA 0 0 @ 0 0 (Bjorck (1990)). 446
(a) Choose an small, so that rank(A) = 3. Then compute Cond2 (A) to check that A is ill-conditioned. 031
BB CC (b) Find the least squares solution to Ax = B BB CCC using @A
(c) (d) (e) (f)
(i) the normal equations method, (ii) the QR factorization method with the Householder, CGS and MGS methods. 031 B 0 CC. Keep A unchanged. Find an upper bound for the relative error Change b to b1 = B @ A 0 in the least squares solution. Change A to A1 = A + A where A = 10;3 A. Keep b unchanged. Find an upper bound for the relative error in the least squares solution. Find the maximum departure from orthogonality of the computed columns of the Q matrix using the CGS and MGS methods. Compute the least squares solution of the problem in (b) using the SVD.
16. (Square-root free Cholesky) Given a symmetric positive de nite matrix A, develop an algorithm for nding the Cholesky decomposition of A without any square-roots:
A = LDLT ; where L is a unit lower triangular matrix and D is a diagonal matrix with positive diagonal entries. Apply your algorithm to solve the full-rank least squares problem based on solving normal equations. 17. (a) Derive a column-oriented version of the modi ed Gram-Schmidt process. (b) Show that the op-count for both the Gram-Schmidt and the modi ed Gram-Schmidt are mn2, where A is m n (m n). (c) Show that the op-count for the Householder method is n2 m ; n3 . 18. (a) Construct an example of the least squares problem where Gram-Schmidt will yield a poor result but modi ed Gram-Schmidt will do fairly well.
447
0 8 21 1 BB 13 34 CC CC, using t = 4. Show that (b) Apply the CGS and MGS methods to the matrix A = B BB @ 21 35 CA
34 39 in both cases the loss of orthogonality between q1 and q2 is unacceptable: q1T q2 = 0:09065.
19. Show that the op-count for QR factorization with column pivoting using the Householder method is 2mr ; r2(m + n) + 2r3=3, where r = rank(A). 20. Show that for the QR method with column pivoting in the rank-de cient case, the basic solution cannot be the minimum-norm solution unless R12 is zero. 21. (a) Show that the minimum-norm solution to Ax = b, with rank(A) = r, obtained by complete QR factorization with column pivoting is given by
T; c 1
x = PW
0
!
;
where c is a vector consisting of the rst r elements of QT b. (b) Find the minimum norm solution to the least squares problem with
01 1 11 B 1 1 1 CC ; A=B @ A 1 1 1
031 B 3 CC : b=B @ A 3
PROBLEMS ON SECTION 7.9 22. Develop an algorithm based on the QR factorization with column pivoting to compute the minimum norm solution to the underdetermined system
Ax = b; where A is m n; m < n. Give a op-count for the algorithm. Apply your algorithm to
0x ! 1 2 3 B Bx 4 5 6 @
1 2
x
3
1 ! 6 CC : A= 15
23. Develop an algorithm similar to the one given in section 7.8.2 for computing the minimum norm solution of the full-rank underdetermined problem. Apply your algorithm to problem #22. 448
24. Using Theorem 7.9.1, prove that the minimum norm solution to an underdetermined system can be obtained by projecting any solution to the system onto R(Ay ). That is, if PAy is the orthogonal projection onto R(Ay ), then the minimum norm solution x is given by
x = PAy y; where y is any solution. Using the above formula, compute the minimum norm solution to the system 0x 1 0 1 10;4 0 1 011 1 0 B C BB 1 0 10;4 0 CC BB x2 CC = BB 1 CC : @ A BB x3 CC @ A ; 4 1 0 0 10 @ A 1
x
4
25. Give a geometrical interpretation of Theorem 7.9.1.
PROBLEMS ON SECTION 7.10 26. Consider the natural algorithm for iterative re nement for improving a computed least squares solution described in the beginning of section 7.10 with x(1) = (0; : : :; 0)T . 001 B C B 0C B (1) 1. x = B C: .. C B @ . CA 0 2. For k = 1; 2; : : : do 2.1 r(k) = b ; Ax(k) . (Compute the residual.) 2.2 Solve the least squares problem: Find c(k) such that
k k
Ac ; r is minimum. ( )
( )
2
2.3 Correct the solution:
xk
( +1)
= x(k) + c(k):
(a) Apply three iterations of the algorithm to each of the following problems: 011 031 B CC B C (i) A = B @ 2 A ; b = B@ 5 CA 03 1 1 9 0 2 1 1 B ;4 C B C (ii) A = B 0 C @ 10 A ; b = B@ 10;4 CA 0 10;4 10;4 449
0 1 B ; (iii) A = B @ 10
1
0 1
1 1 CC BB CC 4 0 A; b = @1A 0 10;4 1 (b) What is the relationship of this algorithm with the iterative algorithm using the augmented system? (Algorithm 7.10.2) 27. Apply Algorithm 7.10.2 to each of the problems of exercise #26, with and without explicitly solving the augmented system in step 2, and compare the results. 28. Apply Algorithm 7.10.2 to the least-squares problem with the 7 7 Hilbert matrix and ; b = 1; 12 ; 13 ; 14 ; 51 ; 16 ; 17 T . Tell how many digits of accuracy per iteration step both in the solution and the residual were obtained.
PROBLEM ON SECTION 7.11 29. In many applications, only the diagonal entries of the Variance-Covariance matrix X = (AT A);1 are needed. Show that these diagonal entries are just the squares of the 2-norms of the rows of R;1, and that these can be computed only with 23 n3 ops.
450
MATLAB PROGRAMS AND PROBLEMS ON CHAPTER 7 You will need the programs housqr, givqr, clgrsch, mdgrsch, lsfrqrh, lsfrmgs, lsfrnme, lsudnme, lsudqrh, lsitrn1, lsitrn2, reganal from MATCOM. 1. Consider the following set of data points: x 0 1 2 3 4 5 6 7 8 9 y 1 2.9 6.8 12 20.5 30.9 42.9 51.5 73 90.5 Using the MATLAB Command vander and the operation `n' compute the least-square t of the data to polynomials of degree 1 through 4. Plot the original data point and the least-squares ts using the MATLAB command plot and polyval, and compare the results. 2. Let
0 1 1 011 1 BB 10; 0 CC BB 1 CC B C B C A=B B@ 0 10; CCA ; b = BB@ 1 CCA : 0 0 1 0 0 ;0:0001 1 0 :0001 1 BB CC BB CC 0 0 : 0009 : 0001 B C B C ; EA = 10; B B@ 0 0:0003 CCA ; b = a0 BB@ :0001 CCA : 3
3
4
4
0 0:0001 :0001 Using MATLAB commands pinv, cond, norm, orth, null, etc., verify the inequalities of Theorems 7.7.1-7.7.4 on di erent sensitivities of least-squares problems. 3. (Implementation of the Least-squares QR Algorithm using Givens Rotations). Using givqr and bcksub, write a MATLAB program to implement the QR algorithm using Givens Rotations for the full rank overdetermined least-square problem: [x] = lsfrqrg (A,b).
451
Test-Data for Problems 3, 4 and 5
For problems 4 and 5 use the following sets of test-data. 1. A randomly generated matrix of order 10. 2. Hilbert matrix of order 10. 0 1 1 1 1 BB 10;3 0 C 0 C B C. 3. B B@ 0 10;3 0 C C A ; 3 0 0 10 For problem #5, generate b so that the least-squares solution x0in each 1 3 BB 10;3 CC case has all entries equal to 1. (e.g. for data matrix in #3, b = BBB ;3 CCC @ 10 A 10;3 .) 4. (The Purpose of this exercise is to compare the accuracy and op-count and timing
for di erent methods for QR factorization of a matrix.)
a. Compute the QR factorization for each matrix A of the data set using i. [Q; R] = qr(A) from MATLAB or [Q; R] = housqr (A) from Chapter 5. (Householder QR). ii. [Q; R] = givqr (A) (Givens QR) iii. [Q; R]= clgrsch(A) from MATCOM (Classical Grass-Schmidt) iv. [Q; R] = mdgrsch(A), from MATCOM (Modi ed Gram-Schmidt) b. Using the results of (a), make the following table for each matrix A. Q^ and R^ stand for the computed Q and R.
452
Table
(COMPARISON OF DIFFERENT QR FACTORIZATION METHODS). METHOD k(Q^ )T Q^ ; I k2 housqr
kA ; Q^R^k kAk
Flop-Count
2
2
Elapsed Time
givqr clgrsch mdgrsch c. Write your conclusions. 5. (The purpose of this exercise is to compare the accuracy, op-count and timing
for di erent least-squares methods for full-rank, overdetermined problems.)
a. Compute the least-squares solution x^ for each data set using i. [x] = lsfrmgs(A,b) (Least-squares using modi ed Gram-Schmidt) ii. [x] = lsfrqrh (A,b) (Least-squares using Householder QR) iii. [x] = lsfrqrg (A,b) (Least-squares using Givens QR) iv. [x] = lsfrnme (A,b) (Least-squares using Normal Equations) v. [x] = pinv (A) * b (Least-squares using generalized inverse) Note: lsfrmgs, lsfrqrh and lsfrnme are all available in MATCOM. pinv is a MATLAB command for computing the generalized inverse of a matrix. b. Using the results of (a) make one table for each data set x^. Note also that the vector x has all entries equal to 1.
453
Table
(COMPARISON OF DIFFERENT METHODS FOR FULL-RANK OVERDETERMINED LEAST-SQUARES PROBLEM) METHOD
kx ; x^k =kxk 2
2
kAx^ ; bk
2
Flop-Count
Elapsed Time
lsfrmgs lsfrqrh lsfrqrg lsfrnme generalizedinverse c. Write your conclusions. 6. Using housqr from MATCOM or [Q; R] = qr(A) from MATLAB, and backsub from Chapter 6, write a MATLAB program, called lsrdqrh(A,b) to compute the minimum norm leastsquares solution x to the rank-de cient overdetermined problem Ax = b, and the corresponding residual r, using Householder QR factorization of A: [x; r] = lsrdqrh (A,b)
(This program should implement Algorithm 7.8.6 of the book.) Test Data: A 20 2 matrix with all entries equal to 1 and b is vector with all entries equal to 2.
7. Using the MATLAB function [U,S,V] = svd (A) to compute the singular value decomposition of A, write a MATLAB program, called lsrdsvd to compute the minimum norm least-squares solution x to the rank-de cient overdetermined system Ax = b: [x] = lsrdsvd (A,b) Use the same test data as in problem #6 and compare the results with respect to accuracy,
op-count and elapsed time. 454
8. Run the programs lsudnme (least-squares solution for the underdetermined full-rank problem using normal equations) and lsudqrh (least-squares solution for the underdetermined full rank problem using Housholder QR factorization) from MATCOM on the following sets of data to compute the minimum norm solution x to the full rank underdetermined problem Ax = b, and compare the results with respect to accuracy, elapsed time, and op-count. ! ! 1 2 3 4 5 6 1 1 1 1 1 1 1 1 A= ; A= ; 0 1 2 3 4 5 6 7
0 10 10 10 10 10 10 10 1 B 0 1 0 0 0 0 0 CC : A=B @ A
0 1 0 0 0 0 0 Construct b for each A so that the minimum norm solution x has all its entries equal to 1. 9. Run the programs lsitrn1 (based on Algorithm 7.10.1) and lsitrn2 (based on Algorithm 7.10.2) from MATCOM on the 88 Hilbert matrix A and constructing b randomly. Compare the algorithms on this data with respect to number of iterations, op-count, and elapsed time, both for the solution and the residual. 10. a. Compute (AT A);1 for each of the following matrices A as follows: i. Compute explicitly (AT A);1 using MATLAB command inv. ii. Run the program reganal from MATCOM. b. Compare the results with respect to accuracy, op-count, and elapsed time.
Test Data:
A = The 0 1 8 8 1 Hilbert1 matrix 1
BB 10; 0 A=B BB @ 0 10;
C
0 C CC 3 0 C A ; 3 0 0 10 A = A 8 8 randomly generated matrix 3
455
8. NUMERICAL MATRIX EIGENVALUE PROBLEMS
8.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 457 8.2 Some Basic Results on Eigenvalues and Eigenvectors : : : : : : : : : : : : : : : : : : 458 8.2.1 Eigenvalues and Eigenvectors : : : : : : : : : : : : : : : : : : : : : : : : : : : 458 8.2.2 The Schur Triangularization Theorem and its Applications : : : : : : : : : : 460 8.2.3 Diagonalization of a Hermitian Matrix : : : : : : : : : : : : : : : : : : : : : : 463 8.2.4 The Cayley-Hamilton Theorem : : : : : : : : : : : : : : : : : : : : : : : : : : 466 8.3 The Eigenvalue Problems Arising in Practical Applications : : : : : : : : : : : : : : 466 8.3.1 Stability Problems for Di erential and Di erence Equations : : : : : : : : : : 467 8.3.2 Vibration Problem, Buckling Problem and Simulating Transient Current of an Electrical Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 473 8.3.3 An Example of the Eigenvalue Problem Arising in Statistics: Principal Components Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 488 8.4 Localization of Eigenvalues : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 490 8.4.1 The Gersgorin Disk Theorems : : : : : : : : : : : : : : : : : : : : : : : : : : : 490 8.4.2 Eigenvalue Bounds and Matrix Norms : : : : : : : : : : : : : : : : : : : : : : 494 8.5 Computing Selected Eigenvalues and Eigenvectors : : : : : : : : : : : : : : : : : : : 495 8.5.1 Discussions on the Importance of the Largest and Smallest Eigenvalues : : : 495 8.5.2 The Role of Dominant Eigenvalues and Eigenvectors in Dynamical Systems : 496 8.5.3 The Power Method, The Inverse Iteration and the Rayleigh Quotient Iteration497 8.5.4 Computing the Subdominant Eigenvalues and Eigenvectors: De ation : : : : 508 8.6 Similarity Transformations and Eigenvalue Computations : : : : : : : : : : : : : : : 515 8.6.1 Eigenvalue Computations Using the Characteristic Polynomial : : : : : : : : 515 8.6.2 Eigenvalue Computations via Jordan-Canonical Form : : : : : : : : : : : : : 520 8.6.3 Hyman's Method : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 520 8.7 Eigenvalue Sensitivity : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 521 8.7.1 The Bauer-Fike Theorem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 521 8.7.2 Sensitivity of the Individual Eigenvalues : : : : : : : : : : : : : : : : : : : : : 524 8.8 Eigenvector Sensitivity : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 527 8.9 The Real Schur Form and QR Iterations : : : : : : : : : : : : : : : : : : : : : : : : : 529 8.9.1 The Basic QR Iteration : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 532 8.9.2 The Hessenberg QR Iteration : : : : : : : : : : : : : : : : : : : : : : : : : : : 535 8.9.3 Convergence of the QR Iterations and the Shift of Origin : : : : : : : : : : : 536
8.10 8.11 8.12 8.13 8.14
8.9.4 The Single-Shift QR Iteration : : : : : : : : : : : : : : : 8.9.5 The Double-Shift QR Iteration : : : : : : : : : : : : : : 8.9.6 Implicit QR Iteration : : : : : : : : : : : : : : : : : : : 8.9.7 Obtaining the Real Schur Form A : : : : : : : : : : : : 8.9.8 The Real Schur Form and Invariant Subspaces : : : : : Computing the Eigenvectors : : : : : : : : : : : : : : : : : : : : 8.10.1 The Hessenberg-Inverse Iteration : : : : : : : : : : : : : 8.10.2 Calculating the Eigenvectors from the Real Schur Form The Symmetric Eigenvalue Problem : : : : : : : : : : : : : : : 8.11.1 The Sturm Sequence and the Bisection Method : : : : : 8.11.2 The Symmetric QR Iteration Method : : : : : : : : : : The Lanczos Algorithm For Symmetric Matrices : : : : : : : : Review and Summary : : : : : : : : : : : : : : : : : : : : : : : Suggestions for Further Reading : : : : : : : : : : : : : : : : :
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CHAPTER 8 NUMERICAL MATRIX EIGENVALUE PROBLEMS
8. NUMERICAL MATRIX EIGENVALUE PROBLEMS Objectives The major objectives of this chapter are to study numerical methods for the matrix eigenvalue problem. Here are some of the highlights of the chapter.
Some practical applications (Section 8.3) giving rise to the eigenvalue problem. The Power method and the Inverse power methods for nding the selected eigenvalues and eigenvectors (Section 8.5).
Eigenvalue and Eigenvector sensitivity (Sections 8.7 and 8.8). The QR iteration algorithm for nding the eigenvalues of a matrix of moderate size (Section 8.9).
Inverse Iteration and Real Schur methods for nding the eigenvectors (Section 8.10). Special methods for the symmetric eigenvalue problem: symmetric QR iteration with Wilkinson shift and the bisection method (Section 8.11).
The Lanczos methods for nding the extremal eigenvalues of a large and sparse symmetric matrix (Section 8.12).
Required Background
The following concepts and tools developed earlier in this book will be required for smooth reading and understanding of the material of this chapter. 1. Norm Properties of Matrices (Section 1.7) 2. The QR factorization of an arbitrary and a Hessenberg matrix using Householder and Givens transformations (Sections 5.4.1 and 5.5.1) 3. Linear system solutions with arbitrary, Hessenberg, and triangular matrices (Section 6.4) 4. The condition number and its properties (Section 6.7) 5. Reduction to Hessenberg and tridiagonal form (sections 5.4.3 and 5.5.3).
456
8.1 Introduction This chapter is devoted to the study of numerical matrix eigenvalue problem. The problem is a very important practical problem and arises in a variety of applications areas including engineering, statistics, economics, etc. Since the eigenvalues of a matrix A are the zeros of the characteristic polynomial det(A ; I ), one would naively think of computing the eigenvalues of A by nding its characteristic polynomial and then computing its zeros by a standard root- nding method. Unfortunately, this is not a practical approach. A standard practical algorithm for nding the eigenvalues of a matrix is the QR iteration method with a single or double shift. Several applications do not need the knowledge of the whole spectrum. A few selected eigenvalues, usually a few largest or smallest ones, suce. A classical method, based on implicit powering of A, known as the power method is useful for this purpose. The symmetric eigenvalue problem enjoys certain remarkable special properties; one such property is that the eigenvalues of a symmetric matrix are well conditioned. A method based on the exploitation of some of the special properties of the symmetric eigenvalue problem, known as the Sturm Sequence-Bisection method, is useful for nding eigenvalues of a symmetric matrix in an interval (note that the eigenvalues of a symmetric matrix are all real). The QR iteration method, unfortunately, cannot be used to compute the eigenvalues and eigenvectors of a very large and sparse matrix. The sparsity gets lost and the storage becomes an issue. An almost classical method by Lanczos has been rejuvenated by numerical linear algebraists recently. The symmetric Lanczos method is useful in particular to extract a few extremal eigenvalues and the associated eigenvectors of a very large and sparse matrix. The organization of this chapter is as follows. Some of the basic theoretical results on eigenvalues and eigenvectors are stated (and proved sometimes) in Section 8.2. Section 8.3 is devoted to the discussions of how the eigenvalue problem arises in some practical applications such as stability analyses of a system of di erential and di erence equations, vibration analysis, transient behaviour of an electrical circuit, buckling problem, principal component analysis in statistics, etc. In Section 8.4 some classical results on eigenvalues locations such as the Gersgorin's disk theorem etc. are stated and proved. Section 8.5 describes the power method, the inverse power method, the Rayleigh-Quotient iteration, etc. for nding a selected number of eigenvalues and the corresponding eigenvectors. In Section 8.6 the diculties of computing the eigenvalues of a matrix via the characteristic 457
polynomial and the Jordan canonical form are highlighted. The eigenvalue and eigenvector sensitivity are discussed in sections 8.7 and 8.8. The most important result in this section is the Bauer-Fike theorem. Section 8.9 is the most important section of this chapter. The QR iteration method with and without shifts and their implementations are described in this section. The Hessenberg-inverse iteration and computations of eigenvectors from the Real Schur form are described in Section 8.10. The methods based on speci c properties of symmetric matrices are discussed in Section 8.11. These include the Sturm sequence-bisection method and symmetric QR iteration method with the Wilkinson shift. The chapter concludes with a brief introduction of the symmetric Lanczos algorithm for nding the extremal eigenvalues of a large and sparse symmetric matrix.
8.2 Some Basic Results on Eigenvalues and Eigenvectors 8.2.1 Eigenvalues and Eigenvectors Let A be an n n matrix. Then is an eigenvalue of A if there exists a nonzero vector x such
that
Ax = x or (A ; I )x = 0: The vector x is a right eigenvector (or just an eigenvector) of A associated with the eigenvalue . The vector y given by y A = y is called a left eigenvector of A associated with the eigenvalue . It is customary to call a right eigenvector just an eigenvector. The homogeneous system (A ; I )x = 0 has a nontrivial solution i det(A ; I ) = 0:
PA () = det(A ; I ) is a polynomial in of degree n and is called the characteristic polynomial
of A.
Thus the n eigenvalues of A are the n roots of the characteristic polynomial. The sum of the eigenvalues of a matrix A is called the trace of A. It is denoted by trace (A) or Tr (A). 458
Recall also from Chapter 1, that if A = (a ), then trace(A) = a11 + + a = sum of the diagonal entries of A. ij
nn
De nition 8.2.1 The eigenvalue with the largest modulus is called spectral radius of A and is denoted by (A) : (A) = max (jij). i De nition 8.2.2 The set of eigenvalues of a matrix A is called the spectrum of A. We now present some basic results on eigenvalues and eigenvectors.
Theorem 8.2.1 The eigenvalues of a triangular matrix are its diagonal entries. Proof. If A = (aij) is triangular, then the matrix (A ; I ) is also triangular with a ; ; a ; ; : : :; ann ; as its diagonal entries. So, n Y det(A ; I ) = (aii ; ); 11
22
i=1
showing that the eigenvalues of A are a11; a22; : : :; ann.
Theorem 8.2.2 The matrices A and TAT ; have the same eigenvalues. In other words, the eigenvalues of two similar matrices are the same. 1
Proof. det(TAT ;1 ; I ) = = = = =
det(T (A ; I )T ;1) det T det(A ; I ) det(T ;1) det T det(T ;1) det(A ; I ) det(T T ;1 ) det(A ; I ) det(A ; I ):
Thus, TAT ;1 and A have the same characteristic polynomial and, therefore, have the same eigenvalues.
459
Note: The converse is not true. Two matrices having the same eigenvalues are not necessarily similar. Here is a simple ! example. 1 0 ! 1 1
Take A = ; B = I22 = . A and B have the same eigenvalues, but they cannot 0 1 0 1 be similar.
8.2.2 The Schur Triangularization Theorem and its Applications Theorem 8.2.3 For any complex matrix A there exists a unitary matrix U such that
UAU = T
is a triangular matrix. The eigenvalues of A are the diagonal entries of T . The theorem is of signi cant theoretical importance. As we shall see below, several important results on eigenvalues and eigenvectors can be derived using this theorem. We shall give a proof of this theorem later in the chapter (Theorem 8.9.1).
Theorem 8.2.4 Let ; : : :; n be the eigenvalues of A, then the eigenvalues of Am m1 ; m2 ; : : :; mn.
1
are In general, if p(x) is a polynomial of degree n,
p(x) = c xn + c xn; + + cn; 0
1
1
then the eigenvalues of p(A) are p(1); p(2); : : :; p(n).
Proof. From U AU = T we have T m = (U AU )(U AU ) (U AU ) (m times) = U Am U: But T m is also a triangular matrix with m1 ; ; mn as its diagonal entries. Since Am and T m are similar, they have the same eigenvalues. 460
In the general case,
U p(A)U = U (c An + c An; + + cnI )U = c U An U + c U An; U + + cn U U = c T n + c T n; + + cnI = T = a triangular matrix. 0
0 0
1
1
1
1
1
1
1
The diagonal entries of the triangular matrix T1 are p(1); p(2); : : :; p(n): Since p(A) and T1 have the same eigenvalues, the theorem is proved.
Theorem 8.2.5 The eigenvalues of a Hermitian matrix A are real. The eigenvec-
tors corresponding to the distinct eigenvalues are pairwise orthogonal.
Proof. Since U AU
= have = = = =
we (U AU ) or U A U or U AU or T
T; T T T T :
Thus, T is also Hermitian and, therefore, diagonal. Since the eigenvalues of A are the diagonal entries of T , and the diagonal entries of a Hermitian matrix must be real, it follows that the eigenvalues of A are real.
Remark: Note that if A is real, then A = A implies that A is symmetric (AT = A). The eigenvalues of a real symmetric matrix are also real.
To prove the second part, let 1 and 2 be two distinct eigenvalues of A, and let x and y be the associated eigenvectors. Then by de nition, we have
Ax = x; x 6= 0; 1
and
Ay = y; y 6= 0; 2
from which we obtain
y Ax = y x 1
461
and
y Ax = y x = y x = y x (since = ): 2
2
2
2
2
Since A = A, we get
yx = yx or ( ; )y x = 0: 1
2
1
2
But, 1 6= 2, so, y x = 0 thus, y and x are orthogonal.
Theorem 8.2.6 The eigenvalues of a Hermitian positive de nite matrix are positive
and conversely, if a Hermitian matrix has all its eigenvalues positive, it must be positive de nite.
Proof. Let be an eigenvalue of A and x be the corresponding eigenvector. Then from Ax = x we obtain
x Ax = xx: Since A is positive de nite, x Ax > 0 and is real. Since A is Hermitian, we then have x x > 0: But, x x is always positive, so > 0. To prove the converse, we note that, since A is Hermitian, by the Schur Triangularization theorem, (Theorem 8.2.3), we have
U AU = diag(i) = diag( ; : : :; n):
P
1
De ne now x = Uy , then x Ax = y U AUy = iyi2 . Since the i's are real and positive, xAx > 0. Again, every nonzero y corresponds to a nonzero x. Thus, for every nonzero x, we have x Ax > 0, proving that A is positive de nite.
Theorem 8.2.7 The eigenvectors associated with the distinct eigenvalues of a ma-
trix are linearly independent.
462
Proof. Let ; : : :; m be the distinct eigenvalues and x ; : : :; xm be the corresponding eigenvectors. Consider
1
1
c x + + cm xm = 0 1
1
where ci ; i = 1; : : :; m are constants. Multiplying by (1I ; A) to the left, we get
c ( I ; A)x + c ( I ; A)x + + cm ( I ; A)xm = 0; 1
or
1
1
2
1
2
1
c ( ; )x + + cm ( ; m )xm = 0: 2
1
2
2
1
(since Ax1 = 1x1):
Multiplying now to the left of the last equation by (2I ; A) we get
c ( ; )( ; )x + + cm( ; s)( ; s)xm = 0: 3
1
3
2
3
3
1
2
Continuing this way (m ; 1) times, we conclude that cm = 0, because i 6= j . Analogously, we can show that cm;1 = cm;2 = = c2 = c1 = 0: Thus, the vectors x1 ; : : :; xm are linearly independent.
8.2.3 Diagonalization of a Hermitian Matrix Theorem 8.2.8 For any Hermitian matrix A, there exists an unitary matrix U such that
U AU = D
is a diagonal matrix. The diagonal entries of D are eigenvalues of A.
Proof. It is an immediate consequence of the Schur Triangularization Theorem. Since A = A from we have
U AU = T; T = T:
Since T is triangular, T = T implies that it is diagonal. 463
Theorem 8.2.9 An arbitrary matrix A is similar to a diagonal matrix if its eigen-
values are distinct.
Proof. Let x ; : : :; xn be the eigenvectors corresponding to the eigenvalues ; : : :; n. De ne 1
1
X = (x ; : : :; xn): 1
Then,
AX = A(x ; : : :; xn0) = ( x ; : : :; nx1n) 0 BB CC B CC = XD = (x ; : : :; xn) B B@ CA 0 n where D = diag( ; : : :; n). Since x ; : : :; xn are the eigenvectors corresponding to the distinct eigenvalues ; : : :; n, they are linearly independent (by Theorem 8.2.7). Therefore, X is nonsingular. So, 1
1 1
1
1
1
1
1
X ; AX = D: 1
De nition 8.2.3 An n n matrix A is said to be nondefective if it is diagonalizable; that is, if it is similar to a diagonal matrix. Otherwise, the matrix is defective. (Compare this de nition with that given in Chapter 1 (De nition 1.3.1))
Thus any complex Hermitian matrix, any real symmetric matrix and a matrix A having distinct eigenvalues are nondefective. It is clear that an arbitrary matrix can not be reduced to a diagonal matrix by similarity. However, it can always be reduced to a block diagonal matrix as the following result shows.
464
Theorem 8.2.10 (The Jordan Theorem) For an n n matrix A, there exists a nonsingular matrix T such that
0J BB 0 BB B T ; AT = B BB BB B@ 0 1
1
where
Jm
1 CC CC CC CC CC CA
0 1 1 i BB 1 0 CC i BB CC BB CC Ji = B C BB C CC BB CA 0 1 @ i
where the i are eigenvalues of A.
Proof. See Horn and Johnson (1985, pp. 121{129). Jordan Canonical Form The diagonal block matrix above is called the Jordan Canonical Form of A. The matrices J ; : : :; Jm are Jordan matrices. If Ji is of degree vi , then 1
v + + vm = n: 1
De nition 8.2.4 The polynomials det(Ji ; I ) = (i ; )v
i
are called the elementary divisors of A. If v1 = v2 = = vm = 1 then the elementary divisors are called linear elementary divisors.
Notes: We note the following: (1) When all the eigenvalues of A are distinct, each Jordan block has order 1 and therefore elementary divisors are linear. In fact, if there are m Jordan blocks, there are m linearly independent eigenvectors, counting the multiplicities of the same block. 465
(2) If there are some nonlinear elementary divisors, then A has eigenvalues whose multiplicities are higher than or equal to the number of independent eigenvectors associated with them. (3) For a real symmetric or complex Hermitian matrix, the elementary divisors are linear.
8.2.4 The Cayley-Hamilton Theorem Theorem 8.2.11 A square matrix A satis es its own characteristic equation; that is, if A = (aij ) is an n n matrix and Pn () is the characteristic polynomial of A,
then
Pn (A) is a ZERO matrix.
Proof. See Matrix Theory by Franklin, pp. 113{114. Example 8.2.1
0 1 0 1 A: Let A = @ P () = 0 P (A) = @ 2
2
2
0 = @
1 2 ; 2 ;1 1: 0 1 0 1 1 2A 0 1 1 0 A;@ A ;2 @ 2 5 1 2 0 1 1 0 0A : 0 0
8.3 The Eigenvalue Problems Arising in Practical Applications The problem of nding the eigenvalues and eigenvectors arise in a wide variety of practical applications. It arises in almost all branches of science and engineering. As we have seen before, the mathematical models of many engineering problems are systems of di erential and di erence equations and the solutions of these equations are often expressed in terms of the eigenvalues and eigenvectors of the matrices of these systems. Furthermore, many important characteristics of physical and engineering systems, such as stability, etc., often can be determined only by knowing the nature and location of the eigenvalues. We will give a few representative examples in this section.
466
8.3.1 Stability Problems for Di erential and Di erence Equations A homogeneous linear system of di erential equations with constant coecients of the form: dx1(t) = a x (t) + a x (t) + + a x (t) n n dt dx (t) = a x (t) + a x (t) + + a x (t) n n dt . 2
or in matrix form
11
1
12
2
1
21
1
22
2
2
.. dxn(t) = a x (t) + a x (t) + + a x (t) n1 1 n2 2 nn n dt
x_ (t) = Ax(t);
where and
(8.3.1)
A = (aij )nn;
0 x (t) 1 B .. CC x_ (t) = dtd B @ . A; 1
xn(t)
arises in a wide variety of physical and engineering systems. The solution of this system is intimately related to the eigenvalue problem for the matrix A. To see this, assume that (8.3.1) has a solution x(t) = vet, where v is not dependent on t. Then we must have vet = Avet; (8.3.2) that is
Av = v;
(8.3.3)
showing that if is an eigenvalue of A and v is corresponding eigenvector. Thus the eigenpair (; v ) of A can be used to compute a solution x(t) of (8.3.1). If A has n linearly independent eigenvectors (which will happen, as we have seen in the last section, when the eigenvalue of A are all distinct), then the general solution of the system can be written as
x(t) = c v e1t + c v e2t + + cnvne t ; 1 1
n
2 2
(8.3.4)
where 1 ; : : :; n are the eigenvalues of A and v1; v2; : : :; vn are the corresponding eigenvectors. In the general case, the general solution of (8.3.1) with x(0) = x0 is given by
x(t) = eAtx ; 0
467
(8.3.5)
where
eAt = I + At + A2!t + : : : 2 2
The eigenvalues i and the corresponding eigenvectors now appear in the computation of eAt : For example, if A has the Jordan Canonical Form
V ; AV0= diag(J ; J ; : : :;1Jk); i 1 0 1
1
(8.3.6)
2
BB . . . . . . CC Ji = B BB .. . . . . . . CCC ; 1A @. 0 i
then
eAt = V diag(eJ1t ; eJ2 t ; : : :; eJ t)V ; ;
(8.3.8)
0 1 p =p! 1 t t = 2 t BB C BB 1 t tp; =(p ; 1)! CCC B CC .. ... ... eJ t = e t B . BB CC : . BB CC .. t @ A 1
(8.3.9)
( )
where
(8.3.7)
k
1
2
1
i
i
Thus, the system of di erential equations (8.3.1) is completely solved by knowing its eigenvalues and eigenvectors of the system matrix A. Furthermore, as said before, many interesting and desirable properties of physical and engineering systems can be studied just by knowing the location or the nature of the eigenvalues of the system matrix A. Stability is one such property. The stability is de ned with respect to an equilibrium solution.
De nition 8.3.1 An equilibrium solution of the system x_ (t) = Ax(t); x(0) = x ; 0
is the vector xe satisfying
(8.3.10)
Axe = 0:
Clearly, xe = 0 is an equilibrium solution and it is the unique equilibrium solution i A is nonsingular. A mathematical de nition of stability is now in order. 468
De nition 8.3.2 An equilibrium solution xe is said to be stable, if, for any t and > 0, there exists a real number (; t ) > 0 such that kx(t) ; xe k < wherever kx ; xe k : 0
0
0
The system (8.3.1) is stable if the equilibration solution xe = 0 is stable.
In other words, the stability guarantees that the system remains close to the origin if the initial position is chosen close to it. In many situations, stability is not enough, one needs more than that.
De nition 8.3.3 An equilibrium solution is asymptotically stable if it is stable and if, for any t , there is a such that 0
whenever
kx(t) ; xek ! 0; as t ! 1 kx ; xek < : 0
De nition 8.3.4 The system (8.3.1) is asymptotically stable if the equilibrium solution xe = 0 is asymptotically stable.
In other words, the asymptotic stability not only guarantees the stability but ensures that every solution x(t) converges
to the origin whenever the initial position is chosen close to it.
Since an asymptotically stable system is necessarily stable, but not conversely, the following convention is normally adopted.
De nition 8.3.5 A system is called marginally stable if it is stable, but not asymptotically stable.
De nition 8.3.6 A system that is not stable is called unstable. Mathematical Criteria for Asymptotic Stability 469
Theorem 8.3.1 (Stability Theorem For a Homogeneous System of Di erential Equations) A necessary and sucient condition for the equilibrium solution
xe = 0 of the homogeneous system (8.3.1) to be asymptotically stable is that the eigenvalues of the matrix A all have negative real parts.
Proof. It is enough to prove that x(t) ! 0 as t ! 1. Since the general solution of the system
x_ (t) = Ax(t) is given by x(t) = eAtx(0), the proof follows from (8.3.6){(8.3.9). Note that if j = j + i j ; j = 1; 2; : : :; n; then e t = e t ei t , and e t ! 0, when t ! 1, i j < 0: j
j
j
j
Stability of a Nonhomogeneous System Many practical situations give rise to mathematical model of the form
x_ (t) = Ax(t) + b;
(8.3.11)
where b is a constant vector. The stability of such a system is also governed by the eigenvalues of A. This can be seen as follows. Let x(t) be an equilibrium solution of (8.3.11). De ne
z(t) = x(t) ; x(t): Then
z_ (t) = x_ (t) ; (x_ )(t) = Ax(t) + b ; Ax(t) ; b = A(x(t) ; x(t)) = Az (t): Thus, x(t) ! x(t) if and only if z (t) ! 0. It, therefore, follows from Theorem 8.3.1 that:
Theorem 8.3.2 (Stability Theorem For a Nonhomogeneous System of Differential Equations) (i) An equilibrium solution of (8.3.11) is asymptotically stable i all the eigenvalues of A have negative real parts. (ii) An equilibrium solution is unstable if at least one eigenvalue has a positive real part.
470
Remarks: Since the stability of an equilibrium solution depends upon the eigenvalues of the matrix A of the system, it is usual to refer to stability of the system itself, or just the stability of the matrix A.
Stability of a System of Di erence Equations Like the system of di erential equations (8.3.11), many practical systems are modeled by a system of di erence equations: xk+1 = Axk + b. A well-known mathematical criterion for asymptotic stability of such a system is given in the following theorem. We leave the proof to the readers.
Theorem 8.3.3 (Stability Theorem for a System of Di erence Equations).
An equilibrium solution of
xk = Axk + b +1
(8.3.12)
is asymptotically stable i all the eigenvalues of A are inside the unit circle. The equilibrium solution is unstable if at least one eigenvalue has magnitude greater than 1.
Summarizing, to determine the stability and asymptotic stability of
a system modeled by a system of rst order di erential or di erence equations, all we need to know is if the eigenvalues of A are in the left half plane or inside the unit circle, respectively. Explicit knowledge of the eigenvalues is not needed.
Example 8.3.1 A European Arms Race Consider the arms race of 1909-1914 between two European alliances. Alliance 1: France and Russia. Alliance 2: Germany and Austria-Hungary. The two alliances went to war against each other. Let's try to explain this historical fact through the notion of stability. 471
First consider the following crude (but simple) mathematical model of war between two countries:
dx = k x ; x + g dt dx = k x ; x + g dt 1
2
where
1
2
1
1
1
2
1
2
2
2
xi(t) = war potential of the country i, i = 1; 2 gi(t) = the grievances that country i has against the other, i = 1; 2: gi , i and ki, i = 1; 2; are all positive constants. ixi denotes the cost of armaments of the country i. This mathematical model is due to L. F. Richardson, and is known as the Richardson model.
Note that this simple model is realistic in the sense that the rate of change of the war potential of one country depends upon the war potential of the other country, and on the grievances that one country has against its enemy-country, and the cost of the armaments the country can a ord. While the rst two factors cause the rate to increase, the last factor has certainly a slowing e ect (that is why we have a ; sign associated with that term). In matrix form, this model can be written as:
x_ (t) = Ax(t) + g where
0 ; A=@
1
1
k
2
0
1
k A x (t) A ; x(t) = @ ; ; x (t) 0 1 g g = @ A: g 1
1
2
2
1
The eigenvalues of A are
2
p ; ( + ) ( ; ) + 4k k : = 2 1
2
1
2
2
1 2
Thus the equilibrium solution x(t) is asymptotically stable if 1 2 ; k1 k2 > 0, and unstable if 1 2 ; k1 k2 < 0. This is because when 1 2 ; k1k2 > 0, both the eigenvalues will have negative real parts; if it is negative, one eigenvalue will have positive real parts. For the above European arms race, the estimates of 1 ; 2 and k1 ; k2 were made under some realistic assumptions and were found to be as follows:
= = 0:2 1
2
472
k = k = 0:9: 1
2
(For details of how these estimates were obtained, see the book by M. Braun (1978)). The main assumptions are that both the alliances have roughly the same strength and 1 and 2 are the same as Great Britain which is usually taken to be the reciprocal of the life-time of the British Parliament ( ve years).) With these values of 1; 2; and k1 ; k2, we have
; k k = ; k = ;:7700: 1
2
1
2 1
2
2 1
Thus the equilibrium is unstable. In fact, the two eigenvalues are: 1.4000 and ;2:2000. For a general model of Richardson's theory of arm races and the role of eigenvalues there, see the book Introduction to Dynamic Systems by David G. Luenberger, John Wiley and Sons, New York, 1979 (pp. 209-214).
Convergence of Iterative Schemes for Linear Systems In Chapter 6 (Theorem 6.10.1) we have seen that the iteration: xk = Bxk + d +1
for solving the linear system
(8.3.13)
Ax = b
converges to the solution x for an arbitrary choice of the initial approximation x1 i the spectral radius (B ) < 1.
We thus see again that only an implicit knowledge of the eigenvalues of B is needed to see if an iterative scheme based upon (8.3.13) is convergent. 8.3.2 Vibration Problem, Buckling Problem and Simulating Transient Current of an Electrical Circuit Analysis of vibration and buckling of structures, simulation of transient current of electrical circuits, etc., often give rise to a system of second-order di erential equations of the form:
B y + Ay = 0; where
0 BB B y = B BB @
1 CC C .. C C: . C A
y (t) y (t) 1 2
yn(t) 473
(8.3.14)
(8.3.15)
The solution of such a system leads to the solution of an eigenvalue problem of the type:
Ax = Bx:
(8.3.16)
This can be seen as follows: Let y = xeiwt be a solution of the system (8.3.14). Then from (8.3.14) we must have
w Bx = Ax: 2
Writing = w2, this becomes
Ax = Bx:
(8.3.17)
Such an eigenvalue problem is called a generalized eigenvalue problem. The number is called a generalized eigenvalue and x is the corresponding generalized eigenvector. Very often in practice the matrices A and B are symmetric and positive de nite. In such a case, an explicit solution of the system can be written down in terms of the generalized eigenvalues and eigenvectors. The eigenvalues of the generalized eigenvalue problem in this case are all real and positive. A solution of (8.3.14) in this case can be written down in terms of the eigenvalues and the eigenvectors.
Eigenvalue-Eigenvector Solution of By + Ay = 0 with A; B Positive De nite Let x1; x2; : : :; xn be the n independent eigenvectors corresponding to the generalized eigenvalues 1; 2; : : :; n, of (8.3.17), then
y=
n X i=1
ci cos(it + di)xi;
where ci and di are arbitrary constants. (See Wilkinson, AEP pp. 34{35.) In vibration problems the matrices B and A are, respectively, called the mass and sti ness matrices, and are denoted by M and K , giving rise to the symmetric generalized eigenvalue problem: Kx = Mx: 474
De nition 8.3.7 The quantities wi = pi; i = 1; : : :; n are called natural frequencies, and x ; : : :; xn are called the amplitudes of vibration of the masses. 1
The frequencies can be used to determine the periods Tp for the vibrations. Thus T = 2 pi
wi
is the period of vibration for the ith mode.
As we will see, the behavior of a vibrating system can be analyzed by knowing the natural frequencies and the amplitudes . We will give a simple example below to illustrate
this. An entire chapter (Chapter 9) will be devoted to the generalized eigenvalue problem later. Other vibration problems such as the buckling problem of a beam gives rise to boundary value problems for the second-order di erential equations. The solutions of such problems using nite di erences lead to also eigenvalue problems. We will also illustrate this in this section . We now describe below how the frequencies and amplitudes can be used to predict the phenomenon of resonance in vibration engineering. All machines and structures such as bridges, buildings, aircrafts, etc., possessing mass and elasticity experience vibration to some degree, and their design require consideration of their oscillatory behavior. Free vibration takes place when a system oscillates due to the forces inherent in the system, and without any external forces. Under free vibration such systems will vibrate at one or more of its natural frequencies, which are properties of the dynamical system and depends on the associated mass and sti ness distribution. For forced vibration, systems oscillate under the excitation of external forces. When such excitation is oscillatory the system is also forced to vibrate at the excitation frequency.
475
Phenomenon of Resonance If the excitation frequency coincides or becomes close to one of the natural frequencies of the system, dangerously large oscillations may result, and a condition of resonance is encountered. This is the kind of situation an engineer would very much like to avoid. The collapse of the Tacoma Narrows (also known as Galloping Gerty) Bridgea at Puget Sound in the state of Washington in 1940 and that of the Broughton Suspension Bridge in England are attributed to such a phenomenon. a For
a complete story of the collapse of the Tacoma bridge, see the book Differential Equations and Their Applications by M. Braun, SpringerVerlag, 1978 (p. 167{169).
In both the above cases, a periodic force of very large amplitude was generated and, the frequency of this force was equal to one of the natural frequencies of the bridge at the time of collapse. In the case of the Broughton Bridge, the large force was set up by soldiers marching in cadence over the bridge. In case of the Tacoma Bridge, it was the wind. Because of what happened
in Broughton, soldiers are no longer permitted to march in cadence over a bridge.
Buildings can be constructed with active devices so that the force due to wind can be controlled. The famous Chicago Sears Tower in the windy city, Chicago, has such a device. Another important property of dynamical systems is damping which is present in all systems due to energy dissipation by friction and other resistances. However, for small values of damping, it has very little e ect on the natural frequencies, and is normally not included in the estimation of natural frequencies. Damping becomes important in limiting the amplitude of oscillation at resonance, major e ect of damping is to reduce amplitude with time. For a continuous elastic body the number of independent co-ordinates or degrees of freedom needed to describe the motion is in nite. However, under many situations, parts of such bodies may be assumed to be rigid, and the system may be treated dynamically equivalent to one having a nite number of the degrees of freedom. In summary, the behavior of a vibrating system can be analyzed by knowing the frequencies and the amplitudes of the masses, and the eigenvalue and the eigenvectors of the matrix of the mathematical model of the system are related to these quantities. Speci cally, the frequencies 476
are the square roots of the eigenvalues and the relative amplitudes are represented by the components of the eigenvectors. Example 8.3.2 Vibration of a Building Consider a two-story building (Figure 8.1(a)) with rigid oor shown below. It is assumed that the weight distribution of building can be represented in the form of concentrated weight at each
oor level, as shown in Figure 8.1(b), and the sti ness of supporting columns are represented by the spring constants ki .
m =m
y
2
2
k =k 2
` y
m =m 1
1
k =k 1
`
(b)
(a)
Figure 8.3
Figure 8.1 The equation of motion for this system can be written as
m y + (k + k )y ; k y = 0 m y ; k y + k y = 0 1 1
1
2 2
or
m
1
0
0 m2
! y !
2
1
2 2
2 1
2 2
k + k ;k + y ;k k 1
1
2
2
2
477
2
2
! y! y
1 2
= 0:
(8.3.18)
Taking m1 = m2 = m; k1 = k2 = k, we have ! 2k ;k m 0 ! y1 ! + ;k k 0 m y2 De ning m the mass matrix M = 0 and 2k the sti ness matrix K =
y ! = 0: y 1 2
0
(8.3.19)
!
m
;k ! ;k k
the above equation becomes
y M y + Ky = 0; where y = y
1
!
2
:
(8.3.20)
Assuming that a solution y is of the form
y = xeiwt; and multiplying the above equation by M ;1 and, setting A = M ;1K and = w2; the equations of motion give rise to the eigenvalue problem or k m ; mk 2
! x ; mk ! x ! = : k x x m 1
1
2
2
(8.3.21)
The eigenvalues 1 and 2 and the corresponding eigenvectors representing two normal modes for this 2 2 problem are easily calculated. The eigenvalues are:
= mk (0:3820);
= mk (2:6180):
1
The corresponding eigenvectors are:
2
!
!
0:5257 0:8507 ; : 0:8507 ;0:5257 The vibration of the building at these two normal modes are shown in the gure:
478
0.851
0.851
rk
! = 0:618 m 1
0.526
rk
! = 1:618 m 2
-0.526
Figure 8.2 Buckling Problem (A Boundary Value Problem) Consider a thin, uniform beam of length l. An axial load P is applied to the beam at one of the ends.
y
P x
`
Figure 8.3 We are interested in the stability of the beam, that is, how and when the beam buckles. We 479
will show below how this problem gives rise to an eigenvalue problem and what role the eigenvalues play. Let y denote the vertical displacement of a point of the beam which is at a distance x from the (de ection) left support. Suppose that both the ends of the beam are simply supported, i.e., d2y and the y(0) = y(l) = 0: >From beam theory the basic relationship between the curvature dx 2 internal moment M for the axes is d2 y = M dx2 EI where E is the modulus of elasticity, and I is the area moment of inertia of column cross-section. Writing the bending moment at any section as M = Py , the equation gives the governing di erential equation which is the so-called bending moment equation:
d y = ;Py: EI dx
(8.3.22)
y(0) = y (l) = 0
(8.3.23)
2
2
The constant EI is called the exural rigidity of the beam. The bending moment equation along with the boundary conditions constitute a boundary value problem. d2y with an appropriate nite We will solve this boundary value problem by approximating dx 2 di erence scheme, as was done in Chapter 6. Let the interval [0; l] be partitioned into n subintervals of equal length h, with x0; x1; : : :; xn as the points of division. That is, 0 = x0 < x1 < x2 < : : : < xj < : : : < xn;1 < xn = l and Let
h = nl :
d y yi ; 2yi + yi; : (8.3.24) dx x x h d y into the bending moment equation (8.3.22) and taking Substituting this approximation of dx 2
+1
2
1
2
= i
2
2
into account the boundary conditions (8.3.23), we obtain the following symmetric tridiagonal matrix eigenvalue problem: 0 2 ;1 0 0 1 0 y 1 0 y 1 BB ;1 2 ;1 0 CC BB y1 CC BB y1 CC BB . . . . CC BB .2 CC BB .2 CC . . . . . . BB . BB .. CC = BB .. CC ; . . . . C (8.3.25) C BB .. . . . . . . . . . CC BB .. CC BB .. CC ;1 A @ . A @ . A @ . 0 0 0 ;1 2 yn yn 480
where
= Ph EI :
(8.3.26)
P = EI h
(8.3.27)
2
Each value of determines a load
2
which is called a critical load. These critical loads are the ones which are of practical interests, because they determine the possible onset of the buckling of the beam.
In general, the smallest value of P is of primary importance, since the bending associated with larger values of P may not be obtained without failure from occurring under the action of the lowest critical value P . Simulating Transient Current for an Electric Circuit ( Chapra and Canale (1988) ).
Given an electric circuit consisting of several loops, suppose we are interested in the transient behavior of the electric circuit. In particular, we want to know the oscillation of each loop with respect to the other. First consider the following circuit of the loop.
;i
_C
L
A
A A A
R
Figure 8.4 Voltage drop across a capacitor is:
VC = Cq ; q = charge in the capacitor C = capacitance
Voltage drop across an inductor:
VL = L di dt ; L is inductor u; Voltage drop across a resistor:
VR = iR; R is resistor u: 481
Kircho 's law of voltage states that the algebraic sum of voltage drops around a closed loop circuit is zero. We then have for this circuit
q L di dt + Ri + C = 0 or
L di + iR + 1 dt C
Z
;1
Z (Because, VC = Cq = C1 i dt:) Now consider the network with four loops. L1
;;;
;
i1
i dt = 0
$ $ $ $ % % % % L2
L3
;;;
+
0
C1
_
L4
;;;
C2
_
i2
;;;
C3
_
i3
i4
C4
_
Figure 8.5 Kircho 's voltage law applied to each loop gives Loop 1:
Zt ; L didt ; C1 (i ; i )dt = 0 ;1 1
1
Loop 2:
(8.3.29)
Zt Zt di 1 1 ; L dt ; C (i ; i )dt + C (i ; i )dt = 0 ;1 ;1
(8.3.30)
Zt Zt di 1 1 ; L dt ; C i dt + C (i ; i )dt = 0: ;1 ;1
(8.3.31)
3
Loop 4:
(8.3.28)
2
Zt Zt 1 1 di (i ; i )dt + C (i ; i )dt = 0 ; L dt ; C ;1 ;1 2
Loop 3:
1
1
2
2
2
3
3
3
4
3
4
4
4
4
482
1
1
2
2
3
2
3
3
4
The system of ODEs given by (8.3.28)-(8.3.31) can be di erentiated and rearranged to give
L ddti + C1 (i ; i ) = 0 L ddti + C1 (i ; i ) ; C1 (i ; i ) = 0 L ddti + C1 (i ; i ) ; C1 (i ; i ) = 0 L ddti + C1 i ; C1 (i ; i ) = 0: 2
1
2
2
2
2
2
3
2
3
2
3
2
4
Assume
1
2
2
3
3
4
4
2
1
1
1
2
4
4
(8.3.32)
2
3
1
2
(8.3.33)
2
3
(8.3.34)
3
(8.3.35)
4
ij = Aj sin(wt); j = 1; 2; 3; 4:
(8.3.36)
(Recall that ij is the current at the j th loop.) >From (8.3.32) ;L1A1w2 sin wt + C1 A1 sin wt ; C1 A2 sin wt = 0 1
or
1
( C1 ; L1w2 )A1 ; C1 A2 = 0: 1
(8.3.37)
1
>From (8.3.33)
;L A w sin wt + C1 A sin wt ; C1 A sin wt ; C1 A sin wt + C1 A sin wt = 0 2
2
2
2
or
2
2
2
1
1
1
2
; C1 A + ( C1 + C1 ; L w )A ; C1 A = 0: 1
1
1
2
2
2
2
2
(8.3.38)
3
>From (8.3.34)
;L A w sin wt + C1 A sin wt ; C1 A sin wt ; C1 A sin wt + C1 A sin wt = 0 3
3
2
3
or
3
3
4
2
2
2
3
; C1 A + ( C1 + C1 ; L w )A ; C1 A = 0: 2
2
2
2
3
3
3
3
4
(8.3.39)
>From (8.3.35)
;L A w sin wt + C1 A sin wt ; C1 A sin wt ; C1 A sin wt = 0 4
or
4
2
4
4
3
3
3
; C1 A + ( C1 + C1 ; L w )A = 0: 3
3
3
4
483
4
2
4
4
(8.3.40)
Gathering the equations (8.3.37){(8.3.40) together, we have ( 1 ; L w 2 )A ; 1 A = 0
C
(8.3.41)
C 1 1 1 ; C A + ( C + C ; L w )A ; C1 A = 0 ; C1 A + ( C1 ; C1 ; L w )A ; C1 A = 0 ; C1 A + ( C1 + C1 ; L w )A = 0 1
1
1
1
1
2
2
1
3
2
3
4
4
2
2
2
3
3
3
2
2
2
2
3
1
3
2
3
(8.3.42)
4
(8.3.43) (8.3.44)
4
or (1 ; L1 C1w2 )A1 ; A2 = 0 ; C2 A1 + ( C2 + 1 ; L2C2w2)A2 ; A3 = 0
(8.3.45) (8.3.46)
C C C ; C A + ( CC + 1 ; L C w )A ; A = 0 ; CC A + ( CC + 1 ; L C w )A = 0: 1
3
2
1
4
3
2
2
3
4
3
3
or
3
2
4
3
4
A ;A ; CC A + CC A + A ; A ; CC A + ( C2 + 1)A ; A ; CC A + ( CC + 1) + A 1
1
3
2
4
3
2
1
2
1
= L2 C2w2A2
3
4
= L3 C3w2A3
(8.3.49)
= L4 C4w2A4
4
3
(8.3.48)
4
3
4
3
2
2
3
2
(8.3.47)
4
= L1 C1w2A1
2
2
3
In matrix form, (8.3.49) is:
0 10 1 ; 1 0 0 BB C C CC BB A BB ; C ( C + 1) ;1 BB A 0 C C BB 0 C BB A ; CC ( CC + 1) ;1 C @ A @ ; CC ( CC + 1) A 0 10 L C 0 0 0 BB CC BB A B 0 LC 0 0 C CC BBB A =w B BB 0 0 LC 0 C @ A B@ A 1
2 1
2 1
2
3 2
3 2
3
4 3
1
4 2
4
1
1
2
2
0
2
2
3
0
0
3
LC 4
4
3
A
1 CC CC CC A 1 CC CC : CC A
4
The above is an eigenvalue problem. To see it more clearly, consider the special case
C = C = C = C = C; 1
2
3
484
4
(8.3.50)
and
L = L = L = L = L; 1
and assume
2
3
4
= LCw ; 2
then the above problem reduces to (1 ; )A1 ; A2 = 0 ;A1 + (2 ; )A2 ; A3 = 0 ;A2 + (2 ; )A3 ; A4 = 0 ;A3 + (2 ; )A4 = 0 or
0 10 (1 ; ) ; 1 BB CC BB BB ;1 (2 ; ) ;1 CC BB BB CC BB ; 1 +(2 ; ) ; 1 @ A@ ;1 +(2 ; )
A A A A
or since ij = Aj sin wt; j = 1; : : :; 4: We have
or
0 10 (1 ; ) ; 1 BB CC BB i BB ;1 (2 ; ) ;1 CC BB i BB CA BB@ i ;1 (2 ; ) ;1 C @ ;1 (2 ; ) i 10 1 0 0 1 ; 1 0 0 CC BB i CC BB i BB BB ;1 2 ;1 0 CC BB i CC = BB i BB 0 ;1 2 ;1 CC BB i CC BB i A@ A @ @ i i 0 0 ;1 2
1 2 3
1 2 3
1 CC CC CC = 0 A
(8.3.51)
4
1 CC CC = 0 CC A
(8.3.52)
4
1
1
2
2
2
3
3
4
1 CC CC : CC A
(8.3.53)
The solution of this eigenvalue problem will give us the natural frequencies (wi2 = i=LC ): Moreover the knowledge of the eigenvectors can be used to study the circuit's physical behavior such as the natural modes of oscillation. These eigenvalues and the corresponding normalized eigenvectors (in four digit arithmetic) are:
= :1206; = 1; = 2:3473; = 3:5321: 1
2
3
485
4
0 :6665 1 0 :5774 1 0 ;:4285 1 0 ;:2280 1 B CB C BB :5774 CC BB :5774 CC B B :5774 C ;0:0000 C B C B CC ; BB CC ; BB CC : ;B B C B C B C B C B @ :4285 A @ ;0:5774 A @ :2289 A @ ;:6565 CA :2280 ;:5774 ;:6565 :4285 From the directions of the eigenvectors we conclude that for all the loops oscillate in the same direction. For the second and third loops oscillate in the opposite directions from the rst and fourth, and so on. This is shown in the following diagram: 1
3
486
= 0:1206 1
+ -
=1 2
+ -
= 2:3473 3
+ -
= 3:5321 4
+ -
Figure 8.6 487
8.3.3 An Example of the Eigenvalue Problem Arising in Statistics: Principal Components Analysis Many practical-life applications involving statistical analysis (e.g. stock market, weather prediction, etc.) involve a huge amount of data. The volume and complexities of the data in these cases can make the computations required for analysis practically infeasible. In order to handle and analyze such a voluminous amount of data in practice, it is therefore necessary to reduce the data. The basic idea then will be to choose judiciously `k' components from a data set consisting of n measurement on p(p > k) original variables, in such a way that much of the information (if not most) in the original p variables is contained in the k chosen components. Such k components are called the rst k \principal components" in statistics. The knowledge of eigenvalues and eigenvectors of the covariance matrix is needed to nd these principal components. Speci cally, if is the covariance matrix corresponding to the random vector
X = (X ; X ; : : :; Xp); 1
2
: : : p 0 are the eigenvalues, and x through xp are the corresponding eigenvectors of the matrix , then the ith principal component is given by 1
2
1
Yi = xTi X; i = 1; 2; : : :; p: Furthermore, the proportion of total population variance due to ith principal component is given by the ratio:
i
i ; i = 1; : : :; p : = n 1 + 2 + : : : + p trace()
(8.3.54)
Note: The covariance matrix is symmetric positive semide nite, and, therefore, its eigenvalues are all nonnegative.
If the rst k ratios constitute the most of the total population variance, then the rst k principal components can be used in statistical analysis. Note that in computing the kth ratio, we need to know only the kth eigenvalue of the covariance matrix; the entire spectrum does not need to be computed. To end this section, we remark that many real-life practices, such as computing the index of Dow Jones Industrial Average, etc., can now be better understood and explained through the principal components analysis. This is shown in the example below.
488
A Stock-Market Example (Taken from Johnson and Wichern (1992)) Suppose that the covariance matrix for the weekly rates of return for stocks of ve major companies (Allied Chemical, DuPont, Union Carbide, Exxon, and Texaco) in a given period of time is given by: 0 1:000 0:577 0:509 0:387 0:462 1 BB 0:577 1:000 0:599 0:389 0:322 CC BB CC B R = B 0:509 0:599 1:000 0:436 0:426 C CC : BB @ 0:387 0:389 0:436 1:000 0:523 CA 0:462 0:322 0:426 0:523 1:000 The rst two eigenvalues of R are:
= 2:857 = :809: 1 2
(8.3.55) (8.3.56)
The proportion of total population variance due to the rst component is approximately 2:857 = 57%: (8.3.57) 5 The proportion of total population variance due to the second component is: :809 = approximately 16%: (8.3.58) 5 Thus the rst two principal components account for 73% of the total population variance. The eigenvectors corresponding to these principal components are:
xT = (:464; :457; :470; :421; :421)
(8.3.59)
xT = (:240; :509; :260; ;:526; ;:582):
(8.3.60)
1
and
2
These eigenvectors have interesting interpretations. >From the expression of x1 we see that the rst component is (roughly) equally weighted sum of the ve stocks. This component is generally called the market component. However, the expression for x2 tells us that the second component represents a contrast between the chemical stocks and the oil-industries stocks. This component will be generally called an industry component. Thus, we conclude that about 57% of total variations in these stock returns is due to the market activity and 16% is due to industry activity. The eigenvalue problem also arises in many other important statistical analysis, for example, in computing the canonical correlations, etc. Interested readers are referred to the book by Johnson and Wichern (1992) for further reading. 489
A nal comment: Most eigenvalue problems arising in statistics,
such as in principal components analysis, canonical correlations, etc., are actually singular value decomposition problems and should be handled computationally using the singular value decomposition to be described in Chapter 10.
8.4 Localization of Eigenvalues As we have seen in several practical applications, explicit knowledge of eigenvalues may not be required; all that is required is a knowledge of distribution of the eigenvalues in some given regions of the complex plane or estimates of some speci c eigenvalues. There are ways such information may be acquired without actually computing the eigenvalues of the matrix. In this section, we shall discuss some of the well-known approaches. We start with a well-known result of Gersgorin (1931).
8.4.1 The Gersgorin Disk Theorems Theorem 8.4.1 (Gersgorin's First Theorem) Let A = (aij )nn. De ne n X ri = jaijj; i = 1; : : :; n: j =1 i=j 6
Then each eigenvalue of A satis es at least one of the following inequalities:
j ; aiij ri; i = 1; 2; : : :; n: In other words, all the eigenvalues of A can be found in the union of disks fz : jz ; aiij ri; i = 1; : : :; ng.
Proof. Let be an eigenvalue of A and x be an eigenvector associated with . Then from Ax = x or from
(I ; A)x = 0; 490
we have
( ; aii)xi =
n X j =1 i=j
aij xj ; i = 1; : : :; n;
6
where xi is the ith component of the vector x. Let xk be the largest component of x (in absolute value). Then, since jxj j=jxkj 1 for j 6= k, we have from above n X
j ; akkj
j =1 j =k
n X
jakj j jjxxj jj k
6
j =1 j =k
jakj j:
6
Thus is contained in the disk j ; akk j rk .
De nition 8.4.1 The disks Ri = jz ; aiij ri, i = 1; : : :; n are called Gersgorin disks in the complex plane.
Example 8.4.1
01 2 31 B 3 4 9 CC A = B @ A 1 1 1
r = 5 r = 12 r = 2: 1 2 3
The Gersgorin disks are:
R : jz ; 1j 5 R : jz ; 4j 12 R : jz ; 1j 2: 1
2
3
(The eigenvalues of A are 7.3067, -0.6533 0:3473i.)
Remark: It is clear from the above example that the Gersgorin rst theorem gives only very
crude estimates of the eigenvalues.
491
Imag
R2
R1 R3
Real
Figure 8.7 Gersgorin's Disks of Example 8.4.1 While the above theorem only tells us that the eigenvalues of A lie in the union of n Gersgorin disks, the following theorem gives some more speci c information. We state the theorem without proof.
Theorem 8.4.2 (Gersgorin's Second Theorem) Suppose that r Gersgorin disks
are disjoint from the rest, then exactly r eigenvalues of A lie in the union of the r disks.
Proof. Horn and Johnson (1985, pp. 344{345). Note: For generalizations of Gersgorin's theorems, see Horn and Johnson (1985) and Brualdi (1993).
492
Example
0 1 0:1 0:2 1 B C A=B @ 0:2 4 0:3 CA 0:4 0:5 8
The Gersgorin disks are:
R : jz ; 1j 0:3 R : jz ; 4j 0:5 R : jz ; 8j 0:9 1 2 3
All the three disks are disjoint from each other. Therefore, by Theorem 8.4.2, each disk must contain exactly one iegenvalue of A. This is indeed true. Note that the eigenvalues A are 0.9834, 3.9671, and 8.0495. Imag
R1
R2
R3
4
1
Figure 8.8 Gersgorin's Disks for Example 8.4.2 493
8
Real
8.4.2 Eigenvalue Bounds and Matrix Norms Simple matrix norms can sometimes be used to obtain useful bounds for the eigenvalues. Here are two examples.
Theorem 8.4.3 For any consistent pair of matrix-vector norms, we have jj kAk; where is an eigenvalue of A. In particular, (A), the spectral radius of A, is bounded by kAk : (A) kAk.
Proof. The proof follows immediately by taking the norm in Ax = x kxk = kAxk kAk kxk or
jj kxk kAk kxk;
or
jj kAk:
Note:
For the above example, kAk1 = 13; (A) = 7:3067. Since the eigenvalues of AT are the same as those of A, we have
Corollary 8.4.1
(A) kAT k:
Combining these two results and taking the in nity norm in particular, we obtain
Theorem 8.4.4 (A) minfmax i
n X j =1
jaijj; max j
494
n X i=1
jaij jg:
Theorem 8.4.5 Let ; ; : : :; n be the eigenvalues of A, then n X jij kAkF : 1
2
2
2
i=1
Proof. The Schur Triangularization Theorem (Theorem 8.2.3) tells us that there exists a unitary
matrix U such that
U AU = T; an upper triangular matrix.
Thus
TT = U AAU:
So, AA is unitarily similar to TT . Since similar matrices have the same traces,
Tr(TT ) = Tr(AA) = kAkF : 2
Again, Tr(TT ) =
n X n X i=1 j =1
jtij j : Also 2
n X n X i=1 j =1
and Thus,
n X n X i=1
jji = 2
n X i=1
jtiij 2
n X n X i=1 j =1
i=1
jtij j = kAkF : 2
jtiij = 2
2
n X i=1
jij : 2
jtij j = kAkF . 2
2
8.5 Computing Selected Eigenvalues and Eigenvectors 8.5.1 Discussions on the Importance of the Largest and Smallest Eigenvalues We have seen before that in several applications all one needs to compute is a few largest or smallest eigenvalues and the corresponding eigenvectors. For example, recall that in the buckling problem, it is the smallest eigenvalue that is the most important one. In vibration analysis of structures, a common engineering practice is just to com-
pute the rst few smallest eigenvalues (frequencies) and the corresponding eigenvectors (modes), because it has been seen in practice that the larger eigenvalues and eigenvectors contribute very little to the total response of the system. The same remarks hold in the case of control problems modeled by a system of second-order di erential equations arising 495
in the nite-element generated reduced order model of large exible space structures (see the book by Inman (1989)). In statistical applications, such as in principal component analysis, only the rst few largest eigenvalues are computed. There are other applications where only the dominant and the subdominant eigenvalues and the corresponding eigenvectors play an important role.
8.5.2 The Role of Dominant Eigenvalues and Eigenvectors in Dynamical Systems Let's discuss brie y the role of dominant eigenvalues and the eigenvectors in the context of dynamical systems. Consider the homogeneous discrete-time system:
xk = Axk ; k = 0; 1; 2; : : :: Let be the dominant eigenvalue of A; that is, j j > j j j j : : : jnj; where ; ; : : :n are the eigenvalues of A. Suppose that A has a set of independent eigenvectors: v ; v ; : : :; vn: Then the state xk at any time k > 0 is given by: xk = kv + k v + : : : + nknvn ; where x = v + : : : + nvn : Since j jk > jijk ; i = 2; 3; : : :; n; it follows that for large values of k, j kj j ikij; i = 2; 3; : : :; n; provided that i = 6 0: This means that for large values of k, the state vector xk will approach the direction of the vector v corresponding to the dominant eigenvalue . Furthermore, the rate at which the state vector approaches v is determined by the ratio of the second to the rst dominant eigenvalue: j = j. (A proof will be presented later.) In the case = 0; the second dominant eigenvalue and the corresponding eigenvector assume +1
1
1
2
3
1
1
1
0
1 1
1 1
2
2
2
2 2
1
1
1
1
1
1
2
1
1
2
the role of the rst dominant eigenpair. Similar conclusions hold for the continuous-time system:
x_ (t) = Ax(t): For details, see the book by Luenberger (1979).
In summary, the long term behavior of a homogeneous dynamical system can essentially be predicted from just the rst and second dominant eigenvalues of the system matrix and the corresponding eigenvectors. 496
The second dominant eigenpair is particularly important in the case 1 = 0: In this case, the long term behavior of the system is determined by this pair.
An Example on the Population Study Let's take the case of a population system to illustrate this. It is well known (see Luenberger (1979), p. 170) that such a system can be modeled by
pk = Apk ; k = 0; 1; 2; : : : +1
where pk is the population-vector. If the dominant eigenvalue 1 of the matrix A is less than 1 (that is, if j1j < 1), then it follows from
pk = k v + : : : + nknvn; 1
1 1
that the population decreases to zero as k becomes large. Similarly, if j1j > 1; then there is long term growth in the population. In the latter case the original population approaches a nal distribution that is de ned by the eigenvector of the dominant eigenvalue. Moreover, it is the second dominant eigenvalue of A that determines how fast the original population distribution is approaching the nal distribution. Finally, if the dominant eigenvalue is 1, then over the long term there is neither growth nor decay in the population.
8.5.3 The Power Method, The Inverse Iteration and the Rayleigh Quotient Iteration In this section we will brie y describe two well-known classical methods for nding the dominant eigenvalues and the corresponding eigenvectors of a matrix. The methods are particularly suitable for sparse matrices, because they rely on matrix vector multiplications only (and therefore, the zero entries in a sparse matrix do not get lled in during the process).
The Power Method The power method is frequently used to nd the dominant eigenvalue and the corresponding eigenvector of a matrix. It is so named because it is based on implicit construction of the powers of A. Let the eigenvalues 1 ; 2; : : :; n of A be such that
j j > j j j j : : : jnj; 1
2
3
that is, 1 is the dominant eigenvalue of A. Let v1 be the corresponding eigenvector. Let max(g ) denote the element of maximum modulus of the vector g . Let be the tolerance and N be the maximum number of iterations. 497
Algorithm 8.5.1 Power Method Step 1. Choose x
0
Step 2. For k = 1; 2; 3; : : : do
x^k = Axk; , xk = x^k = max(^xk). Stop if (max(^xk ) ; max(^xk; )) < or if k > N . 1
1
Theorem 8.5.1 max(^xk) ! , and fxkg ! w , a multiple of v , as k ! 1. 1
1
1
Proof. From above, we have Ak x : xk = max( Ak x ) 0
0
Let the eigenvectors v1 through vn associated with 1; : : :; n be linearly independent. We can then write x0 = 1 v1 + 2 v2 + : : : + n vn, 1 6= 0: So,
Ak x = Ak ( v + v + : : : + nvn ) = k v + k v + : : : + n knvn k k k = [ v + v + : : : + n n vn]: 0
1 1
1
2 2
1 1
1
2
1 1
2
2 2
2
2
1
1
Since 1 is the dominant eigenvalue,
k i
1
Thus,
! 0 as k ! 1; i = 2; 3; : : :; n: Ak x ! cv ; xk = max( Ak x ) 0
and
1
0
fmax(^xk)g ! : 1
Remarks: We have derived the power method under two constraints: 1. 1 6= 0 2. 1 is the only dominant eigenvalue. 498
The rst constraint ( 1 6= 0) is not really a serious practical constraint, because after a few iterations, round-o errors will almost always make it happen. As far as the second constraint is concerned, we note that the method still converges when the matrix A has more than one dominant eigenvalue. For example, let 1 = 2 = : : : = r and j1j > jr+1j > : : : > jnj; and let there be independent eigenvectors associated with 1. Then we have
Ak x
0
=
k 1
k
r X
i=1 r X
1
1
ivi +
n X
i=r+1
i (i=
1
!
)k v
i
i vi
(since (i=1)k is small for large values of k). This shows that in this case the power method converges to some vector in the subspace spanned by v1 : : :; vn.
Example 8.5.1
01 2 31 B 2 3 4 CC A = B @ A
3 4 5 x0 = (1; 1; 1)T :
The eigenvalues of A are 0, ;0:6235, and 9.6235. The normalized eigenvector corresponding to the largest eigenvalue 9.6233 is (:3851; :5595; :7339)T :
k=1:
061 B CC x^ = Ax = B @9A 1
0
12 max(^x10) =112 0 1
k=2:
1
0:50 B C B C x ^ 1 3 C x1 = max(^x ) = B =B 0:75 C @ 4 A @ A 1 1 1 2
1 CC CA ;
0 5:00 B x^ = Ax = B B@ 7:25 2
1
9:50 max(^x2 ) = 9:50 499
1 0 0 : 5263 C B x =B B@ 0:7632 CCA 2
1:0000
k=3:
0 5:0526 1 B 7:3421 CC x^ = Ax = B @ A 3
2
9:6316 x = x0^ = max(^x1 ) 0:5246 B C = B @ 0:7623 CA 1:000 max(^x ) = 9:6316: 3
3
3
3
Thus fmax(^x3)g is converging towards the largest eigenvalue 9.6235 and fxk g is converging towards the direction0of the eigenvector associated with this eigenvalue. (Note that the normalized dominant 1 0:3851 B CC eigenvector B 0 : 5595 @ A is a scalar multiple of x3.) 0:7339
Convergence of the Power Method
The rate of convergence of the power method is determined by the ratio
2
from the following.
k
1
, as is easily seen
k
kxk ; v k = k v + : : : + n n vnk k k j j kv k + : : : + j nj n kvnk k (j jkv k + : : : + j nj kvnk) : 1 1
2
2
1
2
2
2
1
(Since j i j j 2 j; i = 3; 4 : : :; n). 1 1 Thus we have
kxk ; v k
2
1 1
where
1
2
1
2
2
1
1
2
k ; k = 1; 2; 3; : : :;
= (j j kv k + : : : + j nj kvnk): 2
2
This shows that the rate at which xk approaches 1 v1 depends upon how fast j 2 jk goes to zero.
The absolute value of the error at each step decreases by the ratio ( ), that is, if is close to 1
2
2
, then the convergence will be very slow; if this ratio is small, the convergence will 1
1
be fast.
500
The Power Method with Shift In some cases, convergence can be signi cantly improved by using a suitable shift. Thus, if is a suitable shift so that 1 ; is the dominant eigenvalue of A ; I , and if the power method is applied ; to the shifted matrix A ; I , then the rate of convergence will be determined by the ratio 2 2 ; , rather than . (Note that by shifting the matrix A by , the eigenvalues get
shifted by , but the eigenvectors remain unaltered.) By choosing appropriately, in some cases, the ratio ; ; can be made signi cantly smaller than , thus yielding the faster convergence. An optimal choice (Wilkinson AEP, p. 572) of , 1
1
2
1
2
assuming that i are all real, is 21 (1 + n). This simple choice of sometimes indeed yields very fast convergence: but there are many common examples where the convergence can still be slow with this choice of . Consider a 20 20 matrix A with the eigenvalues 20; 19; : : :; 2; 1. The choice of = 1 +2 20 = 10:5 yields the ratio 2 ; = 8:5 , still close to one. Therefore, the rate of convergence will still 1 ; 9:5 be slow. Furthermore, the above choice of is not useful in practice, because the eigenvalues are not known a priori. 1
The Inverse Power Method/Inverse Iteration The following iterative method, known as the inverse iteration, is an e ective method for
computing an eigenvector when a reasonably good approximation to an eigenvalue is known.
Algorithm 8.5.2 Inverse Iteration Let be an approximation to a real eigenvalue 1 such that j1 ; j ji ; j(i 6= 1); that is, is much closer to 1 than to the other eigenvalues. Let be the tolerance and N be the maximum number of iterations.
Step 1. Choose x : Step 2. For k = 1; 2; 3; : : :; do (A ; I )^xk = xk; ; 0
1
xk = x^k = max(^xk): Stop if (kxk ; xk; k=kxk k) < or if k > N . 1
Theorem 8.5.2 The sequence fxkg converges to the direction of the eigenvector corresponding to 1.
501
Remark: Note that inverse iteration is simply the power method applied to (A ; I ); . That is why it is also known as the inverse power method. 1
Proof. The eigenvalues of (A ; I ); are ( ; ); ; ( ; ); ; : : :; (n ; ); and the eigenvectors 1
1
1
1
2
1
are the same as those of A. Thus, as in the case of the power method, we can write
c v + c cn x^k = ( ; )k " ( ; )k : : : + (n ; )k vn ; k ; k # 1 = ( ; )k c v + c ; v : : : + cn ; vn : n Since is closer to than any other eigenvalue, the rst term on the right hand side is the dominating one and, therefore, xk converges to the direction of v . It is the direction of v which 1
1
2
1
2
1 1
1
2
1
1
2
2
1
1
we are trying to compute.
1
An Illustration Let us illustrate the above with k = 1. Suppose that x0 = c1 v1 + c2v2 + + cnvn . Then
x^ = (A ; I ); x = ( ; ); c v + ( ; ); c v + : : : + (n ; ); cnvn: 1
1
0
1
1
1 1
2
1
2 2
1
Since 1 is closer to than any other eigenvalue, the coecient of the rst term in the expansion, namely ( 1; ) , is the dominant one (it is the largest). Thus, x^1 is roughly a multiple of v1, 1 which is what we desire.
Numerical Stability of the Inverse Iteration At rst sight inverse iteration seems to be a dangerous procedure, because if is near 1, the matrix (A ; I ) is obviously ill-conditioned. Consequently, this ill-conditioning might a ect the computed approximations of the eigenvector. Fortunately, in practice the ill-conditioning of the matrix (A ; I ) is exactly what we want. The error at each iteration grows towards the direction of the eigenvector, and it is the direction of the eigenvector that we are interested in. Wilkinson (AEP pp. 620{621) has remarked that in practice x^k is remarkably close to the solution of (A ; I + F )xk = xk;1; where F is small. For details see Wilkinson (AEP pp. 620{621). a a \The
iterated vectors do indeed converge eventually to the eigenvectors of A + F ."
502
Example 8.5.2
01 2 31 B C A=B @ 2 3 4 CA
3 4 5 x0 = (1; 1; 1)T ; = 9:
k = 1:
x^ = (1; 1:5; 2)T 1
x = x^ =kx^ k = (:3714; :5571; :7428)T : 1
k = 2:
1
1 2
x^ = (:619; :8975; 1:1761)T 2
x = x^ =kx^ k = (:3860; :5597; :7334)T : 2
k = 3:
2
2 2
x^ = (:6176; :8974; 1:1772)T 3
x = x^ =kx^ k = (:3850; :5595; :7340): 3
k = 4:
3
3 2
x^ = (:6176; :8974; 1:1772)T 4
x = x^ =kx^ k = (:3850; :5595; :7340)T 4
k = 5:
4
4 2
x^ = (:6177; :8974; 1:1772)T 5
x = x^ =k(^x )k = (:3851; :5595; :7339): 5
5
5
2
Remark: In the above example we have used norm k k as a scaling for the vector xi to empha2
size that the scaling is immaterial since we are working towards the direction of the eigenvector.
Choosing the Initial Vector x
0
To choose the initial vector x0 we can run a few iterations of the power method and then switch to the inverse iteration with the last vector generated by the power method as the initial vector x0 in the inverse iteration. Wilkinson (AEP, p. 627) has stated that if x0 is assumed to be such that
Le = Px
0
503
where P is the permutation matrix satisfying
P (A ; I ) = LU; and e is the vector of unit elements, then only \two iterations are usually adequate", provided that is a good approximation to the eigenvalue. Note: If x0 is chosen as above, then the computations of x^1 involves only solution of the triangular system: U x^1 = e1 :
Example 8.5.3
01 2 31 B C A=B @ 2 3 4 CA :
The eigenvalues of A are: Choose = 9:1
k=1:
3 4 5
0; ;0:6235; and 9:6235:
0 ;8:1 2:0 3:0 1 B C A ; I = B @ 2:0 ;6:1 4:0 CA 3:0 4:0 ;4:1 0 1 1 0 ;8:1 1 0 0 2 3 B C B C L=B @ ;:2469 1 0 CA ; U = B@ 0 ;5:6062 4:7407 CA ;:3704 ;:8456 1 0 0 1:0200 01 0 01 B C P =B @ 0 1 0 CA : 0 0 1 0 1:000 1 B CC : 7531 x0 = P ;1Le = B @ A: ;:2160
0 :4003 1 B :6507 CC x^ = (A ; I ); x = B @ A 1
1
0
:9804
0 :4083 1 B :6637 CC : x = x^ = max(^x ) = B A @ 1
1
1
504
1:000
k=2:
0 :9367 1 B 1:3540 CC x^ = (1 ; I ); x = B @ A 1
2
1
1:7624
0 :5315 1 B :7683 CC ; x =B @ A 2
1:000 which is about 1.3 times the normalized eigenvector.0
:3851 1 B :5595 CC. The normalized eigenvector correct to four digits is B @ A :7339
The Rayleigh Quotient
Theorem 8.5.3 Let A be a symmetric matrix and let x be a reasonably good approximation to an eigenvector. Then the quotient T Rq = = xxTAx x;
is a good approximation to the eigenvalue for which x is the corresponding eigenvector.
Proof. Since A is symmetric there exists a set of orthogonal eigenvectors v ; v ; : : :; vn. Therefore 1
we can write
2
x = c v + : : : + cnvn : Assume that vi ; i = 1; : : :; n are normalized, that is viT vi = 1: Then, since Avi = ivi ; i = 1; : : :; n, and noting that viT vj = 0; i 6= j , we have T T = (c v + : : : + cnvn ) TA(c v + : : : + cn vn) = xxTAx x (c v + : : : + cnvn ) (c v + : : : + cnvn ) T = (c v + : : : + ccnv+n )c (+c :: :v++c : : : + cnnvn ) n = c c ++cc ++: :: :: :++cncn n 2 c n cn 3 6 1 + c + : : : + c 77 = 664 75 : c cn 1+ c +:::+ c 1 1
1 1
1 1
1 1
1 1
1 1
1
2 1
2 1 1
2 1
1
1 1
2 2
2 2 2 2 2
2
2
2
2
2
1
1
2
2
2
1
2
1
1
505
2
1
Because of our assumption that x is a good approximation to v1 , c1 is larger than other ci; i = 2; : : :; n. Thus, the expression within [ ] is close to 1, which means that is close to 1.
De nition 8.5.1 The quotient Rq = xxTAx x is called the Rayleigh Quotient. T
Example 8.5.4 Let
!
1 2 A= : 2 3
x=
Then the Rayleigh Quotient
1
!
:
;:5
T = xxTAx x = ;:2 is a good approximation to the eigenvalue ;:2361.
Note : It can be shown (exercise) that for a symmetric matrix A: n Rq
1
where n and 1 are the smallest and the largest eigenvalue of A respectively.
Rayleigh-Quotient Iteration The above idea of approximating an eigenvalue can be combined with inverse iteration to compute successive approximations of an eigenvalue and the corresponding eigenvector in an iterative fashion, known as Rayleigh-quotient iteration, described as follows. Let N be the maximum number of iterations to be performed.
Algorithm 8.5.3 Rayleigh-Quotient Iteration For k = 0; 1; 2; : : :; do 1. Compute 2. Solve for x^k+1 : 3. Compute
k = xTk Axk =xTk xk (A ; k I )^xk+1 = xk ;
xk = x^k = max(^xk ): +1
+1
+1
4. Stop if the pair ( k ; xk ) is an acceptable eigenvalue-eigenvector pair or if k > N . 506
Convergence: It can be shown (Wilkinson, AEP, p. 630) that the rate of convergence of the
method is cubic.
Choice of x : As for choosing an initial vector x , perhaps the best thing to do is to use the 0
0
direct power method itself a few times and then use the last approximation as x0 .
Remark: Rayleigh Quotient iteration can also be de ned in the nonsymmetric case, where one nds both left and right eigenvectors at each step. We omit the discussion of the nonsymmetric case here and refer the reader to Wilkinson AEP, p. 636. See also Parlett (1974).
Example 8.5.5
01 2 31 B 2 3 4 CC A=B @ A 3 4 5
Let us take
0 :5246 1 B C x =B @ :7622 CA ; 0
1:000 which is obtained after 3 iterations of the power method. Then
k=0:
= xT Ax =(xT x ) = 9:6235 0
0
0
0
0
0 :5247 1 B :7623 CC x =B @ A 1
k=1:
1:000
= xT Ax =(xT x ) = 9:6235 0 1:000 1 B C x =B @ 1:4529 CA ; 1:9059 0 :3851 1 B C The normalized eigenvector associated with 9.6255 is B @ :5595 CA. :7339 Note that .3851 times x is this eigenvector to three digits. Thus two iterations were sucient. 1
1
1
2
2
507
1
1
8.5.4 Computing the Subdominant Eigenvalues and Eigenvectors: De ation Once the dominant eigenvalue 1 and the corresponding eigenvector v1 have been computed, the next dominant eigenvalue 2 can be computed by using de ation. The basic idea behind de ation is to replace the original matrix by another matrix of the same or lesser dimension using a computed eigenpair such that the de ated matrix has the same eigenvalues as the original one except the one which is used to de ate.
Hotelling De ation The Hotelling de ation is a process which replaces the original matrix A = A1 by a matrix A2 of the same order such that all the eigenvalues of A2 are the same as those of A1 except the one which is used to construct A2 from A1 .
Case 1: A is Symmetric First suppose that A = A1 is symmetric. Let (1; x1) be an eigenpair of A1 . De ne
A = A ; x xT ; 2
1
1
1
1
where xT1 x1 = 1: Then
A xi = A xi ; x xT xi ; i = 1; 2; : : :; n: Since x is orthogonal to the other eigenvectors, and assuming xT x = 1; we have For i = 1 : A x = A x ; x xT x = x ; x = 0 = 0: For i = 6 1: A xi = A xi ; 0 = ixi: Thus the eigenvalues of A are 0; ; ; : : :; n, and x through xn are the corresponding eigenvec2
1
1
1
1
1
1
2
1
1
1
1
1
1
2
2
2
1
1
1
1
1
1
1
3
1
tors.
Case 2: A is Nonsymmetric The idea above can be easily generalized to a nonsymmetric matrix; however, we need both left and right eigenvectors here. Let (x1 ; y1) be the pair of right and left eigenvector of A = A1 corresponding to 1. Then de ne A2 = A1 ; 1x1 y1T where
y T x = 1: 1
1
508
Then using the bi-orthogonality conditions of the eigenvectors xi and yi we have For i = 1 : A2x1 = A1x1 ; 1x1y1T x1 = 1x1 ; 1x1 = 0 For i 6= 1 :
A xi = A xi ; x y T xi = ixi ; 0 = ixi : 2
1
1
1 1
Again, we see that 1 = 0; and, 2 through n are the eigenvalues of A2 corresponding to the eigenvectors x1 through xn.
Remarks: Though the Hotelling de ation is commonly recom-
mended in the engineering literature as a practical process for computing a selected number of eigenvalues and eigenvectors and is wellsuited for large and sparse matrices, the method both in the
symmetric and nonsymmetric cases is numerically unstable (see Wilkinson AEP, p. 585).
Householder De ation We will now construct a de ation using similarity transformation on A1 with Householder matrices. Of course, the technique will work with any similarity transformation; however, we will use Householder matrices for reasons of numerical stability. The method is based upon the following result:
Theorem 8.5.4 Let ( ; v ) be an eigenpair of A and H be a Householder matrix such that Hv is a multiple of e1 , then
1
1
1
0 BB 0 A = HAH = B BB .. @. 1
1
0
A2
::: 1 0 C
1 T CC b A; CC = @ 0 A A 1
2
where A2 is (n ; 1) (n ; 1), and the eigenvalues of A2 are the same as those of A except for 1; in particular, if j1j > j2j j3j : : : jnj, then dominant eigenvalue of A2 is 2, which is the second dominant (subdominant) eigenvalue of A. 509
Proof. From Av = v we have 1
1 1
HAHHv = Hv (since H = I ): 1
1
2
1
That is, HAH (ke1) = 1ke1 (since Hv1 = ke1 ) or HAHe1 = 1e1. (This means that the rst column of HAH is 1 times the rst column of the identity matrix.) Thus HAH must have the form 0 1
BB 0 HAH = B BB .. @.
1
:::
A2
CC CC : CA
0 Since det(HAH ; I ) = 1 det(A2 ; I ), it follows that the eigenvalues of HAH are 1 plus (n ; 1) eigenvalues of A2 . Moreover, if
j j > j j > j j j j : : : jnj; 1
2
3
4
the dominant eigenvalue of A2 is 2, which is the second dominant eigenvalue of A.
Algorithm 8.5.4 Householder De ation to Compute the Subdominant Eigenvalue 1. Compute the dominant eigenvalue 1 and the corresponding eigenvector v1 using the power method and the inverse power method. 2. Find a Householder matrix H such that
01 BB CC 0 Hv = B BB .. CCC : @.A 1
0
3. Compute HAH . 4. Discard the rst row and the rst column of HAH and nd the dominant eigenvalue of the (n ; 1) (n ; 1) matrix thus obtained.
Example 8.5.6
0 :2190 :6793 :5194 1 B :0470 :9347 :8310 CC : A=B @ A :6789 :3835 :0346
The eigenvalues of A are: .0018, -0.3083, 1.4947.
510
0 ;:5552 1 B C 1. = 1:4947; v = B @ ;:7039 CA : ;:4430 0 ;1:5552 1 B ;0:7039 CC 2. u = v ; kv k e = B @ A ;0:4430 1
1
1
1 2 1
H=I;
2uuT uT u
0 ;:5552 ;:7039 ;:4430 1 B C =B @ ;:7039 :6814 ;:2005 CA ;:4430 ;:2005 ;:8738 011 B CC Hv = B @0A: 1
0
1 0 1:4977 ;:3223 ;:3331 1 0 1 : 4977 ; : 3223 ; : 3331 CC B 0 CC = BB 3. HAH = B : 1987 : 2672 B CA : A @ A @ 0 0 :3736 ;:5052 0 4. The dominant eigenvalue of A is ;0:3083 which the subdominant eigenvalue of A. Computing the Subdominant Eigenvector 2
2
Once the subdominant eigenvalue 2 has been computed using the above procedure, the corresponding eigenvector v2 can be found from the inverse iteration. We, however, show below that this eigenvector can be computed directly from A2 without invoking inverse iteration. Let v2(2) be the eigenvector of A2 corresponding to 2 . Then it can be shown that the eigenvector v1(2) of A1 corresponding to 1 is given by
v = v (2) 1
where is determined from
!
(2) 2
;
(1 ; 2 ) + bT v2(2) = 0: Once v1(2) is found, the subdominant eigenvector v2 of A corresponding to 2 is obtained from
v = Hv : (2) 1
2
(Note that HAHv1(2) = 2v1(2)). Thus to compute the subdominant eigenvector v2 of A: Given
1 0 T b A: HAH = A = @ 1
0 A2
1
511
Algorithm 8.5.5 Computing the Subdominant Eigenvector 1. Compute the eigenvector v2(2) of A2 corresponding to 2. 2. Compute given by
T = b ;v : (2) 2
2
3. Compute the eigenvector v1(2) of A1 :
! v = : v (2) 1
4. Compute the vector v2 of A:
(2) 2
v = Hv : 2
Example 8.5.7
1
(2) 1
0 :2190 :6793 :5194 1 B C A=B @ :0470 :9347 :8310 CA :6789 :3835 :0341
0 1:4947 ;:3223 ;:3333 1 B 0 C ; = 1:4947 = ;:3083 HAH = B :1987 :2672 C @ A 0 :3736 ;:5052 1
2
bT = (;:3223 ; :3331)
!
:1987 :2672 A = : ;:3736 ;:5052 2
1. 2. 3. 4.
;:4662 ! v = :8847 T = b ;v = :0801 0 :0801 1 B ;:4662 CC v =B @ A ;:8847 0 ;:1082 1 B ;:5514 CC : v = Hv = B @ A 0:8314 Computing the other largest Eigenvalues and Eigenvectors (2) 2
(2) 2
2
1
(2) 1
2
(2) 1
512
Once the pair (2; v2) is computed, the matrix A2 can be de ated using this pair to compute the pair (3; v3). From the vector pair (3; v3), we then compute the pair (4; v4) and so on. Thus by repeated applications of the process we can compute successively all the n eigenvalues and eigenvectors of A.
Remarks: However, if more than a few eigenvalues are needed, the QR iteration method to be described a little later should be used, because in that case, the QR iteration will be more cost-e ective, and it is a stable method. Computing the Smallest Eigenvalues It is easy to see that the power method applied to A;1 gives us the smallest eigenvalue in magnitude (the least dominant one) of A. Let A be nonsingular and let the eigenvalues of A be ordered such that
j j > j j j j jn; j > jnj > 0: 1
2
3
1
Then the eigenvalues of A;1 (which are the reciprocals of the eigenvalues of A) are arranged as: 1 1 1 > 1 > 0:
n n; n; 1
2
j j 1
That is, 1 is the dominant eigenvalue of A;1 . This suggests that the reciprocal of the smallest n eigenvalue can be computed by applying the power method to A;1.
Algorithm 8.5.6 Computing the Smallest Eigenvalue in Magnitude 1. Apply the power method to A;1 . 2. Take the reciprocal of the eigenvalue obtained in step 1.
Note: Since the power method is implemented by matrix-vector multiplication only, the inverse of A does not have to be computed explicitly. This is because computing A; x, 1
where x is a vector, is equivalent to solving the linear system: Ay = x. To compute the next least dominant eigenvalue (the second smallest eigenvalue in magnitude), we compute the smallest eigenvalue and the corresponding eigenvector and then apply de ation. The inverse power method applied to the (n ; 1) (n ; 1) matrix at bottom right hand 513
corner will yield the reciprocal of the second smallest eigenvalue of A. Once the required eigenvalues are computed, we can always use the inverse power method to compute the corresponding eigenvectors or use de ation as shown earlier.
Remark: To accelerate the convergence, a suitable shift should be made. Example 8.5.8
01 4 51 B C A=B @ 2 3 3 CA :
1 1 1 The power method (without shift) applied to A;1 with the starting vector x0 = (1; ;1; 1)T gave = 9:5145. Thus the smallest eigenvalue of A is: 1 = :1051:
(Note the eigenvalues of A are 6.3850, 01.4901 and .1051.)
Summary of the Process for Finding the Selected Eigenvalues and Eigenvectors To compute a selected number of eigenvalues (mainly the rst few largest or smallest eigenvalues and the corresponding eigenvectors), the following combination of the power method, inverse iteration and de ation is recommended to be used in the following sequences: 1. Use the power method (the power method applied to A;1) to compute a reasonably good approximation of the largest (smallest) eigenvalue in magnitude and of the corresponding eigenvectors. 2. Use the inverse iteration with the approximate eigenvalue (keeping it xed) and the approximate eigenvector obtained in step 1 as the starting vector. 3. Apply now de ation to compute the next set of eigenpairs.
514
8.6 Similarity Transformations and Eigenvalue Computations Recall (Theorem 8.2.2) that two similar matrices A and B have the same eigenvalues, that is, if X is a nonsingular matrix such that X ;1AX = B; then the eigenvalues of A are the same as those of B . One obvious approach to compute the eigenvalues of A, therefore, will be to reduce A to a suitable \simpler" form B by similarity so that the eigenvalues of B can be more easily computed. However, extreme caution must be taken here. It can be shown (Golub and Van Loan MC 1984, p. 198) that
Theorem 8.6.1 (X ; AX ) = X ; AX + E where 1
1
kE k kX k kX ; k kAk : 2
2
1
2
2
Remark: Since the error matrix E clearly depends upon Cond (X ), the above theorem tells us that in computing the eigenvalues and eigenvectors we should avoid ill-conditioned transforming matrices to transform A by similarity to a \simpler" form. Primarily because of this the eigenvalues of a matrix A are not computed using the characteristic polynomial of A or by transforming A to the Jordan Canonical form. Below we will discuss them in 2
some detail.
8.6.1 Eigenvalue Computations Using the Characteristic Polynomial Why should eigenvalues not be computed via the characteristic polynomial?
Since the eigenvalues of a matrix are the zeros of the characteristic polynomial, it is natural to think of computing the eigenvalues of A by nding the zeros of its characteristic polynomial. However, this approach is NOT numerically e ective.
515
Diculties with Eigenvalue Computations Using the Characteristic Polynomial First, the process of explicitly computing the coecients of the characteristic polynomial may be numerically unstable. Second, the zeros of the characteristic polynomial may be very sensitive to perturbations on the coecients of the characteristic polynomial. Thus if the coecients of the characteristic polynomial are not computed accurately, there will be errors in the computed eigenvalues. In Chapter 3 we illustrated the sensitivity of the root- nding problem by means of the Wilkinsonpolynomial and other examples. We will now discuss the diculty of computing the characteristic polynomial in some detail here. Computing the characteristic polynomial explicitly amounts to transforming the matrix to a block-companion (or Frobenius) form. Every matrix A can be reduced by similarity to
0C 1 0 B . . . CC C=B @ A; 1
0 Ck where each Ci is a companion matrix. The matrix C is said to be in Frobenius form. If k = 1, the matrix A is nonderogatory. Assume that A is nonderogatory and let's see how A can be reduced to a companion matrix by similarity. This can be achieved in two stages:
Reduction of a Matrix to a Companion Matrix Stage 1: The matrix A is transformed to an unreduced Hessenberg
matrix H by orthogonal similarity using the Householder or Givens method.
Stage 2: The transformed unreduced Hessenberg matrix H is further reduced to a companion matrix by similarity.
516
We have already seen in Chapter 5 how a nonderogatory matrix A can be transformed to an unreduced Hessenberg matrix by orthogonal similarity in a numerically stable way using the Householder or Given method. Consider now stage 2, that is, the transformation of the unreduced Hessenberg matrix H to a companion matrix C . Let X be the nonsingular transforming matrix, that is HX = XC where 00 0 : 0 c 1 BB 1 0 : 0 c1 CC 2C BB C: B C = B 0 1 0 0 c3 C BB .. .. .. . . . .. .. CCC . .A @. . . 0 0 : 0 cn If x1 ; x2; : : :; xn are the n successive columns of X , then from
HX = XC; we have
Hxi = xi ; i = 1; : : :; n ; 1; +1
and
Hxn = c x + c x + + cnxn: 1
Eliminating x2 through xn we have
1
2
2
(H )x = 0; 1
where (x) is the characteristic polynomial of C . Since the two similar matrices have the same characteristic polynomial, we have by the CayleyHamilton Theorem (H ) = 0: This means that x1 can be chosen arbitrarily. Once x1 is chosen, x2 through xn can be determined by the recursion: xi+1 = Hxi ; i = 1; : : :; n ; 1: Thus it follows that if x1 is chosen so that
X = (x ; Hx ; : : :; Hxn; ) 1
1
is nonsingular, then X will transform H to C by similarity. 517
1
Choose x1 = (1; 0; : : :; 0), then the matrix 01 BB 0 h BB 21 X = (x1; Hx1; : : :; Hxn;1) = B BB 0. 0. h21. h32 .. B@ .. .. 0 0 0 is nonsingular because hi+1;i 6= 0; i = 1; : : :; n ; 1.
...
.. .
0 h21h32 hn;n;1
1 CC CC CC CC A
So, if one or more subdiagonal entries hi ;i of H are signi cantly small, then the inverse of X will have large entries and consequently X will be ill-conditioned. Thus in +1
such a case the transformation of an unreduced Hessenberg matrix H to a companion matrix will be unstable
Thus, the rst stage, in which A is transformed to H using the Householder or the Givens method, is numerically stable, while the second stage, in which H is further reduced to C , can be highly unstable. Example.
0 1 2 31 B 0:0001 1 1 CC H=B @ A 0
2 3 x1 = (1; 0; 0)T ; x2 = Hx1 = (1; 0:0001; 0)T ;
x = Hx = (1:0002; 0:0002; 0:0002)T ; 0 1 1 1:0002 1 B 0 0:0001 0:0002 CC X=B @ A 0 0 0:0002 3
2
00 0 1 1 B 1 0 ;4:9998 CC X ; AX = C = B @ A 1
0 1 5 Cond2(X ) = 3:1326 104 :
(Note that the existence of a small subdiagonal entry of H ; namely h21 , made the transforming matrix X ill-conditioned.) 518
There are also other equivalent methods for reducing H to C . For example, Wilkinson, (AEP, p. 406) describes a pivoting method for transforming an unreduced Hessenberg matrix H to a companion matrix C using Gaussian elimination, which also shows that small subdiagonal entries can make the method highly unstable. The subdiagonal entries are used as pivots and we have seen before that small pivots can be dangerous. Note that there are other approaches for nding the characteristic polynomial of a matrix. For example, LeVerrier's method (Wilkinson, AEP pp. 434{435) computes the coecients of the characteristic polynomial using the traces of the various powers of A. Here Wilkinson has shown that in LeVerrier's method, severe cancellation can take place while computing the coecients from the traces using Newton's sums. The Newton's sums determining the coecients ci ; i = 0; : : :; n, of the characteristic polynomial det(A ; I ) = n + cn;1n;1 + + c1 + c0; are given by
cn; = ;trace(A); kcn;k = ;(trace(Ak ) + cn; trace(Ak; ) + + cn;k trace(A)); k = 2; : : :; n: 1
1
1
+1
For details, see (Wilkinson AEP, p. 434). Having emphasized the danger of using the Frobenius form in the eigenvalue-computations of a matrix, let's point out some remarks of Wilkinson about Frobenius forms of matrices arising in certain applications such as mechanical and electrical systems. \Although we have made it clear that we regard the use of the Frobenius form as dangerous, in that it may well be catastrophically worse-conditioned than the original matrix, we have found the program based on its use surprisingly satisfactory in general for matrices arising from damped mechanical or electrical systems. It is common for the corresponding characteristic polynomial to be well-conditioned. When this is true methods based on the use of the explicit characteristic polynomial are both fast and accurate." Quotation from Wilkinson AEP p. 482 519
Remarks: The above remarks of Wilkinson clearly support a long tradition by engineers of computing the eigenvalues via the characteristic polynomial.
8.6.2 Eigenvalue Computations via Jordan-Canonical Form Let us now discuss the use of some other suitable canonical forms in eigenvalue computations. In this connection, of course, the Jordan-Canonical Form is the one that comes to mind rst. Recall (Theorem 8.2.10) that given an n n matrix A, there exists a nonsingular matrix X such that
X ; AX = diag(J ; : : :; Jk); 1
1
0 1 1 0 i BB . . . . . . CC C: Ji = B BB ... 1 C CA @
where
If Ji is of order i , then
i
0
+ + : : : + k = n: 1
2
The matrix on the right hand side is the Jordan Canonical Form (JCF) of A, i 's are the eigenvalues. Thus, the eigenvalues of A are displayed as soon as the JCF is computed.
Unfortunately, this computation can also be highly unstable.
Whenever A is close to a nondiagonalizable matrix, the transforming matrix X will very ill-conditioned (see Golub and Van Loan MC 1984, pp. 196{197). It then follows from
Theorem 8.6.1 that the computed JCF in that case will be inaccurate.
8.6.3 Hyman's Method Before we leave this section, we wish to point out that there is a method for implicitly evaluating the characteristic polynomial of a Hessenberg matrix and its derivative at a certain given point, without explicitly computing the coecients of the characteristic polynomial. This method, known as Hyman's method, can in turn be used in a root- nding procedure (such as Newton's method) to compute a zero of the characteristic polynomial, and hence an eigenvalue of the associated matrix. Let H = (hij ) be an unreduced upper Hessenberg matrix. Set pn() = 1. 520
Compute pn;1() through p1 () using the recurrence:
13 20 n X 66 @ hi ;j pj () ; pi ()A 77 6 77 pi () = 66 j i 77 ; i = n ; 1; : : :; 1: hi ;i 64 5 +1
+1
= +1
+1
Then the characteristic polynomial det(A ; I ) of A is given by det(H ; I ) = h21 h32 hn;n;1
0n 1 X @ hij pj () ; p ()A : j =1
1
Wilkinson (AEP, p. 429) has shown that Hyman's method has favorable numerical properties. Thus Hyman's method can be combined with Newton's method for nding the zeros of a polynomial to obtain an isolated eigenvalue of a matrix. However, if all the eigenvalues are needed, one
should use the QR iteration method to be described later.
8.7 Eigenvalue Sensitivity In the previous two sections we have cautioned the readers about the danger of computing the eigenvalue via Jordan Canonical Form or the Frobenius form of a matrix. The danger was mainly the possibility of the transforming matrix X being ill-conditioned. In this section we will see now what speci c role the condition number of the transforming matrix: Cond(X ) = kX k kX ;1k, plays in the eigenvalue-sensitivity. We have illustrated the eigenvalue-sensitivity phenomenon by means of some examples in Chapter 3. Here is a more speci c result, known as the Bauer-Fike Theorem, on the eigenvalue sensitivity of a diagonalizable matrix.
8.7.1 The Bauer-Fike Theorem Theorem 8.7.1 Let A be diagonalizable, that is, the Jordan Canonical Form of A is a diagonal matrix D. Then for an eigenvalue of A + E , we have min ji ; j kX k kX ;1k kE k; where k k is a subordinate matrix norm, and 1; 2; : : :; n are the eigenvalues of A.
Proof. Consider two cases: 521
Case 1: = i for some i. The theorem is trivially true. Case 2: 6= i for any i. Then the diagonal entries of the diagonal matrix I ; D are di erent from zero. Since the determinant of a matrix is equal to the product of its eigenvalues, the matrix (I ; D) is nonsingular. Now from (A + E )x = x we have
Ex = (I ; A)x = (I ; XDX ; )x = X (I ; D)X ; x:
(Note that A = XDX ;1:)
1
1
Set X ;1x = y . Then from the last equation, we have, by multiplying the equation by X ;1 to the left (I ; D)y = X ;1Ex or
y = (I ; D); X ; EXy (note that x = Xy ). 1
1
Taking a subordinate norm on both sides, we have
kyk = k(I ; D); X ; EXyk k(I ; D); k kX ; kkE k kX k kyk: 1 1
1
1
Dividing both sides by y , we get 1 k(I ; D);1k kX ;1kkE k kX k: Now for a subordinate norm,
1 = max i ; i
k(I ; D);1k So, or,
= min(1 ; ) : i i
1 min(1 ; ) kX ;1k kX k kE k: i
i
min( ; i ) kX ;1k kX k kE k: 522
Implications of the Theorem The above theorem tells us that if Cond(X ) = kX k; kX ; k 1
1
is large, then an eigenvalue of the perturbed matrix A + E can be signi cantly di erent from an eigenvalue i of A. In general, the more ill-conditioned the eigenvector matrix X is, the more ill-conditioned will be the eigenproblem for A. Remark: In case A is not diagonalizable, a similar result also holds. (For details, see Golub
and Van Loan, MC 1984 p. 209.)
Example 8.7.1
0 1 ;0:2113 0:0957 1 B C A = B 2 ;0:0756 C @0 A 0
0
3
0 0 ;0:2113 :1705 1 B C X = 10 B 1 ;1:0636 C @0 A 5
0
0
1:2147
X ; AX = diag(1; 2; 3): 1
The eigenvalues of A are: 1, 2, 3
01 0 01 B CC E = 10; B 0 1 0 @ A 4
0 0 1 0 1:0011 ;0:2113 0:0957 1 B C A+E = B 2:0001 ;0:8756 C @ 0 A: 0 0 3:0001 The eigenvalues of A + E are: 1:0001; 2:0001 3:001 Cond2(X ) = 1:8294 105 523
kE k = 10; Cond (X ) kE k = 18:2936: 4
2
2
2
8.7.2 Sensitivity of the Individual Eigenvalues The condition number kX k kX ; k gives an overall assessment of the changes in eigenvalues with 1
respect to changes in the coecient of the matrix. However, as we have seen from the examples in Chapter 3, some eigenvalues of A may be more sensitive than the others. In fact, some may be very well-conditioned while others are ill-conditioned. Similarly, some eigenvectors may be well-conditioned while others are not. It is therefore more appropriate to talk about conditioning of the individual eigenvalues, rather than conditioning of the eigenvalue problem. Recall that in Chapter 3 an analysis of the illconditioning of the individual eigenvalues of the slightly perturbed Wilkinson-matrix was given in terms of the numbers si . In general, this can be done for any diagonalizable matrix. Let X ;1AX = diag(1; : : :; n). Then the normalized right-hand and left-hand eigenvectors corresponding to an eigenvalue i are given by Xei ; y = (X ;1)T ei : xi = kXe i k(X ;1)T eik2 i k2 De nition 8.7.1 The number 1 , where s is de ned by
si
i
si = yiT xi is called the condition number of the eigenvalue i . If this number is large then i is an ill-conditioned eigenvalue. This typically happens when A
is close to a matrix having some nonlinear elementary divisors (see Wilkinson, AEP p. 183). Notes: 1. There are n condition numbers associated with the n eigenvalues of A. 2. If xi and yi are real, then si is the cosine of the angle between xi and yi .
Example 8.7.2
01 2 31 B C A=B @ 0 0:999 1 CA 0
0
524
2
x x x y y y
= = = = = =
1 2 3 1 2 3
(1; 0; 0)T (1; ;0:005; 0)T (:9623; :1923; :1925)T (0:0004; :7066; ;0:7076)T (0; ;0:7075; :7068)T (0; 0; 1)T
s = yT x = 3:5329 10; s = yT x = 3:5373 10; s = yT x = :1925: 1
1
1
2
2
2
3
3
3
4 4
Thus, the above computations clearly show that 1 = 1; 2 = :999 are ill-conditioned, while 3 = 2 is well-conditioned. Indeed, when a(3; 1) was perturbed to 0.000001 and the eigenvalues of the perturbed matrix were computed, the rst two eigenvalues of the perturbed matrix (those corresponding to 1 and of the .999 of the original matrix) become complex. The computed eigenvalues of the perturbed matrix were (to three digits): 0:999 + 0:001i; 0:999 ; 0:001i; and 2: For yet another nontrivial example, see exercise #23.
A Relationship Between s and Cond(X ) i
It is easy to see that the condition numbers si and Cond2(X ) with respect to the 2-norm are related. T ; Xeij si = jyiT xi j = kXejeki X i k(X ; )T ei k 1
2
Now
1
1
2
= kXe k k(X ;1)T e k : i 2 i 2
kXeik kX k keik = kX k ; 2
2
2
2
and
k(X ; )T eik k(X ; )T k keik = k(X ; )T k = kX ; k : 1
1
2
1
525
2 2
2
1
2
So,
1 kX k kX ;1k = Cond (X ): 2 2 2
si Thus, for each i,
1
si Cond (X ):
Example 8.7.3
2
01 2 31 B C A=B @ 0 0:999 1 CA 0
0
2
1 = 2:8305 103 s1 1 = 2:8270 103 s2 1 = 5:1940:
s
3
Cond2(X ) = 6:9603 103 : Thus,
1 < Cond (X ); 2
si
i = 1; 2; 3:
The Condition Numbers and Linear Dependence of Eigenvectors Since for a diagonalizable matrix the columns of the matrix X are the eigenvectors of A, Cond2 (X ) gives us an indication of how linearly independent the eigenvectors are:
If Cond (X ) is large it means that the eigenvectors are nearly dependent. 2
For the above matrix A, the matrix of eigenvectors is: 0 ;0:4675 + :8840i ;0:4675 ; :8840i 0:9623 1 B ;0:0005 ; 0:0005i ;0:0005 + 0:0005i 0:1923 CC : X=B @ A 0:0000 ; 0:0000i 0:0000 + 0:000i 0:1925 Note the linear dependence of the rst two eigenvectors. Cond2(X ) = 2:5965 103 : 526
The Eigenvalue-Sensitivity of a Normal Matrix A matrix A is called normal if
AA = A A;
where A = (A)T : A Hermitian matrix is normal. Normal matrices are diagonalizable. A remarkable property of a normal matrix A is that if X is the transforming matrix that transform A to a diagonal matrix, thus Cond2 (X ) = 1: Thus an immediate consequence of the Bauer-Fike theorem is:
Corollary to the Bauer-Fike Theorem: Let A be a normal matrix, and ; : : :; n be the
eigenvalues of A. Then for an eigenvalue of A + E we have
1
min ji ; j kE k2:
Eigenvalue Sensitivity of a Normal Matrix In other words, the eigenvalues of a normal matrix are perfectly well-conditioned.
Remark: The normal matrices most commonly found in practical applications are symmetric (or Hermitian, if complex) matrices. Thus, by the Corollary above, the eigenvalues of a symmetric (or Hermitian) matrix are well-conditioned. We will discuss the symmetric eigenvalue problem in more detail in Section 8.11.
8.8 Eigenvector Sensitivity We shall not go into details in this discussion on the sensitivity of eigenvectors. We will just state a theorem (in somewhat crude form) that will highlight the main di erences between eigenvalue and eigenvector sensitivities. For an exact statement and proof, see Watkins (FMC pp. 332{333).
527
Theorem 8.8.1 Let A be a very small perturbation of A and let the eigenvalue
k of A be perturbed by k ; that is k + k is an eigenvalue of A + A: Let xk + xk be the eigenvector corresponding to k + k . Then, assuming that the eigenvalues of A are all distinct, we have X jk xk + xk = xk + xj + 0(kAk ); j 6 k (k ; j )sj 2
=
where
jk = yj (A)xj :
Implications of the theorem The above theorem tells us that if A is perturbed by a small amount, then the amount of perturbation an eigenvector x experiences is determined by k
1. condition numbers of all the eigenvalues other than k , and 2. the distance of k from the other eigenvalues. An immediate consequence of this theorem is that if there is a multiple eigenvalue or
an eigenvalue near another eigenvalue, then there are some ill-conditioned eigenvectors. This is signi cant especially for a Hermitian or a symmetric matrix, because we know that the eigenvalues of a Hermitian matrix are all well-conditioned, but the eigenvectors could be ill-conditioned. If the eigenvalues are well-separated and well-conditioned, then the eigenvectors are well-conditioned.
Example 8.8.1
01 0 01 B 0 :99 0 CC A=B @ A
0 0 2 0 1 :0001 0 1 B :0001 :99 0 CC : A0 = A + A = B @ A 0 0 2 The eigenvalues of A+A are 1, .99, 2. (No change, since are 0 A;is1 symmetric 1 0 :01 1 the 0eigenvalues 1 0 well-conditioned.) However, the eigenvectors of A0 are BB@ ;0:01 CCA, BB@ ;1 CCA and BB@ 0 CCA while those 0 0 1 528
011 001 001 B CC B C B C of A are B @0A; B @1C A and B @ 0 CA. Note that the eigenvector corresponding to = 2 has 0
0
3
1
not changed; while the other two eigenvectors have changed; because of the proximity of the eigenvalues 1 and .99.
8.9 The Real Schur Form and QR Iterations In the preceding discussions we have seen that computing eigenvalues of A via reduction of A to the Frobenius or to the Jordan Canonical Form is not numerically e ective. If the transforming matrix is ill-conditioned, then there may be large errors in the computed canonical form and this in turn will introduce large errors in the eigenvalues.
A question therefore arises as to whether we can obtain a similarity reduction of A to a suitable canonical form using a well-conditioned transforming matrix.
A perfectly well-conditioned matrix, for example, is an orthogonal matrix (or unitary, if it is complex) the condition number (with respect to 2-norm and F-norm) of such a matrix, being 1. Indeed, if a matrix A is transformed to a matrix B using unitary similarity trans-
formation, then a perturbation in A will result in a perturbation in B of the same magnitude. That is, if B = U AU
and
U (A + A)U = B + B;
then
kBk kAk : 2
Example 8.9.1
2
0 ;:5774 ;:5774 ;:5774 1 01 2 31 BB ;:5774 :7887 ;:2113 CC B CC 3 4 5 ; U = A=B @ A @ A ;:5774 ;:2113 :7887 6 7 8 0 13 ;:6340 ;2:3660 1 B ;:9019 C B=B 0 0 C @ A ;6:0981 0 0 A = 10; I 0 1:00001 2 1 3 B C A = A + A = B 4:00001 5 C @ 3 A 5
1
6
529
3
3
7
8:00001
0 13:00001 ;:633974 ;2:3660 1 B C B = U (A + A)U = B 0 C @ ;:9019 :00001 A ;6:0981 0 0:00001 B = B ; B = 10; I 1
5
1
3 3
kAk = kBk = 10; : 5
A perfect canonical form displaying the eigenvalues is a triangular form (the diagonal entries are the eigenvalues). In this context we now recall a classical result due to Schur (Theorem 8.2.3). We restate this important result below and give a proof.
Theorem 8.9.1 (The Schur Triangularization Theorem; restatement of Theorem 8.2.3) If A is an n n matrix, then there exists a unitary matrix U such that
U AU = T
where T is a triangular matrix with the eigenvalues 1; 2; : : :; n as the diagonal entries.
Proof. We will prove the theorem using induction on n.
If n = 1; the theorem is trivially true. Next assume the theorem is true for n = k ; 1, then we will show that it is also true for n = k: Let u be a normalized eigenvector of A associated with an eigenvalue 1. De ne
U = (u; V ); 1
where V is k k ; 1 and is unitary. Then U1 is unitary; and,
0 1 BB CC 0 B CC A = U AU = B BB ... CC ; ^ A @ A 1
1
1
1
0
where A^ is (k ; 1) (k ; 1). By our hypothesis there exists unitary matrix V1 of order (k ; 1) such that T^ = V1 (A^)V1 530
is triangular. Then, de ning
0 1 1 0 0 BB CC BB 0 CC U = B .. CC ; B@ . V A 2
1
0 we see that U2 is unitary (because V1 is so), and
U A U = U U AU U = U AU 2
So,
1
2
2
1
1
2
1 0 CC BB 0 CC B U AU = B CC : BB ... V AV ^ ^ = T A @ 1
1
0
2
Since T^ is triangular, so is U AU: Since the eigenvalues of a triangular matrix appear on the diagonal, we are done. Since a real matrix can have complex eigenvalues (occurring in complex conjugate pairs), even for a real matrix A, U and T in the Schur Theorem above can be complex. However, we can choose U to be real orthogonal if T is replaced by a quasi-triangular matrix R, known as the Real Schur Form of A (RSF).
Theorem 8.9.2 (The Real Schur Triangularization Theorem) Let A be an n n real matrix. Then there exists an n n orthogonal matrix Q such that 0R R R 1 k BB C 0 R R kC B C T Q AQ = R = B . . . ... C B@ ... CA 0 0 Rkk where each Rii is either a scalar or a 2 2 matrix. The scalars diagonal entries correspond to real eigenvalues and 2 2 matrices on the diagonal correspond to 11
complex conjugate eigenvalues.
531
12
1
22
2
Proof. The proof is similar to Theorem 8.9.1. De nition 8.9.1 The matrix R in Theorem 8.9.2 is known as the Real Schur Form (RSF) of A.
Notes: 1. The 2 2 matrices on the diagonal are usually referred to as \bumps". 2. The columns of Q are called Schur vectors. For each k(1 k n), the rst k columns
of Q form an orthonormal basis for the invariant subspace corresponding to the rst k eigenvalues.
Remark: Since the proofs of both the theorems are based on the knowledge of eigenvalues
and eigenvectors of the matrix A, they can not be considered to be constructive. They do not help us in computing the eigenvalues and eigenvectors. We present below a method known as the QR iteration method, for computing the Real-Schur form of A. A properly implemented QR-method is widely used nowadays for computing the eigenvalues of an arbitrary matrix. As the name suggests, the method is based on the QR factorization and is iterative in nature. The QR iteration method was proposed in algorithmic form by J. G. Francis (1961), though its roots can be traced to a work of Rutishauser (1958). The method was also independently discovered by the Russian mathematician Kublanovskaya (1961).
Note: Since the eigenvalues of a matrix A are the n zeros of the char-
acteristic polynomial and it is well-known (proved by Galois more than a century ago) that the roots of a polynomial equation of
degree higher than four cannot be found in a nite number of steps, any numerical eigenvalue-method for an arbitrary matrix has to be iterative in nature.
8.9.1 The Basic QR Iteration We rst present the basic QR iteration method. Set A0 = A: 532
Compute now a sequence of matrices (Ak ) de ned by
A = QR A = R Q =Q R A = R Q =Q R ; 0
0
0
1
0
0
1
1
2
1
1
2
2
and so on. In general,
Ak = Qk Rk = Rk; Qk; k = 1; 2; : : : The matrices in the sequence fAk g have a very interesting property. Each matrix in the se1
1
quence is orthogonally similar to the previous one and is therefore orthogonally similar to the original matrix. It is easy to see this. For example, A = R Q = QT A Q (sinceQT A = R ) A = R Q = QT A Q : 1
0
0
0
0
0
2
1
1
1
1
1
0
0
0
Thus A1 is orthogonally similar to A, and A2 is orthogonally similar to A1 . Therefore, A2 is orthogonally similar to A, as the following computation shows:
A = QT A Q = QT (QT A Q )Q = (Q Q )T A (Q Q ): 2
1
1
1
1
0
0
0
1
0
1
0
0
1
Since each matrix is orthogonally similar to the original matrix A, and therefore, has the same eigenvalues as A; then if the sequence fAk g converges to a triangular or quasi-triangular matrix, we will be done. The following result shows that under certain conditions, indeed this happens (see Wilkinson AEP, pp. 518{519).
A Condition for Convergence Theorem 8.9.3 (A Convergence Theorem for Basic QR Iteration) Let the eigenvalues ; : : :; n be such that j j > j j > : : : > jnj; and let the eigenvector 1
1
2
matrix X of the left eigenvectors (that is, X ;1) be such that its leading principal minors are nonzero. Then fAk g converges to an upper triangular matrix or to the Real Schur Form. In fact, it can be shown that under the above conditions, the rst column of Ak approaches a multiple of e1 . Thus, for suciently large k we get
0 B Ak = B @
1
u1
0 Ak
533
CC A:
We can apply the QR iteration again to Ak and the process can be continued to see that the sequence converges to an upper triangular matrix.
Example 8.9.2
!
1 2 A= : 3 4 Eigenvalues of A are: 5.3723, and -0.3723. j1j > j2j.
k=0: A = A=Q R ;0:3162 ;0:9487 ! Q = ;0:9487 0:3162 ! ;3:1623 ;4:4272 R = : 0 ;0:6325 0
0
0
0
0
k=1:
k=2:
!
5:2 1:6 A1 = R0 Q0 = =Q R :6 ;:2 ! 1 1 ;0:9934 ;0:1146 Q1 = ;0:1146 ;:9934 ! ;5:2345 ;1:5665 R1 = : 0 ;0:3821
!
5:3796 ;0:9562 A2 = R1Q1 = = Q2R2: ;0:0438 ;0:3796
(Note that we have already made some progress towards obtaining the eigenvalues.) ;1 ;0:0082 ! Q = ;0:0081 1 ;5:3797 0:9593 ! R = : 0 ;0:3718 2
2
k=3:
!
5:3718 1:0030 A3 = R2Q2 = = Q3 R3 0:0030 ;0:3718 1 ;0:0006 ! Q3 = ;0:0006 1 ;5:3718 ;1:0028 ! R3 = : 0 ;0:3723 534
k=4:
!
5:3723 ;0:9998 A4 = R3Q3 = : ;0:0002 ;0:3723
8.9.2 The Hessenberg QR Iteration The QR iteration method as presented above is not ecient if the matrix A is full and dense. We have seen before that the QR factorization of such a matrix A requires 0(n3) ops and thus n iterations of QR method will require 0(n4 ) ops, making the method impractical. Fortunately, something simple can be done:
Reduce the matrix A to a Hessenberg matrix by orthogonal similarity before starting the QR iterations. An interesting
practical consequence of this is that if A = A0 is initially reduced to an upper Hessenberg matrix and is assumed to be unreduced, then each member of the sequence fAk g is also upper Hessenberg. This can be seen as follows: Suppose Ak is an unreduced upper Hessenberg matrix, and Givens rotations are used to factorize Ak into QkRk. Then Qk = J (2; 1; )J (3; 2; ) J (n; n ; 1; ) is also upper Hessenberg. Thus, since Rk is upper triangular, Ak+1 = Rk Qk is also upper Hessenberg. Since the QR factorization of a Hessenberg matrix requires only 0(n2 ) ops, the QR iteration method with the initial reduction of A to Hessenberg will be bounded by 0(n3), making the method ecient.
Example 8.9.3
0 0:2190 ;0:5651 ;0:6418 1 B ;0:6805 0:1226 0:4398 CC A=A =B @ A ;0:0000 0:8872 0:8466 0 ;0:3063 ;0:4676 0:8291 1 0 ;0:7149 0:2898 0:6152 1 B 0:9519 ;0:1505 0:2668 CC ; R = BB 0:0000 1:0186 0:9714 CC : Q =B @ A @ A 0:0000 0:8710 0:4913 ;0:0000 0 0:0011 0
0
0
535
k=1:
0 0:4949 0:8265 ;0:2132 1 B C A =R Q =B @ 0:9697 0:6928 0:7490 CA 1
0
0
0:0000 0:0010 0:0006 0 0:4546 ;0:8907 ;0:0021 1 0 1:0886 0:9928 0:5702 1 B C B C Q1 = B @ 0:8907 0:4546 0:0011 CA ; R1 = B@ 0:0000 ;0:4213 0:5303 CA : 0:0000 ;0:0023 1:0000 0:0000 ;0:0000 0:0018
k=2:
0 1:3792 B A =R Q =B @ ;0:3752 ;0:0000 0 0:9649 0:2625 ;0:0000 1 B CC Q =B ; 0 : 2625 0 : 9649 ; 0 : 0000 @ A; ;0:0000 0:0000 1:0000 2
1
1
2
;0:5197 0:5690 1 C ;0:1927 0:5299 C A ;0:0000 0:0018 0 1:4293 ;0:4509 0:4099 1 B CC R =B 0 ; 0 : 3224 0 : 6607 @ A: 0 0 0:0018 2
(Note that each Q and each A is upper Hessenberg.) i
i
8.9.3 Convergence of the QR Iterations and the Shift of Origin Once it is known that the QR iteration method can be made ecient by initial reduction of A to a Hessenberg matrix, it is natural to wonder about the rate of convergence of the iterations. By rate of convergence we mean how fast the subdiagonal entries of the transformed Hessenberg matrix converge to zero. Let A be initially reduced to an upper Hessenberg matrix H and the QR iteration is performed on H to obtain the sequence Hk . Then it can be shown (Golub and Van Loan, MC 1984 p. 228) that the subdiagonal entry h(i;ik);1 of Hk converges to zero at a rate determined by the ratio
k i : i; Thus, the rate of convergence will be very slow if the moduli of two eigenvalues and ;1 are very close to each other. Fortunately, the rate of convergence can signi cantly be improved by shifting the origin. 1
i
i
Let ^ i be an approximation of an eigenvalue i of H . Let the QR iteration be applied to the matrix H^ = H ; ^ i I: 536
The eigenvalues of H^ are i ; ^ i; 2 ; ^ i; : : :; n ; ^i: Let these eigenvalues be ordered so that
j ; ^ij j ; ^ij jn ; ^ij: 1
2
Then, in this case, the ith subdiagonal entry of H^ k will converge to zero at a rate determined by the ratio ; ^ k i i ; i;1 ; ^ i
rather than by the ratio j i jk : The former is usually smaller than the latter. i;1 Consider the convincing example from Ortega and Poole (INMD, p. 227). Let i = :99; i;1 = 1:1; ^ i = 1: Then ; ^ i i = :1 i;1 ; ^ i while
i i;
1
= :9:
This observation tells us that if we apply the QR iteration to the shifted matrix H^ rather than to the original matrix H , then the rate of convergence will be faster. Of course, once an eigenvalue of H^ is found, the corresponding eigenvalue of H can be computed just by adding the shift back.
8.9.4 The Single-Shift QR Iteration The above argument suggests the following modi ed QR iteration method, known as the singleshift QR iteration method. 1. Transform A to an upper Hessenberg matrix H . 2. Set H0 = H . k) becomes computationally zero. For k = 0; 1; 2; : : : do until h(n;n ;1 (k ) Hk ; hnn I = QkRk , Hk+1 = Rk Qk + h(nnk)I .
In the above h(ijk) denotes the (i; j )th entry of Hk . Of course, each of the matrices
can overwrite by H .
fH g k
To implement the single shift QR iteration, we need to have an approximate value ^ i of the eigenvalue i . Experimentally it has been observed that if we let the unshifted QR iteration (the 537
Basic QR) run for a few iterations (say s), then h(nns) can be taken as a reasonably good approximation to an eigenvalue. Thus starting with h(nns) as a shift, we can continue the iterations using the (n; n)th element of the current matrix as the next shift.
Convergence: The rate of convergence of the single-shift QR iteration method is ultimately
cubic to the eigenvalue of minimum modulus (see Wilkinson AEP, pp. 548{549).
Remark: An implicit version of the single shift iteration known as the implicit QR iteration,
can be worked out, where one does not subtract the shifts but implicitly constructs the matrix Hk+1. This is explained in the exercise #29.
Example 8.9.4 Single-shift QR Iteration
01 1 11 B C H=H =B @ 1 2 3 CA : 0
k=0:
0 1 1
00 1 11 B 1 1 3 CC = Q R H ;h I =B @ A 0 1 0 0 0 0:7071 ;0:7071 1 B C Q =B 0 C @ ;1 0 A (0) 33
0
0
0
0
0 0:7071 0:7071
0 ;1:0000 ;1:0000 ;3:000 1 B C R =B 0:4142 0:7071 C @ 0 A 0 0 ;0:7071 0 2:000 ;2:8284 ;1:4142 1 B C H =R Q +h I =B @ ;1:4142 1:5000 0:5000 CA : 0 ;0:5000 0:5000 0
1
k=1:
0
(0) 33
0
H ;h I = Q R 0 ;0:7276 ;0:6342 ;0:2615 1 B C Q =B @ 0:6860 ;0:6727 ;0:2774 CA 0 ;0:3812 0:9245 0 ;2:0616 2:7440 1:3720 1 B C R =B 1:3117 0:5606 C @ 0 A 0 0 0:2311 1
(1) 33
1
1
538
1
1
0 3:8824 ;1:0613 1:0464 1 B C H = R Q +h I = B @ 0:8998 ;0:5960 0:1544 CA : 0 ;0:0881 0:7137 2
k=2:
1
1
(1) 33
H ;h I = Q R 0 ;0:9620 0:2721 0:0247 1 B ;0:2732 ;0:9580 ;0:0870 CC Q =B @ A 0 ;0:0905 0:9959 0 ;3:2940 1:3787 ;1:0488 1 B C R =B 0:9740 0:1367 C @ 0 A 0 0 0:0124 0 3:5057 ;2:1221 ;1:2459 1 B C H =R Q +h I =B @ ;0:2661 ;0:2318 ;:0514 CA : 0 ;0:0011 0:7260 2
(2) 33
2
2
2
2
3
k=3:
2
(2) 33
2
H ;h I = Q R 0 ;0:9955 ;0:0953 ;0:0001 1 B C Q =B @ 0:0953 ;0:9955 ;0:0010 CA 0 ;0:0010 1:000 0 ;2:7924 2:0212 1:2451 1 B C R =B 1:1556 0:0675 C @ 0 A 0 0 0:0001 0 3:6983 ;1:7472 1:2434 1 B C H = R Q +h I = B @ 0:1101 ;0:4244 0:0664 CA : 0 0:0000 0:7261 3
(3) 33
3
3
3
3
4
3
3
(3) 33
The iteration is clearly converging towards the eigenvalue .7261. (The eigenvalues of H , in four digit arithmetic, are: :7261; 3:6511; ;0:3772:)
8.9.5 The Double-Shift QR Iteration If the desired eigenvalue is real, the above single-shift QR iteration works well. However, a real matrix can have complex eigenvalues. In such a case: i) at some stage of iteration, we could encounter a trailing 22 submatrix on the bottom right-hand corner corresponding to a complex conjugate pair of eigenvalues, if there are any, 539
ii) the (n; n)th entry of the trailing 2 2 matrix, which is real, will not be a good approximation, iii) it is natural to use the eigenvalues of that 2 2 matrix as shift parameters yielding the double-shift QR iteration.
One Iteration-step of the Double-Shift QR (Complex) Let the eigenvalues of the 2 2 bottom right hand corner of the Hessenberg matrix Hs be k1 and k2 = k1 . Then one iteration-step of the double-shift QR iteration is:
Hs ; k I = QsRs ; Hs = Rs Qs + k I 1
+1
1
Hs ; k I = Qs Rs ; Hs = Rs Qs + k I: +1
2
+1
Example 8.9.5
+1
+2
+1
+1
2
01 2 21 B 0 0 1 CC H =H=B @ A: 0 ;1 0 k = i; k = ;i: 0
1
H ; kiI = Q R : 0
0
2
0 ;1 0 1 0 B C Q =B @ 0 ;:7071 ;0:7071i CA
0
0
0
:7071i
:7071
0
0
0
0 ;1 + i ;2 ;2 1 B C R =B @ 0 1:4142i ;1:4142 CA 0
0 1 1:4142 ; 1:4142i ;1:4142 + 1:4142i 1 B0 CC H =R Q +k I =B ;i 0 @ A 1
0
0
1
H ;k I = Q R : 1
2
1
0
0
0 ;1 0 1 0 B C Q =B @ 0 ;:9124 ;:2717i CA
1
1
0
:2717i
i
:9624
0 ;1 ; i ;1:4142 + 1:4142i 1:4142 ; 1:4142i 1 B CC R =B 0 :5434 @ 0 A 1
0
0
1:9248
0 1 1:7453 ; :9717i 1:7453 ; :9717i 1 B CC H = R Q +k I = B ;:8523i :5230 @0 A: 0 ;:5230 :8523i 2
1
1
2
540
;:8523i :5230 ! Note that the eigenvalues of the 2 2 bottom right hand corner matrix ;:5230 :8523i are ;i and i. Thus the eigenvalues of H are 1; i and ;i. 2
Avoiding Complex Arithmetic in Double-Shift QR Iteration Since k1 and k2 are complex, the above double-shift QR iteration step will require complex arithmetic for implementation, even though the starting matrix Hs is real. However, with a little manipulation complex arithmetic can be avoided. We will discuss this aspect now.
We will show that the matrix H +2 is orthogonally similar to H through a real transforming matrix, and can be formed directly from H without computing H +1. s
s
s
Consider the matrix
s
N = (Hs ; k I )(Hs ; k I ) = Hs ; (k + k )Hs + k k I: 2
2
1
1
2
1
2
Since k2 = k1 ; the matrix N is real. Next, we show that (QsQs+1 )(Rs+1Rs) is the QR factorization of N .
N = (Hs ; k I )(Hs ; k I ) = (Hs ; k I )QsRs = Qs Qs (Hs ; k I )QsRs = Qs (Hs ; k I )Rs = Qs Qs Rs Rs: 2
1
2
2
+1
2
+1
+1
Since N is real and (Qs Qs+1)(Rs+1Rs) is the QR factorization of N , the matrix Qs Qs+1 can be chosen to be real. Finally, we show that Hs+2 is orthogonally similar to Hs through this real transforming matrix Qs Qs+1.
Hs
+2
= = = = = =
Rs Qs + k I Qs (Hs ; k I )Qs + k I Qs (RsQs + (k ; k )I )Qs + k I Qs [Qs (Hs ; k I )Qs + (k ; k )I ]Qs + k I Qs Qs HsQs Qs (Qs Qs ) Hs (QsQs ): +1
+1
+1
+1
2
2
+1
+1
1
+1
1
+1
+1
+1
2
2
+1
1
2
2
+1
2
+1
Since QsQs+1 and Hs are real, Hs+2 is also real. From the above discussions we conclude that it is possible to obtain Hs+2 directly from Hs through real orthogonal transformations. 541
Eigenvalues k and k need not be computed explicitly 1
2
Though computing the eigenvalues of a 2 2 matrix is almost a trivial job, we note that k1 and k2 need not be computed explicitly. To form the matrix
N = (Hs ; k I )(Hs ; k I ) = Hs ; (k + k )Hs + k k I; 1
2
2
1
2
1
2
all we need to compute is the trace and the determinant of the 2 2 matrix. Let !
hn; ;n; hn; ;n hn;n; hnn 1
1
1
1
be the 2 2 right hand corner matrix of the current matrix Hs: Then
t = k + k = sum of the eigenvalues = trace = hn; ;n; + hnn is real; 1
2
1
1
d = k k = product of the eigenvalues = determinant =hn; ;n; hnn ; hn;n; hn; ;n is real: 1 2
1
1
1
1
This allows us to write the one-step of double-shift QR iteration in real arithmetic as follows:
One Step of Double Shift QR Iteration (Real Arithmetic) 1. Form the matrix N = Hs2 ; tHs + dI . 2. Find the QR factorization of N : N = QR. 3. Form Hs+2 = QT Hs Q. We will call the above computation Explicit Double Shift QR iteration for reasons to be stated in the next section.
Example 8.9.6
01 2 31 B 1 0 1 CC H=H =B @ A 0 ;2 2 0
t = 2; d = 2:
0 3 ;8 5 1 B C N = H ; tH + dI = B @ ;1 2 3 CA : ;2 0 0 2
542
Find the QR Factorization of N :
0 ;:8018 ;:5470 ;:2408 1 B :2673 :0322 ;0:9631 CC Q=B @ A :5345 ;0:8365 0:1204 0 ;:8571 1:1007 2:5740 1 B ;1:1867 3:0455 ;0:8289 CC : H = QT H Q = B @ A 2
0
0:0000 1:8437 0:8116
8.9.6 Implicit QR Iteration After all this, we note, with utter disappointment, that the above double-shift (explicit) QR iteration is not practical. The reason for this is that forming the matrix N itself in step 1 requires 0(n3) ops. Fortunately, a little trick again allows us to implement the step in 0(n2) ops.
One Iteration of the Double-Shift Implicit QR 1. Compute the 1st column n1 of the matrix N . 2. Find a Householder matrix P0 such that P0n1 is a multiple of e1 . 3. Form Hs0 = P0T HsP0 . 4. Transform Hs0 to an upper Hessenberg matrix by orthogonal similarity using Householder matrices: (PnT;2 P2T P1T )Hs0 (P1 P2 Pn;2 ) = Hs0 +2 :
Using the implicit Q-theorem (Theorem 5.4.3) we can show that the matrix H +2 of the explicit QR and H 0+2 of the implicit QR are both unreduced upper Hessenberg and are essentially the same matrix. The above four steps constitute one iteration of the double-shift implicit QR. To unders
s
stand how these four steps can be implemented in 0(n2) ops, we must give a close look into the structures of the above computations. First, since Hs is Hessenberg, computing the 1st column of N = (Hs ; k1 I )(Hs ; k2 I ) is almost trivial. It contains only three nonzero elements. They can be written down explicitly. 543
0n BB n BB Bn Ne = n = B BB BB 0. B@ ..
Let
11 21 31
1
1
0
then
n n n
11 21 31
1 CC CC CC CC ; CC CA
= h211 ; th11 + d + h12h21 = h21 (h11 + h22 ; t) = h21 h32:
Here hij refers to the (ij )th entry of Hs . Second, because only three elements of n1 are nonzero, the Householder matrix P0 has the form ! P^0 0 P0 = ; 0 In;3 where P^0 is a 3 3 Householder matrix. Because of this form of P0 , and Hs being Hessenberg, the matrix Hs0 = P0 Hs P0 is not a full matrix. It is a Hessenberg matrix with a bulge. For example, when n = 6, we have 0 1
BB 0 BB B0 Hs0 = P Hs P = B BB BB 0 B@ 0
CC C C C CC : C CC C A 0 0 0 0 A bulge will be created at each step of the reduction of Hs to Hessenberg form and the constructions of Householder matrices P through Pn; amount to chasing these bulges systematically. Each Pk ; k = 1; 2; : : :; n ; 3 has the form 0I 1 0 k B ^ CC Pk = B @ Pk A; 0 In;k; where P^k is a 3 3 Householder matrix. The last Householder matrix Pn; has the form ! In; 0 Pn; = : 0 P^n; 0
1
0 0
0
2
3
2
2
2
2
544
Taking into consideration the above structures of computations, it can be shown that one step of the implicit QR iteration requires only o(n ) ops. 2
For details of this 0(n2) computations of one iteration of the double-shifted implicit QR, see the book by Stewart (IMC pp. 375{378) and the recent book by Watkins (FMC, pp. 277{278).
Algorithm 8.9.1 One Iteration-Step of the Double-Shift Implicit QR Let H be an n n unreduced upper Hessenberg matrix. Then the following algorithm, which constitutes one step of the double-shift implicit QR iteration, produces orthogonal matrices P0 ; P1; : : :; Pn;2 such that
QT HQ; where Q = P P ; : : :; Pn; ; 0
1
2
is an unreduced upper Hessenberg matrix. The algorithm overwrites H with QT HQ. 1. Compute the shifts:
t = hn; ;n; + hnn d = hn; ;n; hnn ; hn;n; hn; ;n: 1
1
1
1
1
1
2. Compute the 1st three nonzero entries of the 1st column of N = H 2 ; tH + dI :
x = n = h ; th + d + h h y = n = h ( h + h ; t) z = h =h h : 11
2 11
21
21
32
21
11
11
12
21
12
32
3. Compute the Householder matrices P0P1 : : :Pn;2 such that the nal matrix is upper Hessenberg. For k = 0; 1; 2; : : :; n ; 3 do. (a) Find a 3 3 Householder matrix P^k such that
0x1 01 B y CC = BB 0 CC : P^k B @ A @ A z
Form
0I k B B Pk = @ P^k 0
545
0
1 CC A:
0
In;k;
3
Form PkT HPk and overwrite with H :
H PkT HPk : Update x; y and z :
x hk y hk z hk
+2
;k+1
+3
;k+1
+4
;k+1
(if k < n ; 3):
(b) Find a Householder matrix P^n;2 of order 2 such that
P^n;
2
Form
Pn; = 2
!
!
x = : y 0 In; 0
2
!
0
P^n;
2
:
Flop-count. One step of the double-shift implicit QR iteration takes about 6n ops. If the 2
transferring matrix Q is needed and accumulated, then another 6n2 ops will be needed (see Golub and Van Loan MC 1983, p. 234).
01 2 31 Example 8.9.7 1. t = 2; d = 2; H = BB@ 1 0 1 CCA 0 ;2 2 2. x = n = 3; y = n = ;1; z = n = ;2 11
3. k = 0:
21
31
031 T BB CC P = I ; 2uuu T u ; where u = @ ;1 A ;2 0 ;:8018 :2673 :5345 1 CC B : 2673 : 9604 ; 0 : 0793 P =B A @ :5345 ;:0793 0:8414 0 ;:8571 ;2:6248 ;0:9733 1 B :0581 0:8666 1:9505 CC : H = P T HP = B @ A 1:1852 ;0:7221 2:9906 0
0
0
0
Update x and y :
546
x = h = :0581 y = h = 1:1852 21
31
Find P1 :
!
0:0490 ;0:9988 ; ;0:9988 0:0490 01 0 B P1 = B @ 0 0:0490 0 ;0:9988 0 ;0:8571 B ;1:1867 H = P1T HP1 = B @ 0
P^1 =
!
!
;1:1866 P^1 xy = ; 0 1 0 C; ;0:9988 C A 0:0490 1 1:1008 2:5739 C: 3:0456 ;0:8290 C A 1:8436 0:8116
Note that the matrix H obtained by the implicit QR is the same as H2 obtained earlier in section 8.9.5 by the explicit QR. (Example 8.9.6)
8.9.7 Obtaining the Real Schur Form A 1. Transform the matrix A to Hessenberg form. 2. Iterate with the implicit double step QR method. Typically, after two to three steps of the doubly-shift implicit QR iteration, one or two (and sometime more) subdiagonal entries from the bottom of the Hessenberg matrix converge to zero. This then will give us a real or pair of complex conjugate eigenvalues. Once a real or a pair of complex conjugate eigenvalues is computed, the last row and the last column in the rst case, or the last two rows and the last two columns in the second case, can be deleted and computation of the other eigenvalues can be continued with the submatrix. This process is also known as de ation. Note that the eigenvalues of the de ated submatrix are also the eigenvalues of the original matrix. For, suppose immediately before de ation, the matrix has the form:
!
A0 C 0 Hk = ; 0 B0 where B 0 is the 2 2 trailing submatrix or is a 1 1 matrix. Then the characteristic equation of Hk : det(I ; Hk ) = det(I ; A0) det(I ; B 0 ): 547
Thus, the eigenvalues of Hk are the eigenvalues of A0 together with those of B 0 . But Hk is orthogonally similar to the original matrix A and therefore has the same eigenvalues as A.
When to Accept a Subdiagonal Entry as Zero A major decision that we have to make during the iteration procedure is when to accept a subdiagonal entry as zero so that the matrix can be de ated. It seems that there are no clear-cut conventions here; however, we have given a commonly used criterion above. For a good discussion on this matter, see the book \The Symmetric Eigenvalue Problem" by B. N. Parlett (1980).
Accept a subdiagonal entry hi;i;1 to be zero if
jhi;i; j (jhiij + jhi; ;i; j): 1
1
Example 8.9.8 Find the Real Schur Form of 2 66 1 H = 64 1
1
3
2 37 0 1 775 : 0 ;2 2
The RSF is
Iteration 1 2 3 4
h ;1:1867 21
0.3543 0.0129 0.0000
2 3 ; 1 : 1663 ; 1 : 3326 ; 2 : 0531 66 7 64 0 1:2384 1:6659 775 : 0 ;1:9409 2:9279
The eigenvalues of the 2 2 right-hand lower corner submatrix are 2:0832 1:5874i.
Beresford Parlett is a professor of mathematics at the University of California at Berkeley. He has made some outstanding contributions in the area of numerical matrix eigenvalue problem, especially for large and sparse problems. His book \The Symmetric Eigenvalue Problem" is an authoritative book in this area."
548
Example 8.9.9 Find the Real Schur Form of 2 3 0 : 2190 ; 0 : 0756 0 : 6787 ; 0 : 6391 66 7 66 ;0:9615 0:9032 ;0:4571 0:8804 777 h=6 64 0 ;0:3822 0:4526 ;0:0641 775 0 0 ;0:1069 ;0:0252 Iteration h21 1 0.3860 2 ;0:0672 3 0.0089 4 ;0:0011 5 0.0001 .
h h ;0:5084 ;0:0084 ;0:3773 0.0001 ;0:3673 0 ;0:3590 0 ;0:3905 0 32
43
The RSF is
2 3 1.4095 0.7632 ; 0 : 1996 0.8394 66 7 66 0.0001 0.1922 0.5792 0.0494 777 h=6 : 64 0 ;0:3905 0.0243 ;0:4089 775 0 0 0 ;0:0763 2 3 0:1922 0:5792 5 The eigenvalues of 4 are 0:1082 0:4681i. ;0:3905 0:0243
Balancing As in the process of solving linear system problems, it is advisable to balance the entries of the original matrix A, if they vary widely, before starting the QR iterations. The balancing is equivalent to transforming the matrix A to D;1AD, where the diagonal matrix D is chosen so that a norm of each row is approximately equal to the norm of the corresponding column. The EISPACK routine BALANC that balances the entries of the matrix A is applied to A before the other routines are used to compute the eigenvalues. In general, preprocessing the matrix by balancing improves the accuracy of the QR iteration method. Note that no round-o error is involved in this computation and it takes only 0(n2) ops.
Flop-count. Since QR iteration method is an iterative method, it is hard to give an exact
op-count for this method. However, empirical observations have established that it takes about two 549
QR iterations per eigenvalue. Thus, it will require about 8n3 ops to compute all the eigenvalues (Golub and Van Loan, MC 1984 p. 235). If the transforming matrix Q and the nal quasitriangular matrix T are also needed, then the cost will be about 15n3 ops.
Round-o Property The QR iteration method is quite stable. An analysis of the round-o property of the
algorithm shows that the computed real Schur form (RSF) is orthogonally similar to a nearby matrix A + E , where kE kF (n)kAkF
(n) is a slowly growing function of n. The computed orthogonal matrix Q is also orthogonal.
8.9.8 The Real Schur Form and Invariant Subspaces De nition 8.9.2 Let S be a subspace of the complex plane C n. Then S will be called an invariant subspace (with respect to premultiplication by A) if x 2 S implies that Ax 2 S . Thus, since
Ax = x
for each eigenvalue , each eigenvector is an invariant subspace of dimension 1 associated with the corresponding eigenvalue.
The Real Schur Form of A displays information on the invariant subspaces. Basis of an Invariant Subspace from RSF Let
QT AQ = R =
R
11
R R
12
!
0 22 and let us assume that R11 and R22 do not have eigenvalues in common. Then the rst p columns of Q, where p is the order
of R11, forms a basis for the invariant subspace associated with the eigenvalues of R11.
In many applications, such as in the solution of algebraic Riccati equations (see Laub (1979)), one needs to compute the orthonormal bases of an invariant subspace associated with a selected 550
number of eigenvalues. Unfortunately, the transformed Real Schur Form obtained by QR iteration may not give the eigenvalues in some desired order. Thus, if the eigenvalues are not in a desired order, one wonders if some extra work can be done to bring them into that order. That this can indeed be done is seen from the following simple discussion. Let A be 2 2. Let ! 1 r12 T Q1 AQ1 = ; 1 6= 2: 0 2 If 1 and 2 are not in right order, all we need to do to reverse the order is to form a Givens rotation J (1; 2; ) such that ! ! r12 J (1; 2; ) = : 2 ; 1 0 Then Q = Q1 J (1; 2; )T is such that
QT AQ =
Example 8.9.10
r 0 2
12 1
!
!
:
1 2 A= 2 3 :8507 :5257 ! Q1 = ;:5257 :8507 ! ; 0 : 2361 0 : 0000 QT1 AQ1 = 0:0000 4:2361 ! 0 ;1 J (1; 2; ) = ;1 0 ! 4:4722 ! 0 J (1; 2; ) = ;4:4722 0 ;0:5257 ;0:8507 ! T Q = Q1J (1; 2; ) = ;0:8507 0:5257 ! 4:2361 0:00 T Q AQ = : 0:00 ;0:2361 The above simple process can be easily extended to achieve any desired ordering of the eigenvalues in the Real Schur Form. For details see (Golub and Van Loan, MC 1984 p. 241). The process is quite inexpensive. It requires only k(8n) ops, where k is the number of interchanges required to achieve the desired order. Stewart (1976) has provided useful Fortran routines for such an ordering of the eigenvalues. 551
8.10 Computing the Eigenvectors 8.10.1 The Hessenberg-Inverse Iteration As soon as an eigenvalue is computed by QR iteration, we can invoke inverse iteration (algorithm 8.5.2) to compute the corresponding eigenvector. However, since A is initially reduced to a Hessenberg matrix H for the QR iteration, it is natural to take advantage of the structure of the Hessenberg matrix H in the solutions of the linear system that need to be solved in the process of inverse iteration. Thus the Hessenberg-Inverse iteration can be stated as follows:
Algorithm 8.10.1 The Hessenberg-Inverse Iteration 1. Reduce the matrix A to an upper Hessenberg matrix H :
P T AP = H: 2. Compute an eigenvalue ; whose eigenvector x is sought, using the implicit QR iteration. 3. Apply the inverse iteration For k = 1; 2; : : : do Solve (H ; I )z (k) = y (k;1) y(k) = z(k)= max(z (k)): Stop if k(y (k) ; y (k;1))=y (k)k < or if k > N , the maximum number of iterations. 4. Recover the eigenvector x:
x = Py k ; ( )
where y (k) is the approximation of the eigenvector y obtained from Step 3.
Note: From Ax = x we have or That is,
P T APPx = Px Hy = y: x = Py:
552
8.10.2 Calculating the Eigenvectors from the Real Schur Form The eigenvectors can also be calculated directly from the Real Schur Form without invoking the inverse iteration. The process is described as follows: Let A be transformed to the RSF T by the implicit QR iteration:
QAQ = T: Then Ax = x can be written as
QAQQx = Q x:
That is, writing Q x = y , we have
Ty = y:
Thus, after A has been transformed to the RSF T , an eigenvector x corresponding to an eigenvalue can be computed as follows: 1. Solve the homogeneous triangular system Ty = y . 2. Compute x = Qy: We now show how the solution of Ty = y can be simpli ed assuming that T is triangular and that all the eigenvalues of A (that is, the diagonal entries of T ) are distinct. Let = tkk : That is we are trying to nd a y such that (T ; tkk I )y = 0: Write
T ; tkkI =
T
T 0 T 11
y y= y
1
!
2
;
;
(T ; tkk I )y = 0
gives
T
T 0 T 11
that is,
!
22
where T11 is k k. Partition y accordingly:
where y1 has k-entries. Then
12
12
! y!
22
y
1 2
=0
T y + T y = 0; T y = 0: 11
1
12
2
22
553
2
Now, T22 is nonsingular, because its diagonal entries are tjj ; tkk ; j = k + 1; : : :; n; which are di erent from zero. So, the homogeneous system
T y =0 22 2
has only the trivial solution y2 = 0. From above, we have
T y = 0: 11 1
Since the ith diagonal entry of T11 is zero, T11 is singular; therefore, y1 6= 0: Again, note that T11 has the form 0 1 ^11 s T A; T11 = @ 0 0 where T^11 is (k ; 1) (k ; 1) and nonsingular. Thus T11y1 = 0 reduces to
T^ y^ + sz = 0; 11 1
where
!
y^ y = ; z z can be chosen to be any nonzero number. Since T^ is upper triangular, y^ can be computed by 1
1
11
1
back substitution.
Algorithm 8.10.2 Computing an Eigenvector Directly from the RSF 1. Transform A to RSF by the implicit QR iteration:
Q AQ = T (Assume that T is triangular and that the diagonal entries of T are di erent.) 2. Select the eigenvalue = tkk whose eigenvector x is to be computed: 3. Partition 4. Partition
(T ; tkk I ) =
T = 11
5. Solve by back substitution:
T ! : 0 T
T
11
12 22
T^
11
0
s
!
0
T^ y^ = ;sz; 11 1
choosing z as a nonzero number. 554
:
!
y^ 6. Form y = : z ! y 7. Form y = : 1
1
1
0 8. Compute x = Qy:
Example 8.10.1
01 1 11 B 2 3 4 CC A = B @ A 4 6 7
0 10:8998 2:5145 2:7563 1 B CC T = B 0 0 : 3571 : 2829 @ A 0 0 ;0:2569 0 ;:1358 ;:9606 ;:2423 1 B ;:4747 :2778 ;:8352 CC : Q = B @ A ;:8696 ;:0016 :4937 Suppose we want to compute the eigenvector corresponding to = t22 = :3571: Then ! 10:5428 2:5145 T11 = : 0 0
T^
10:5428 2:5145 1 ;(T^11);1sz = ;:2385
= s = Choose z = y^1 = 11
!
!
y^ ;:2385 y = = : z 1 1
1
Choose y2 = 1. Then
0 ;:2385 1 ! y B 1 CC y = =B A @ y 0 1 0 ;0:9283 B C x = Qy = B @ 0:3910 CA : 1 2
0:2058
555
It is easily veri ed that
0 ;0:3315 1 B C Ax = B @ 0:1396 CA 0:0734
0 ;0:3315 1 B 0:1396 CC : x=B @ A
and
2
0:0734
8.11 The Symmetric Eigenvalue Problem The QR iteration can, of course, be applied to nd the eigenvalues of a symmetric matrix. Indeed, in the symmetric case the method simpli es to a large extent. We will discuss below the symmetric QR iteration brie y. However, since the eigenvalues and eigenvectors of a symmetric matrix enjoy certain special remarkable properties over those of the nonsymmetric case, some special methods exploiting these properties can be developed for the symmetric problem. One such method is based on the Sturm sequence property of the characteristic polynomial of the matrix.
Some Special Properties A. The eigenvalues of a real symmetric matrix are real. The eigenvectors associated with the distinct eigenvalues are orthogonal (Theorem 8.2.5). B.
The Real Schur form of a real symmetric matrix is a diagonal matrix. Proof. >From the Real Schur Triangularization Theorem (Theorem 8.9.2) we have
QT AQ = R
where R is in Real Schur Form, that is, R is a triangular matrix with each diagonal entry as either a scalar or a 2 2 matrix. Now, each 2 2 matrix on the diagonal corresponds to a pair of complex conjugate eigenvalues. Since a real symmetric matrix cannot have a complex eigenvalue, it follows that R can not have a 2 2 matrix on the diagonal; therefore R is a diagonal matrix. C. General Perturbation Property. 556
Let A be a n n real symmetric matrix. Let A0 = A + E; where E is a real symmetric perturbation of the matrix A, and let 1 2 n and 01 02 0n be the eigenvalues of A and A0 , respectively. Then it follows from the Bauer-Fike Theorem (Theorem 8.7.1) that i ; kE k2 0i i + kE k2; i = 1; 2; : : :; n: This result is remarkable.
Perturbation Result on the Eigenvalues of a Symmetric Matrix The eigenvalues of a real symmetric matrix are wellconditioned, that is, small changes in the elements of A cause only small changes in the eigenvalues of A. In fact, since
kE k = maxflargest and smallest of the eigenvalues of E g; 2
then the eigenvalues of the perturbed matrix A0 cannot di er
from the eigenvalues of the original matrix A by more than the largest eigenvalue of the perturbated matrix E . (See also the corollary of the Bauer-Fike Theorem given earlier.) (Section 8.7.2) Example 8.11.1
01 2 31 B 2 3 4 CC ; A=B @ A
E = 10; I : 4
3
3
3 4 6 The eigenvalues of A are ;0:4203, 0.2336, and 10.1867. The eigenvalues of A + E are ;0:4203, .2337, 10.1868. Note that kE k2 = 10;4. D. Rank-one Perturbation Property. In this section we state a theorem that shows how are the eigenvalues shifted if E is a rank-one perturbation. The result plays an important role in the divide and conquer algorithm (Dongarra and Sorensen (1987)) for the symmetric eigenvalue problem. 557
Eigenvalues of a Rank-One Perturbed Matrix Theorem 8.11.1 Suppose B = A + bbT , where A is an n n symmetric matrix, is a scalar and b is an n-vector. Let ::: n be the eigenvalues of A and 0 : : : 0n be the eigenvalues of B . Then 1
2
1
0i 2 [i; i; ]; i = 2; :::; n; if 0; 1
0i 2 [i ; i]; i = 1; :::; n ; 1; if < 0; +1
Proof. See Wilkinson APE, pp. 97-98. Example 8.11.2
01 2 31 B C A=B @ 2 4 5 CA ; = ;1; b = (1; 2; 3)T :
3 5 6 The eigenvalues of B are: 03 = ;3:3028; 02 = 0; 01 = 0:3023. The eigenvalues of A are: 3 = ;0:5157; 2 = 0:1709; 1 = 11:3443. It is easily veri ed that 2 < 01 < 1; and 3 < 02 < 2:
8.11.1 The Sturm Sequence and the Bisection Method In this section we describe a method to nd the eigenvalues of a symmetric matrix. The method is particularly useful if eigenvalues are required in an interval. In principle, it can be used to nd all eigenvalues. However, in practice, if all the eigenvalues are required explicitly,
the symmetric QR iteration method is preferred over the method of this section for matrices of moderate sizes. First, the symmetric matrix A is transformed to a symmetric tridiagonal matrix T using Householder's method described earlier:
0 B B 0 B B . . . .. .. .. B PAP T = T = B B ... ... ... B B B B n; n; n; @ 0 1
1
1
2
2
2
558
n;
1
1
n
1
1 CC CC CC CC : CC CA
Let pi() denote the characteristic polynomial of the i i principal submatrix of T . Then these polynomials satisfy a three term recursion:
pi() = ( i ; )pi; () ; i; pi; (); i = 2; 3; : : :; n 1
with
2
1
2
p () = 1 and p () = ; : 0
1
1
Without loss of generality, we may assume that i 6= 0; i = 1; 2; : : :; n ; 1: Recall that the matrix T with this property is called unreduced. If a subdiagonal entry of T is zero, then T is a block diagonal and its eigenproblem can thus be reduced to that of its submatrices. For unreduced symmetric tridiagonal matrix T , the following interlacing property is very important.
Theorem 8.11.2 (Interlacing property) Let T be an unreduced symmetric tridiagonal matrix and ik the ith smallest eigenvalue of its k k principle sub( )
matrix. Then
k
( +1) 1
<k <k ( ) 1
( +1) 2
< < ik
( +1)
< ik < ik ( )
( +1) +1
< < kk
( +1)
< kk < kk : ( )
( +1) +1
(See Golub and Van Loan, MC 1989 p. 438.) We shall show that this interlacing property leads to an interesting result on the number of eigenvalues of T. It is clear that pk () = (;1)k k + . Thus pk () is positive if is negative with large magnitude. The zeros (1k) < (2k) < < (kk) of pk separate the real line into k + 1 intervals (;1; (1k)); ((1k); (2k)); : : :; ((kk;)1; (kk)); ((kk); 1):
pk () is positive in the rst interval and takes alternative signs in consecutive intervals. The interlacing property and the sign changes of pk () can be illustrated in the following gure.
559
p (λ)
+
+
+
+
+
+
+
+
+
+
p (λ)
+
+
+
+
+
-
-
-
-
-
p (λ)
+
+
+
+
+
+
+
p (λ)
+
+
+
+
-
-
p (λ)
+
+
+
+
+
p (λ)
+
+
-
-
-
0 1
-
-
+
-
-
+
+
-
-
+
+
-
-
+
+
-
2
-
3
- +
4
+
-
+
5
Let be a real number. There are two cases for the signs of p0() and p1 (): µ
µ p (λ)
+
+
+
+
+
+
p (λ)
+
+
+
-
-
-
+
+
+
+
+
+
+
+
+
-
-
-
0 1
(1)
λ
(1)
λ
1
1
Case 2: signs agree
Case 1: no sign agreement
Clearly, a sign agreement between p0() and p1 () \pushes" (1) to the right of . By interlacing 1 property, this implies
< < < < nn : (1) 1
(2) 2
( )
Thus, it \pushes" one eigenvalue (nn) of A (or T ), as well as one eigenvalue of each k k submatrix, to the right of . Suppose we have counted the sign agreements between consecutive two terms of p0(); p1(); : : :; pk (). Let's consider the signs of pk () and pk+1 () and interlacing property:
560
µ
µ +
p (λ) k
p (λ)
-
+ λ
+
λ
(k+1) i-1
λ
i-1
-
+
k+1
(k)
(k+1)
λ
- +
+
+
p (λ)
- +
(k)
k
(k)
-
+
(k)
λ
λ
i-1
i
+ + λ
i
-
p (λ)
(k+1)
+
k+1
λ
i+1
Case 1: no sign agreement
-
+ (k+1) i-1
+
i
+ +
-
(k+1)
λ
i
λ
(k+1) i+1
Case 2: signs agree
Let be between (i;k)1 and (ik). It is clear from the above gure that a sign agreement between pk () and pk+1() \pushes" ki +1 to the right of , or it means that pk+1 has one more zeros than pk in [; 1). There is no zero of p0 in [; 1). We know the number of zeros of p1 in [; 1) by the sign agreement or disagreement of p0 () and p1(). Generally, when we know the number of zeros of pk in [; 1), then we know the number of zeros of pk+1 in [; 1) by checking the sign agreement between pk () and pk+1(). This recursion can be continued until we know the number of zeros of pn, i.e., the number eigenvalues of T . The rule is clear:
The number of sign agreements between consecutive terms of p (); p (); : : :; pn() equals to the number of eigenvalues of T larger than . 0
1
Note : It is also clear that pk () = 0 should be considered an sign agreement with pk() because one more zero is \pushed" into [; 1). +1
Example 8.11.3 Let
02 1 01 B C T =B @ 1 2 1 CA : 0 1 2 561
p () = 1; p () = 2 ; ; p () = (2 ; ) ; 1 p () = (2 ; )p () ; p () = (2 ; )[(2 ; ) ; 1] ; (2 ; ) = (2 ; ) ; 2(2 ; ): 0
1
2
2 3
2
1
2
3
Let = 0. Then the sequence p0; p1(); p2() and p3 () is 1, 2, 3, 4. There are three agreements in sign. Thus all the eigenvalues of T are greater than or equal to zero. In fact, since p3(0) = 4 6= 0, it follows that all the eigenvalues of T are positive. The eigenvalues of T are easily seen to be p p 2; 2 + 2; 2 ; 2: Let = 2 then, the sequence p0(); p1(); p2(); and p3() is: 1; 0; ;1; 0: We count the agreements in sign of
+ + ;;
There are two agreements; con rming that T has two eigenvalues greater than or equal to 2. The Sturm-sequence and Bisection method for computing a speci ed zero of pn () now can be stated as follows:
Algorithm 8.11.1 The Sturm Sequence { Bisection Algorithm Let 1 < 2 < < n be the eigenvalues of T , that is, the zeros of pn(). Suppose the desired eigenvalue is n;m+1 for a given m n. Then I. Find an interval [s1 ; s2] containing n;m+1 : Since i kT k, initially, we can take
s = ;kT k1 ; s = kT k1: 1
2
II. Compute s3 = s1 +3 s2 .
III. Compute N (s3) = the number of agreements in sign in the sequence 1; p1(s3); p2(s3); : : :; pn(s3 ): If N (s3) < m, set s2 = s3 otherwise, set s1 = s3 .
Let > 0 be a preassigned small positive number. Test if js2 ; s1j < . If so accept s3 = s1 + s2 2 as an approximate value of n;m+1. Otherwise go to II. 562
Note: After k steps, the desired zero is located in an interval of width (s 2;k s ) . Example 8.11.4 Let 02 1 01 B C A=B @ 1 2 1 CA : 0 1 2 p Suppose we want to approximate = 2 ; 2 2
1
m = 3:
Iteration 0. Initially, s = ;4; s = 4 s =0 N (s ) = N (0) = 3: Set s = s : 1
2
3
3
1
3
Iteration 1. s = 0; s = 4 s = 0 +2 4 = 2 N (s ) = N (2) = 2 < 3: Set s = s 1
2
3
3
2
Iteration 2.
3
s = 0; s = 2; s = 0 +2 2 = 1 1
2
3
N (s ) = number of agreements of sign in the sequence: 1; ;1; 0; ;1: (+ ; ;;) 3
N (s ) = 2 < 3 Set s = s : 3
2
Iteration 3. s = 0; s = 1; s = :5 N (s ) = 3: Set s = s : 1
2
3
3
1
563
3
3
1
The eigenvalue 1 is clearly in the interval [:5; 1], which is, in fact, the case. We can continue our iterations until the length of the interval js2 ; s1 j < :
8.11.2 The Symmetric QR Iteration Method When A is symmetric, the transformed Hessenberg matrix PAP T = H is also symmetric; therefore, it is symmetric tridiagonal. So, in this case we have to deal with the eigenvalue problem of a symmetric tridiagonal matrix rather than of a Hessenberg matrix. To apply the QR iteration method to a symmetric tridiagonal matrix, we note that, if the starting matrix is a symmetric tridiagonal matrix T , then so is each matrix Tk in the sequence
Tk ; k I = Qk Rk and, furthermore, we need only 0(n) ops to generate each Tk (note that the QR factorization of a symmetric tridiagonal matrix requires only 0(n) ops). Thus, the symmetric QR iteration is much more ecient than the nonsymmetric case. Also, since the eigenvalues of a symmetric matrix are all real and the Real Schur Form of a symmetric matrix is a diagonal rather than triangular matrix, the double-shift strategy discussed for the general eigenvalue problem is not needed in this case. However, in this case a popular shift, known as the Wilkinson-shift, is normally used. Instead of taking the (n; n)th entry at every iteration as the shift, the eigenvalue of the trailing 2 2 matrix that is closer to the (n; n)th entry, is usually chosen as the shift. This is known as the Wilkinson-shift. Thus if a trailing 2 2 submatrix of Tk is given by
tnk; ;n; k tn;n ; ( )
1
( )
!
k tn;n ; ; k tnn ( )
1
1
( )
1
then the Wilkinson-shift is
r
= tnnk + r ; sign(r) ( )
k r + tn;n ; 2
( )
2
1
where
(t(nk;) 1;n;1 ; t(nnk) ) r= : 2 Thus the symmetric method comes in two stages:
Algorithm 8.11.2 Symmetric QR Iteration I. Transform A to a symmetric tridiagonal matrix T using orthogonal similarity transformations:
PAP T = T: 564
II. Apply single-shift QR iteration to T with the Wilkinson-shift : Set T = T1 For k = 1; 2; : : : do until convergence occurs Tk ; I = QkRk Tk+1 = Rk Qk + I .
Remark: It is possible to compute Tk from Tk without explicitly forming the matrix Tk ; k I . This is known as Implicit Symmetric QR. For details, see Golub and Van Loan, pp. 278{281. +1
Also see exercise #28 in this chapter.
Convergence of the Symmetric QR Iteration k) to It can be shown (see Lawson and Hanson, SLP p. 109) that the convergence of t(n;n ;1 zero is quadratic, that is, there exists e > 0 depending upon A such that for all k k j ejt k j : jtn;n ; n;n; ( +1) 1
( )
1
2
Remark: In practice, it has been seen that the convergence is almost always cubic; but the quadratic convergence is all that has been proved (for a proof, see Lawson and Hanson SLP, pp. 240{244). Flop-Count: The symmetric QR algorithm requires only about 23 n ops. However if the 3
transforming orthogonal matrix Q is accumulated the ops count becomes 5n3.
Round-o error property. As in the general nonsymmetric case, the symmetric QR with implicit shift is stable. It can be shown that, given a symmetric matrix A, the symmetric QR
algorithm with implicit shift generates an orthogonal matrix Q and a diagonal matrix D such that
QT AQ = D + E; where
kE kF (n)kAkF and (n) is a slowly growing function of n.
565
Furthermore, each computed eigenvalue ^ satis es the inequality: j ; ;1j p(n)kAk2 : Thus, the eigenvalues with modulus close to kAk2 are computed accurately; the other ones may not be. i
i
i
Note: If the starting matrix itself is a symmetric tridiagonal matrix, then the eigenvalues and the eigenvectors can be computed much more accurately.
8.12 The Lanczos Algorithm For Symmetric Matrices There are areas of applications such as power systems, space science, quantum physics and chemistry, nuclear studies, etc., where the eigenvalue problems for matrices of very large order are commonly found. Most large problems arising in applications are sparse. Research in this area is very active. The symmetric problem is better understood than the nonsymmetric problem. There are now wellestablished techniques to compute the spectrum, at least the extremal eigenvalues of very large and sparse symmetric matrices. A method originally devised by Lanczos has received considerable attention in this context. We will discuss this method and its applications to eigenvalue computations in this section. The QR iteration method described in the last section is not practical
for large and sparse eigenvalue problem. The sparsity gets destroyed and the storage becomes an issue. The Symmetric Lanczos Algorithm Given an n n symmetric matrix A and a unit vector v , the Lanczos algorithm constructs 1
simultaneously a symmetric tridiagonal matrix T and an orthonormal matrix V such that
T = V T AV:
(8.12.1)
The algorithm can be deduced easily as follows: Cornelius Lanczos (1893-1974), a Hungarian mathematician and physicist, made signi cant contributions to problems in physics and mathematics that are considered to be of vital interest in current scienti c computing. The algorithm, which is now known as the symmetric Lanczos algorithm, is one of the rst algorithms for large and sparse eigenvalue problems. The symmetric Lanczos algorithm has served as a \seed" algorithm for further research in this area. There was a conference at Northern Carolina State University honoring Lanczos 100th birthday in December, 1993.
566
0 0 BB . . . T =B BB .. . . . . . . n; @.
Let
1
1
1
2
n;
0 and V = (v1; v2; : : :; vn). Then the equation
n
1
1
1 CC CC CA
V T AV = T or or
AV = V T
0 0 BB . . . . . . A(v ; v ; : : :; vn) = (v ; v ; : : :; vn) B BB .. . . . . . . n; @. 0 n; n 1
1
1
1
2
1
2
1
gives
1 CC CC CA
1
Avj = j vj + j ; vj ; + j vj ; j = 1; 2; : : :; n ; 1 1
1
+1
(8.12.2)
(where we assume that 0v0 = 0). Multiplying both sides of the above equation by vjT to the left, and observing that the orthonormality condition gives
vjT vj = 1 vjT vk = 0; j 6= k; we obtain Also, if we write
j = vjT Avj ; j = 1; 2; : : :; n:
(8.12.3)
rj = Avj ; j vj ; j ; vj ; ;
(8.12.4)
1
1
then from (8.12.2) and (8.12.4), we get
vj = rj = j ; +1
provided that
j 6= 0:
The nonvanishing of j can be assured if we take j = krj k . The above discussion gives us the following basic Lanczos algorithm: 2
567
Algorithm 8.12.1 The Basic Symmetric Lanczos Given an n n symmetric matrix A and a unit vector v1 , the following algorithm constructs simultaneously the entries of a symmetric tridiagonal matrix T and an orthonormal matrix V = (v1 ; : : :; vn) such that: V T AV = T: Set vo = 0; o = 1; ro = v1: For j = vj = j = rj = j =
1; 2; : : :; n do
rj ; = j ; vjT Avj (A ; j I )vj ; j ; vj ; k rj k 1
1
1
1
2
Notes: 1. The vectors v1 ; v2; : : :; vn are called Lanczos vectors. 2. Each Lanczos vector vj +1 is orthogonal to all the previous ones; provided j 6= 0. 3. The whole algorithm can be implemented just by using a subroutine that can perform matrixvector multiplication, and thus the sparsity of the original matrix A can be maintained. In
fact, in contrast with Householder's method or Givens' method, the matrix A is never altered during the whole procedure. Such a feature makes the Lanczos algorithm so attractive for sparse computations.
4. The vectors fv1 ; : : :; vj g from an orthonormal basis of the Krylov subspace fv1; Av1; : : :; Aj ;1v1 g. 5. The Arnoldi method described in chapter 6 is the same as the symmetric Lanczos method described above. We now show the Lanczos algorithm can be reformulated using only three n{vectors. This is important when n is large.
Algorithm 8.12.2 The Reformulated Lanczos Set v0 = 0; 0 = 1; r0 = v1: For j = 1; 2; : : :; n do 568
1. vj = rj ;1= j ;1 2. uj Avj 3. rj uj ; j ;1vj ;1 4. j = vjT uj 5. rj = rj ; j vj 6. j = krj k:
Example 8.12.1
01 2 31 B C A=B @ 2 3 4 CA ;
011 B CC r =v =B @0A: 0
3 4 5
j=1:
0
011 B CC = 1; = 3:6056; v = B @0A: 1
j=2:
1
1
0
0 0 1 B C = 8:0769; = :6154; v = B @ :5547 CA :8321 ! ! 1 3:6056 2
2
T =
1
1
1
2
2
j=3:
1
2
=
3:6056 8:0769
0 0 1 B :8321 CC = 0:0769; = 1:5466 10; ; v = B @ A ; 0 : 5547 0 1 3:6056 1 0 B 3:6056 :0769 :6154 CC T =T =B @ A 0 :6154 ;0:0769 01 0 1 0 B CC V =B ; : 5547 : 8321 @ A: 0 :8321 ;0:5547 3
14
3
3
Note that V T AV = T: 569
3
Round-o Properties of the Symmetric Lanczos Algorithm It is clear from (8.12.2) that if the symmetric Lanczos algorithm is run from j = 1 to k(k < n), then we have AVj = Vj Tj + rj eTj ; where Vj = (v1 ; : : :; vj ) and Tj is the j j principal submatrix of T . If V~j , T~j and r~j denote the respective computed quantities, then it can be shown (see Paige 1980) that
AV~j = V~j T~j + r~j eTj + Ej where kEj k2 = kAk2; which shows that the Lanczos algorithm has very favorable numerical properties as far as the equation AVj = Vj Tj + rj eTj is concerned. However, the loss of orthogonality
among the Lanczos vectors is the real concern as explained in the following. Loss of Orthogonality
The Lanczos algorithm clearly breaks down when any of the j = 0. However, as we will see later, computationally this is a blessing in disguise. We immediately obtain an invariant subspace. This, however, seldom happens in practice. The real diculty is that the computed ~j can be very small due to the cancellation that takes place during the computations of rj , and a small ~j can cause a severe loss of orthogonality among the computed vectors v~j , as can be seen from the following result (Golub and Van Loan, MC 1984 p. 333):
jv~jT v~j j jr~j vj j +~ kAk : j j j T
2
+1
Thus, if ~j is small, then v~j and v~j +1 will be far from being orthogonal. Since the Lanczos algorithm plays an important role in sparse matrix computations, a signi cant amount of research has been devoted to making the algorithm a practical one. There now exist procedures such as Lanczos with complete orthogonalization which produce the Lanczos vectors that are orthogonal to working precision, or Lanczos with selective orthogonalization that is used to enforce orthogonality only in selective vectors, whenever needed. For details, we refer the readers to the well-known books in the area by Parlett (SEP 1980), and by Cullum and Willoughby (1985a), (1985b). Several papers of Paige (1970, 1971, 1972, etc.), whose pioneering work in the early 1970's rejuvenated the interests of the researchers in this area, are also very useful reading. Procedures for complete and selective orthogonalization have been described in some detail in Golub and Van Loan MC, 1984 pp. 334{337. 570
Computing the Eigenvalues of A The sole purpose of presenting the Lanczos algorithm here was to show that the Lanczos matrices Tj can be used to compute certain eigenvalues of the matrix A. We remarked earlier, when a j is exactly equal to zero, we have an invariant subspace. This is indeed good news. Unfortunately, this happens very rarely in practice. In practice, for large enough values of j , the eigenvalues of Tj provide very good approximations to the extremal eigenvalues of A. The question arises: Can we give posteriori bounds? To this end, we introduce the following de nitions.
De nition 8.12.1 Let (j ; zj ) be an eigenpair of Tj . Then the pair (j ; yj ); where yj is de ned by yj = VjT zj , is called the Ritz pair; the j 's are known as the Ritz values and yj 's are called the Ritz vectors. Now, returning to the question of how well a Ritz pair approximates an eigenpair of A, we state the following result (for a proof, see Golub and Van Loan, MC 1984 p. 327).
Theorem 8.12.1 Let Ri = Ayi ; yi i; i = 1; : : :; j . Then in each interval [i ; kRik ; i + kRik ], there is an eigenvalue of A. 2
2
Thus, it follows from the above theorem that kRi k2 is a good measure of how accurate is the Ritz pair (i ; yi).
How do we compute kRik ? Fortunately, we can compute kRik from Tj without computing i and yi at every step. 2
Let
SjT Tj Sj = diag( ; ; : : :; j ); 1
and
2
Yj = (y ; y ; : : :; yj ) = Vj Sj = Vj (s ; s ; : : :; sj ): 1
2
1
2
Then
kRik = kAyi ; yiik = kAVjsi ; Vj siik = k(AVj ; Vj Tj )si k (because Tj si = si i ) = k( j vj eTj )si k (Note that AVj ; Vj Tj = j vj eTj ) = j j j keTj si k (since kvj k = 1): = j j j jsjij; +1
+1
+1
571
where sij is the (j; i)th entry of Sj . The above discussion can be summarized in the following theorem:
Theorem 8.12.2 (Residual Theorem for Ritz Pair) Let Tj denote the j j
symmetric tridiagonal matrix obtained after j steps of the Lanczos algorithm. Let SjT Tj Sj denote the Real Schur Form (RSF) of Tj ;
SjT Tj Sj = diag( ; : : :; j ); 1
that is, 1 ; : : :; j are the eigenvalues of Tj . Let
Yj = (y ; y ; : : :; yj ) = Vj Sj : 1
2
Then for each i from 1 to j , we have
kRik = kAyi ; iyik = j j j jsjij; where
sji = eTj si ; si is the ith column of Sj :
Example 8.12.2
01 2 31 B 2 3 4 CC : A=B @ A
3 4 5 j=2: The eigenvalues of A are: 0, -.6235, 9.6235. ! 1 3:6056 T2 = ; 2 = :6154 3:6056 8:0769 ! :9221 :3870 S2 = ;:3970 :9211 1 = ;:5133; 2 = 9:5903: (Note that 2 = 9:5903 is a good approximation of the largest eigenvalue 9.6235 of A).
R = j j js j = :6154 :3870 = :2382; 1
2
21
R = j j js j = :5674: 2
2
22
572
j=3:
0 1 3:6056 B T =B @ 3:6056 8:0769
1
0 C :6154 C 3 A 0 :6154 ;0:0769 0 :8277 ;:4082 :3851 1 B C S3 = B @ ;:2727 :1132 :9210 CA :4196 :9058 :0584
R = j j js j = 6:487 10; R = j j js j = 1:4009 10; R = j j js j = 9:0367 10; : 1
3
31
2
3
32
3
3
33
15
14 16
(Note the smallness of R1 ; R2, and R3 in this case.)
8.13 Review and Summary This chapter has been devoted to the study of the eigenvalue problem, the problem of computing the eigenvalues and eigenvectors of a matrix: Ax = x. Here are the highlights of the chapter. 1. Applications of the Eigenvalues and Eigenvectors. The eigenvalue problem arises in a wide variety of practical applications. Mathematical models of many of the problems arising in engineering applications are systems of di erential and di erence equations, and the eigenvalue problem arises principally in the analysis of stability of these equations. Maintaining the stability of a system is a real concern for engineers. For example, in the study of vibrations of structures, the eigenvalues and eigenvectors are related to the natural frequencies and amplitude of the masses, and if any of the natural frequencies becomes equal or close to a frequency of the imposed periodic force on the structure, resonance occurs. This is a situation engineers would like to avoid. In this chapter we have included examples on the European arms race, buckling of a beam, simulating transient current of an electric circuit, vibration of a building, and principal components analysis in statistics with a reference to a stock market analysis. We have attempted just to show how important the eigenvalue problem is in practical applications. 2. Localization of Eigenvalues. In several applications explicit knowledge of the eigenvalues is not required; all that is needed is a knowledge of the distribution of the eigenvalues in a region of the complex plane or estimates of some speci c eigenvalues. 573
The Gersgorin disk theorems (Theorems 8.4.1 and 8.4.2) can be used to obtain a region
of the complex plane containing all the eigenvalues, or in some cases, a number of the eigenvalues in a region. The estimates are, however, very crude. Also, jj kAk. (Theorem 8.4.3) This result says that the upper bound of any
eigenvalue of A can be found by computing its norm.
Recall that this result is important in convergence analysis of iterative methods for linear systems. 3. The Power Method and The Inverse Iteration. There are applications such as analysis of dynamical systems, vibration analysis of structures, buckling of a beam, principal component analysis in statistics, etc., where only the largest and the smallest (in magnitudes) eigenvalues or only the rst or last few eigenvalues and their corresponding eigenvectors are needed. The power method and the inverse power method based on implicit construction of powers of A can be used to compute these eigenvalues and the eigenvectors. The power method is extremely simple to implement and is suitable for large and sparse matrices, but there are certain numerical limitations. In practice, the power method should be used with a suitable shift. The inverse power method is simply the power method applied to (A ; I );1; where is a suitable shift. It is widely used to compute an eigenvector when a reasonably good approximation to an eigenvalue is known. 4. The Rayleigh Quotient Iteration. The quotient T Rq = xxTAx x;
known as the Rayleigh quotient, gives an estimate of the eigenvalue of a symmetric matrix A for which x is the corresponding eigenvector. This idea, when combined with the inverse iteration method, can be used to compute an approximation to an eigenvalue and the corresponding eigenvector. The process is known as the Rayleigh quotient iteration. 5. Sensitivity of Eigenvalues and Eigenvectors.
The Bauer-Fike Theorem (Theorem 8.7.1) tells us that if A is a diagonalizable matrix, then the condition number of the transforming matrix X , Cond(X ) = kX k kX ; k, plays 1
574
the role of the condition number of the eigenvalue problem. If this number is large, then a small change in A can cause signi cant changes in the eigenvalues. Since a symmetric matrix A can be transformed to a diagonal matrix by orthogonal similarity and the condition number of an orthogonal matrix (with respect to 2-norm) is 1, it immediately follows from the Bauer-Fike Theorem that the eigenvalues of a symmetric matrix are insensitive to small perturbations. If an eigenvalue problem is ill-conditioned, then it might happen that some eigenvalues are more sensitive than the others. It is thus important to know the sensitivity of the individual eigenvalues. Unfortunately, to measure the sensitivity of an individual eigenvalue, one needs the knowledge of both left and right eigenvectors corresponding to that eigenvalue. The condition number of the eigenvalue i is the reciprocal of the number jyiT xi j, where xi, and yi are, respectively, the normalized right and left eigenvectors corresponding to i . The sensitivity of an eigenvector xk corresponding to an eigenvalue k depends upon (i) the condition number of all the eigenvalues other than k and (ii) the distance of k from the other eigenvalues (Theorem 8.8.1).
Thus, if the eigenvalues are well-separated and well-conditioned, then the eigenvectors are well-conditioned. On the other hand, if there is a multiple eigen-
value or there is an eigenvalue close to another eigenvalue, then there are some illconditioned eigenvectors. This is especially signi cant for a symmetric matrix. The
eigenvalues of a symmetric matrix are well-conditioned, but the eigenvectors can be quite ill-conditioned.
6. Eigenvalue Computation via the Characteristic Polynomial and the Jordan Canonical Form. A similarity transformation preserves the eigenvalues and it is well known that a matrix A can be transformed by similarity to the Jordan Canonical Form (JCF) and to the Frobenius form (or a comparison form if A is nonderogatory). The eigenvalues of these condensed forms are rather easily computed. The JCF displays the eigenvalues explicitly, and with the companion or Frobenius form, the characteristic polynomial of A is trivially computed and then a root- nding method can be applied to the characteristic polynomial to obtain the eigenvalues, which are the zeros of the characteristic polynomial. However, computation of eigenvalues via the characteristic polynomial or the JCF is not recommended in practice. Obtaining these forms may require a very ill-conditioned transforming 575
matrix, and the sensitivity of the eigenvalue problem depends upon the condition number of this transforming matrix.
In general, ill-conditioned similarity transformation should be avoided in eigenvalue computation. The use of well-conditioned transforming matrices, such as orthogonal matrices, is desirable. 7. The QR Iteration Algorithm. The most widely used algorithm for nding the eigenvalues of a matrix is the QR iteration algorithm. For a real matrix A, the algorithm basically constructs iteratively the Real Schur Form (RSF) of A by orthogonal similarity. Since the algorithm is based on repeated QR factorizations and each QR factorization of an n n matrix requires 0(n3 ) ops, the n steps of QR iteration algorithm, if implemented naively (which we call the basic QR iteration), will require 0(n4)
ops, making the algorithm impractical.
The matrix A is, therefore, initially reduced to a Hessenberg matrix H by orthogonal similarly before the start of the QR iteration. The key observations here are: the
reduction of A to H has to be made once for all, and the Hessenberg form is preserved at each iteration.
The convergence of the Hessenberg-QR iteration algorithm, however, can be quite slow
in the presence of a near-multiple or a multiple eigenvalue. The convergence can be accelerated by using suitable shifts. In practice, double shifts are used. At each iteration, the shifts are the eigenvalues of the 2 2 submatrix at the bottom right hand corner. Since the eigenvalues of a real matrix can be complex, complex arithmetic is usually required. However, computations can be arranged so that complex arithmetic can be avoided. Also, the eigenvalues of the 2 2 bottom right hand corner matrix at each iteration do not need to be computed explicitly. The process is known as the double shift implicit QR iteration. With double shifts, the eigenvalues are computed two at a time. Once two eigenvalues are computed, the matrix is de ated, and the process is applied to the de ated matrix.
The double shift implicit QR iteration seems to be the most practical algorithm for computing the eigenvalues of a dense matrix of modest size.
8. Ordering the Eigenvalues. The eigenvalues appearing in RSF obtained by the QR iteration algorithm do not appear in the desired order, although there are some applications which need 576
this. However, with a little extra work, the eigenvalues can be put in the desired order. There is an excellent Fortran routine, designed by Stewart, which will accomplish this. 9. Computing the Eigenvectors. Once an approximation to an eigenvalue is obtained for the QR iteration, inverse iteration can be invoked to compute the corresponding eigenvector. Since the matrix A is initially reduced to a Hessenberg matrix for practical implementation of the QR iteration algorithm, advantage can be taken of the structure of a Hessenberg matrix in computing an eigenvector using inverse iteration. Alternatively, one can compute the eigenvectors directly from the RSF of A. 10. The Symmetric Eigenvalue Problem.
The QR iteration algorithm, of course, can be used to compute the eigenvalues of a sym-
metric matrix. A shift called the Wilkinson shift is normally used here. The convergence of the symmetric QR iteration algorithm with the Wilkinson-shift has been proven to be quadratic; however, in practice very often it is cubic. The eigenvalues and eigenvectors of a symmetric matrix A enjoy some remarkable special properties, and there are methods for the symmetric problem that can exploit these properties. One such method is the bisection method. The given symmetric matrix A is rst transformed to a symmetric tridiagonal matrix T using orthogonal similarity and then the well-known bisection algorithm for the root nding problem is applied to the characteristic polynomial of T , obtained by a simple recursion (Section 8.10.1). This recursion not only gives the characteristic polynomial of T , but also gives the characteristic polynomials of all the leading principal submatrices. A remarkable fact is that these polynomials have the Sturm sequence property, which is used in the implementation of the bisection method. The bisection method is especially useful for nding eigenvalues of a symmetric matrix in a given interval of the real line. There are other methods, such as the divide and conquer method, the Jacobi method, etc., for the symmetric eigenvalue problem. We have not discussed these methods in this chapter. These methods are important primarily for parallel computations. 11. Large and Sparse Eigenvalue Problem. The eigenvalue problem for large and sparse matrices is an active area of research. The state-of-the-art techniques using Lanczos or Arnoldi 577
type of methods with some sort of reorthogonalization and proper preconditioning can compute only a few extremal eigenvalues. The techniques for symmetric eigenvalue problems are more well-established and betterunderstood than those for the nonsymmetric problem. For the sake of completeness and to give the readers an idea of how the Lanczos-type methods are used in eigenvalue computations, we have included only a very brief description of the symmetric Lanczos method (Section 8.12).
8.14 Suggestions for Further Reading Most books on vibration discuss eigenvalue problems arising in vibration of structures. However, almost all eigenvalue problems here are generalized eigenvalue problems; as a matter of fact, they are symmetric de nite problems. For references of the well-known books in vibration, see section 9.10 in the next chapter. The books by Inman and by Thompson are, in particular, very useful and important books in this area. For learning more about how the eigenvalue problem arises in other areas of engineering, see the books on numerical methods in engineering by Chapra and Canale, and by Peter O'Neil, referenced in Chapter 6. There are other engineering books (too numerous to list here), especially in the areas of electrical, mechanical, civil and chemical engineering, containing discussions on eigenvalue problems in engineering. The Real Schur Form (RSF) of a matrix is an important tool in numerically e ective solutions of many important control problems, such as solutions of the Lyapunov matrix equation (Bartels and Stewart (1972)), the Sylvester Matrix Equation (Golub, Nash, and Van Loan (1979)), Algebraic Riccati equations (Laub (1979), Van Dooren (1982)), Byers (1984), etc. For details, see the book Computational Methods for Linear Control Systems, by Petkov, Christov and Konstantinov, Prentice-Hall, 1991, the IEEE Reprint, Volume by Patel et al. (1994), and the book Numerical Methods in Control Theory, by B. N. Datta (in preparation). Some references of how eigenvalue problems (especially large and sparse eigenvalue problems) arise in other areas of sciences and engineering such as power systems, physics, chemistry, structural mechanics, oceanography, etc., are given in the book Lanczos Algorithms for Large Symmetric Eigenvalue Computations, vol. 1, by Jane Cullum and Ralph Willoughby, Birkhauser, Boston, 1985. For some generalizations of the Gersgorin disk theorems, see the recent paper by Brualdi (1993). This paper contains results giving a region of the complex plane for each eigenvalue; for a 578
full description of the Gersgorin disk theorems and applications, see Matrix Analysis by Roger Horn and Charles Johnson, Cambridge University Press, Cambridge, 1985 (Chapter 6). A nice description of stability theory in dynamic systems is given in Introduction to Dynamic Systems: Theory, Models, and Applications by David Luenberger, John Wiley & Sons, New York, 1979. For results on eigenvalue bounds, see the paper by Varah (1968). For computation of Jordan Canonical Form, see the papers by Golub and Wilkinson (1976), Kagstrom and Ruhe (1980a and 1980b), and Demmel (1983). Descriptions of the usual techniques for eigenvalue and eigenvector computations|the power method, the inverse power method, the Rayleigh-Quotient iteration method, and the QR iteration method, etc.|can be found in all numerical linear algebra books: Golub and Van Loan (MC, 1983 and 1989), Stewart (IMC), Hager (ANLA), Watkins (FMC), Wilkinson (AEP). The Wilkinson
AEP is, of course, the most authoritative book in this area. The papers by Varah (1968), (1970), etc. and Peters and Wilkinson (1979) are important
in the context of inverse iteration. A Fortran program for ordering the eigenvalues of a real upper Hessenberg matrix appears in Stewart (1976). An important book in the area of symmetric eigenvalue problem is the book The Symmetric Eigenvalue Problem by Beresford Parlett, Prentice-Hall, Englewood Cli s, NJ, 1980. For a proof of the global convergence of the symmetric QR iteration with Wilkinson-shifts, see the book SLP by Lawson and Hanson, pp. 240{247. For a description of the divide and conquer method, see Dongarra and Sorensen (1987) and the original paper of Cuppen (1981). Watkins (FMC) contains a nice description of the Jacobi method (Chapter 6). Again, the book by Golub and Van Loan (MC 1984 and 1989) is a rich source of references. A basic discussion on Lanczos methods with and without reorthogonalizations, and their applications to solutions of positive de nite linear systems and eigenvalue computations, is contained in the book by Golub and Van Loan (MC 1984, Chapter 9). Two authoritative books in this area are: Lanczos Algorithms for Large Symmetric Eigenvalue Computations, volumes I and II, by J. K. Cullum and R. A. Willoughby, Birkhauser, 1985, and The Symmetric Eigenvalue Problem by B. N. Parlett, Prentice-Hall. Another recent book in this area is Eigenvalue Problem for Large Matrices, by Y. Saad (1993). Since the pioneering work by Paige in the 1970's, many papers in this area have been published, and this is an active area of research. See the list of references given in Golub and Van Loan (MC 1989, Chapter 9). 579
The doctoral thesis of Chris Paige (1971) and several of his follow-up papers (Paige (1976), Paige (1980), etc.) are also well worth reading. These are considered to be the \seed papers" for further recent active research in this area.
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Exercises on Chapter 8 (Use MATLAB whenever needed and appropriate)
PROBLEMS ON SECTION 8.3 1. Consider the following model for the vertical vibration of a motor car: car body mass m1
y
2
d (suspension)
k1
1
m2
y
d (tire)
1
2
k2
Road (a) Formulate the equation of motion of the car, neglecting the damping constants d1 of the shock absorber and d2 of the tire:
M y + Ky = 0 where M = diag(m1; m2);
;k ! k +k ! y y= : y
k K= ;k 1
and
1
1
1
2
1 2
Determine the stability of motion when
m = m = 1200kg 1
2
N: k = k = 300 cm 1
581
2
(b) Formulate the equation of motion when just the damping d2 of the tire is neglected:
M y + Dy_ + Ky = 0; M , and K are the same as in part (a), and d D= ;d 1
1
;d ! : d 1
1
Investigate the stability of motion in this case when
d = 4500 Ns m: 1
Hints: Show that the system
M y + Dy_ + Ky = 0 is equivalent to the rst order system
x_ (t) = Ax(t); where and
A=
!
0
;M ; K 1
I ; ;M ; D 1
!
y (t) x(t) = : y_ (t)
2. Write the solution of the equation M y + Ky = 0 with M and K as given in #1(a), using initial conditions y (0) = 0 and y_ (0) = (1; 1; : : :; 1)T . 3. Develop an eigenvalue problem for an LC network similar to the case study given in section 8.2.2, but with only three loops. Show that in this case the natural frequencies are given by
8 p > LC : 4451 j > < p w = > 1:2470j LC > : 1:8019jpLC
Find the modes; and illustrate how the currents oscillate in these modes.
582
PROBLEMS ON SECTION 8.4 4. Apply the Gersgorin disk theorems to obtain bounds of the eigenvalues for the following matrices: 0 10 1 1 1 B 2 10 1 CC ; (a) A = B @ A 2 2 10 01 0 01 B2 5 0C C (b) B @ A; 0 12 1 ;61 0 0 1 BB C ;1 2 ;1 0 C B C (c) B B@ 0 ;1 2 ;1 CCA ; 0 01 ;01 ;01 20 1 BB ;1 2 ;1 0 CC CC ; (d) B BB CA 0 ; 1 2 ; 1 @ 2 0 10:000 0 :577;1 :509 :387 :462 1 BB :577 1:000 :599 :389 :322 CC BB CC B (e) B :509 :599 1:000 :436 :426 C CC ; BB @ :387 :389 :436 1:000 :523 CA :322 :426 :523 1:000 0 1:452 1 ;i 0 1 B 1 C: (f) B 1 1 C @ A 0 1;i 1+i 5. Using a Gersgorin disk theorem, prove that a diagonally dominant matrix is nonsingular. 6. Let x be an eigenvector corresponding to an isolated eigenvalue in the Gersgorin disk Rk . Prove that jxk j > jxij for i 6= k. 7. Let A = (aij be an n n symmetric matrix. Then using a Gersgorin disk theorem prove that each eigenvalue of A will lie in one of the intervals: [aij ; ri ; aij + ri]. Find an interval where all the eigenvalues of A must lie.
583
PROBLEMS ON SECTION 8.5 8. Applying the power method and inverse power method nd the dominant eigenvalue and the corresponding eigenvector for each of the matrices in the exercise #4. 9. Prove that if 1; : : :; n are the eigenvalues of A and v1; : : :; vn are the corresponding eigenvectors, then 1 ; ; : : :; n ; are the eigenvalues of A ; I , and the corresponding eigenvectors are v1; : : :; vn. 10. Explain the slow rate of convergence of the power method with the following matrices: 03 2 31 B C (a) A = B @ 0 2:9 1 C A; 0 01 00 10 1 B C (b) A = B @ 1 10 0 CA ; 1 1 9:8 Choose a suitable shift and then apply the shifted power method to each of the matrices and observe the improvement of the rates of convergence. 11. (Orthogonal Iteration) The following iterative procedure generalizes the power method and is known as the orthogonal iteration process. The process can be used to compute p(p > 1) largest eigenvalues and the corresponding eigenvectors. Let Q1 be an n p orthonormal matrix. Then For k = 2; 3; : : : do 1) Compute Bk = AQk;1 2) Factorize into QR: Bk = Qk Rk : Apply the above method to compute the rst two dominant eigenvalues and eigenvectors for each of the matrices in Exercise #4. 12. (Inverse Orthogonal Iteration) The following iteration, called the Inverse Orthogonal iteration, generalizes the inverse power method; and, can be used to compute the p smallest eigenvalues. Let Q1 be an n p orthonormal matrix. 584
For k = 2; 3; : : : 1) Solve for Bk : ABk = Qk;1. 2) Factorize into QR: Bk = Qk Rk : Apply the Inverse Orthogonal Iteration to compute the 2 smallest (least dominant) eigenvalues of each of the matrices in Exercise #4. 13. Let T be a symmetric tridiagonal matrix. Let the Rayleigh-Quotient Iteration be applied to T with x0 = en , then prove that x1 = qn, where qn is the last column of Q in (T ; 0I ) = QR: 14. Compute the subdominant eigenvalue of each of the matrices in Exercise #4 using Householder de ation. Then compute the corresponding eigenvector without invoking the inverse iteration. 15. Compute the smallest eigenvalue of each of the matrices A in Exercise #4 by applying the power method to A;1 , without explicitly computing A;1. 16. De ation Using Invariant Subspace. Suppose that we have an n m matrix X with independent columns and an m m matrix M such that
AX = XM: Consider the QR factorization of X :
QT X = Then show that (a)
QAQT
A A = 0 A 1
2 3
!
R 0
!
:
;
(b) the eigenvalues of A are those of A1 and A3 , (c) the eigenvalues of A are those of M .
PROBLEMS ON SECTIONS 8.6{8.8 17. Construct a simple example to show that an arbitrary similarity transformation can worsen the conditioning of the eigenvalues of the transformed matrix. 18. If B = U AU , where U is unitary, and U (A + A)U = B + B; then show that kB k2 = kAk2. 585
19. (a) Prove that A is normal if there exists an unitary matrix U such that U AU = D, where D is diagonal. (b) Prove that a matrix A has a set of n orthonormal eigenvectors i A is normal. (c) Prove that a normal matrix A is unitary i its eigenvalues are on the unit circle, that is, for each eigenvalue of A, we have jj = 1. (d) How does the Real Schur Form of a normal matrix look? (e) Using the eigenvector-sensitivity theorem (Theorem 8.8.1), show that if A is normal, then the eigenvector xk corresponding to the eigenvalue k is well-conditioned if k is well-separated from the other eigenvalues. 20. Show both theoretically and experimentally that the eigenvectors of the matrix
A = diag(1 + ; 1 ; ; 2); where is a very small positive number, are ill-conditioned. However, the eigenvalues are well-conditioned. 21. Show that the unitary similarity transformation preserves the condition number of an eigenvalue. 22. Prove the Bauer-Fike Theorem using Gersgorin's rst theorem. 23. (a) Given
!
1 1 A= ; 0 1+ nd the eigenvector-matrix X such that X ;1AX is diagonal; hence show that the eigenvalues of A are ill-conditioned. Verify the ill-conditioning of the eigenvalues of A computationally by constructing a small perturbation to A and nding the eigenvalues of the perturbed matrix. (b) Consider 0 12 11 10 2 1 1 BB 11 11 10 2 1 CC BB C BB 0 10 10 2 1 CCC A=B BB 2 1 CCC :
BB C @ CA
0 0 0 0 11 1 Show that the largest eigenvalues of A are well-conditioned while the smallest ones very ill-conditioned, by computing the eigenvalues using MATLAB. 586
PROBLEMS ON SECTIONS 8.9 AND 8.10 24. Prove that if R is upper triangular and Q is upper Hessenberg, then RQ is upper Hessenberg. 25. Construct an example to show that for an arbitrary upper Hessenberg matrix H , the Q-matrix in H = QR need not be an upper Hessenberg. However, if R has nonzero diagonal entries, then Q in H = QR, is always upper Hessenberg. 26. Prove that (a) the QR factorization of an n n Hessenberg matrix requires 0(n2) ops, (b) the QR factorization of a symmetric tridiagonal matrix requires only 0(n) ops. 27. Apply 3 iterations of the single-shift QR iteration to each of the following matrices and observe the convergence or nonconvergence of the subdiagonal entries. (a)
01 2 01 B 2 3 4 CC A=B @ A 0 4 1
(b)
04 5 61 B C A=B @ 1 0 1 CA 0 ;2 2
28. Perform one step of the single-shift QR iteration the matrix 01 1 11 B C H=B @ 2 10 6 CA 0 :1 1 and observe the quadratic convergence. 29. Implicit single-shift QR. Consider one step of the single-shift QR iteration:
Hk ; hnnk I = Qk Rk; Hk = RkQk + hnnk I ( )
or (simply)
+1
( )
H ; I = QR; H = RQ + I; is real: 587
(a) Prove that the rst column of Q is a multiple of the rst column of H ; I , and therefore contains only two nonzero entries. (b) Denote the rst column of H ; I by h1 = (h11 ; ; h21 ; 0; : : :; 0)T : Find a Givens rotation P0 such that P0 h1 is a multiple of e1 . Show that the rst column of P0 is the same as the rst column of Q, except possibly for the sign. (c) Form H 0 = P0T HP0 . Find Givens rotations J32 ; J43; : : :; Jn;n;1 such that
H 0 = (J J ; : : :; Jn;n; )T H 0(J ; : : :; Jn;n; ) 1
32
43
1
32
1
is upper Hessenberg. Show that the matrix
Q~ = P J ; : : :; Jn;n; 0
32
1
has the same rst column as P0 and hence the same rst column as Q. Conclude nally from the Implicit Q Theorem that the Hessenberg matrix H10 is essentially the same as
H 0:
The steps (a) to (c) constitute one step of the implicit single-shift QR iteration. Apply one step of the implicit single-shift QR iteration to the symmetric tridiagonal matrix 0 2 ;1 0 1 BB ;1 . . . CC BB . . . C B@ .. . . . . ;1 CCA : 0 ;1 2 30. Construct one step of the explicit double-shift QR iteration (real arithmetic) and one step of the implicit double-shift QR iteration for the matrix 01 2 3 41 BB 3 4 5 6 CC CC A=B BB CA 0 1 0 1 @ 0 0 ;2 2 and show that the nal Hessenberg matrices are (essentially) the same. 31. LR Iteration. In analogy with QR iteration algorithm, develop a LR iteration algorithm, based on LU decomposition of A, making the necessary assumptions. Set A = A1 2) Compute Ak = Lk Rk ; Ak+1 = Rk Lk ; k = 1; 2; : : : Why is this algorithm not to be preferred over the QR iteration algorithm? 588
32. Considering the structures of the matrices Pi ; i = 0; 1; : : :; n ; 2 in the implicit double-shift QR iteration step (Exercise #29), show that it requires only 4n2 ops to implement this step. 33. Show that the matrices Hs and Hs+2 in the double-shift QR iteration have the same eigenvalues. 34. Prove the following: Let H = H0 . Generate the sequence fHk g:
Hk ; k kI = Qk Rk; Hk l = Rk Qk + k I: +
Then
ni=1 (H ; i I ) = (Q1 ; : : :; Qn)(Rn; : : :; R1):
PROBLEMS ON SECTIONS 8.11 AND 8.12 35. (a) Let
0 1 ;1 0 0 1 BB ;1 2 ;1 0 CC CC : A=B BB @ 0 ;1 2 ;1 CA 0 0 ;1 2 Without computing the eigenvalues show that jj < 4 for each eigenvalue of A. Show
that there are exactly two eigenvalues greater than 2 and two less than 2. Apply Sturm Sequence -Bisection Algorithm (Algorithm 8.11.1). to compute the eigenvalue close to 2. (b) Apply the inverse Hessenberg iteration algorithm to A to compute the eigenvector associated with the eigenvalue close to 2. (c) Compute the eigenvalues of A by applying the symmetric QR iteration with the Wilkinsonshift. 36. (a) Prove that the eigenvalues of an unreduced real symmetric tridiagonal matrix are real and distinct. (b) Prove that if is an eigenvalue of multiplicity k of an unreduced symmetric tridiagonal matrix T , then at least (k ; 1) subdiagonal entries of T must be zero. 37. (a) Develop a QR-type algorithm to compute the eigenvalues of a symmetric positive de nite matrix A, based upon the Cholesky decomposition. (Section 6.4.6). 589
(b) Test your algorithm with the matrix A of problem #35.
!
38. Let
A= : Prove that the eigenvalue of A closest to is given by () ; p = ; jjsign + + where = ; . 2
2
2
2
A ;A 39. Let A = A + iA be a Hermitian matrix. Then prove that B = A A How are the eigenvalues and eigenvectors of A related to those of B ? 1
1
2
2
2
1
!
is symmetric.
40. Take the 50 50 symmetric tridiagonal matrix T with 2 along the diagonal and -1 along the sub diagonal and super diagonal. Apply Algorithm 8.12.3 to T with j = 1; 2; : : :; 20. Find pproximations to a few extreme eigenvalues of T using Theorems 8.12.1 and 8.12.2.
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MATLAB PROGRAMS AND PROBLEMS ON CHAPTER 8 1. Write a MATLAB program to compute the dominant eigenvalue of a matrix using the power method. [lambda1] = power(A,x0,epsilon,n) (a) Modify the program power to incorporate a shift sigma [lambda1] = powershift(A,x0,sigma,epsilon,n) (b) Apply power and powershift to the following matrices and compare the speed of convergence
Test Data:
03 2 3 1 B 0 0:99 1 CC ; A=B @ A
0 0 2:9 A = A randomly generated matrix of order 5. A = The Wilkinson bidiagonal matrix of order 20. 2. (a) Using linsyspp from Chapter 6 or the MATLAB command `n', write a MATLAB program called invitr to implement the inverse iteration algorithm x = invitr(A,x0,sigma,epsilon,n) (b) Using linsyspp from Chapter 6 or the MATLAB command `n', write a MATLAB program called powersmall to compute the smallest eigenvalue (in magnitude) of a matrix A lambdan = powersmall(A,x0,epsilon,n)
Test Data and Experiment: (a) Take the 20 20 symmetric tridiagonal matrix appearing in the buckling problem of section 8.3.2. Apply power to compute the dominant eigenvalue lambda1 by choosing x0 arbitrarily. (b) Now compute the smallest eigenvalue in magnitude, lambdan, by using (i) powershift with sigma = lambda1 (ii) powersmall with the same x0 as used to implement power. 591
(c) Compare the speed of (i) and (ii) in (b). (d) Find the smallest critical load that will buckle the beam (e) Taking sigma = lambdan, nd the eigenvector corresponding to the smallest eigenvalue n using invitr. 3. Using power, invitr, housmat (from Chapter 5) and housmul (from Chapter 4), write a MATLAB program to implement the Householder de ation algorithm that computes the subdominant eigenvalue of a matrix: lambda2 = housde t(a)
Test data: A single sex, chohort population model can be represented by the following system of di erence equations :
p (k + 1) = p (k) + p (k) + ::: npn (k) 0
0 0
1 1
pi (k + 1) = i pi(k); i = 0; 1; :::; n ; 1; +1
or in matrix form
pk = Apk : +1
Here i is the birth rate of the ith age group and i is the rate of survival of that group (see Luenberger (1979), p 170). Taking 0 = 0; 1 = 2 = ::: = n = 1, and i = 1; i = 0; 1; :::; n ; 1, determine by using power, invitr, housde t, if there is a long term growth in the population; if this is so, what is the nal population distribution; and how fast is the original distribution approaching the nal distribution. Consult the example on population study in Section 8.4.2. 4. (The purpose of this exercise is to study how are the eigenvalues of a matrix A
are a ected by conditioning of the transforming matrix.)
(a) For each of the following matrices construct a matrix X of appropriate order which is upper triangular with all the entries equal to 1 except for a few very small diagonal
592
entries. Then compute the eigenvalues of A and those of (X );1AX using MATLAB commands eig and inv : 01 1 0 01 00 0 2 1 BB 0 1 0 0 CC CC ; A = BB 1 0 ;5 CC ; A=B BB @ A @ 0 0 0:9 1 CA 0 1 4 0 0 0 1 A = The Wilkinson bidiagonal matrix of order 20. (b) Repeat part (a) by taking X as a Householder matrix of appropriate order. (c) Compare the results of (a) and (b). 5. (a) Compute the eigenvalues of the following matrices using: 1) MATLAB commands poly and roots. 2) MATLAB command eig.
00 0 2 1 01 1 B C B A=B @ 1 0 ;5 CA ; A = B@ 0 :19
1
1 C 1 C A; 0 1 4 0 0 0:9999 A = The Wilkinson bidiagonal matrix of order 20. (b) Compare your results of (1) and (2) for each matrix. 6. (The purpose of this exercise is to study the sensitivities of the eigenvalues of
some well-known matrices with ill conditioned eigenvalues).
Perform the following on each of the matrices in the test-data: (a) Using the MATLAB command [V; D] = eig(A), nd the eigenvalues and the matrix of right eigenvectors. Then nd the matrix of left eigenvectors W as follows W = (inv(V))' /norm(inv(V)') (b) Compute si = wiT vi ; i = 1; :::; n: where Wi and Vi are the ith columns of W and V . (c) Compute ci = the condition number of the ith eigenvalue = 1=si, i = 1; 2; :::; n. (d) Perturb the (n,1)th entry of A by = 10;10. Then compute the eigenvalues ^ i ; i = 1:::; n of the perturbed matrix using the MATLAB command eig. (e) Make the following table for each matrix.
593
i
^i
ji ; ^ij
Cond (V) ci
(f) Write your conclusions.
Test-Data: (1) A = The Wilkinson bidiagonal matrix of order 20 (2) A = The transpose of the Wilkinson Bidiagonal matrix of order 20. (3) 0 12 11 10 3 2 1 1 BB 11 11 10 3 2 1 CC C B
BB A=B BB BB B@
2 0 0 0
C C CC C CC 2 1C A
1 1 7. Study the sensitivites of the eigenvectors of the following matrices by actually computing the the eigenvectors of the original and perturbed matrices using the MATLAB command [V; D] = eig(A), [V; D] = eig(A^), where A^ is the matrix obtained from A by perturbing (n; 1)th entry by = 10;5.
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01 0 0 BB 0 :999 0 A=B BB @0 0 0 0
0
1) A = diag(1; :9999; 1; :9999; 1) 2) A randomly generated matrix of order 5.
1
0 C 0 C CC 2 C A 0 0:0005
8. Write a MATLAB program called qritrb to implement the basic QR iteration using givqr from Chapter 5 : (a) [A] = qritrb(A; num); num is the maximum number of iterations. (b) Modify the program now to implement the Hessenberg QR iteration [A] = qritrh(A; num) num is the number of iterations (c) Compare the speed of the programs by actually computing the op-count and elapsed time . 9. (a) Write a MATLAB program to compute one step of the explicit double shift QR iteration: [A] = qritrdse(A) (b) Write a MATLAB program to compute one step of the implicit QR iteration with double shift: [A] = qritrdsi(A). (c) Compare your results of (a) and (b) and conclude that they are essentially the same. 10. Write a MATLAB program to de ate the last k rows and k columns of a Hessenberg matrix. hprime = de at(H; k). Test your program with a randomly generated matrix with di erent values of k. Note that for k = 1, hprime will be of order n ; 1, for k = 2 hprime will be of order n ; 2, and so on. 11. Using Qritrdsi and de at, write a MATLAB program to determine the Real Schur form of a Hessenberg matrix A; [h] = rsf(h; eps); where eps is the tolerance. Test: Generate randomly a 20 20 Hessenberg matrix and make the following table using rsf. 595
Iteration h21 h32 h43 h54 h20;1
TABLE
12. (a) Write a MATLAB program, called polysymtri to compute the characteristic polynomial pn() of an unreduced symmetric tridiagonal matrix T , based on the recursion in Section 8.11.1 [valpoly] = polysymtri(A; lambda): (b) Using polysymtri, write a MATLAB program called signagree that nds the number of eigenvalues of T greater than a given real number , based on Theorem 8.11.1: [number] = signagree (T; meu). (c) Using polysmtri and signagree, implement the Bisection algorithm of Section 8.11.1: [lambda] = bisection (T; m; n). Compute n;m+1 for m = 1; 2; 3; ::: , using bisection, and then compare your results with those obtained by using eig(T).
Test Data:
A = The dymmetric tridiagonal matrix arising in Buckling Problem in Section 8.3.2, with n = 20. 13. Write a MATLAB program, called lanczossym, to implement the reformulated symmetric Lanczos algorithm of Section 8.11.3. Using lanczossym nd the rst ten eigenvalues of a randomly generated symmetric matrix of order 50. (To generate a symmetric matrix B, generate A rst, then take B = A + AT ). (Note : The programs power, invitr, qritrb, qritrh, qritrdsi, etc. are in MATCOM. But it is a good idea to write your own codes).
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9. THE GENERALIZED EIGENVALUE PROBLEM
9.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 597 9.2 Generalized Schur Decomposition : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 599 9.3 The QZ algorithm : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 601 9.3.1 Reduction to Hessenberg-Triangular Form : : : : : : : : : : : : : : : : : : : : 602 9.3.2 Reduction to the Generalized Schur Form : : : : : : : : : : : : : : : : : : : : 604 9.4 Computations of the Generalized Eigenvectors : : : : : : : : : : : : : : : : : : : : : 611 9.5 The Symmetric-De nite Generalized Eigenvalue Problem : : : : : : : : : : : : : : : 612 9.5.1 The QZ Method for the Symmetric-De nite Pencil : : : : : : : : : : : : : : : 614 9.5.2 The Cholesky-QR Algorithm : : : : : : : : : : : : : : : : : : : : : : : : : : : 614 9.5.3 Diagonalization of the Symmetric-De nite Pencil : : : : : : : : : : : : : : : : 616 9.6 Symmetric-De nite Generalized Eigenvalue Problems Arising in Vibrations of Structures : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 619 9.6.1 A Case Study on the Vibration of a Spring-Mass System : : : : : : : : : : : : 619 9.6.2 A Case Study on the Vibration of a Building : : : : : : : : : : : : : : : : : : 622 9.6.3 Forced Harmonic Vibration : : : : : : : : : : : : : : : : : : : : : : : : : : : : 625 9.7 Applications to Decoupling and Model Reduction : : : : : : : : : : : : : : : : : : : : 629 9.7.1 Decoupling of a Second-Order System : : : : : : : : : : : : : : : : : : : : : : 630 9.7.2 The Reduction of a Large Model : : : : : : : : : : : : : : : : : : : : : : : : : 637 9.7.3 A Case Study on the Potential Damage of a Building Due to an Earthquake : 639 9.8 The Quadratic Eigenvalue Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : 643 9.9 The Generalized Symmetric Eigenvalue Problems for Large and Structured Matrices 646 9.9.1 The Sturm Sequence Method for Tridiagonal A and B : : : : : : : : : : : : : 646 9.9.2 The Lanczos Algorithm for the Pencil A ; B : : : : : : : : : : : : : : : : : 647 9.9.3 Estimating the Generalized Eigenvalues : : : : : : : : : : : : : : : : : : : : : 649 9.10 The Lanczos Method for Generalized Eigenvalues in an Interval : : : : : : : : : : : : 651 9.11 Review and Summary : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 652 9.12 Suggestions for Further Reading : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 655
CHAPTER 9 THE GENERALIZED EIGENVALUE PROBLEM
9. THE GENERALIZED EIGENVALUE PROBLEM Objectives The objectives of this chapter is to study engineering applications and numerical methods for the generalized eigenvalue problem: Ax = Bx. Some of the highlights of this chapter are:
The QZ algorithm for generalized Schur form (Section 9.3). The Cholesky-QR algorithm and simultaneous diagonalization techniques for a symmetric de nite pencil (Section 9.5).
Engineering vibration problems giving rise to generalized symmetric de nite eigenvalue problems (Section 9.6).
Applications of simultaneous diagonalization techniques to decoupling and model reduction (Section 9.7).
The Lanczos algorithm for large and sparse symmetric de nite problems (Section 9.9).
The Lanczos algorithm for generalized eigenvalues in an interval (Section 9.10).
Background Material Needed for this Chapter
We need the following major tools developed earlier to understand the material in this chapter. 1. The Householder and Givens methods to create zeros in a vector and the corresponding QR factorization algorithms (Sections 5.4, 5.4.1, 5.4.2, 5.5, 5.5.1). 2. The Cholesky algorithm for solving a symmetric positive de nite system (Section 6.4.7). 3. The Inverse iteration algorithm (Chapter 8, Section 8.5). 4. The Rayleigh quotient iteration algorithm (Chapter 8, Section 8.5).
9.1 Introduction In this chapter we consider the following eigenvalue problems known as the generalized eigenvalue problem.
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Statement of the Generalized Eigenvalue Problem Given n n matrices A and B , nd scalars and nonzero vectors x such that Ax = Bx: Note that the standard eigenvalue problem for the matrix A is a special case of this problem (take B = I ).
De nition 9.1.1 In the problem AX = BX , is called a generalized eigenvalue and the vector x is a generalized eigenvector associated with . It is easy to see that is a root of the characteristic equation det(A ; B ) = 0:
De nition 9.1.2 The matrix A ; B is called a matrix pencil. If B is nonsingular, then the pencil is called a regular pencil. A Note on the Use of the Word `Pencil' \The rather strange use of the word \pencil" comes from optics and geometry. An aggregate of (light) rays converging to a point does suggest the sharp end of a pencil and, by a natural extension, the term came to be used for any one parameter family of curves, spaces, matrices, or other mathematical objects." (Parlett, The Symmetric Eigenvalue Problem (1980), p. 302). Note that when B is nonsingular, a generalized eigenvalue is also an eigenvalue of B ;1 A. If B is singular, then zero is an eigenvalue of B, in that case the corresponding generalized eigenvalue is 1. Many engineering applications give rise to generalized eigenvalue problems. In fact, the ma-
jority of eigenvalue problems arising from real physical systems are generalized eigenvalue problems. For example, the eigenproblems for vibrations of structures such as buildings 598
and bridges are the so-called symmetric de nite generalized eigenvalue problems for the mass and sti ness matrices: Kx = Mx, where M is the mass matrix and K is the sti ness matrix. M and K are usually real symmetric, and furthermore, M is symmetric positive de nite.
De nition 9.1.3 If A and B are real symmetric matrices and furthermore, if B is positive de nite, then the generalized eigenvalue problem
Ax = Bx is called the symmetric de nite generalized eigenvalue problem. This chapter is devoted to the study of the generalized eigenvalue problem with particular attention to the symmetric de nite problem. The chapter is organized in the following manner. In Section 9.2 we present a result that shows how the generalized eigenvalues, and eigenvectors, can be extracted once the matrices A and B are reduced to triangular forms. In Section 9.3 we describe the QZ algorithm for the generalized eigenvalue problem. It is a natural generalization of the QR iteration algorithm described in Chapter 8. In Section 9.4 we show how to compute a generalized eigenvector when a reasonable approximation of a generalized eigenvalue is known using the inverse iteration. Sections 9.5, 9.6, and 9.7 are dedicated to the study of the symmetric de nite generalized eigenvalue problems. Several case studies on the problems arising in vibration of structures are presented. A popular algorithm widely used in engineering practice|namely, the simultaneous diagonalization algorithm is presented and several engineering applications of this technique are discussed. Finally, in Section 9.8, we present the Sturm sequence and the Lanczos methods for large and sparse generalized eigenvalue problems.
9.2 Generalized Schur Decomposition As a generalization of the procedure for the standard eigenvalue problem, it is natural to think of solving the generalized eigenvalue problem by simultaneously reducing the matrices A and B to some convenient forms from which generalized eigenvalues can be easily extracted. Theorem 9.3.1 shows the existence of such convenient forms, while Theorem 9.2.1, illustrates the extraction of the generalized eigenvalues once the matrices A and B have been reduced to such forms.
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Theorem 9.2.1 Let A and B be n n matrices and B be nonsingular. Let U1 and U2 be unitary matrices such that
0t BB 11 0 U1AU2 = T1 = B BB .. @. 0
0 t0 BB 011 U1BU2 = T2 = B BB .. @. 0
... ... ... ...
tnn
t0nn
1 CC CC CA 1 CC CC : CA
Then the generalized eigenvalues i ; i = 1; : : :; n of the regular pencil det(A ; B ) are given by i = tii=t0ii:
Proof. From Ax = Bx, we have U1AU2U2 x = U1 BU2U2 x: De ne y = U2 x. Then from above we have
U1 AU2 y = U1 BU2y ; that is, T1y = T2y . Since B is nonsingular, so is T2. Thus, we have
T2;1T1y = y; which shows that is an eigenvalue of
T2;1 T1:
Since T1 and T2;1 are both upper triangular matrices, so is their product T2;1 T1. Thus, the eigenvalues of T2;1 T1 are tii =t0ii; i = 1; : : :; n.
Remarks: 1. Note that U2 x = y is an eigenvector associated with . 2. If A and B are real matrices, then U1 can be chosen such that U1 AU2 = T1 is in Real Schur Form. 600
9.3 The QZ algorithm We will describe an analog of the QR iteration, known as the QZ iteration algorithm for computing the generalized eigenvalues, developed by Cleve Moler and G. W. Stewart.Assume that B is nonsingular. Then the basic idea is to apply the QR iteration algorithm to the matrix C = B ;1 A (or to AB ;1 ), without explicitly forming the matrix C , because if B is nearly singular, then it is not desirable to form B ;1 . In this case the elements of C will be much larger than those of A and B , and the eigenvalues of C will be computed inaccurately. Note that the eigenvalues of B;1 A are the same as those of AB ;1 , because AB ;1 = B (B ;1 A)B ;1 . Thus, they are similar. If AB ;1 or B ;1 A is not to be computed explicitly, then the next best alternative, of course, is to transform A and B to some reduced forms and then extract the generalized eigenvalues i from these reduced forms. The simultaneous reduction of A and B to triangular forms by equivalence is guaranteed by the following theorem:
Theorem 9.3.1 (The Generalized Real Schur Decomposition) Given two n n real matrices A and B , there exist orthogonal matrices Q and Z such that QT AZ is an upper Real Schur matrix and QT BZ is upper triangular: A0 = QT AZ : B0 = QT BZ :
A0 = an upper Real Schur matrix. B0 = an upper triangular matrix.
The pair (A0; B 0 ) is said to be in generalized Real Schur Form. We shall now give a constructive proof of this theorem. The reduction to the generalized Real Schur Form is achieved in two stages.
Stage I. A and B are reduced to an upper Hessenberg and an upper triangular matrix, respec-
tively, by simultaneous orthogonal equivalence:
A QT AZ : an upper Hessenberg matrix B QT BZ : an upper triangular matrix Dr. Cleve Moler is the developer of the popular software package MATLAB. He was a professor and head of the Computer Science Department at the University of New Mexico. He is currently with MathWorks, Inc. He was also one of the developers of another popular mathematical software \LINPACK".
601
Stage II. The Hessenberg-Triangular pair (A; B) is reduced further to the generalized Real Schur Form by applying implicit QR iteration to AB;1 . This process is known as the QZ Algorithm. We will now brie y sketch these two stages in the sequel.
9.3.1 Reduction to Hessenberg-Triangular Form Let A and B be two n n matrices. Then 1. Find an orthogonal matrix U such that
B U T B : upper triangular: Form
A UTA
(in general, A will be full). 2. Reduce A to Hessenberg form while preserving the triangular structure of B . This step is achieved as follows: To start with, we have
0 BB BB . T AU A = B BB .. B@ 0 BB 0 BB T BU B = B BB 0. B@ ..
1 CC C CC ; CC C A 1 C C CC . . . C 0 C: . . . ... C CA 0 0 0 0 First, the (n; 1)th entry of A is made zero by applying a rotation Qn;1;n in the (n ; 1; n) plane: 0 1 BB CC CC BB . . CC : . A Qn;1;nA = B BB B@ CCA 0 602
This transformation, when applied to B to the left, will give a ll-in in the (n; n ; 1) position:
0 BB 0 BB B Qn;1;nB = B BB 0. 0 . .. B@ .. 0 0 0
1 CC C C: C C .. C .C A The rotation Zn;1;n = J (n ; 1; n; ) is now applied to the right of B to make the (n; n ; 1) entry of B zero. Fortunately, this rotation, when applied to the right of A, does not destroy the zero produced earlier. Schematically, we have
0 1 BB 0 CC BB CC B B B Zn;1;n = B 0 0 C CC BB .. . . . . CA . . @. 0 0 0 0 1 BB CC BB C BB CCC A AZn;1;n = B : .. C BB ... ... ... .C C BB C @ 0 CA 0
The entries (n ; 1; 1); (n ; 2; 1); : : :; (3; 1) of A are now successively made zero, each time applying an appropriate rotation to the left of A, followed by another appropriate rotation to the right of B to recover the undesirable ll-in in B . At the end of the rst step, the matrix A is Hessenberg in its rst column, while B remains upper triangular:
0 BB BB A=B BB 0. B@ ..
0 1 1 BB CC CC C BB . . CC C .. C . . B . . C; B = B0 .C CC : B .. . . . .. C . . . . . . . . B@ . . . . CA . .C A 0 0 0 The zeros are now produced on the second column of A in the appropriate places while retaining the triangular structure of B in the analogous manner. The process is continued until the matrix A is an upper Hessenberg matrix and B is upper triangular.
603
Example 9.3.1
01 2 31 B 1 3 4 CC ; A=B @ A
01 1 11 B 0 1 2 CC : B=B @ A
1 3 3
0 0 2
1. Form Q23 to make a31 zero:
01 0 1 0 B 0 :7071 :7071 CC Q23 = B @ A 0 ;:7071 :7071 0 1 1 2 3 B C A A(1) = Q23A = B @ 1:4142 4:2426 4:9497 CA : 0 0 ;0:7071
2. Update B :
01 1 1 1 B C B B (1) = Q23B = B @ 0 0:7071 2:8284 CA 0 ;0:7071 0 3. Form Z23 to make b32 zero: 01 0 0 1 B C Z23 = B @ 0 0 ;1 CA
0 1 0 01 1 ;1 1 B CC B B(1) Z23 = Q23 BZ23 = B 0 2 : 8284 ; 0 : 7071 @ A: 0 0 0:7071
4. Update A:
0 1 3 ;2 1 B C A A(1) Z23 = Q23AZ23 = B @ 1:4142 4:9497 ;4:2426 CA : 0 ;0:7071 0
A is in upper Hessenberg and B is in upper triangular form.
9.3.2 Reduction to the Generalized Schur Form At the beginning of this process, we have A and B as an upper Hessenberg and an upper triangular matrix, respectively, obtained from stage 1. We can assume without loss of generality that the matrix A is an unreduced upper Hessenberg matrix. The basic idea now is to apply an
implicit QR step to AB;1 without ever completely forming this matrix explicitly. Thus a QZ step, analogous to an implicit QR step, will be as follows: 604
Compute the rst column of N = (C ; 1I )(C ; 2I ), where C = AB;1 , and 1 and 2 are suitably chosen shifts.
Find a Householder matrix Q1, such that Q1Ne1 is a multiple of e1. Form Q1A and Q1B. Transform simultaneously Q1A to an upper Hessenberg matrix A1, and Q1B to an upper triangular matrix B1 . A1 Q0(Q1A)Z : an upper Hessenberg; B1 Q0(Q1B )Z : an upper triangular. Using the implicit Q theorem we can show (Exercise) that the matrix A1B1;1 is essentiatlly the same as that we would have obtained by applying an implicit QR step directly to AB ;1 . Application of a few QZ steps sequentially will then yield a quasi-triangular matrix R = QT AZ and an upper triangular T = QT BZ , from which the generalized eigenvalues can be easily computed. The above four steps constitute one single step of the QZ algorithm. We now show how to implement these steps.
Computation of the rst column of (C ; 1I )(C ; 2I ) The real bottleneck in implementing the whole algorithm is in computing the rst column of (C ; 1I )(C ; 2 I ), without forming C = AB ;1 explicitly. Fortunately, this can be done. First, note that because A is upper Hessenberg and B is upper triangular, this rst column contains at most three nonzero entries in the rst three places:
0x1 BB y CC BB CC B z CC n1 = Ne1 = (C ; 1I )(C ; 2 I )e1 = B BB CC : BB 0. CC B@ .. CA
0 To compute x; y; z; all we need to know is the rst two columns of C , which can be obtained just by inverting the 2 2 leading principal submatrix of B ;1 ; the whole B ;1 does not need to be computed. Thus, if c1 and c2 are the rst two columns of C = AB;1, then
b b (c1; c2) = (a1; a2) 11 12 0 b22 605
!;1
;
where ai ; i = 1; 2 are the rst two columns of A. Note that c1 has at most two nonzero entries and c2 has at most three. Let 0 1
c 0c 1 BB c12 CC 11 B C BB 22 CC B c21 C B C BB c32 CC B CC c1 = B 0 and c = CC : 2 B B CC B B 0 . B B BB .. CCC @ .. CA @ .A 0
Then it is easy to see that
Example 9.3.2 Let
0
0 x 1 0 (c ; )(c ; ) + c c 1 11 1 11 2 12 21 BB y C B C C B @ A = @ c21(c11 ; 2) + c21(c22 ; 1) CA : z
c21c32
01 1 1 11 B C B 2 1 4 1C B CC ; A=B B @ 0 1 1 1 CA 0 0 1 1
01 2 3 41 BB 0 1 1 1 CC CC : B=B BB @ 0 0 1 2 CA 0 0 0 3
The 2 2 leading principal submatrix of B =
011 BB 2 CC c1 = B BB CCC ; @0A
0 ;1 1 BB ;3 CC c2 = B BB CCC : @1A
!
1 2 : 0 1
0 0 Choose 1 = 1; 2 = 1;
then x = ;2; y = ;8; z = 2:
Computation of A1 and B1 Since the rst column n1 of N = (C ; 1 I )(C ; 2I ) has at most three nonzero entries, the Householder matrix Q1 that transforms n1 to a multiple of e1 has the form ! Q^ 1 0 Q1 = 0 In;3 606
where Q^ 1 is a 3 3 Householder matrix. Therefore
0 BB BB BB B
A Q1A = B BB 0 BB BB ... @
0
.. . 0 0
0 BB BB
BB BB
B0 B Q1B = B BB BB ... BB BB ... @
.. . 0
.. .
1 C CC C C C CC CC C .. C .C CA
1 C CC C C C CC CC : C .. . . . . . . .. C CC . .C .. . . . . . . .. C . .C CA 0 0 0 That is, both the Hessenberg form A and the triangular form of B are now lost, in that there are now an unwanted nonzero in the (3,1) position of Q1 A and the unwanted nonzeros in the (2,1), (3,1) and (3,2) position of Q1 B . The job now will be to cleverly chase the unwanted nonzero entries
0 0
and make them zero using orthogonal transformations. To do this, Householder matrices Z1 and Z2 are rst constructed to make B triangular, that is, to make the nonzero entries b31; b32 and b21 zero. We then have
0 BB 0 BB BB 0 0 BB ... B B Z1Z2 = B BB .. BB .. BB .. BB .. @. 0 0 0
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0
1 CC CC CC CC CC ; CC CC CC CA
0 BB BB BB
BB
A A Z1Z2 = B BB BB 0 BB 0 BB .. @.
1 CC C CC C CC C C: C CC C CC CA 0 0 0
0
0 0
(Note that we now have unwanted zeros in the (3,1), (4,1) and (4,2) positions of A.) Next, a Householder matrix Q2 is created to introduce zeros in the (3,1) and (4,1) positions of A; B is then updated. We now have
0 B B B B B B 0 B B B A Q2 A = B 0 B B B 0 B B .. B B @.
1 C 0 1 CC B ... C CC BB CC B CC ... ... ... C CC BB 0 CC BB C ... ... ... C CC = BB 0 A0 CC ; CC C BB ... CC 0 0 ... ... ... C CC BB . CC BB .. . . . . . . . . . . . . . . . ... C CA @ A 0 0 0 0 0 0 1 0 0 1 CC BB CC BB BB 0 CC BB 0 CC BB 0 CC BB 0 CC C B BB CC BB ... C C CC ; 0 = B Q2 B = B 0 0 CC BB BB CC BB 0 0 . . . . . . CC BB 0 CC CC BB .. BB . . . CC . . 0 . . . . .. C B . B B@ .. .. A @ A 0 0 0 0 0 The process is now repeated on the submatrices A0 and B 0 which have the same structures of A and B to start with. The continuation of this process will nally yield A1 , and B1 in the desired
forms. In view of the above discussion, let's now summarize one step of the QZ iteration.
Algorithm 9.3.1 One step of the QZ Algorithm Given A, an unreduced upper Hessenberg matrix, and B , an upper triangular matrix, the 608
following steps constitute one step of the QZ iteration; that is, these steps construct orthogonal matrices Q and Z such that QAZ is an upper Hessenberg and QBZ is upper triangular. 1. Choose the shifts 1 and 2. 2. Compute the rst column of N = (C ; 1 I )(C ; 2I ), where C = AB ;1 , without explicitly !
b11 b12 Let (c1; c2) be the rst two columns of C . Then (c1; c2) = (a1; a2) 0 b22 The three nonzero entries of the rst column of N are given by: forming B ;1 :
;1
.
x = (c11 ; 1 )(c11 ; 2) + c12c21 y = c21(c11 ; 2 ) + c21(c22 ; 1) z = c21c32. 0x1 BB y CC BB CC BB z CC
The rst column of N = n1 = B BB 0 CCC. BB .. CC @.A 0
01 BB 0 CC 3. Find a Householder matrix Q1 such that Q1 n1 = B BB .. CCC : @.A 0
4. Form Q1 A and Q1 B .
5. Transform Q1A and Q1 B , respectively, to an upper Hessenberg matrix A1 and a triangular matrix B1 by orthogonal equivalence in the way shown above, by taking advantage of the special structures of the matrices, creating orthogonal matrices Q2 through Qn;2 and Z1 through Zn;2 .
Obtaining the Transforming Matrices The transforming matrices Q and Z are obtained as follows: The matrix Q: Q = Q1Q2 Q3 Qn;2; The matrix Z : Z = Z1 Z2 Zn;2: Note that Q has the same rst row as Q1 . 609
Flop Count: One QZ step requires 13n2 op. If Q and Z are to be accumulated, then an
additional 5n2 and 8n2 ops, respectively, will be required.
Choosing the Shifts The double shifts 1 and 2 at a QZ step can be taken as the eigenvalues of the lower 2 2 submatrix of C = AB ;1 : The 2 2 lower submatrix of C can be computed without
explicitly forming B;1.
Algorithm 9.3.2 The QZ Algorithm Given A and B are n n real matrices. I. Transform (A; B ) to a Hessenberg-Triangular pair by orthogonal equivalence:
A QT AZ : an upper Hessenberg, B QT BZ : an upper Triangular. II. Apply a sequence of the QZ steps to the Hessenberg-Triangular pair (A; B ) to produce fAk g and fBk g, with properly chosen shifts. III. Watch the convergence of the sequences fAk g and fBk g:
fAk g ;! R, Quasi-Triangular fBk g ;! T , Triangular.
Flop Count: The implementation of I to III requires about 15n3 ops. The formation of Q
and Z , if required, needs, respectively, another 8n3 and 10n3 ops (from experience it is known that about two QZ steps per eigenvalue are adequate).
Round-o Properties: The QZ iteration algorithm is as stable as the QR iteration algo-
rithm. It can be shown that the computed R^ and S^ satisfy QT0 (A + E )Z0 = R^ QT0 (B + F )Z0 = S^, where Q0 and Z0 are orthogonal and
kE k = kAk and kF k = kB k; is the machine precision. 610
9.4 Computations of the Generalized Eigenvectors Once an approximate generalized eigenvalue is computed, the corresponding generalized eigenvector v can be computed using the inverse iteration as before.
Algorithm 9.4.1 Computations of the Generalized Eigenvectors 1. Choose an initial eigenvector v0 . 2. For k = 1; 2; : : : do Solve (A ; B )^vk = Bvk;1; vk = v^k=kv^k k1.
3. Stop if kvk k;v vkk;1k , for a prescribed small . k
A Remark on Solving (A ; B)vk = Bvk;1. In solving (A ; B )vk = Bvk;1, substantial savings can be made by exploiting the HessenbergTriangular structure to which the pair (A; B ) is reduced as a part of the QZ algorithm. Note that for a given , the matrix A ; B is a Hessenberg matrix. Thus, at each iteration only a Hessenberg system needs to be solved.
611
Example 9.4.1 0 3 ;1:5 0 1 B CC A = 109 B ; 1 : 5 3 ; 1 : 5 @ A; 0 ;1:5 1:5 02 0 01 B C B = 103 B @ 0 3 0 CA : 0 0 4
1 = a generalized eigenvalue of(A ; B) = 1950800:
011 B 1 CC : v0 = B @ A 1
k=1: Solve for v1: Solve:
(A ; 01B )^v1 = Bv 0 :0170 1 B C v^1 = B @ ;0:0102 CA 0:0024 0 :8507 1 B CC v^1 = v^1 =kv^1k = B ; : 5114 @ A: 0:1217
9.5 The Symmetric-De nite Generalized Eigenvalue Problem In this section, we study the symmetric de nite eigenvalue problem:
Ax = Bx; Here A and B are symmetric and B is positive de nite. As said before, the symmetric de nite generalized eigenvalue problem arises in a wide variety of engineering applications. It is routinely solved in vibration analysis of structures.
612
Frequencies, Modes, and Mode Shapes The eigenvalues are related to the natural frequencies, and \the size and sign of each element of an eigenvector determines the shape of the vibration at any instant of time". The eigenvectors are therefore referred to as mode shapes or simply as modes. \The language of modes, mode shapes, and natural frequencies form the basis for discussing vibration phenomena of complex systems. An entire industry has been formed around the concept of modes" (Inman (1994)). We start with an important (but not surprising) property of the symmetric de nite pencil.
Theorem 9.5.1 The symmetric-de nite pencil (A ; B) has real eigenvalues and
linearly independent eigenvectors.
Proof. Since B is symmetric positive de nite, it admits the Cholesky decomposition: B = LLT : So, from we have whence or
Ax = Bx Ax = LLT x L;1A(LT );1LT x = LT x; Cy = y;
where y = LT x.
The matrix C = L;1A(LT );1 is symmetric; therefore is real. The assertion about the eigenvectors is obvious, since a symmetric matrix has a set of n independent eigenvectors. 613
An Interval Containing the Eigenvalues of a Symmetric De nite Pencil The eigenvalues of the symmetric de nite pencil A ; B lie in the interal [;kB ;1 Ak; kB ;1 Ak]. (Exercise #26.)
Computational Algorithms 9.5.1 The QZ Method for the Symmetric-De nite Pencil The QZ algorithm described in the previous section for the regular pencil A ; B can, of course, be applied to a symmetric-de nite pencil. However, the drawback here is that both the symmetry and de niteness of the problem will be lost in general. We describe now a specialized algorithm for a symmetric-de nite pencil.
9.5.2 The Cholesky-QR Algorithm The proof of Theorem 9.5.1 immediately reveals a computational method for nding the generalized eigenvalues of a symmetric-de nite pencil:
614
Algorithm 9.5.1 The Cholesky-QR Algorithm For the Symmetric-De nite Pencil 1. Find the Cholesky factorization of B :
B = LLT : 2. Form C = L;1 A(LT );1, by taking advantage of the symmetry of A. 3. Compute the eigenvalues and the eigenvectors yi ; i = 1; : : :; n, of the symmetric matrix C , using the symmetric QR iteration. 4. Compute the generalized eigenvectors xi by solving: LT xi = yi , i = 1; : : :; n.
Stability of the Algorithm When B is well-conditioned, there is nothing objectionable in the algorithm. However, if B is ill-conditioned or nearly singular, so will be L;1 and then the matrix C cannot be computed accurately. Therefore, in this case, the eigenvalues and the eigenvectors will be inaccurate. Speci cally, it can be shown (Golub and Van Loan MC 1984, p. 310) that a computed eigenvalue ~ obtained by the algorithm is the exact eigenvalue of the matrix (L;1A(LT );1 + E ), where kE k2 kAk2 kB ;1 k2. Thus, ill-conditioning of B will severely a ect the computed eigenvalues,
even if they are themselves well-conditioned.
As we will see in the next section, in many practical applications only a few of the smallest generalized eigenvalues are of interest. These eigenvalues sometimes can be computed reasonably accurately, even when B is ill-conditioned, by using the ordered RSF of B in place of the Cholesky decomposition, as follows:
Algorithm 9.5.2 Computing the Smallest Eigenvalues with Accuracy 1. Compute the Real Schur Form of B and order the eigenvalues of B from smallest to largest:
U T BU = D = diag(d1; : : :; dn): 2. Form
p
p
L = UD; 12 = U diag( d1 ; : : :; dn);1 :
3. Form
C = L;1AT (LT );1
4. Compute the eigenvalues of C . 615
Example 9.5.1
Cond(B ) = 105: 1.
01 2 31 B C A=B @ 2 3 4 CA ;
01 0 01 B C B=B @ 0 :00001 0 CA :
3 4 5
0
0 :00001 0 0 1 B 0 1 0 CC ; U T BU = D = B @ A 0
2.
0
1
00 1 01 B 1 0 0 CC : U =B @ A
0 1
0 0 1
0 0 1 1 0 B C L = UD; = B @ 316:2278 0 0 CA : 1 2
0
3.
0 1
0 0:0003 0:0063 0:0126 1 B C C = L;1 AT (LT );1 = B 3 C @ 0:0063 1 A:
0:0126 4. The eigenvalues of C are: 0; ;0:6055; 6:6055.
3
5
5. The two smallest generalized eigenvalues are: 0; ;0:6665.
Remark: Two smallest generalized eigenvalues have been computed by the procedure rather
accurately. However, the largest one is completely wrong. This example shows how ill-conditioning of B can impair the occuracy of the computed eigenvalues.
9.5.3 Diagonalization of the Symmetric-De nite Pencil Simultaneous Diagonalization of A and B The Cholesky-QR iteration algorithm of the symmetric-de nite pencil gives us a method to nd a nonsingular matrix P that transforms A and B simultaneously to diagonal forms by congruence. This can be seen as follows: Let Q be an orthogonal matrix such that
QT CQ = diag(c1; : : :; cn): 616
Set P = (L;1 )T Q. Then
P T AP = QT L;1A(L;1)T Q = QT CQ = diag(c1; c2; : : :; cn); and
P T BP = QT L;1B (L;1)T Q = QT L;1LLT (L;1)T Q = I:
Algorithm 9.5.3 Simultaneous Diagonalization of a Symmetric De nite Pencil Given a symmetric matrix A and a symmetric positive de nite matrix B , the following algorithm computes a nonsingular matrix P such that P T BP is the identity matrix and P T AP is a diagonal matrix. 1. Compute the Cholesky factorization of B :
B = LLT : 2. Form C = L;1 A(LT );1, by taking advantage of the symmetry of A (C is symmetric). 3. Applying the symmetric QR iteration algorithm to C , nd an orthogonal matrix Q such that QT CQ is a diagonal matrix. 4. Form P = (L;1)T Q.
Flop Count: The algorithm for simultaneous diagonalization requires 7n3 ops. Example 9.5.2 Consider
01 2 31 B C A=B @ 2 3 4 CA
0 10 1 1 1 B C B=B @ 1 10 1 CA ;
1 1 10 A is symmetric and B is symmetric positive de nite.
3 4 5
1. The Cholesky decomposition of B = LLT : 1 0 3:1623 0 0 B :3162 3:1464 0 CC L=B A @ :3162 :2860 3:1334 617
2. Form C = L;1A(L;1 )T
0 0:1000 0:1910 :2752 1 B C C=B @ 0:1910 0:2636 0:3320 CA
3. Find an orthogonal Q such that
0:2752 0:3320 0:3864
QT CQ = diag( c1; : : :; cn) 0 :4220 ;0:8197 ;0:3873 1 B :5684 ;0:0936 0:8174 CC Q = B @ A 0:7063 0:5651 ;0:4262 4. Form
5. Verify:
0 :09409 ;0:2726 ;0:1361 1 B C P = (L;1)T Q = B @ :1601 ;0:0462 0:2722 CA 0:2254 0:1803 ;0:1361 P T AP = diag(:8179; ;0:0679; 0): P T BP = Identity = diag(1; 1; 1):
Modal Matrix In vibration problems involving the mass and sti ness matrices, that is, when K is the sti ness matrix and M is the mass matrix in the symmetric-de nite generalized eigenvalue problem:
Kx = Mx the matrix P that simultaneously diagonalize M and K , respectively, to the identity and a diagonal matrix, is called the modal matrix, and the columns of the matrix P are called the normal modes.
Orthogonality of the Eigenvectors Note that if p1; : : :; pn are the columns of P , then it is easy to see that
pTi Bpi = 1; i = 1; : : :; n pTi Bpj = 0; i = 6 j: and
pTi Api = ci ; i = 1; : : :; n pTi Apj = 0; i = 6 j: 618
Generalized Rayleigh-Quotient The Rayleigh-Quotient de ned for a symmetric matrix A in Chapter 8 can easily be generalized for the pair (A; B ).
De nition 9.5.1 The number
T Ax = xxT Bx (kxk2 = 1);
is called the generalized Rayleigh quotient.
It can be used to compute approximations to generalized eigenvalues k and eigenvectors xk for the symmetric de nite generalized eigenvalue problem, as shown in the following algorithm.
Algorithm 9.5.4 Generalized Rayleigh Quotient Iteration Choose x0 such that kx0 k = 1: For k = 0; 1; : : : do until convergence occurs
Axk : 1. Compute k = xxTk Bx T
2. Solve for x^k+1:
3. Normalize x^k+1 :
k
k
(A ; k B )^xk+1 = Bxk
xk+1 = kx^x^k+1k : k+1 2
9.6 Symmetric-De nite Generalized Eigenvalue Problems Arising in Vibrations of Structures As we stated earlier, almost all eigenvalue problems arising in vibrations of structures are the symmetric-de nite generalized eigenvalue problems. We will now give some speci c examples.
9.6.1 A Case Study on the Vibration of a Spring-Mass System Consider the system of three spring masses shown in Figure 9.1 with spring constants k1; k2 and k3. We would like to study the con guration of the system corresponding to the di erent modes of vibration.
619
k
1
m
1
y
1
k
2
m
2
y
2
k
3
m
3
y
Figure 9.1
3
3 Degrees of Freedom
1. Formulate the problem as a Symmetric De nite Generalized Eigenvalue Problem Let the displacements of the masses from static equilibrium positions be de ned by the generalized coordinates y1 ; y2; and y3 , respectively. The sti ness matrices are very often determined using physical characteristics of the systems. The generalized eigenvalues and eigenvectors are then used to study the behavior of the system. The equations of motion for the system can be written as:
m1y1 + (k1 + k2)y1 ; k2y2 = 0 620
m2y2 ; k2y1 + (k1 + k2)y2 ; k3y3 = 0 m3y3 ; k3y2 + k3y3 = 0 or, in matrix form,
0 m 0 0 1 0 y 1 0 k + k ;k 10y 1 001 0 1 1 1 2 2 1 B C B C B C B C B C B @ 0 m2 0 C AB @ y2 CA + B@ ;k2 k2 + k3 ;k3 CA B@ y2 CA = B@ 0 CA ; 0 0 m3 y3 0 ;k3 k3 y3 0
that is,
M y + Ky = 0;
where M = diag(m1 ; m2; m3); and
0 k + k ;k 1 0 1 2 2 B C K=B @ ;k2 k2 + k3 ;k3 CA : 0 ;k3 k3
Assuming harmonic motion, we can write
y1 = x1 ei!t y2 = x2 ei!t y3 = x3 ei!t where x1 ; x2 and x3 are, respectively, the amplitudes of the masses, m1 , m2 and m3, and ! denotes the natural angular frequency. Substituting these expressions for y1 ; y2 and y3 into the equations of motion, and noting that
yk = ;!2 xk ei!t; k = 1; 2; 3; we have
;m1x1!2 + (k1 + k2)x1 ; k2x2 = 0 ;m2x2!2 ; k2x1 + (k1 + k2)x2 ; k3x3 = 0 ;m3x3!2 ; k3x2 + k3x3 = 0 which can be written in matrix form as: 0m 0 0 10x 1 10x 1 0 k + k ;k 0 1 1 2 2 BB ;k k + k ;k CC BB x CC = !2 BB 01 m 0 CC BB x1 CC 2 A@ 2A @ @ 2 1 2 3A@ 2A x3 0 0 m3 x3 0 ;k3 k3 621
or
Kx = Mx
where = ! 2 . The eigenvalues i = !i2 ; i = 1; 2; 3 are the squares of the natural frequencies
of the rst, second, and third modes of vibration, respectively. 2. Solve the Generalized Eigenvalue Problem Using the Cholesky-QR Algorithm
Let's take m1 = 20000; m2 = 30000; m3 = 40000; k1 = k2 = k3 = 109 1:5. Then
M = diag(20000 ; 40000)1 0 3 ; 30000 ;1:5 0 BB C 9 K = 10 @ ;1:5 3 ;1:5 C A 0 ;1:5 1:5 0 141:4214 1 0 0 B C L = B 173:2051 0 C @ 0 A 0 0 200:000 0 1:5 ;:6124 0 1 B C C = 105 B @ ;:6124 1 ;:4330 CA 0 ;:4330 :3750 The generalized eigenvalues = the eigenvalues of C are: 105(1:9508; :8382; :0860). The natural frequencies are: 102 (4:4168; 2:8951; 0:9273). The generalized eigenvectors are: 0 0:0056 ;0:0040 0:0017 1 BB ;0:0034 ;0:0035 0:0031 CC : @ A 0:0008 0:0028 0:0040
These eigenvectors can be used to determine di erent con gurations of the system for di erent modes. That is, they can be used to see how the structures vibrates under di erent modes. 9.6.2 A Case Study on the Vibration of a Building Consider a four-story reinforced concrete building as shown in the gure below. The oors and roofs, which are fairly rigid are represented by lumped masses m1 to m4 having a horizontal motion caused by shear deformation of columns, and k1 to k4 are equivalent spring constants of columns that act as springs in parallel. 622
m
4
y 4
k
4
m
3
y 3
k
3
m
2
y 2
k
2
m
1
y 1
k
1
Figure 9.2 Schematic of 4-story building
We would like to study the con guration when the building is vibrating in its rst two modes (corresponding to two smallest eigenvalues). 1. Formulate the problem as a symmetric-de nite generalized eigenvalue problem in terms of mass and sti ness matrices:
Kx = Mx: As in the previous example, M is diagonal and K is tridiagonal:
M = diag(m1; m2; m3; m4) 0 k + k ;k 1 0 0 1 2 2 BB ;k k + k ;k C 0 C 2 2 3 3 B CC : K = B B@ 0 ;k3 k3 + k4 ;k4 C A 0 0 ;k4 k4 Taking m1 = 5 107; m2 = 4 107; m3 = 3 107; m4 = 2 107, and k1 = 10 1014; k2 = 623
8 1014, k3 = 6 1014; k4 = 4 1014, we have 0 18 ;8 0 0 1 BB ;8 14 ;6 0 CC CC K = 1014 B BB @ 0 ;6 10 ;4 CA 0 0 ;4 4 and M = 107diag(5; 4; 3; 2): 2. Find the generalized eigenvalues and eigenvectors using the Cholesky-QR algorithm. 0 7:0711 0 1 0 0 BB 0 6:3246 0 CC 0 B CC L = 103 B B@ 0 0 5:4772 0 C A 0 0 0 4:4721 0 3:6 ;1:7889 1 0 0 BB C ;1:7889 3:5000 ;1:7321 0 C B CC : 7 C = 10 B B@ 0 ;1:7321 3:3333 ;1:6330 C A 0 0 ;1:6330 2:000 The eigenvalues of C which are the generalized eigenvalues, are: 0 6:1432 1 BB 1:8516 CC CC : 7 10 B BB @ 0:3435 CA 4:0950 The corresponding generalized eigenvectors are: 0 :0666 1 0 ;0:0785 1 0 0:0370 1 0 0:0896 1 B BB ;0:0858 CC BB 0:0753 CC BB ;0:02777 CC B C ;0:1058 C B C B C B C CC : ; 3 ; 3 ; 3 ; 3 10 B ; 10 B C CC ; 10 BB CC ; 10 BBB B C B CA 0 : 0977 0 : 0104 0 : 1091 ; 0 : 1085 @ A @ A @ A @ ;0:0472 0:1403 0:1318 0:1036 The eigenvectors corresponding to the two smallest eigenvalues 107(:3435)and 107(1:8516) are: 0 0:0370 1 0 ;0:0785 1 BB 0:0753 CC BB ;0:0858 CC B C B C ; 3 ; 3 10 B B@ 0:1091 CCA ; 10 BB@ 0:0104 CCA : 0:1318 0:1403 The rst two modes of vibration are shown in Figures 9.3 and 9.4, respectively. 624
m
x
x
4
4
k
m
x
x
k
m
4
3
x 3
k
m2
2
m
k
k
m
3
2
k
m
2
1
1
k
x
1
Figure 9.3 First Mode of Vibration of 4-Story Building
3
2
x x1
4
4
3
3
m 4
1
2
k
1
Figure 9.4 Second Mode of Vibration of 4-Story Building
9.6.3 Forced Harmonic Vibration In the previous two examples, we considered vibration of systems without any external force. Consider now a system with two degrees of freedom excited by a harmonic force F1 sin !t, as shown in Figure 9.5.
625
k
F1 sin ω t m
1
y
1
k
m2
y2
k
Figure 9.5 Forced vibration of a 2 degree of freedom system
Then the di erential equations of motion for the system become
m1y1 = ;k(y1 ; y2 ) ; ky1 + F1 sin !t m2y2 = k(y1 ; y2 ) ; ky2 or
"m
# ( y ) " 2k ;k # ( y ) ( F ) 1 1 + = 1 sin !t: 0 m2 y2 ;k 2k y 2 0 Assuming the solution to be ! x! y 1
0
1
y2
=
and substituting into the equations we get
1
x2
(9.6.1)
sin !t
(2k ; m1! 2 )x1 ; kx2 = F1 ;kx1 + (2k ; m2!2)x2 = 0: In matrix form,
" (2k ; m !2) ;k
1
;k # ( x1 ) ( F1 ) = : (2k ; m2 ! 2) x2 0
These equations can be easily solved to obtain explicit expressions for the amplitudes x1 and x2 : 2 x1 = m m(2(!k2;;m!22!)(!)F2 1; !2) (9.6.2) 1 2 1 2 (9.6.3) x2 = m m (! 2 ;kF!12)(!2 ; !2) 1 2
1
626
2
where !1 and !2 are the modal frequencies. For the special case when m1 = m2 = m, !1 and !2 are given by s s and x1 and x2 are given by
!1 = mk
and
!2 = 3 mk
)F1 x1 = m2(!(22k;;!m! 2 )(! 2 ; ! 2 ) ; 1 2 (9.6.4) kF 1 x2 = m2(!2 ; ! 2)(! 2 ; ! 2) : 1 2 From above, it follows immediately that whenever ! is equal to or close to !1 or !2 , the amplitude 2
becomes arbitrarily large, signaling the occurrence of resonance. Note that in this case, the denominator is zero or close to it. This situation is very alarming to an engineer.
In other words, if the frequency of the imposed periodic force is equal or nearly equal to one of the natural frequencies of a system, resonance results, a situation which is quite alarming. Example 9.6.1 To demonstrate the consequences of excitation at or near resonance, let us consider an airplane landing on a rough runway. The fuselage and engine are assumed to have a combined mass m1. The wings are modeled by lumped masses m2 and m3 , and sti ness k2 and k3 ; k1 represents the combined sti ness of the landing gear and tires. The masses of the wheels and landing gear are assumed negligible compared to the mass of the fuselage and the wings. The system is modeled in Figure 9.6.
627
y
v
y0 l = 20 m Figure 9.6 Airplane landing on a runway
The runway is modeled by a sinusoidal curve as shown in the Figure 9.6. Let the contour be described by y = y0 sin !t, and let the airplane be subjected to a forcing input of f1 sin !t, where f1 = k1 y0. The equations of motion for the 3 degree of freedom system so described are given by: 9 2 k + k + k ;k ;k 3 8 y 9 8 f sin !t 9 2m 0 0 38 y > > > > > > 1 1 = 66 77 < = 66 1 2 3 2 3 77 < 1 = < 1 0 m 0 y + ; k k 0 y = 0 : 2 2 2 2 2 4 5> 4 5> > > > > : ; : ; : ; 0 0 m3 y3 ;k3 0 k3 y3 0 The airplane is shown schematically in Figure 9.7, where m1 represents the mass of the fuselage, m2 and m3 represent the lumped masses of the wings. The combined sti ness of the landing gear and tires is denoted by k1, and k2 and k3 are the sti nesses of the wings. Finally, y1 , y2 , and y3 represent motion relative to the ground.
628
y
1
y3
y m
k3
k m1
3
k
2
2
m2
1
y
Figure 9.7 Model of the airplane
Let
k1 k3 = k2 m1 m2
= = = =
1:7 105 N/m 6 105 N/m 1300kg m3 = 300kg:
The natural frequencies obtained by solving the generalized eigenvalue problem: Kx = Mx = ! 2 Mx are given by:
!1 = 9:39 rad/sec !2 = 44:72 rad/sec !3 = 54:46 rad/sec: The forcing frequency ! is related to the landing velocity v by v = !`=2 . So if ! = !1 , v = !21` = 9:39 220m/sec = 29:8m/sec = 107km/hr: Thus, if the landing velocity is 107 km/hr, or close to it, there is danger of excitation at or near the resonance.
9.7 Applications to Decoupling and Model Reduction We have already seen how generalized eigenvalue problems arise in applications. In this section we will mention a few more engineering applications. These include decoupling of a second-order system of di erential equations, and model reduction. 629
9.7.1 Decoupling of a Second-Order System Case 1. The Undamped System As we have seen before, vibration problems in structural analysis give rise to a homogeneous system of di erential equations of second order of the form:
M y + Ky = 0; where
(9.7.1)
y = ddty2 2
0y 1 BB y1 CC y=B BB ..2 CCC @.A
and
yn
is an n-vector. The matrices M and K are, as usual, the mass and sti ness matrices. Assuming that these matrices are symmetric and M is positive de nite, we will now show how the simultaneous diagonalization technique described earlier can be pro tably employed to solve this system of second-order equations. The idea is to decouple the system into n uncoupled equations so that each of these uncoupled equations can be solved using a standard technique. Let P be the modal matrix such that:
P T MP = I; P T KP = = diag(!12; : : :; !n2 ): Let y = Pz , so the homogeneous system
M y + Ky = 0 becomes Next premultiplying by P T , or
MP z + KPz = 0: P T MP z + P T KPz = 0 z + z = 0:
The homogeneous system therefore decouples:
zi + !i2zi = 0;
i = 1; 2; : : :; n: 630
(9.7.2) (9.7.3)
or
z1 + !12z1 = 0 z2 + !22z2 = 0 .. .
where
;
(9.7.4)
zn + !n2 zn = 0
0z 1 1 B C B z2 C B z=B CC ; = diag(!12; !22; : : :; !n2 ): .. C B @.A zn
The solution of the original system now can be obtained by solving these n decoupled equations using standard techniques and then recovering the original solution y from
y = Pz: Thus, if the solutions of the transformed system (9.7.4) are given by
zi = Ai cos !i t + Bi sin !it; i = 1; 2; ; n; then the solutions of the original system (9.7.1) are given by
0 y 1 2 A cos ! t + B sin ! t 1 1 66 1 1 BB y1 C BB .2 C C = P 66 A2 cos !2t + B2 sin !2t 6. B@ .. C C A 64 .. yn
An cos !nt + Bn sin !n t
3 77 77 77 : 5
(9.7.5)
The constants Ai and Bi are to be determined from the initial conditions such as:
yijt=0 = displacement at time t = 0 y_ijt=0 = initial velocity.
Example 9.7.1 Consider the system with three masses in the example of the Section 9.6.1. Suppose that the system, when released from rest at t = 0, is subjected to the displacement as given below: We would like to nd the undamped time response of the system. The initial conditions are:
y1 = 1 y2 = 2 y3 = 3
3 77 y_1 = 0 75 y_2 = 0 : y_3 = 0
631
Since the initial velocities are zeros, we obtain
y_1 = PB1 !1 = 0 y_2 = PB2 !2 = 0 y_3 = PB3 !3 = 0: These give B1 = B2 = B3 = 0. Again, at t = 0, we have
Recall that for this problem
0y 1 0A 1 011 1 B CC BB 1 CC BB CC B @ y2 A = P @ A2 A = @ 2 A : y3
A3
3
0 0:0056 ;0:0040 0:0017 1 B ;0:0034 ;0:0035 0:0031 CC : P =B @ A
0:0008 Then the solution of the linear system
0:0028 0:0040
0A 1 011 BB 1 CC BB CC P @ A2 A = @ 2 A A3
3
is
A1 = 2:5154 A2 = 55:5358 A3 = 710:6218: Substituting these values of A1 ; A2; A3 and the values of !1 ; !2; and !3 obtained earlier in
y = Pz or
0 y 1 0 z 1 0 A cos ! t 1 1 1 1 1 B C B C B C B @ y2 C A = P B@ z2 CA = P B@ A2 cos !2t CA ; y3
z3
A3 cos !3 t
we obtain the values of y1 ; y2; and y3 that give the undamped time response of the systems subject to the given initial conditions.
632
Case 2. The Damped System Space Some damping, such as that due to air resistance, uid and solid friction, etc., is present in all real systems. Let us now consider damped homogeneous systems. Let D be the damping matrix. Then the equations of motion of the damped system becomes
M y + Dy_ + Ky = 0:
(9.7.6)
Assume that D is a linear combination of M and K ; that is
D = M + K;
(9.7.7)
where and are constants. Damping of this type is called proportional or Rayleigh damping. Let P be the modal matrix. Then we have
P T DP = P T MP + P T KP = I + : Let z = P T y . Then the above homogeneous damped equations are transformed to n decoupled equations: zi + ( + !i2)z_i + !i2 zi = 0; i = 1; 2; : : :; n: (9.7.8)
In engineering practice it is customary to assume modal damping. In modal damping, the damping is proportional to mass or sti ness or a combination of them in a certain special manner. Let and be chosen so that + !i2 = 2i !i: i is called the modal damping ratio of the ith mode. i is usually taken as a small number between 0 and 1. The most common values are 0 0:05. (See Inman (1994), pp. 196.) However, in some applications, such as in the design of exible structures, fig are taken to be as low as 0.005. On the other hand, for an automatic shock absorber, a value as high as = 0:5 is possible. Assuming modal damping, the decoupled equations (9.7.8) become:
zi + 2i!i z_i + !i2 zi = 0; i = 1; 2; : : :; n: The solutions of these equations are then given by
zi = e;i !i t
Ai cos !i
q
1 ; 2t + B i
i sin !i
q
1 ; 2t i
;
i = 1; 2; : : :; n;
where the constants Ai and Bi are to be determined from the given initial equations. The original system can now be solved by solving these n uncoupled equations separately, and then recovering the original solution y from y = Pz . 633
Example 9.7.2 Usefulness of Proportional Damping Consider the following system with two degrees of freedom (DOF). We will show here why
proportional damping is useful.
y
y
1
k
2
2k
k m
2m
d
d
1
d
2
3
The equations of motion of the system are developed by considering a free body diagram for each mass. For mass m:
d1y_1 ky1
m
-
d2(y_2 ; y_1 ) k(y2 ; y1 )
)
-
my1
The equation of motion for this mass is:
;d1y_1 + d2(y_2 ; y_1) ; ky1 + k(y2 ; y1) = my1 or
my1 + (d1 + d2)y_ 1 ; d2y_2 + 2ky1 ; ky2 = 0:
634
For mass 2m:
d2(y_2 ; y_1 ) k(y2 ; y1 )
2m
)
d3y_2 2ky2
-
2my2
The equation of motion for this mass is:
;d2(y_2 ; y_1) ; d3y_2 ; k(y2 ; y1) ; 2ky2 = 2my2 or
2my2 ; d2y_1 + (d2 + d3 )y_ 2 ; ky1 + 3ky2 = 0:
Thus, for the whole system we have " m 0 # ( y ) " d + d ;d # ( y_ ) " 2k ;k # ( y ) ( 0 ) 1 2 1 1 + 1 2 + = : 0 2m y2 ;d2 d2 + d3 y_2 ;k 3k y2 0 Let k = 2 and m = 5, and assume that D = M + K: Consider now two cases. For this 2 DOF system, the mass matrix "m 0 # "5 0 # M= = ; 0 2m 0 10 the sti ness matrix " 2k ;k # " 4 ;2 # K= = ; ;k 3k ;2 6 and the damping matrix " d + d ;d # 2 D= 1 2 : ;d2 d2 + d3
Case 1: d1 = 1; d2 = 1; d3 = 2 " 2 ;1 # " 5 0 # " 4 ;2 # = + : ;1 3 0 10 ;2 6 2 = 5 + 4 ;1 = ;2 3 = 10 + 6
(9.7.9) (9.7.10) (9.7.11)
Equations (9.7.9), (9.7.10) and (9.7.11) are uniquely satis ed with = 1 and = 0. So, this is a 2 case of proportional damping. 635
Case 2: d1 = 2; d2 = 4; d3 = 1 " 6 ;4 # " 5 0 # " 4 ;2 # = + ;4 5 0 10 ;2 6 6 = 5 + 4 ;4 = ;2 5 = 10 + 6 :
(9.7.12) (9.7.13) (9.7.14)
Now equations (9.7.12), (9.7.13) and (9.7.14) cannot all be satis ed with any set of values of and . This is a case of nonproportional damping. In the rst case, the equations of motion can be decoupled using y = Pz, obtaining
real mode shapes and real natural frequencies. However, in the second case, such decoupling is not possible. This type of damping will lead to complex natural frequencies and mode shapes. Damped Systems Under Force Excitation When a damped system is subject to an external force F , the equations of motion are given by 8 F (t) 9 > 1 > > > > < F2(t) > = M y + Dy_ + Ky = F (t) = > .. > : (9.7.15) . > > > : Fn(t) > ; Assuming that M is symmetric positive de nite, K is symmetric, and damping is proportional, it is easy to see from our previous discussion that the above equations can be decoupled using simultaneous diagonalization. Let P = (pij ) be the modal matrix. Then
0p p 11 21 B B p12 p22 B B .. .. B . . B T P F =B B p1i p2i B B . . B @ .. .. p1n p2n
pn1 1 0 F1 1 C BB F CC pn2 C CB 2 C
B.C .. C . C CC BBB .. CCC : BF C pni C CC BBB ..i CCC .. C . A@ . A
pnn
Fn
The decoupled equations will then be given by
zi + 2i!i z_i + !i2 zi = p1iF1 + p2iF2 + + pniFn ; 636
or where Ei(t) =
n X j =1
zi + 2i!i z_i + !i2zi = Ei(t);
(9.7.16)
pji Fj ; i = 1; 2; : : :; n:
The function Ei (t) is called the exciting or forcing function of the ith mode. If each force Fi is written as Fi = fis(t); then
Ei(t) = s(t)
De nition 9.7.1 The expression
n X j =1
n X j =i
pjifj :
pjifi
is called the mode participation factor for the ith mode. Once the decoupled equations are solved for zi we can recover the solutions of the original equations from: y = Pz or
0y 1 0z 1 BB y1 CC BB z1 CC BB .2 CC = P BB .2 CC : B@ .. CA B@ .. CA
The solutions of the decoupled equations
yn
zn
zi + 2i !i z_i + !i2zi = Ei(t) depend upon the nature of the force F (t). For example, when the force is a shock type force, such as an earthquake, etc., one is normally interested in maximum responses, and the maximum values of z1; z2; : : :; zn can be obtained from the responses of a single equation of one degree of freedom (see the example below).
9.7.2 The Reduction of a Large Model Many applications give rise to a very large system of second order di erential equations: M y + Dy_ + Ky = F (y; t): 637
(9.7.17)
For example, the large space structure (LSS) is a distributed parameter system. It is therefore in nite dimensional in theory, but of very large dimension in practice. A nite element generated reduced order model can be a large second order system. The dimension of the problem in such an application can be several thousand. Naturally, the solution of a large system will lead to a solution of a very large generalized eigenvalue problem. Unfortunately, e ective numerical
techniques for computing a large number of generalized eigenvalues and eigenvectors are virtually nonexistent and not very well developed (see the section on the Generalized Symmetric Eigenvalue Problem for Large and Structure Matrices). It is therefore natural to think of solving the problem by constructing a reduced-order model with the help of a few eigenvalues and eigenvectors which are feasible to compute. Such a thought is based on an assumption that in many instances the response of the structure depends mainly on a rst couple of eigenvalues (low frequencies); usually the higher modes do not get excited. We will now show how the computations can be simpli ed by using only the knowledge of a few eigenvalues and eigenvectors. Suppose that, under the usual assumption that M and K are symmetric and of order n and M is positive de nite, we were able to compute only the rst few normal modes, perhaps m of them where m n. Let the matrix of these normal modes be Pnm : Then
P T MP = Imm P T KP = mm = diag(!12; : : :; !m2 ): Setting y = Pz and assuming that the damping is proportional to mass or sti ness, the system of n di erential equations (9.7.17) then reduces to m equations:
zi + 2i!i z_i + !i2zi = Ei(t); i = 1; 2; : : :; m; where Ei is the ith coordinate of the vector P T F . Once this small number of equations is solved, the displacement of any masses under the external force can be computed from:
yi = Pzi;
i = 1; : : :; n:
Sometimes only the maximum value of the displacement is of interest. Several vibration groups in industry and the military use the following approximation to obtain the maximum value of yi (see Thompson (1988)):
v uX m t jpj zj(max)j2; jyijmax = jp1z1(max)j + u j =2
where pi is the ith column of P evaluated at zi . 638
9.7.3 A Case Study on the Potential Damage of a Building Due to an Earthquake Suppose we are interested in nding the absolute maximum responses of the four-story building considered in the example of section 9.6.2, when the building will be subjected to a strong earthquake, using the known response of the building due to a previous earthquake. We will use only
the rst two modes (the modes corresponding to the two lowest frequencies) in our calculations. In the gure, the yi ; i = 1; : : :; 4 denote the displacement of each oor relative to the moving support, whose motion is denoted by y0 . y k
4
4
y k
3
3
y k
2
y k
2
1
1
y
0
Figure 9.8 Building subject to an earthquake
The decoupled normal mode equations in modal form in this case can be written as:
zi + 2i!i z_i + !i2zi = ;Eiy0; i = 1; 2; where
Ei =
4 X
j =1
y0 = absolute acceleration of the moving support, pjimj = mode participation factor of the chosen mode pi due to support existence. 639
pji are the coordinates of the participating mode Pi, that is, 20 p 1 0 p 13 66BB p11 CC BB p12 CC77 P = (p1; p2) = 666B BB 21 CCC ; BBB 22 CCC777 ; 4@ p31 A @ p32 A5 p41 p42 where p1 and p2 are the two chosen participating modes. Let R1 and R2 denote the maximum relative responses of z1(max) and z2(max) obtained from previous experience. Then we can take
z1(max) = E1R1 z2(max) = E2R2: This observation immediately gives:
0 BB y1 BB y2 BB y3 @ y4
1 CC CC CC A
0 1 0 p BB 11 CC BB p12 B p21 CC B p22 = E1R1 B BB p31 CC + E2R2 BBB p32 @ A @
max
p41
p42
1 CC CC CC : A
Using now the data of the example in section 9.6.2, we have
m1 = 107(5);
m2 = 107(4);
m3 = 107(3);
m4 = 107(2):
p1 = the eigenvector corresponding to the smallest eigenvalue 0 0:0370 1 BB CC 0 : 0753 B C = 10;3 B B@ 0:1091 CCA 0:1318
p2 = the eigenvector corresponding to the second smallest eigenvalue 0 ;0:0785 1 BB 0:0858 CC CC ; 3 = 10 B BB CA 0 : 0104 @ 0:1403 E1 = m1p11 + m2p21 + m3 p31 + m4p41 = 104(1:0772) E2 = m1p12 + m2 p22 + m3p32 + m4 p42 = 103(;4:2417) 640
Assume that R1 = 1:5 inches, R2 = :25 inches. We obtain
0y 1 BB y1 CC BB 2 CC B@ y3 CA y4
= max
+
=
=
0 0:0370 1 BB 0:0753 CC CC 10;3:104(1:0772)(1:5) B BB @ 0:1091 CA 00:;1318 1 0:0785 BB C 0:0858 C B C ; 3 3 10 :10 (;4:2417)(:25) B B@ 0:0104 CCA 0 0:5979 1 0 0:0833 1 0:1403 BB 1:2169 CC BB 0:0910 CC BB C B C B@ 1:7636 CCA + BB@ ;0:0110 CCA 0 20::1293 1 ;0:1488 6812 in. BB 1:3079 in. CC BB C B@ 1:7526 in. CCA
1:9805 in.: Thus, the absolute maximum displacement (relative to the moving support) of the rst oor is 0.6812 in., that of the second oor is 1.3079 in., etc. The absolute maximum relative displacements are obtained by adding the terms using their absolute values: 0y 1 0 0:6812 in. 1 1 B C BB 1:3079 in. CC B y2 C B C CC =B B C BB B C @ y3 A @ 1:7747 in. CA y4 (abs. max.) 2:2781 in.
641
m
y4
4
k
m
y3
4
3
k
m2
y2
m
y1
3
k
2
1
k
1
Figure 9.9 Absolute maximum displacement
Note: The contribution to the second participating mode to responses is small in comparison with the contribution of the rst mode.
Average Maximum Relative Displacement of the Masses The absolute maximum relative displacement of the masses provides us with an upper bound for the largest relative displacements the masses can have, and thus help us to choose design parameters. Another practice for such a measure in engineering literature has been to use the root sum square of the same terms, giving the \average" maximum relative displacement values:
q
(yi)average max = (E1R1pi1 )2 + (E2R2pi2 )2 + + (Ek Rk Pik )2 For the above example, k = 2 and the average maximum relative displacements are given by (y1 )average max = (y2 )average max =
q
(E1R1p11)2 + (E2R2 p12)2 = :9975 inches;
q
(E1R1p21)2 + (E2R2 p22)2 = 1:5610 inches;
and so on.
642
9.8 The Quadratic Eigenvalue Problem So far we have considered the second-order system:
M y + Dy_ + Ky = 0
(9.8.1)
under the assumption of proportional or modal damping. This assumption let us use the simultaneous diagonalization on M; K; and D (see Section 9.7.1, Case 2), which, of course, as we have seen in Section 9.5, amounts to solving a generalized eigenvalue problem. In case of general damping, we will, however, have the quadratic eigenvalue problem: (2M + D + K )x = 0: This problem has 2n eigenvalues and there are 2n eigenvectors corresponding to them. We describe two existing approaches to solve this quadratic eigenvalue problem and show how to extract frequencies and from these eigenvalues and eigenvectors. We assume, as before, M is symmetric positive de nite.
Approach 1: Reduction to a standard Eigenvalue Problem. Multiplying both sides of the equation (9.8.1). by M ;1 we have
y + M ;1 Dy_ + M ;1 Ky = 0:
(9.8.2)
y_ = z_1 = z2
(9.8.3)
y = z_2 = ;M ;1 Dy_ ; M ;1 Ky (from 9.8.2) = ;M ;1 Dz2 ; M ;1 Kz1 (from 9.8.3)
(9.8.4)
Write Then
The equations (9.8.3) and (9.8.4) can now be combined into the single matrix equation: 0 I ! z1 ! z_1 ! = ;M ;1K ;M ;1D z2 z_2 That is z_ = Az where
!
(9.8.5) (9.8.6)
!
z 0 I z = 1 ; and A = : (9.8.7) z2 ;M ;1K ;M ;1D Assuming a solution of the equation (9.8.6) of the form z = xet , we have the standard eigenvalue problem:
Ax = x; 643
(9.8.8)
where A is 2n 2n as given by 9.8.7. There are 2n eigenvalues i ; i = 1; :::; 2n of the above problem and 2n corresponding eigenvectors. If the second-order system is the model of a vibrating structure, then the natural frequencies and the modal damping ratios can be computed from the eigenvalues and eigenvectors as follows.
Computing Frequencies and Damping Ratios Let k = k + i k , that is, k and k are, respectively, the real and imaginary part of the complex eigenvalue k . Then the natural frequencies !k are given
q
!k = 2k + k2; k = 1; 2; :::; n:
(9.8.9)
(See Inman (1994), pp. 462). Note that the eigenvalues k occur in complexconjugate pairs. The modal damping ratios are given by:
k = p ;2 k 2 : k + k
(9.8.10)
Approach 2: Reduction to a Symmetric Generalized Eigenvalue Problem Let's substitute z1 = y and z2 = y_ into
M y + Dy_ + Ky = 0; then
M z_2 = ;Dz2 ; Kz1:
Again, since
(9.8.11)
z_1 = y_ = z2
we can write
; K z_1 = ;Kz2
(9.8.12)
Combining (9.8.11) and (9.8.12) we get
! ! ;K 0 ! z_1 ! 0 ;K z1 = 0 M z_2 ;K ;D z2 644
(9.8.13)
That is
B z_ = Az
(9.8.14)
! ;K ! ;K 0 ! z1 A= ; B= z= (9.8.15) ;K ;D 0 M z2 Assuming again a solution of (9.8.14) of the form z = xet, the equation (9.8.14) yields the generalized eigenvalue problem: Ax = Bx; (9.8.16) where A and B are both symmetric and are given by (9.8.15). Since A and B are both 2n 2n, the solution of (9.8.16) will give 2n eigenvalues and eigenvectors. The natural frequencies and modal damping ratios can be extracted from the eigenvalue as in Approach 1. where
0
Summary: The quadratic eigenvalue problem (2M + D + K )x = 0 can be either reduced to the standard nonsymmetric eigenvalue problem:
Ax = x where
I A= ; 1 ;M K ;M ;1D 0
!
or to the generalized symmetric eigenvalue problem:
Ax = Bx where
;K ! ;K 0 ! A= ; B= : ;K ;D 0 M 0
Once the eigenvalues and eigenvectors are computed, the natural frequencies and the modal damping ratios of a vibrating structure whose mathematical model is the second-order system (9.8.1) can be computed using (9.8.9) and (9.8.10), respectively. A Word of Caution About Using Approach 1 If M is ill-conditioned, the explicit formation of the matrix A given by (9.8.7) by actually computing M ;1 will be computationally disastrous. A will be computed inaccurately and so will be the eigenvalues and eigenvectors. Also note that A is nonsymmetric. 645
9.9 The Generalized Symmetric Eigenvalue Problems for Large and Structured Matrices As we have seen in several previous case studies, in many practical situations the matrices A and B are structured: tridiagonal and banded cases are quite common. Unfortunately, the Cholesky algorithm when applied to such structured problems, will very often destroy the sparsity. Even though A and B are banded, the matrix C = L;1 A(LT );1 will in general be full. Thus,
simultaneous diagonalization techniques is not practical for large and sparse matrices. 9.9.1 The Sturm Sequence Method for Tridiagonal A and B We now present a method when A and B are both symmetric tridiagonal and B is positive de nite. The method is a generalization of the Sturm-sequence algorithm for the symmetric eigenvalue problem described earlier in Chapter 8, and takes advantage of the tridiagonal forms of A and B . Let 0 0 1 0 0 0 0 1 1 1 BB . . . . . . CC BB 10 . .1. . . . CC 1 0 C 1 B B C A=B B@ ... . . . . . . n;1 CCA ; B = BB@ ... . . . . . . n0 ;1 CCA : 0 n;1 n 0 n0 ;1 0n De ne the sequence of polynomials fp()g given by:
p0() = 1; p1() = 1 ; 01; pr () = ( r ; r 0r )pr;1() ; ( r ; r0 )2pr;2 (); r = 2; 3; : : :; n:
(9.9.1) (9.9.2) (9.9.3)
Then it can be shown (exercise #22) that these polynomials form a Sturm sequence. The generalized eigenvalues of the pencil (A ; B ) are then given by the zeros of pn (). The zeros can be found by bisection or any other suitable root- nding method for polynomials. For a proof of the algorithm, see Wilkinson AEP, pp. 340-341.
Example 9.9.1
01 0 01 01 1 0 1 B 0 2 0 CC ; B = BB 1 4 1 CC : A=B @ A @ A 0 0 3
0 1 5
p0() = 1; p1() = 1 ; 646
p2() = (2 ; 4)(1 ; 2) ; (0 ; )2 1 = 32 ; 6 + 2 p3() = (3 ; 5)(32 ; 6 + 2) ; (;)2(1 ; ) = ;143 + 382 ; 28 + 6: The roots of p3() = 0 are: 1.6708, 0.6472, and 0.3964. It can be easily checked that these are the generalized eigenvalues of the pencil (A ; B ).
Computing a Generalized Eigenvector Once a generalized eigenvalue is computed, the corresponding eigenvector can be computed using the inverse iteration by taking full advantage of the tridiagonal forms of A and B as follows: Let y0 be the initial approximation of an eigenvector corresponding to a computed eigenvalue . Then For i = 0; 1; 2; : : :; do until convergence occurs. 1. Solve for xi+1 : (A ; B )xi+1 = yi 2. Form yi+1 = Bxi+1 :
Remarks: 1. About two iterations per eigenvector are adequate. 2. Note that the system (A ; B )xi+1 = yi can be solved by taking advantage of the tridiagonal forms of A and B .
9.9.2 The Lanczos Algorithm for the Pencil A ; B In this section, we shall show how the symmetric Lanczos method described in Chapter 8 for a matrix A can be extended to the symmetric de nite generalized eigenvalue problem Ax = Bx.
The Lanczos method to be described here is useful for estimating a few extremal eigenvalues of a large and sparse symmetric de nite generalized eigenvalue problem. Let's repeat that large symmetric de nite generalized eigenvalue problems are common in practice and like most practical problems, they are sparse. The Lanczos method, which relies on matrix-vector multiplications only, is suitable to preserve sparsity. Let B = LLT be the Cholesky factorization of B . 647
Algorithm 9.9.1 The Generalized Symmetric Lanczos Algorithm Given a symmetric matrix A and a symmetric positive de nite matrix B , the following generalized Lanczos algorithm constructs a symmetric tridiagonal matrix
0 BB 1 Tj = B BB ..1 @.
1 CC ... ... CC .. .. . . j ;1 C A 1
j ;1 j
and an orthonormal matrix Vj = fv1 ; v2; : : :; vj g such that
VjT AVj = Tj ; VjT BVj = Ijj : 1. Choose v1 such that v1T Bv1 = 1. 2. Set 0 = 1; r0 = v1; v0 = 0: 3.
For i = 1; 2; : : :; j do Solve: Bvi = ri;1= i;1
i = viT (Avi ; Bvi;1 ) ri = Avi ; i Bvi ; i;1Bvi;1 i = kL;1rik2 :
NOTES: 1. The Lanczos vectors in this case are B-orthonormal:
viT Bvi = 1 viT Bvj = 0; i 6= j: 2. For j = n:
V T BV = Inn V T AV = Tnn; a symmetric tridiagonal matrix. 3. If (; s) is an eigenpair of Tj , then (; Vj s) is the corresponding Ritz pair for the pencil (A; B ). 648
Computational Remarks Note that L;1ri can be computed by solving the lower triangular system:
Lyi = ri : Also, is equivalent to
Bvi = ri;1= i;1 LT vi = L;1 (ri;1= i;1);
since B = LLT .
Thus, to implement the scheme, we need (i) a Cholesky-factorization routine for a sparse matrix, (ii) routines for sparse triangular systems, and (iii) a routine for matrixvector multiplication. 9.9.3 Estimating the Generalized Eigenvalues Fact: For large enough j the extreme eigenvalues of the tridiagonal matrices 0 1 1 2 BB . . . . . . CC Tj = B BB ..2 . . . . . . CCC j A @. j j will approximate the generalized eigenvalues of the pencil (A ; B ).
Example 9.9.2 01 2 31 B 2 3 4 CC A = B A @
3 4 5 B = diag(1; 2; 3)
L = diag(1; 1:4142; 1:7321) 0 = 1; v1 = (1; 0; 0)T ; r0 = v1: (Note that v1T Bv1 = 1).
649
i=1: 1 = v01T (Av 1 1 ; Bv0) = 1 0 B 2 CC r1 = B @ A 3
1 = 2:2361:
i=2:
0 0 1 B 0:4472 CC v2 = B @ A
0:4472 2 = v2T (Av2 ; !Bv1) = 3:2000 ! 1 1 1 2:2361 T2 = = : 1 2 2:2361 3:2000 The eigenvalues of T2 are ;0:3920; 4:5919. The generalized eigenvalues of (A; B ) are: 4.6013, ;0:4347, 0. (Note that the largest eigenvalue of T2, 4.5919, is a reasonable approximation of the largest generalized eigenvalue 4.6013 of the pencil (A ; B).)
i=3:
0 0 B ;:5477 v3 = B @
1 CC A
= 0:3651 3 = v3T (Av3 ; Bv2 ) = ;0:0333: Thus,
0 0 1 0 1 2:2361 1 0 1 1 B CC = BB C T3 = T = B @ 1 2 2 A @ 2:2361 3:2000 0:2449 CA : 0 2 3 0 0:2449 ;0:0333 The eigenvalues of T are: ;0:4347; 0; 4:6013. The eigenpairs of T2 are: !! !! 0 : 8489 0 : 5285 ;0:3920; ;0:5285 and 4:5919; 0:8489 : The Ritz pairs are: 0 0:5285 11 0 0 0:8489 11 0 CC CC B B B B B @;0:3920; B@ ;0:2363 CACA ; B@4:5919; B@ 0:3796 CACA : 0:3796 ;0:2363 650
9.10 The Lanczos Method for Generalized Eigenvalues in an Interval In several important applications, such as in vibration and structural analysis, one is frequently interested in computing a speci ed number of eigenvalues in certain parts of the spectrum. We give a brief outline of a Lanczos algorithm for the generalized symmetric eigenvalue problem to do this. The algorithm was devised by Ericsson and Ruhe (1980).
Algorithm 9.10.1 The Lanczos Method for Generalized Eigenvalues in an Interval Given (A; B ), A symmetric and B symmetric positive de nite, the following algorithm computes the generalized eigenvalues k of the pencil (A ; B ) in a speci ed interval [ ; ]. Denote
S (; ) = the set of eigenvalues in (; ); C (; ) = the set of converged eigenvalues in (; ): Set
1 = ; C = (null set).
For i = 1; 2; : : : do until jC ( ; )j = jS ( ; )j 1. Factorize (A ; i B ) :
A ; i B = LDLT :
Count the number of negative diagonal entries of D. Set that count equal to jS ( ; i)j. 2. Choose a random unit vector v1. 3. Apply the symmetric Lanczos algorithm to the matrix
W T (A ; iB );1 W; where
B = WW T ;
with v1 as the starting Lanczos vector.
4. Let 1 ; 2; : : :; r be the converged eigenvalues from Step 3. Then add i + 1 to C; s = s 1; : : :; r. 5. Determine a new shift i+1 . 651
Remarks: Note that in applying the symmetric Lanczos to W T (A ; i B);1 W; we need a matrix-vector multiply of the form
y = W T (A ; iB );1 Wx which can be obtained as follows: 1. Compute z = Wx 2. Solve for p:
(A ; i B )p = z
which can be done by making use of the factorization in Step 1 of the algorithm. 3. Compute y = W T p. For details of this algorithm, its implementation and proof, see the paper by Ericsson and Ruhe (1980).
9.11 Review and Summary This chapter has been devoted to the study of the most commonly arising eigenvalue problem in engineering, namely, the generalized eigenvalue problem involving two matrices:
Ax = Bx: We now review and summarize the most important results: 1. Existence Results. There exist Schur and Real Schur analogs of the ordinary eigenvalue problem for the generalized eigenvalue problem (Theorems 9.2.1 and 9.3.1). 2. The QZ Algorithm. The most widely used algorithm for the generalized eigenvalue problem is the QZ algorithm, which constructs the generalized Real Schur Form of (A; B ). The algorithm comes in two stages. In stage I, the pair (A; B ) is reduced to a Hessenberg-triangular pair. In stage II, the Hessenberg-triangular pair is further reduced to the generalized Real Schur Form using implicit QR iteration to AB ;1 . B ;1 is never formed explicitly. The algorithm requires about 15n3 ops and is backward stable. 652
3. The Generalized Symmetric Eigenvalue Problem. Almost all eigenvalue problems arising in structural and vibration engineering are of the form
Mx = Kx; where M is symmetric positive de nite, and K is symmetric positive semide nite. This is called the symmetric-de nite generalized eigenvalue problem. Because of the importance of this problem, it has been studied in some depth here. Several case studies from vibration engineering have been presented in section 9.6 to show how this problem arises in important practical applications. These include: (i) vibration of a free spring-mass system, (ii) vibration of a building, (iii) and forced harmonic vibration of a spring-mass system. The natural frequencies and amplitudes of a vibrating system are related, respectively, to the generalized eigenvalues and eigenvectors. If the frequency of the imposed periodic
force becomes equal or nearly equal to one of the natural frequencies of the system, then resonance occurs, and the situation is quite alarming. The fall of Tacoma Bridge in the state of Washington, in the USA, and of Boughton Bridge in England are related to such a phenomenon. (See Chapter 8.)
The QZ method can, of course, be used to solve a symmetric de nite generalized eigen-
value problem. However, both symmetry and de niteness will be lost in general. A symmetry-preserving method is the Cholesky-QR algorithm. This has been described in section 9.5.1. The accuracy obtained by this algorithm can be severely impaired if the matrix B in Ax = Bx is ill-conditioned. A variation of this method due to Wilkinson that computes the smallest eigenvalues with reasonable accuracy has been described in this section. The method constructs an ordered real Schur form of B rather than the Cholesky factorization. 4. Simultaneous Diagonalization and Applications. The Cholesky-QR algorithm for the symmetric de nite problem Ax = Bx 653
basically constructs a nonsingular matrix P that transforms A and B simultaneously to diagonal forms by congruence:
P T AP = a diagonal matrix P T BP = I: This is called simultaneous diagonalization of A and B . In vibration and other engineering applications, this decomposition is called modal decomposition and the matrix P is called a modal matrix. The technique of simultaneous diagonalization is a very useful technique in engineering practice. Its applications include (i) decoupling of a second order system of di erential equations
M y + Dy_ + ky = 0 to n independent equations
zi + ( + !i2 )z_i + !i2 zi = 0;
i = 1; 2; : : :; n;
where D = M + K , and (ii) making of a reduced order model from a very large system of second-order systems. Decoupling and model reduction are certainly very useful approaches for handling a large second-order system of di erential equations. Unfortunately, however, the simultaneous diagonalization technique is not practical for large and sparse problems. On the other hand, many practical problems, such as the design of large space structures, etc., give rise to very large and sparse symmetric de nite eigenvalue problems. The simultaneous diagonalization technique preserves symmetry, but destroys the other exploitable properties o ered by the data of the problem, such as sparsity. (Note that most
practical large problems are sparse and maintaining sparsity is a major concern to the algorithm developers to economize the storage requirements.) 5. The Quadratic Eigenvalue Problem: The eigenvalue problem (2M + D + K )x = 0 is
discussed in Section 9.8. It is shown that the problem can be reduced either to a standard 2n 2n nonsymmetric problem (9.8.8) or to a 2n 2n symmetric generalized eigenvalue problem (9.8.16). It is also shown how to extract the frequencies and modal damping ratios of a vibrating system governed by a second-order system once the eigenvalues are obtained (Equations 9.8.9 and 9.8.10). 654
6. The Sturm-sequence and Lanczos Methods. We have given an outline of the Sturmsequence method for the generalized eigenvalue problem with tridiagonal matrices A and B . The symmetric de nite tridiagonal problems arise in several applications. We have also given a very brief description of the symmetric Lanczos method for a generalized eigenvalue problem. The chapter concludes with a Lanczos-algorithm due to Ericsson and Ruhe for nding a speci ed number of generalized eigenvalues in an interval of a symmetric de nite generalized eigenvalue problem (section 9.9).
9.12 Suggestions for Further Reading Almost all books in vibration and structural engineering discuss implicitly or explicitly how the generalized eigenvalue problems arise in these applications. Some of the well-known books in the literature of vibration are: 1. Vibration with Control, Measurement, and Stability, by Daniel J. Inman, Prentice Hall, Englewood Cli s, NJ 07632. 2. Theory of Vibrations with Applications, (Third Edition) by W. T. Thomson, Prentice Hall, Englewood Cli s, NJ 1988. 3. An Introduction to Mechanical Vibrations, (Second Edition), by Robert F. Steidel, Jr., John Wiley & Sons, New York, 1979. 4. Linear Vibrations, by P. C. Muller and W. O. Schiehlen, Martinus Nijho Publications, Dordrecht, 1985. 5. Vibration of Mechanical and Structural Systems, by M. L. James, G. W. Smith, J. C. Wolford, and P. W. Whaley. Harper & Row, New York, 1989. 6. Engineering Vibration, by Daniel J. Inman, Prentice Hall, New York, 1994. The QZ iteration algorithm has been discussed in detail in the books by Golub and Van Loan (MC 1984 Chapter 7), Stewart (IMC), and Watkins (FMC Chapter 5). (The original paper of Moler and Stewart (1973) is worth reading in this context.) For further reading on simultaneous diagonalization technique, see Golub and Van Loan (MC 1984 Chapter 8) and the references therein. For results on perturbation analysis of the generalized eigenvalue problem, see Stewart (1978). 655
For applications of the symmetric de nite generalized eigenvalue problem to earthquake engineering, see the book Introduction to Earthquake Engineering, (Second Edition), by S. Okamoto, University of Tokyo Press, 1984. A technique more ecient than the Cholesky-QR iteration method for computing the generalized eigenvalues of a symmetric de nite pencil for banded matrices has been proposed by Crawford (1973). See also the recent work of Wang and Zhao (1991) and Kaufman (1993). For a look-ahead Lanczos algorithm for the quadratic eigenvalue problem see Parlett and Chen (1991). Laub and Williams (1992) have considered the simultaneous triangularizations of matrices M; D; and K of the second-order system M y + Dy_ + Ky = 0. Chapter 15 of Parlett's book \The Symmetric Eigenvalue Problem" (1980) is a rich source of knowledge in this area.
656
Exercises on Chapter 9 PROBLEMS ON SECTION 9.2 1. Show that when A and B have a common null vector, then the generalized characteristic equation is identically zero: det(A ; B ) = 0: 2. (a) Prove that if A and B are n n matrices, then det(A ; B ) is a polynomial of degree at most n. (b) The degree of det(A ; B ) is equal to n i A is nonsingular.
PROBLEMS ON SECTIONS 9.3 AND 9.4 3. Show that the matrix Q1 in the initial QZ step can be computed just by inverting the 2 2 leading principal submatrix of the triangular matrix B . 4. Show that the shifts 1 and 2 in a QZ step, which are the eigenvalues of the lower 2 2 principal submatrix of C = AB ;1 , can be computed without forming the complete B ;1 (Hint: Computation depends only on the lower right 3 3 submatrix of B ;1 ). 5. Using the Implicit Q Theorem, show that the matrix Q in the QZ step has the same rst row as Q1 . 6. Give op-count for (a) Hessenberg-Triangular reduction with and without accumulations of the transforming matrices. (b) One step of the QZ iteration. (c) Reduction of (A; B ) to the generalized Schur form. 7. Work out an algorithm based on Gaussian elimination with partial pivoting for the Hessenbergtriangular reduction. What is the op-count? 8. How are the generalized eigenvectors of the pair (A; B ) related to the orthogonally equivalent ~ B~ ), where pair (A;
A~ = UAV B~ = UBV: 657
PROBLEMS ON SECTION 9.5 9. Rework the Cholesky-QR algorithm (Algorithm 9.5.1) of the symmetric de nite pencil by using the Schur decomposition of B :
V T BV = D: 10. Consider the Hessenberg-triangular reduction of (A; B ) with B singular. Show that in this case, if the dimension of the null space AB is k, then the Hessenberg-Triangular structure takes the form: ! A11 A12 ! 0 B12 A= ; B= ; 0 A22 0 B22 where A11 is a k k upper triangular matrix, A22 is upper Hessenberg, and B22 is (n ; k) (n ; k) upper triangular and nonsingular. How does this help in the reduction process of the generalized-Schur form? 11. Work out the op-count for the Cholesky-QR algorithm of the symmetric de nite pencil. 12. Let A and B be positive de nite. Show that the algorithm for simultaneous diagonalization requires about 7n3 ops. 13. Generalized Orthogonal Iteration. Consider the following iterative algorithm known as the generalized orthogonal iteration: (a) Choose an n m orthonormal matrix Q0 such that QT0 Q0 = Imm . (b) For k = 1; 2; : : : do i. Solve for Zk : BZk = AQk;1. ii. Find QR factorization of Zk : Zk = Qk Rk . Apply the above algorithm to 01 BB 1 A=B BB @1 1
1
1 2 3 4
14. Given
01 1 11 0 10 1 0 1 B C B C (a) A = B @ 1 1 1 CA ; B = B@ 1 10 1 CA : 1 1 1
0
1 10 C B 3 4C CC ; B = BBB 1 B@ 1 4 5C A 5 6 1
0
1 10
658
1 10 1 1
1 1 10 1
1
1 C 1C CC : 1C A 10
0 10 1 1 1 01 B C B (b) A = B @ 1 10 1 CA ; B = B@ 12
1 2 1 3 1 4
1 3 1 4 1 5
1 CC A:
1 1 1 1 3 Find the generalized eigenvalues and eigenvectors for each of the above pair using i. the QZ iteration followed by inverse iteration, ii. the generalized Rayleigh-Quotient iteration, iii. techniques of simultaneous diagonalization.
PROBLEMS ON SECTIONS 9.6 AND 9.7 15. Suppose that a bridge trestle has a natural frequency of 5 Hz (known from an earlier test). Suppose that it de ects about 2mm at midspan under a vehicle of 90000 kg. What are the natural frequencies of the bridge and the vehicle? 16. For the equation of motion under the force F (t) = F1 ei!t:
my + ky = F1 ei!t; nd the amplitude and determine the situation which can give rise to resonance. 17. Consider the following spring-mass problem:
659
3k
y1 2m k
y
2
m k
y
3
2m 3k
(a) Determine the equations of motion. (b) Set up the generalized eigenvalue problem in the form
Kx = Mx; and then determine the natural frequencies and modes of vibration. (c) Find the con guration of the system in each mode.
660
18. Consider the four story building as depicted by the following diagram:
m
4
m
3
m
2
m
1
Given
m1 m2 m3 m4 k1 k3
= = = = = =
1:0 105kg 0:8 105kg 0:5 105kg 0:6 105kg 15 108N/m; k2 = 12 108N/m; 15 108N/m; k4 = 10 108N/m:
Find the maximum amplitude of each oor for a horizontal displacement of 3 mm with a period of .25 second.
661
19. Consider the following diagram of a motor car suspension with vibration absorber (taken from the book Linear Vibrations by P. C. Muller and W. O. Schiehlen, p. 226).
car body
y1 (t)
y3 (t) m1 m3 k
y2 (t)
d
1
k
d
3
axle & wheel suspension
m2
k
3
1
absorber
tire
2
guide way
ye (t) Motor car suspension with vibration absorber
Given
m1 m2 m3 k1
1200kg; 80kg; 20kg; 300 N ; cm N; k2 = 3200 cm N: k3 = 600 cm = = = =
Find the various amplitude frequency responses with di erent damping values of the absorber d3 = 0; 300; 600; 1000 Ns/m. (The response of a system is measured by the amplitude ratios.)
662
PROBLEMS ON SECTIONS 9.8 AND 9.9 20. Find the eigenvalues of the quadratic pencil: (M2 + K + D)x = 0 using both approach 1 and approach 2 of Section 9.8, and compare the results. 0 2 ;1 0 1 01 1 11 B ;1 2 ;1 CC ; D = BB 1 1 1 CC : M = I33; K = B @ A @ A 0 ;1 2 1 1 1 21. Deduce the equations (9.8.9) and (9.8.10) for frequencies and the mode shapes. 22. Show that the polynomials de ned by (9.9.1){(9.9.3) form a Sturm sequence. 23. Find the generalized eigenvalues for the pair (A; B ), where
01 1 01 0 10 1 0 1 B C B C A=B @ 1 1 1 CA ; B = B@ 1 10 1 CA
0 1 1 0 1 10 using the Sturm sequence method described in section 9.9.1.
24. Estimate the generalized eigenvalues of the pair (A; B ) of problem #14 using the generalized symmetric Lanczos algorithm. 25. Find the generalized eigenvalues of the pair (A; B ) of problem #23 in the interval (0; 0:3). 26. Prove that the eigenvalues of the symmetric de nite pencil A;B lie in the interval [;kB ;1 Ak; kB ;1 Ak]. in exercise #23.
663
MATLAB PROGRAMS AND PROBLEMS ON CHAPTER 9 You will need backsub from MATCOM. 1. Write a MATLAB program, called hesstri, to reduce a pair of matrices (A; B ) to a (Hessenbergtriangular) pair: [H; T ] = hesstri(A; B ): Test your program by randomly generating A and B , each of order 15. 2. (a) Write a MATLAB program, called qzitri, to implement one step of QZ iteration algorithm: [A1; B 1] = qzitri (A; B ): (b) Now apply one step of qritrdsi. (Double-shift implicit QR iteration from Chapter 8) to C = AB;1 : [C ] = qritrdsi (C ): (c) Compute D = A1 B 1;1 , where (A1; B 1) is the Hessenberg-triangular pair obtained in step (a). (d) Compare C and D to verify that they are essentially the same.
Test-Data: A = A 15 15 randomly generated unreduced upper Hessenberg matrix. B = A 15 15, upper triangular matrix with all entries equal to 1, except two diagonal entries each equal to 10;5.
3. Using lynsyspp from Chapter 6 or the MATLAB Command `n' and hesstri, write a MATLAB program, called eigenvecgen to compute a generalized eigenvector u corresponding to a given approximation to a generalized eigenvalue : [u] = eigvecgen(A; B; ): Test your program using randomly generated matrices A and B , each of order 15, and then compare the result with that obtained by running the MATLAB command : [U; D] = eig(A; B ): 4. (The purpose of this exercise is to compare the accuracy of di erent ways to
nd the generalized eigenvalues and eigenvectors of a symmetric de nite pencil (K ; M )). 664
(a) Using the MATLAB commands chol, inv, eig, and `n' (or bascksub from Chapter 3), write a MATLAB program, called geneigsymdf, to implement the Cholesky algorithm for the symmetric-de nite pencil (K ; M ) (Algorithm 9.5.1). [V 1; D1] = geneigsymdf (K; M ); where V 1 is a matrix containing the generalized eigenvectors as its columns and D1 is a diagonal matrix containing the generalized eigenvalues of the symmetric-de nite pencil K ; M . (b) Run eig(K; M ) from MATLAB to compute (V 2; D2): [V 2; D2] = eig(K; M ): (c) Run eig(inv(M ) K ) from MATLAB to compute (V 3; D3) : [V 3; D3] = eig (inv (M ) K ): (d) Compare the results of (a), (b), and (c).
Test-data: M = diag( 0 m; m; ; m)2525 BB k ;k 0 0 BB ;k 2k ;k 0 B .. . . . . . . K = B . . . B.
BB .. B@ .
0 m = 100; k = 108 :
0 0 .. .
. . . . . . . . . . . . ;k 0 ;k 2k
1 CC CC CC CC CC A 2525
5. Using chol, inv, eig from MATLAB, write a MATLAB program, called diagsimul, to simultaneously diagonalize a pair of matrices (M; K ), where M is symmetric positive de nite and K is symmetric (Algorithm 9.5.2): (I; D) = diagsimul (M; K ):
Test-data:
Use M and K from the case-study on the vibration of a building in Section 9.6.2. 665
6. Using diagsimul, solve example 9.7.1 from the book completely. 7. (The purpose of this exercise is to compare di erent approaches for solving the quadratic eigenvalue problem: (2M + D + K )x = 0.)
0 1 0 I A by (i) Approach 1 of Section 9.8. Form the matrix A = @ ; 1 ; 1 ;M K ;M D using MATLAB commands inv, eig, and zeros. Then compute the eigenvalues and
eigenvectors of A using MATLAB command: [V1; D1] = eig (A): (ii) (Approach 2 of Section 9.8): Use the program geneigsymdf to compute the eigenvalues: [V 2; D2] = geneigsymdf (A; B );
0 1 0 1 0 ; K ; K 0 A; B = @ A: where A = @ ;K D 0 M
(iii) Use the MATLAB command [V 3; D3] = eig (A; B ), where the matrices A and B are given as in part (ii). Compare the results of the above three approaches with respect to accuracy, op-count, and elapsed time. (The eigenvalues and the eigenvectors obtained by (iii) can be taken as the accurate ones).
Test-Data:
Use the same M and K as in problem 4 and D = 10;5 M . 8. Find the natural frequencies and modal ratios using the results (9.8.9) and (9.8.10) of the book with the same M; K; and D as in problem #7. 9. Write a MATLAB program called lansymgen to transform the symmetric de nite pair (A; B ) to (TJ; IJ ), where TJ is a j j symmetric tridiagonal matrix and IJ is the j j identity matrix, j n. (alpha; beta) = langsymgen (A; B ); where alpha is a vector containing the diagonal entries of TJ and beta is a vector containing the o -diagonal entries of TJ . Test-data and the experiment: Take B = M; A = K; where M and K are the same as given in exercise #7. Run lansymgen using j = 10, 20, 30, 40, and 50. Tell how the extreme eigenvalues of TJ for each value of j approximate the generalized eigenvalues of the pencil (A ; B ). Find the generalized eigenvalues using the MATLAB 666
command [U; D] = eig(A; B )].
667
10.THE SINGULAR VALUE DECOMPOSITION (SVD)
Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 669 The Singular Value Decomposition Theorem : : : : : : : : : : : : : : : : : : : : : : : 670 A Relationship between the Singular Values and the Eigenvalues : : : : : : : : : : : 673 Applications of the SVD : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 675 Sensitivity of the Singular Values : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 678 The Singular Value Decomposition and the Structure of a Matrix : : : : : : : : : : : 679 10.6.1 The Norms and the Condition Number : : : : : : : : : : : : : : : : : : : : : 680 10.6.2 Orthonormal Bases for the Null Space and the Range of a Matrix : : : : : : : 681 10.6.3 The Rank and the Rank-De ciency of a Matrix : : : : : : : : : : : : : : : : : 682 10.6.4 An Outer Product Expansion of A and Its Consequences : : : : : : : : : : : 686 10.6.5 Numerical Rank : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 687 10.7 Computing the Variance-Covariance Matrix : : : : : : : : : : : : : : : : : : : : : : : 688 10.8 The Singular Value Decomposition, the Least Squares Problem, and the Pseudo-Inverse689 10.8.1 The SVD and Least Squares Problem : : : : : : : : : : : : : : : : : : : : : : 689 10.8.2 Solving the Linear System Using the Singular Value Decomposition : : : : : 692 10.8.3 The SVD and the Pseudoinverse : : : : : : : : : : : : : : : : : : : : : : : : : 692 10.9 Computing the Singular Value Decomposition : : : : : : : : : : : : : : : : : : : : : : 695 10.9.1 Computing the SVD from the Eigenvalue Decomposition of AT A : : : : : : : 695 10.9.2 The Golub-Kahan-Reinsch Algorithm : : : : : : : : : : : : : : : : : : : : : : 696 10.9.3 The Chan SVD : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 703 10.9.4 Computing the Singular Values with High Accuracy: The Demmel-Kahan Algorithm : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 705 10.9.5 The Di erential QD Algorithm : : : : : : : : : : : : : : : : : : : : : : : : : : 709 10.9.6 The Bisection Method : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 709 10.10Generalized SVD : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 710 10.11Review and Summary : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 711 10.12Some Suggestions for Further Reading : : : : : : : : : : : : : : : : : : : : : : : : : : 712
10.1 10.2 10.3 10.4 10.5 10.6
CHAPTER 10 THE SINGULAR VALUE DECOMPOSITION (SVD)
10. THE SINGULAR VALUE DECOMPOSITION (SVD) Objectives The major objectives of this chapter are to study theory, methods, and applications of the Singular Value Decomposition (SVD).Here are the highlights of the Chapter.
The SVD existence theorem (Section 10.2) A relationship between singular values and the eigenvalues(Section 10.3) An application of the SVD to the fetal ECG (Section 10.4) Sensitivity of singular values (Section 10.5) Applications of the SVD to rank-related problems, computing orthonormal bases, matrix approximations, etc. (Section 10.6)
Use of the SVD in least squares and the pseudoinverse problems (Section 10.8) The Golub-Kahan-Reinsch algorithm (Section 10.9) The Demmel-Kahan algorithm (Section 10.9)
Background Material Needed for this Chapter
The following background material will be needed for smooth reading of this chapter: 1. Norm concepts and related results (Section 1.7) 2. Rank concept (Section 1.3) 3. Orthonormal bases (Sections 1.3 and 5.6) 4. Creating zeros with Householder and Givens matrices (Sections 5.4 and 5.5) 5. QR factorization of a nonsquare matrix (Section 5.4.2) 6. Statements and methods for least-squares problems and the pseudoinverse (Chapter 7) 7. The Bauer-Fike Theorem (Theorem 8.7.1)
668
10.1 Introduction So far we have studied two important decompositions of a matrix A, namely, the LU and the QR decompositions. In this chapter we will study another very important decomposition of a complex matrix A: the singular value decomposition or in short, the SVD de ned by
A = U V where U and V are unitary and is a diagonal matrix. If A is real, U and V are real orthogonal.
We will assume that A is real throughout this chapter.
The SVD has a long and fascinating history. The names of at least ve classical and celebrated mathematicians|E. Beltrami (1835-1899), C. Jordan (1838-1921), J. Sylvester (1814-1897), E. Schmidt (1876-1959), and H. Weyl (1885-1955)|can be associated with the development of the theory of the SVD. Some details of the contributions of these mathematicians to the SVD can be found in an interesting recent paper by G. W. Stewart (1993). Also the book by Horn and Johnson (1991) contains a nice history of the SVD. In recent years the SVD has become a computationally viable tool for solving a wide variety of problems arising in many practical applications. However, this has only been possible due to the pioneering work of some contemporary numerical analysts such as: G. H. Golub, W. Kahan, etc., who showed us how to compute the SVD in a numerically ecient and stable way. In this chapter we will discuss the important properties, some applications, and the existing computational methods for the SVD. The organization of this chapter is as follows. In section 10.2 we prove the existence theorem on the SVD. There are several proofs available in the literature: Golub and Van Loan (MC, 1984, p. 17) Horn and Johnson (1991), Pan and Sigmon (1992), etc. We will, however, give a more traditional and constructive proof that relies on the relationship between the singular values and singular vectors of A with the eigenvalues and eigenvectors of AT A: The later aspect is treated fully in Section 10.3. In section 10.4 we give a real-life application of the SVD on separating out the ElectroCardiogram (ECG) of a fetus from that of the mother, taken from the work of Callaerts, De Moor, Van Dewalle and Sansen (1991). In section 10.5 we discuss the sensitivity of the singular values. The singular values are insensitive to perturbations, and this is remarkable. In section 10.6 we outline several important applications of the SVD relating to the structure of a matrix: nding the numerical rank, orthonormal bases for the range and null space of A, approximation of A by another matrix of lower rank, etc. The ability of the SVD to perform the above computations in a numerically reliable way makes the use of the SVD so 669
attractive in practical applications. In section 10.8 we discuss the use of the SVD in least squares solutions and computation
of the pseudoinverse. The SVD is the most numerically reliable approach to handle these problems.
Finally, section 10.9 is devoted to computing the SVD. Here, besides describing the widely-used Golub-Reinsch method, we brie y outline the recent improvements to the method by Demmel and Kahan (1990). A brief mention of the very recent method of Fernando and Parlett (1992) will be mentioned too.
10.2 The Singular Value Decomposition Theorem Theorem 10.2.1 Let A be a real m n matrix. Then there exist orthogonal
matrices U and V such that
A = Umm mnVnTn;
(10.2.1)
where is an m n \diagonal" matrix. The diagonal entries of are all nonnegative and can be arranged in nonincreasing order. The number of nonzero diagonal entries equals the algebraic rank of A.
Remark: Notice that when m n, has the form 1 0 1 CC BB ... CC : B =B BB CA n C @
0(m;n)n Proof. Consider the matrix AT A. It is an n n symmetric positive semide nite matrix; therefore its eigenvalues are nonnegative. Denote the eigenvalues of AT A by 1 = 12 , 2 = 22; : : :; n = n2 . Assume that these have been ordered such that 1 2 r > 0; and
r+1 = r+2 = = n = 0: We have seen before that a symmetric matrix has a set of orthonormal eigenvectors. Denote the set of orthonormal eigenvectors of AT A corresponding to 1 through n by v1; : : :; vn; that is, v1 through vn are orthonormal and satisfy:
AT Avi = i2vi; i = 1; : : :; n: 670
Then, and
viT AT Avi = i2;
(10.2.2)
viT AT Avj = 0; if i 6= j .
(10.2.3)
Write
V1 = (v1; : : :; vr ); V2 = (vr+1; : : :; vn); where v1 through vr are the eigenvectors associated with the nonzero eigenvalues 1 through r , and vr+1 ; : : :; vn correspond to the zero eigenvalues. De ne now a set of vectors fui g by u = 1 Av ; i = 1; : : :; r: (10.2.4) i
i
i
The ui 's, i = 1; : : :; r; form an orthonormal set, because uT u = 1 (Av )T 1 (Av ) i j
i
j j = 1 (viT AT Avj ) 8i j < 0 when i =6 j = : 1 when i = j i
(10.2.5)
(by (10.2.2) and (10.2.3)). De ne U1 = (u1 ; : : :; ur ), and choose U2 = (ur+1 ; : : :; um ) such that uTj A = 0; j = r + 1; : : :; m. The set fu1 ; : : :; ur ; ur+1;:::; um g forms an orthonormal basis of the m-space C m . De ne now V = (V1; V2); and U = (U1; U2). Then U and V are orthonormal, and
0 1 vT AT 1 BB 1 1T T CC BB v2 A CC 0 uT 1 1 BB ... CC BB uT CC B C U T AV = B BB ..2 CCC A(v1; : : :; vn) = BBB 1r vrT AT CCC A(v1; : : :; vn) BB uT CC @ . A T BB r.+1 CC um B@ .. CA 1 2
(Using 10.2.4) (10.2.6)
uTm
(10.2.7)
671
0 1 2 1 B B 0 B B .. B . B B = B 0 B B B B @ 0
0
0 0 ...
1
0 0 C 1 2 0 0C CC 2 2 .. ... CC . C= 1r r2 0 0 C CC C 0 ... 0 C A 0 0 The statement about the rank follows immediately from the fact that the rank is invariant under unitary matrix multiplications: 1
rank (A) = rank (U V T ) = rank (); and that the rank of a diagonal matrix is equal to the number of nonzero diagonal entires. The decomposition (10.2.6) is known as the Singular Value Decomposition (SVD) of A.
De nition 10.2.1 The scalars 1; 2; : : :; r are called the nonzero-Singular Values of A. Note that the other (n ; r) singular values are zero singular values. De nition 10.2.2 The columns of U are called the left singular vectors and those of V are called the right singular vectors. A Convention For the remainder of this chapter we will assume, without any loss of gener-
ality, that m n, because if m < n, we consider the SVD of AT and if the SVD of AT is U V T , then the SVD of A is V U T . We will follow the convention that the singular values are ordered as above. Thus 1 = max is the largest singular value and = min the smallest singular value. Also, by (A), we will denote the set of singular values of A. This is also n
referred to as the singular spectrum.
Notes: 1. When m n, we have n singular values. 2. The singular values of A are uniquely determined while the matrices U and V are not unique (Why?) 672
Example 10.2.1
01 21 B2 3C A=B @ C A 3 4
0 6:5468 0 1 B 0 0:3742 CC =B @ A
0 0:3381 0:8480 0:4082 1 B 0:5506 0:1735 ;0:8165 CC U =B @ A 0:7632 ;0:5009 0:4082 33
0
0
32
!
0:5696 ;0:8219 V= 0:8299 0:5696 22
There are two singular values; they are 6.5458, 0.3742.
10.3 A Relationship between the Singular Values and the Eigenvalues In the proof of Theorem 10.2.1 we have used the fact that the singular values of A are the nonnegative square roots of the eigenvalues of AT A (in fact, that's how they have been de ned). We put this important fact in the form of a theorem and give an independent proof.
Theorem 10.3.1 The singular values of an m n matrix A are the nonnegative square roots of the eigenvalues of AT A.
Proof. AT A = (V T U T )(U V T ) = V T V T = V 0V T ;
(10.3.1)
where 0 is an n n diagonal matrix with 12; : : :; n2 as its diagonal entries. Thus, V T AT AV = 0 = diag(12; 22; : : :; n2 ): Since the singular values are nonnegative numbers, the theorem is proven.
Corollary 10.3.1 Let A be a symmetric matrix with its eigenvalues 1; : : :; n. Then the singular values of A are jij; i = 1; : : :; n. Proof. Since A = AT ; AT A = A2: 673
(10.3.2) (10.3.3)
Therefore, from Theorem 10.3.1 above we have that the n singular values of A are the square roots of the n eigenvalues of A2 . Since the eigenvalues of A2 are 21 ; : : :; 2n, we have the result.
Corollary 10.3.2 An n n matrix A is nonsingular i all its singular values are di erent from zero.
Proof. We know that det(AT A) = (det(A))2: Thus A is nonsingular if and only if AT A is non-
singular. Since a matrix is nonsingular i its eigenvalues are all di erent from zero, proof is now immediate from Theorem 10.3.1.
Theorem 10.3.2 The nonzero singular values of an m n (m n) matrix A are positive eigenvalues of ! 0 A C=
Proof. Let
where and Then it is easy to see that
0nn
A = U V T :
Write
De ne
mm AT
U =
! U1 U2 m n m (m ; n)
=
1(nn) : 0(m;n)n
!
!
U~ ;U~ U2 P = ~1 ~ 1 ; V V 0n(m;n) U~1 = p1 U1 2
V~ = p1 V: 2
0 0 01 BB 1 C T P CP = @ 0 ;1 0 C A; 0
674
0
0
which shows that the nonzero eigenvalues of C are 1; : : :; k , ;1 ; : : :; ;k , where 1 through k are the nonzero singular values of A.
10.4 Applications of the SVD The SVD has become an e ective tool in handling various important problems arising in a wide variety of applications areas, such as control theory, signal and image processing identi-
cation and estimation, speech synthesis, pattern recognition, time series analysis, electrical network, biomedial engineering, etc.
The aspects of control theory and identi cation problems requiring use of the SVD include problems on controllability and observability, realization of state space models, the H-in nity control, balancing, robust feedback stablization, model reduction, etc. In signal and speech processing the SVD can be regarded as a lter that produces an estimate of a signal from noisy data. For example, when a person speaks, speech is absent about 50% of the time. Thus, if there are background noises coming from, say, a fan, a vibrating machine, etc., then these disturbances dominate in the microphone signal, when speech is absent. In such a situation the ratio of speech signal to the background noise has to be enhanced and the SVD can be used e ectively to do so. In image processing applications the SVD is routinely used for image compression, to remove noise in a picture, etc. For details see Jain (1989). In biomedial engineering the SVD plays an important role in obtaining a meaningful fetal ECG from that of the mother, see Van Dewalle and DeMoor (1988). The crux of using the SVD in these applications is in the fact that these applications require: determination of the \numerical" rank of a matrix, nding an approximation of a matrix using matrices of lower ranks, computations of the orthonormal bases for the row and column spaces of a matrix as well as the bases for their orthogonal complements, etc. In most cases these computations need to be done under the presence of impurities in the data (called \noises"), and as we will see in later sections in this chapter, the SVD is very e ective for these computations. Furthermore, the SVD is the most e ective tool in solving least squares and the generalized least-square problems. (See Golub and Van Loan MC, 1984 and 1989.) We now present in this section a real-life application of the SVD in biomedial engineering.
Fetal ECG (Taken from (Van Dewalle and DeMoor (1988).)
Consider the problem of taking the electrocardiogram of a fetus by placing the cutaneous electrodes at the heart and the abdomen of the pregnant mother. The maternal ECG (MECG) clearly 675
disturbs the observation of the fetal ECG (FECG), because the contributions of the material heart signals are much stronger than those of the fetal heart. The objective then will be to detect the FECG while simultaneously suppressing the MECG respective to noise. Suppose there are p measurements signals m1 (t); : : :; mp (t). Let these measurements be arranged in a vector called the measurement vector m(t):
m(t) = (m1(t); m2(t); : : :; mp(t))T :
(10.4.1)
Let there be r source signals s1 (t); s2(t); : : :; sr (t) arranged in the source signal vector s(t):
s(t) = (s1(t); s2(t); : : :; sr (t))T :
(10.4.2)
Obviously, the measurement signals are corrupted by an additive noise signal, and there exists a relationship between the measurement signals and the source signals. It can be assumed (see Van Dewalle and de Moor (1988)) that this relationship is linear, and, indeed, each measurement signal mi (t) can be written as a linear combination of r source signals si (t) and additive noise signal ni (t). This leads to the following equations:
m1(t) = t11s1(t) + t12 s2 (t) + + t1r sr (t) + n1 (t) m2(t) = t21s1(t) + t22 s2 (t) + + t2r sr (t) + n2 (t)
.. . mp(t) = tp1 s1(t) + tp2 s2(t) + + tpr sr (t) + nr (t):
(10.4.3)
m(t) = T s(t) + n(t);
(10.4.4)
or where
T = (tij );
and
n(t) = (n1(t); n2(t); : : :; nr (t))T :
(10.4.5)
The matrix T is called the transfer matrix and depends upon the geometry of the body, the positions of the electrodes and sources, and the conductivities of the body tissues. The problem now is to get an estimate of the source signals s(t) knowing only m(t), and from that estimate, separate out the estimate of fetal source signals. Let each measurement consist of q samples. Then the measurements can be stored in a matrix M of order p q. We now show that the SVD of M can be used to get estimates of the source signals. Let
M = U V T be the SVD of M . 676
(10.4.6)
Then the p q matrix S^ de ned by
S^ = U T M
(10.4.7)
will contain p estimates of the source signals. Next, we need to extract the estimates of the fetal source signals from S^; let this be called S^F : Partition the matrix of singular values of M as follows: 0 0 0 1 m B C; B = @ 0 F 0 C (10.4.8) A 0 0 0 where M contains rm large singular values, associated with the maternal heart, F contains rf singular values, those smaller ones associated with the fetal heart, and 0 contains the remaining singular values associated with noise, etc. Let U = (UM ; UF ; U0) be a conformable partitioning of U . Then, obviously, we have
0 UT 1 M ^S = U T M = B B@ UFT CCA M 0 U T M 1 U0T0 S^ 1 BB MT CC BB ^M CC = @ UF M A = @ SF A ;
(10.4.9)
S^0
U0T M
Thus S^F = UFT M . Once S^F is determined, we can also construct a matrix F containing only the contributions of fetus in each measured signal, as follows:
F = UF S^F =
rmX +rf i=rm +1
i uiviT ;
where ui and vi are the ith column of U and V , and i is the ith singular value of M . The signals in S^F are called the principal fetal signals. The above method has been automated and an on-line adaptive algorithm to compute the U matrix has been designed. For the details of the method and test results, see the paper by Callaerts, DeMoor, Van Dewalle and Sansen (1990). Note that if the SVD of M is given by
M = U V T ! VT ! 1 0 1 = (U1; U2) ; 0 2 V2T then S can be estimated by S^ = U1T M: 677
10.5 Sensitivity of the Singular Values Since the squares of the singular values of A are just the eigenvalues of the real symmetric matrix AT A, and we have seen before that the eigenvalues of a symmetric matrix are insensitive to small perturbations, the same thing should be expected of the singular values as well. Indeed, the following important result holds. The result is analogous to a result on the sensitivity of the eigenvalues of a symmetric matrix stated in Chapter 8. The result basically states that, like the eigenvalues of a symmetric matrix, the singular values of a matrix are well-conditioned.
Theorem 10.5.1 (Perturbation Theorem for Singular Values). Let A and B = A + E be two m n matrices (m n). Let i ; i = 1; : : :; n and ~i, i = 1; : : :; n be, respectively, the singular values of A and A + E , in decreasing order. Then j~i ; ij kE k2; for each i.
Proof. The proof is based on the interesting relationship between the singular values of the matrix A and the eigenvalues of the symmetric matrix A~ =
0
A
AT 0
!
;
displayed in Theorem 10.3.2. It has been shown there that the nonzero eigenvalues of A~ are 1 ; : : :; k ; ;1; : : :; ;k, where 1 through k are the nonzero singular ! values of 0A. The !remaining 0 B E eigenvalues of A~ are, of course, zero. De ne now B~ = ; E~ = T ; we have BT 0 E 0 B~ ; A~ = E~ . The eigenvalues of B~ and E~ are related respectively, to the singular values of B and E in the same way the eigenvalues of A~ are related to the singular values of A. The result of Theorem ?? ~ A; ~ and E~ . now follows immediately by applying the Bauer-Fike Theorem (Theorem 8.7.1) to B;
Example 10.5.1 The singular values of A:
01 2 31 B 3 4 5 CC ; A=B @ A 6 7 8
00 0 B0 0 E=B @
1
0 C: 0 C A 0 0 :0002
1 = 14:5576 2 = 1:0372 3 = 0:0000 678
(10.5.1)
(10.5.2)
The singular values of A + E :
~1 ~2 ~3 kE k2 j~1 ; 1j j~2 ; 2j j~3 ; 3j
= = = = = = =
14:5577 1:0372 0:0000 :0002 1:1373 10;4 0 0:
(10.5.3)
We now present a result (without proof) on perturbation of the singular values that uses Frobenius norm instead of 2-norm. The result of Theorem 10.5.2 will be used later in deriving a result on the closeness of A to a number of lower rank. (Theorem 10.6.2.)
Theorem 10.5.2 Let A; E; i and ~i, i = 1; :::; n be the same as in Theorem 10.5.1. Then v u n u tX(~i ; i)2 kE kF : i=1
10.6 The Singular Value Decomposition and the Structure of a Matrix The SVD can be very e ectively used to compute certain important properties relating to the structure of a matrix, such as its rank (in fact, nearness to rank-de ciency, which is an
important practical issue), Euclidean norm, condition number and orthonormal basis for the null space and range, etc. We will discuss these applications in this section.
679
10.6.1 The Norms and the Condition Number Theorem 10.6.1 Let 1 2 n be the n singular values of an m n matrix A(m n). Then 1. kAk2 = 1 = max , 2. kAkF = (12 + 22 + + n2 ) 21 , 3. kA;1k = 1 , when A is n n and nonsingular, 2
n 4. Cond2(A) = kAk2 kA;1k2 = 1 = max . n min
Proof. 1. kAk2 = kU V T k2 = kk2 = max (i ) i = 1 : (Note that the 2-norm and F -norm are invariant under unitary matrix products.) 2. kAkF = kU V T kF = kkF = (12 + 22 + + n2 ) 12 : 3. To prove 3 we note that the largest singular value of A;1 is 1 . (Note that when A is n invertible, n 6= 0). Then the result follows from 1. 4. Follows from the de nition of Cond2 (A) and 1 and 3.
Remark: When A is rank-de cient, min = 0, and we say that Cond(A) is in nite. Example 10.6.1
01 2 31 B CC A = B 3 4 5 @ A
6 7 8 1 = 14:5576; 2 = 1:0372; 680
3 = 0:0000
1. kAk2 = 1 = 14:5576
p
2. kAkF = 12 + 22 + 32 = 14:5945
10.6.2 Orthonormal Bases for the Null Space and the Range of a Matrix In Chapter 5 we have shown how to construct orthonormal bases for the range and the null-space of a matrix A using QR decomposition. In this section we show how the singular vectors can be used to construct these bases. Let r be the rank of A, that is,
1 2 r > 0; r+1 = = n = 0:
(10.6.1) (10.6.2)
Let uj and vj be the columns of U and V in the SVD of A. Then the set of columns vj
corresponding to the zero singular values of A form an orthonormal basis for the nullspace of A. This is because, when j = 0; vj satis es Avj = 0 and is therefore in the null-space of A. Similarly, the set of columns uj corresponding to the nonzero singular values is an orthonormal basis for the range of A. Orthogonal Projections Once the orthonormal bases for the range R(A) and the null-space N (A) of A are obtained, the orthogonal projections can be easily computed. Thus if we partition U and V as
U = (U1 ; U2);
V = (V1; V2);
where U1 and V1 consist of the rst r columns of U and V , then
Computing Projections Using the SVD 1. Projection onto R(A) = U1U1T . 2. Projection onto N (A) = V2V2T . 3. Projection onto the orthogonal complement of R(A) = U2U2T . 4. Projection onto the orthogonal complement of N (A) = V1 V1T .
681
Example 10.6.2
01 2 31 B 3 4 5 CC A = B @ A
6 7 8 1 = 14 ; 2 = 1:0372 13 = 0: 0 0:5576 :2500 0:8371 0:4867 B C U = B @ 0:4852 0:3267 ;0:8111 CA 0:8378 ;0:4379 0:3244 0 0:4625 ;0:7870 0:4082 1 B 0:5706 ;0:0882 ;0:8165 CC V = B @ A 0:6786 ;0:6106 0:4082
0 0:4082 1 B ;0:8165 CC. An orthonormal basis for the null-space of A = B @ A 0:4082 0 0:2500 1 0:8371 B C An orthonormal basis for the range of A = B @ 0:4852 0:3267 CA. (Compute now the four 0:8370 ;0:4379 orthogonal projections yourself.)
10.6.3 The Rank and the Rank-De ciency of a Matrix Finding the rank of an m n matrix and, in particular, determining the nonsingularity of a square
matrix are very important and frequently arising tasks in linear algebra and many important applications. The most obvious and the least expensive way of determining the rank of a matrix is, of course, to triangularize the matrix using Gaussian elimination and then to nd the rank of the reduced upper triangular matrix. Theoretically, nding the rank of a triangular matrix is trivial; one can just read it o from the diagonal. Unfortunately, however, this is not a very reliable approach in oating point computations. In practice, it is more important, as we will see below, to determine if the given matrix is near a matrix of a certain rank and in particular, to know if it is near a rank-de cient matrix. The Gaussian elimination method which uses elementary transformations, may transform a rank-de cient matrix rank into one having full rank, due to numerical round-o error. Another approach, which is more reliable than Gaussian elimination, to determine the rank and rank-de ciency, is the QR factorization with column pivoting. As we have seen before in 682
Chapter 5 this method constructs a permutation matrix P , and orthogonal matrix Q such that
QT AP
R1 R2 !
; (10.6.3) 0 0 R1 is upper triangular and nonsingular, and the dimension r of R1 is the rank of A. This result above, however, does not determine reliably the nearness of A to a rank-de cient matrix. For details, see again Golub and Van Loan (MC, 1984, pp. 166-167) and our discussions on rank-revealing QR factorization in Chapter 5. =
The most reliable way to determine the rank and nearness to rank-de ciency is to use the SVD. Since the number of nonzero singular values determines the rank of a matrix, we can say that a matrix A is arbitrarily near a matrix of full rank: just change each zero singular value by a small number . It is, therefore, more meaningful to know if a matrix is near a matrix of a certain rank, rather than knowing what the rank is. The singular value decomposition exactly answers this question. Suppose that A has rank r, that is, 1 2 r > 0 and r+1 = = n = 0. Then the question is how far A is from a matrix of rank k < r. The following theorems can be used to answer the question. Theorem 10.6.2 is generally known as the Eckart-Young Theorem (see Eckart and Young (1939)).
Theorem 10.6.2 Let A = U V T be the SVD of A, and let rank(A) be r > 0. Let k r. De ne Ak = U k V T ;
where
1 0 0 1 BB CC ... B 0C CC : k = B BB 0 CA k @
0 0 Then (i) rank(Ak ) = k, (ii) out of all the matrices of rank k, it is closest to A in Frobenius norm.
Proof. The statement about the rank is obvious, because, rank(Ak ) = rank(U k V T ) = rank(k ) = k: 683
To prove the second part, we rst prove that if A has (n ; k) small singular values, then A is close to Ak . Indeed, kAk ; Ak2F = kU (k ; )V T k2F = kk ; k2F = k2+1 + + n2 : Thus if the (n ; k) singular values k+1; : : :; n are small, then it is close to Ak . Next, we show that if B is any other matrix of rank k,
kB ; Ak2F kAk ; Ak2F : To see this, let's denote the singular values of B by 1 2 n . Then since B has rank k, k+1 = k+2 = = n = 0. By Theorem 10.5.2, we then have n X kB ; Ak2F j i ; ij2 k2+1 + + n2 = kAk ; Ak2F : i=1
We now present a similar result (without proof) using 2-norm.
Theorem 10.6.3 Out of all the matrices of rank k(k < r), the matrix B, closest to A is given by where
B = U k V T ;
0 1 0 0 1 BB CC ... BB CC k = B : B@ 0 CA k C 0
0
Furthermore, the distance of B from A : kB ; Ak2 = k+1:
Proof. See Watkins (FMC, pp. 413-414). Corollary 10.6.1 If the distance of the matrix B from A is less than the smallest singular value of A, then B must have rank at least r. That is, if B is such that kB ; Ak2 < r , then rank(B ) = k r:
Corollary 10.6.2 The relative distance of a nonsingular matrix A to the nearest singular matrix B is Cond1 (A) . That is, 2
kB ; Ak2 = 1 : kAk2 Cond2 (A) 684
Implication of the above results Distance of A Matrix From the Nearest Matrix of Lower Rank The above result states that the smallest nonzero singular value gives the distance from A to the nearest matrix of lower rank. In particular, for a nonsingular n n matrix A, n gives the measures of the distance of A to the nearest singular matrix. Thus, in order to know if a matrix A of rank r is close enough to a matrix of lower rank, look into the smallest nonzero singular value r . If this is very small, then the matrix is very close to a matrix of rank r ; 1, because there exists a perturbation of size small as jr j which will produce a matrix of rank r ; 1. In fact, one such perturbation is ur r vrT .
Example 10.6.3
01 0 B0 2 A = B @ rank(A) =
U = V = A0 = rank(A0) =
1
0 C 0 C A 0 0 0:0000004 3; 3 = 0:0000004 00 1 01 BB 1 0 0 CC ; @ A 0 0 1 00 1 01 BB 1 0 0 CC @ A 0 0 1 01 0 01 B C A ; u3 3v3T = B @ 0 2 0 CA : 0 0 0 2:
685
The required perturbation u3 3 v3T to make A singular is very small:
00 0 0 1 B 0 0 0 CC : 10;6 B @ A 0 0 :4
10.6.4 An Outer Product Expansion of A and Its Consequences In many applications the matrix size is so large that the storage of A may require a major part of total available memory space. The knowledge of nonzero singular values and the associated left and right singular vectors can help store the matrix with a substantial amount of savings in storage space. Let A be m n (m n), and 1 ; : : :; r be the nonzero singular values of A. Let the associated right and left singular vectors be v1; : : :; vr , and u1 ; : : :; ur , respectively. Then the following theorem shows that A can be written as the sum of r matrices each of rank 1. Since a rank-one matrix of order m n can be stored using only (m + n) storage locations rather than mn locations needed to store an arbitrary matrix, this will result in a very substantial savings, if r is substantially less than n.
Theorem 10.6.4
A=
r X j =1
j uj vjT :
Proof. The proof follows immediately from the relation A = U V T ; noting that r+1 through n are zero.
De nition 10.6.1 The representation of A = expansion of A.
Example 10.6.4
r X j =1
j uj vjT is called the dyadic or the outerproduct
01 2 31 B CC A=B 3 4 5 @ A 6 7 8
686
1 = 14:5576 2 = 1:0372 3 = 0; r = 2:
0 :2500 1 0 :8371 1 B C B C u1 = B @ :4852 CA ; u2 = B@ :3267 CA 8379 1 :43891 0 ::4625 0 ;;:07870 B C B C v1 = B @ :5706 CA ; v2 = B@ ;:0882 CA :6786 0:6106 01 2 31 B C 1u1v1T + 2 u2v2T = B @ 3 4 5 CA : 6 7 8
An Important Consequence Theorem 10.6.4 has an important practical consequences. In practical applications, the matrix A generally has a large number of small singular values. Suppose there are (n ; k) small singular values of A which can be neglected. Then the matrix Ak = 1u1 v1T + a2 u2v2T + + k uk vkT is a very good approximation of A, and such an approximation can be adequate in applications. For example, in digital image processing, even when k is chosen much less than n, the digital image corresponding to Ak can be very close to the original image. For details see Andrews and Hunt (1977). If A is n n, the storage of Ak will require 2nk locations compared to n2 locations needed to store A, resulting in a large amount of savings.
10.6.5 Numerical Rank As we have just seen in the last section in practical applications that need singular values we have to know when to accept a computed singular value to be \zero". Of course, if it is of an order of \round-o zeros" (machine epsilon), we can declare it to be zero. However, if the data shares large relative error, it should also be taken into consideration. A practical criterion would be
Accept a computed singular value to be zero if it is less than or equal to 10;tkAk1; where the entries of A are correct to t digits. Having de ned a tolerance = 10;tkAk1 for a zero singular value, we can have the following convention (see Golub and Van Loan (MC (1989)), p. 247): 687
A has \numerical rank" r if the computed singular values ~1; ~2; : : :; ~ n
satisfy
~1 ~2 ~r > ~r+1 ~n
(10.6.4)
Thus to determine the numerical rank of a matrix A, count the \large" singular values only. If this number is r, then A has numerical rank r.
Remark: Note that nding the numerical rank of a matrix will be \tricky" if there is no
suitable gap between a set of singular values.
10.7 Computing the Variance-Covariance Matrix In Chapter 7 we have described a method to compute the variance-covariance matrix (AT A);1 using the QR factorization of A. We note here this matrix can also be computed immediately once the singular values and the left singular vectors of A have been computed.
Computing (AT A);1 Using the SVD Let
A = U V T
be the SVD of A, then the entries of the matrix C = (AT A);1 = (cij ) are given by n v v X ik jk , where n = rank(A). (Note that C = (AT A);1 = V ;2V T :) cij = 2 k=1
Example 10.7.1
k
01 21 B 2 3 CC : A=B @ A 3 4
2 2 c11 = v112 + v122 = 4:833; 1 2 2 v 11v21 v12v22 c12 = 2 + 2 = ;3:3333; 1 2
688
c22 = v212 + v222 = 2:3333: 2
2
1
2
10.8 The Singular Value Decomposition, the Least Squares Problem, and the Pseudo-Inverse In Chapter 7 we have seen that the full-rank linear least squares problem can be e ectively solved using the QR decomposition, but pivoting will be needed to handle the rank-de cient case. We have just seen that the singular values of A are needed to determine the reliably nearness to rank de ciency. We will now see that the SVD is also an e ective tool to solve the least squares problem, both in the full rank and rank-de cient cases.
10.8.1 The SVD and Least Squares Problem We have brie y discussed how to solve a least-squares problem using the SVD in Chapter 7. Here we give a full treatment. Consider the least squares problem once more: Find x such that krk2 = kAx ; bk2 is minimum. Let A = U V T be the SVD of A. Then we have
krk2 = k(U V T x ; b)k2 = kU (V T x ; U T b)k2 = ky ; b0k2 ; where V T x = y and U T b = b0 . Thus, the use of the SVD of A reduces the least squares problem for a full matrix A to one with a diagonal matrix : Find y such that
ky ; b0k2
is minimum. The reduced problem is trivial to solve. We have k m X X ky ; b0k2 = jiyi ; b0ij2 + jb0ij2; i=1
i=k+1
689
where k is the number of nonzero singular values of A. Thus the vector
0y 1 BB y1 CC y=B BB ..2 CCC @.A yn
that minimizes ky ; b0k2 is given by:
0 yi = bi ; when i 6= 0 i yi = arbitrary; when i = 0. (Note that when k < n; yk+1 through ym do not appear in the above expression and therefore do not have any e ect on the residual.) Of course, once y is computed, the solution can be recovered from x = V y: Since corresponding to each \zero" singular value i ; yi can be set arbitrarily, in the rank-
de cient case, we will have in nitely many solutions to the least squares problem. There are instances where this rank de ciency is actually desirable because it provides a rich family of solutions which might be used for optimizing some other aspect of the original problem. In the full rank case the least squares solution is, of course, unique. The above discussion can be summarized in the following algorithm.
Algorithm 10.8.1 Least Squares Solutions Using the SVD 1. Find the SVD of A:
A = U V T :
0 b0 1 1 B C B b02 C B 0 T 2. Form b = U b = B C. .. C B @ . CA b0m
0y 1 BB ..1 C 3. Compute y = @ . C A choosing yn
8 b0 1 > i; < when = 6 0 i CA : yi = > i : arbitrary, when i = 0:
4. Then the family of least squares solutions is
x = V y: (Note that in the full-rank case, the family has just one number.) 690
Flop-count: Using the Golub-Kahan-Reinsch method to be described later, it takes about 2mn2 + 4n3 ops to solve the least squares problem, when A is m n and m n. (In deriving this op-count, it is noted that the complete vector b does not need to be computed; then only the columns of U that correspond to the nonzero singular values are needed in computation.)
An Expression for the Minimum Norm Least Squares Solution It is clear from Step 3 above that in the rank-de cient case, the minimum 2-norm least squares solution is the one that is obtained by setting yi = 0, whenever i = 0. Thus, from above, we have the following expression for the minimum 2-norm solution:
Minimum Norm Least-Squares Solution Using the SVD x= where r = rank(A).
Example 10.8.1
1. 1 = 7:5358; 2 = 0:4597; 3 = 0:
r uT b0 X i i i=1
i vi;
01 2 31 B C A=B @ 2 3 4 CA ; 1 2 3
(10.8.1)
061 B CC b=B @9A: 6
A is rank-de cient.
0 :4956 :5044 :7071 1 B :7133 ;:7008 0:0000 CC ; U =B @ A 0:4956 0:5044 ;0:7071 2. b0 = U T b = (12:3667; ;0:2547; 0)T :
0 :3208 ;0:8546 0:4082 1 B 0:5470 ;0:1847 ;0:8165 CC : V =B @ A 0:7732 0:4853
3. y = (1:6411; ;0:5541; 0): 4. The minimum 2-norm least-squares solution is V y = (1; 1; 1)T : 691
0:4082
10.8.2 Solving the Linear System Using the Singular Value Decomposition Note that the idea of using the SVD in the solution of the least squares problem can be easily applicable to determine if a linear system Ax = b has a solution, and, if so, how to compute it. Thus if A = U V T ; then is equivalent to
Ax = b y = b0
where y = V T x; and b0 = U T b: So, the system Ax = b is consistent i the diagonal system y = b0 is consistent (which is trivial to check), and a solution of Ax = b can be computed by solving the diagonal system y = b0 rst, and then recovering x from x = vy . However, this approach will be much more expensive than the Gaussian elimination and QR methods. That is why the SVD is not
used in practice, in general, to solve a linear system. 10.8.3 The SVD and the Pseudoinverse
In Chapter 7 we have seen that when A is an m n (m n) matrix having full rank, the pseudoinverse of A is given by: Ay = (AT A);1AT : A formal de nition of the pseudoinverse of any matrix A (whether it has full rank or not) can be given as follows:
692
Four Properties of the Pseudoinverse The pseudoinverse of an m n matrix A is an n m matrix X satisfying the following conditions: 1. AXA = X . 2. XAX = X: 3. (AX )T = AX: 4. (XA)T = XA:
The pseudoinverse of a matrix always exists and is unique. We now show that the SVD provides a nice expression for the pseudoinverse. Let A = U V T be the SVD of A; then it is easy to verify that the matrix
V yU T ; where y = diag( 1 ); j (if j = 0, use 1 = 0);
(10.8.2)
j
satis es all the four conditions, and therefore is the pseudoinverse of A. Note that this expression for the pseudoinverse coincides with A;1 when A is nonsingular, because
A;1 = (AT A);1AT = (V T U T U V T );1V T U T = V ;1(T );1V T V T U T = V ;1U T
(10.8.3)
(Note that in this case, y = ;1 ): The process for computing the pseudoinverse Ay of A using the SVD of A can be summarized as follows.
Algorithm 10.8.2 Computing the Pseudoinverse Using the SVD 1. Find the SVD of A:
A = U V T : 693
2. Compute
0 BB BB y = diag B BB BB @
1
1
1
2
...
1
r
1 CC CC C; 0C CC CA
0 0 where 1 ; : : :; r are the r nonzero singular values of A.
3. Compute Ay = V y U T .
Example 10.8.2 Find the pseudoinverse of 0 0 0 ;1 1 0 1 0 0 1 0 1 ; 6 ; 6 1 3 9 9 C B C B C B 1 6 6 B C B C B A = @ ;1 0 0 A @ 0 2 0 A @ ; 9 3 ; 9 C A 6 6 1 ;9 ;9 3 0 ;1 0 0 0 0 A = U V T ; Ay = V yU T 01 0 01 B 1 CC y = B @0 2 0A 0 0 0
0 0 ;1 0 1 B 0 0 ;1 CC UT = B @ A ;1 0 0 V T = V: Thus
0 1 ; 6 ; 6 1 0 1 0 0 1 0 0 ;1 0 1 0 0 ; 1 ; 1 1 B 36 19 ; 96 C BB 0 0 ;1 CC = BB 23 ; 31 CC B 1 0C C C B Ay = B 0 ; @ 9 3 9 A@ 2 A@ A @0 3 6A 6 6 1 2 1 ;9 ;9 3 0 0 0 ;1 0 0 0 3 3
The Pseudoinverse and the Least Squares Problem >From (10.8.1) and (10.8.2), it follows that the minimum 2-norm solution of the linear least squares problem obtained through the SVD in the previous section is
Ayb; where Ay is the pseudoinverse of A.
694
Example 10.8.3
01 2 31 061 B 2 3 3 CC ; b = BB 8 CC A=B @ A @ A
1 2 0 ;0:9167 B Ay = B @ ;0:1667 0:5833 The minimum 2-norm least squares solution is
3
6 1 1:3333 ;0:9167 C 0:3333 ;0:1667 C A ;0:6667 0:5833
011 B 1 CC : Ay b = B @ A 1
10.9 Computing the Singular Value Decomposition As mentioned before, the ability of the singular values and the singular vectors to compute e ectively the numerical rank, the condition number, the orthonormal bases of the range and the null spaces, the distance of a matrix A from a matrix of lower rank, etc., is what makes the singular value decomposition so attractive in many practical applications. Therefore, it is important to know how the SVD can be computed in a numerically e ective way. In this section we will discuss this most important aspect of the SVD.
10.9.1 Computing the SVD from the Eigenvalue Decomposition of AT A We have seen (Theorem 10.3.1) that the singular values of a matrix A are just the nonnegative square roots of the eigenvalues of the symmetric matrix AT A. It is thus natural to think of computing the singular values by nding the eigenvalues of the symmetric matrix AT A. However, this is not a numerically e ective approach because, as we have seen in Chapter 7, some vital information may be lost due to round-o error in the process of computing AT A. The following simple example will illustrate the phenomenon:
Example 10.9.1
!
1:0001 1:0000 A= 1:0000 1:0001 The singular values of A are 2.0010 and 0.0001. Now ! 2:0002 2:0002 T A A= 2:0002 2:0002 695
(10.9.1)
(to four signi cant digits). The eigenvalues of AT A are 0.0000 and 4.0004.
10.9.2 The Golub-Kahan-Reinsch Algorithm The following algorithm is nowadays a standard computational algorithm for computing the singular values and the singular vectors. The LINPACK SVD routine is based on this algorithm. The ALGOL codes for this procedure has been described in a paper by Golub and Reinsch (1970). A Fortran program implementing the algorithm appears in Businger and Golub (1969). The algorithm comes in two stages:
Stage I: The m n matrix A(m n) is transformed to an upper bidiagonal matrix by orthogonal equivalence: ! B U0Tmm Amn V0nn =
where B is an n n bidiagonal matrix given by
0b BB 011 BB . B@ .. 0
b12
0
(10.9.2)
1
0 .. C . C C
... ... C: ... ... b n;1;n C A 0 0 bnn
Stage II: The bidiagonal matrix B is further reduced by orthogonal equivalence to a diagonal
matrix :
by
U1T BV1 = = diag(1; : : :; n):
(10.9.3)
Of course, is the matrix of singular values. The singular vector matrices U and V are given
U = U0U1 V = V0V1
(10.9.4) (10.9.5)
Remark: In the numerical linear algebra literature, Stage I is known as the Golub-Kahan bidiagonal procedure, and Stage II is known as the Golub-Reinsch algorithm. We will call the combined two-stage procedure by the Golub-Kahan-Reinsch method. High relative accuracy of the singular values of bidiagonal matrices. The following result due to Demmel and Kahan (1990) shows that the singular values of a bidiagonal matrix can be computed with very high accuracy. 696
Theorem 10.9.1 Let B = (bij) be an n n bidiagonal matrix. Let B = (bij ). Suppose that bii + bii = 2i;1bii, and bi;i+1 + bi;i+1 = 2i bi;i+1; j = 6 0. 2 n ; 1 ; 1 Let = i=1 max(j ij; j i j): Let 1 n be the singular values of B and let 10 n0 be the singular values of B + B . Then i 0 ; i i
i = 1; : : :; n:
(10.9.6)
Reduction to Bidiagonal Form The matrices U0 and V0 in Stage I are constructed as the product of Householder matrices as follows: U0 = U01U02 U0n (10.9.7) and
V0 = V01V02 V0;n;2:
(10.9.8)
Let's illustrate the construction of U01 ; V01 and U02; V02, and their role in the bidiagonalization process with m = 5 and n = 4: First, a Householder matrix U01 is constructed such that
0 BB 0 BB (1) A = U01A = B BB 0 B@ 0
0
Next, a Householder matrix V01 is constructed such that
0 BB 0 BB A(2) = A(1) V01 = B BB 0 B@ 0
0
1 C C CC C CC C A
1
0 01 B 0 0C C CC B C CC BB 0 CC BB 0 0 C = A C CC B CC B CA C B 0 A @ 0
0
The process is now repeated with A(2); that is,
697
Householder matrices U02 and V02 are constructed so that 0 0 01 BB 0 0 CC BB CC (2) B U02A V02 = B 0 0 C CC BB @ 0 0 CA 0 0 Of course, in this step we will work with the 4 3 matrix A0 rather than the matrix A(2) . Thus, the orthogonal matrices U020 and V020 will be constructed rst such that 0 01 BB CC 0 B C U020 A0V020 = B B@ 0 CCA ; 0 then U02 and V02 will be constructed from U020 and V020 in the usual way, that is, by embedding them in identity matrices of appropriate orders. The process is continued until the bidiagonal matrix B is obtained.
Example 10.9.2
01 2 31 B C A=B @ 3 4 5 CA : 6 7 8
Step 1. U01 A(1)
Step 2. V01 A(2)
0 ;0:1474 ;0:4423 ;0:8847 1 B ;0:4423 0:8295 ;0:3410 CC = B @ A ;0:8847 ;0:3410 0:3180 0 ;6:7823 ;8:2567 ;9:7312 1 B 0 C = U01A = B 0:0461 0:0923 C @ A 0 ;0:9077 ;1:8154 01 1 0 0 B 0 ;0:6470 0:7625 CC = B @ A 0 ;0:5571 0:6470 1 0 ;6:7823 12:7620 0 C B 0 = A(1) V01 = B ;1:0002 0:0245 C A @ 0 1:9716 ;0:4824 698
Step 3. U02
01 1 0 0 B C = B @ 0 ;0:0508 0:9987 CA 0 0:9987 0:0508
0 ;6:7823 12:7620 1 0 B C B = U02A(2) = U021A(1)V01 = U02U01AV01 = B ;1:0081 ;1:8178 C @ 0 A 0
0
0
Note that from the above expression of B , it immediately follows that zero is a singular value of A.
Finding the SVD of the Bidiagonal Matrix The process is a variant of the QR iteration. Starting from the n n bidiagonal matrix B obtained in Stage 1, it successively constructs a sequence of bidiagonal matrices fBk g such that each Bi has possibly smaller o -diagonal entries than the previous one. The ith iteration is equivalent to applying the implicit symmetric QR, described in Chapter 8, with Wilkinson shift to the symmetric tridiagonal matrix BiT Bi without, of course, forming the product BiT Bi explicitly. The e ective tridiagonal matrices are assumed to be unreduced (note that the implicit symmetric QR works with unreduced matrices); otherwise we would work with decoupled SVD problems. For example, if bk;k+1 = 0, then B can be written as the direct sum of two bidiagonal matrices B1 and B2 and (B ) = (B1) [ (B2). The process has guaranteed convergence and the rate of convergence is quite fast. The details of the process can be found in the book by Golub and Van Loan (MC 1984 pp. 292-293). We outline the process brie y in the following. In the following just one iteration step of the method is described. To simplify the notation, let's write 0 1 1 2 BB . . . . . . CC C B=B (10.9.9) BB ... C nC @ A
n be an n n bidiagonal matrix. Then the Wilkinson-shift for the symmetric matrix B T B is the eigenvalue of 2 2 right-hand corner submatrix ! 2n;1 + n2;1 n;1 n (10.9.10) n n;1 2n + n2 that is closer to 2n + n2 : 699
1. Compute a Givens Form J1 =
J10
rotation J 0 0
!
1
such
that J 0
1
!
!
21 ; = : 1 2 0
:
0 In;2 2. Apply J1 to the right of B , that is form
B BJ1
This will give a ll-in at the (2,1) position of B. That is, we will have 0 1 BB + . . . . . . CC BB CC . . . . B . . C B BJ1 = B C BB ... C CA @
(10.9.11)
where + indicates a ll-in.
(The ll-in is at the (2,1) position.) The idea is now to chase the nonzero entry `+' down the subdiagonal to the end of the matrix by applying the Givens rotations in an appropriate order as indicated by the following: 0 + 1 BB . . . CC BB CC . . T . . B . . C 3. B J2 B = B C BB ... C CA @
(Fill-in at 0the (1,3) position) 1 BB . . . CC BB CC . . . . B B BJ3 = B + . . C C BB ... C CA @ (Fill-in at (3,2) position)
0 1 BB + CC BB CC . . T . . B . . C 4. B J4 B = B C BB ... C CA @ (Fill-in at (2,4) position) 700
0 1 BB CC BB CC . . B B BJ5 = B . C C BB ... C CA + @
(Fill-in at the (4,3) position.) And so on.
At the end of one iteration we will have a new bidiagonal matrix B orthogonally equivalent to the original bidiagonal matrix B :
B = (J2Tn;2 J4T J2T )B(J1J3 Jn;1 ):
Example 10.9.3
01 2 01 B C B=B @ 0 2 3 CA
1. The Wilkinson shift = 15:0828
0 0 1
0 ;:9901 :1406 0 1 B C J1 = B @ ;:1406 ;:9901 0 CA 0
2.
0
0 ;1:2713 ;1:8395 0 1 B ;0:2812 ;1:9801 3 CC B BJ1 = B @ A 0
(The ll-in at the (2,1) position.) 3.
0
0
1
0 ;:9764 ;:2160 0 1 B C J2 = B @ 0:2160 ;:9764 0 CA 0 0 1 0 1:3020 2:2238 ;:6480 1 B 0 1:5361 ;2:9292 CC B J2B = B @ A (The ll-in at (1,3) position)
0
0
01 0 1 0 B C J3 = B @ 0 0:9601 :2797 CA 0 ;:2797 :9601 701
1
0 1:3020 2:3163 1 0 B C B BJ3 = B @ 0 2:2942 ;2:3825 CA 0 ;:2797 :9601
(Fill-in at the (3,2) position) 4.
01 0 1 0 B 0 :9926 ;:1210 CC J4 = B @ A 0 :1210 :9926
0 1:3020 2:3163 1 0 B C B J4B = B @ 0 2:3112 ;2:4812 CA 0
0
Stopping Criterion
:6646
The algorithm typically requires a few iteration steps before the o -diagonal entry n becomes negligible. A criterion for o -diagonal negligibility given in Golub and Van Loan (MC 1984 p. 434) is:
A Criterion for Neglecting an O -Diagonal Entry Accept an o -diagonal i to be zero if
j ij (j ij + j i;1j); where is a small multiple of the machine precision .
Flop-Count: The cost of the method is determined by the cost of Stage I. Stage II is iterative and is quite cheap. The estimated op-count is: 2m2n + 4mn2 + 92 n3 (m n). This count includes the cost of U; ; and V . There are applications (e.g., least squares) where all three matrices are not explicitly required. A nice table of di erent op-count of the Golub-Kahan-Reinsch SVD and the Chan SVD (to be described in the next section) for di erent requirements of U; , and V appears in Golub and Van Loan (MC 1984, p. 175).
Round-o Property: It can be shown (Demmel and Kahan (1990)) that the computed SVD,
U^ ^ (V^ )T , produced by the Golub-Kahan-Reinsch algorithm is nearly the exact SVD of A + E , that is,
A + E (U^ + U^ )^ (V^ + V^ ); 702
where U^ + U^ and V^ + V^ are orthogonal, (kE k2jkAk2) p(m; n); k U^ k p(m; n); kV^ k p(m; n); and p(m; n) is a slowly growing function of m and n.
Entrywise Errors of the Singular Values Furthermore, let i be a computed singular value, then
ji ; ij p(n)kAk2 = p(n)max; where p(n) is a slowly growing function of n. The result says that the computed singular values can not di er from the true singular values by an amount larger than = p(n)max . Thus, the singular values which are not much smaller than max will be computed
by the algorithm quite accurately, and the small ones may be inaccurate. 10.9.3 The Chan SVD
T. Chan (1982) observed that the Golub-Kahan-Reinsch algorithm for computing the SVD described in the last section, can be improved in the case m > 13 8 n, if the matrix A is rst factored into QR and then the bidiagonalization is performed on the upper triangular matrix R. The improvement naturally comes from the fact that the work required to bidiagonalize the triangular matrix R is much less than that required originally to bidiagonalize the matrix A. Of course, once the SVD of R is obtained, one can easily retrieve the SVD of A. Thus the Chan-SVD can be described as follows: 1. Find the QR factorization of A:
QT A =
R
!
:
0 2. Find the SVD of R using the Golub-Kahan-Reinsch algorithm:
R = X Y T :
(10.9.12)
(10.9.13)
Tony Chan, a Chinese-American mathematician is a professor of mathematics at the University of California, Los Angeles. He developed this improved algorithm for SVD when he was a graduate student at Stanford University.
703
Then the singular values of A are just the singular values of R. The singular vector matrices U and V are given by: U = Q[X j0]; V = Y: (10.9.14)
Flop-Count: The Chan-SVD requires about 2m2n +11n3 ops to compute ; U and V , compared
to 2mn + 4mn2 + 92 n3 ops required by the Golub-Kahan-Reinsch SVD algorithm. Clearly, there will be savings with the Chan-SVD when m > n.
Remark: The Chan-SVD should play an important role in several applications such as in control theory, where typically m n: Example 10.9.4
01 21 B CC A=B @2 3A:
1. The QR factorization of A:
4 5
0 ;0:2182 ;0:8165 ;0:5345 1 B ;0:4364 ;0:4082 0:8018 CC Q=B @ A ;0:8729 0:4082 ;0:2673 0 ;4:5826 ;6:1101 1 B C R=B ;0:8165 C @ 0 A 0
0
(Q T A = R) 2. The SVD of R:
R = X Y T 0 ;0:9963 0:0856 B X = B @ ;0:0856 ;0:9963 0 0 ! 0:5956 ;0:8033 Y = 0:8033 0:5956 0 7:6656 0 1 B C = B @ 0 0:4881 CA 0 0 704
1
0 C 0 C A 1:0000
3. The Singular Value Decomposition of A = U V T The Singular Values of A are 7.6656, 0.4881. 0 0:2873 0:7948 ;0:5345 1 B 0:4698 0:3694 0:8018 CC U = Q[X j0] = B @ A 0:8347 0:4814 ;0:2673 V = Y
Flop-Count for the Least-Squares Problem Using the SVD The number of ops required to solve the linear least squares problem (m n) using the SVD computed by the Golub-Kahan-Reinsch algorithm is about 2mn2 + 4n3 . If the Chan-SVD is used, 3 the op-count is about mn2 + 17 3n. Recall that the op-count for the normal equations approach for the least-squares problem is mn2 + n3 and that using QR factorization (with Householder matrices) is mn2 ; n3 =3. Again a 2 6 nice table of di erent least squares methods in regards to their eciency appears in Golub and Van Loan (MC 1984, p. 177).
10.9.4 Computing the Singular Values with High Accuracy: The Demmel-Kahan Algorithm The round-o analysis of the Golub-Kahan-Reinsch algorithm tells us that the algorithm can compute the singular values not much smaller than kAk2 quite accurately; however, the smaller singular values may not be computed with high relative accuracy. Demmel and Kahan (Demmel and Kahan, 1990) in an award-winning paper presented a new algorithm that computes all the singular values, including the smaller ones, with very high relative accuracy. They call this algorithm \QR iteration with a zero shift". The reason for this is that the new algorithm corresponds to the Golub-Kahan-Reinsch algorithm when the shift is zero. The
new algorithm is based on a remarkable observation that when the shift is zero, no cancellation due to subtraction occurs, and therefore, the very small singular values can be found (almost) as accurately as the data permits. James Demmel is a professor of Computer Science at the University of California-Berkeley. He won several awards including the prestigious Householder Award (jointly with Ralph Byers of the University of Kansas) in 1984, the Presidential Young Investigator Award in 1986, the SIAM (Society for Industrial and Applied Mathematics) Award in 1988, and jointly with William Kahan, in 1991, for the best linear algebra paper and the Wilkinson Prize in 1993.
705
Indeed the e ect of the zero shift is remarkable. This is demonstrated in the following. Let J10 be the Givens rotation such that ! ! 21 0 (J1) = : 1 2 0 ! J10 0 Compute J1 = . Then 0 In;2 0 0 1 BB + CC BB CC ... ... CC ; BJ1 = B BB . .. C B@ CA
where there is a ll-in at the (2,1) position as before, but note that the (1,2) entry now is zero instead of being nonzero. This zero will now propagate through the rest of the algorithm and is indeed the key to the e ectiveness of this algorithm. Let us now apply J2 to the left of BJ1 to zero the nonzero entry at the (2,1) position. Then 0 + 1 BB . . . CC C B J2 BJ1 = B BB ... C CA @
There is now a ll-in at the (1,3) position as in the Golub-Kahan-Reinsch algorithm. Since the rank of the 2 2 matrix !
b12 b13 b22 b23 is 1, it follows that when J3 is applied to J2 BJ1 to the right to zero out (1,3) entry, it will zero out
the (2,3) entry as well; that is, we will have
0 0 1 BB 0 CC BB CC ... BB CC B J2 BJ1 J3 = B ... ... C BB CC BB ... C CA @
Thus, we now have one extra zero on the superdiagonal compared to the same stage of the GolubKahan algorithm: This phenomenon continues. Indeed, the rotation J4 zeros out the (3,4) entry as well as the (2,4) entry, and so on. 706
Example 10.9.5
1.
01 2 01 B C B=B @ 0 2 3 CA 0 0 1
0 0:4472 ;0:8944 0 1 B 0:8944 0:4472 0 CC J1 = B @ A 0
2.
0
1
0 2:2361 0 0 1 B C B BJ1 = B @ 1:7889 0:8944 3 CA :
0 0 1 (Note that there is a ll-in at the (2,1) position, but the (1,2) entry is zero.) 3.
0 0:7809 0:6244 0 1 B C J2 = B @ ;0:6247 0:7809 0 CA : 0
0
1
0 2:8636 0:5588 1:8741 1 B C B J2 B = B @ 0 0:6984 2:3426 CA : 0
0
1
Note that the rank of the 2 2 matrix ! 0:5588 1:8741 0:6984 2:3476 is one. 01 0 1 0 B C J3 = B @ 0 0:2857 ;0:9583 CA 0 0:9583 0:2857 0 2:8636 1:9557 0 1 B C B BJ3 = B @ 0 2:4445 0 CA 0 0:9583 0:2857 (Note that both (1,3) and (2,3) entries are zero.) 4.
01 1 0 0 B C J4 = B @ 0 0:9310 0:3650 CA 0 ;0:3650 0:9310 707
0 2:8636 1:9557 0 1 B C B J4B = B @ 0 2:6256 0:1043 CA
0 0 0:2660 Note that the singular values of B are: 3.8990, 1.9306, and 0.2657.
Stopping Criterion: Let the nonzero diagonal entries of B be denoted by 1; : : :; n, and the nonzero o -diagonal entries by 2 ; : : :; n. De ne the two sequences fk g and fk g as follows: 1st sequence: n = j nj For j = n ; 1 1 to (n-1) do j = j j j (j+1/(j+1 + j j +1j)) and
2nd sequence: 1 = f 1g For j = n ; 1 to 1 do j+1 = j j+1j(j /(j + j j+1j)): A Criterion for Setting an O -Diagonal Entry to Zero Set j = 0 if
j j/j+1j or j j /j j ; where < 1 is the desired relative accuracy, and fj g and fj g are computed as
above. Remark: For details of how to apply these two di erent criteria in di erent situations in practical implementations, see Demmel and Kahan (1990).
Rate of convergence: The convergence of the last o -diagonal element to zero is linear with
constant factor n2 ;1=n2 : If there is a cluster of singular values of multiplicity m di erent from the remaining ones then the convergence will be linear with constant factor n2 ;m+1 /n2 ;m :
Round-o error: Let i; i = 1; : : :; n be the singular values of the computed bidiagonal matrix B obtained by applying one implicit zero shift QR to B. Let 1 2 n. It has been shown by Demmel and Kahan (1990) that, if
w 69n2 < 1; 708
then
ji ; ij 1 ;w w i; i = 1; : : :; n:
Moreover, if after k iterations, the singular values of the then bidiagonal matrix Bk are k1 kn, then when w 69n2 < 1, we have 1 ji ; kij (1 ; w)k ; 1 i 69kn2 i: The above result basically states that \the relative error in the computed singular values can only grow with the squares of the dimension". This is a rather pessimistic result. The authors have given another round-o result that states that with the approach of the convergences of the algorithm, \errors do not accumulate at all and the error in the computed i and i is bounded by c ; c another modest constant". An error bound on the computed singular values is of the form 0
0
c0n ; (1 ; c0 n)
where c0 is another modest constant. For details and the proofs of these results, the reader is referred to the paper by Demmel and Kahan (1990).
10.9.5 The Di erential QD Algorithm Recently Fernando and Parlett (1992) have reported an algorithm based on the di erential form of the quotient-di erence that computes all the singular values with maximal relative accuracy. This algorithm is claimed to be at least four times faster than the Golub-Kahan-Reinsch algorithm. Unfortunately, we can not describe the algorithm here, as the machinery needed to explain the algorithm has not been developed in this book.
10.9.6 The Bisection Method To conclude the section, we remark that the bisection method for the symmetric eigenvalue problem discussed in Chapter 8 can be adapted to compute the singular values of a matrix. Note that the singular values of 2b b 3 0 11 12 66 77 ... ... 6 77 B = 66 ... b n;1;n 7 4 5 0 bnn are positive eigenvalues of " 0 BT # : B 0 709
The last matrix can be permuted to obtain 0 0 b11 B B b11 . . . b12 B B B b12 . . . b22 B B B b22 . . . b23 B T =B B b23 . . . B B ... B B
B B @
... ...
bn;1;n
bn;1;n ...
bnn
1 CC CC CC CC CC CC CC CC bnn C A
0 which is a symmetric tridiagonal matrix with 0 diagonal entries. Thus the bisection method described earlier in Chapter 8 can now be used to nd the positive eigenvalues of T , i.e. the singular values of B . The advantage of the bisection method is its simplicity, accuracy and inherent parallelism, see a recent paper by Li, Rhee and Zeng (1993). However, it is inecient in sequential computation compared to QR based algorithms.
10.10 Generalized SVD The SVD Theorem (Theorem 10.2.1) can be generalized for a pair of matrices A and B , and this generalized SVD is useful in certain applications such as constrained least squares problems. The generalization was rst performed by Van Loan (1976). We merely state the theorem here without proof. For a proof, see Golub and Van Loan (MC, 1984, pp. 318-319).
Theorem 10.10.1 (Generalized Singular Value Decomposition Theorem) Let A and B be, respectively, matrices of order m n and p n (m n). Then there exist orthogonal matrices U and V and a nonsingular matrix W such that
U T AW = C = diag(c1; : : :; cn); ci 0 V T BW = D = diag(d1; : : :; dq ); di 0 where q = min(p; n); d1 dr > dr+1 = = dq = 0, r = rank(B ). The elements dc1 ; dc2 ; : : :; dcr are called the generalized singular values of A and B . 1
2
r
Charles F. Van Loan is a professor of Computer Science at Cornell University. He is the co-author of the celebrated book \Matrix Computations".
710
10.11 Review and Summary In this section we summarize, for easy reference, the most important results discussed in this chapter. 1. Existence and Uniqueness of the SVD: The singular value decomposition (SVD) of a matrix A always exists (Theorem 10.2.1).
A = U V T : The singular values (the diagonal entries of ) are unique, but U and V are not unique. 2. Relationship of the singular values and singular vectors with the eigenvalues: The singular values of A are the nonnegative square roots of the eigenvalues of AT A. (Theorem 10.3.1, see also Theorem 10.3.2.) 3. Sensitivity of the singular values: The singular values are insensitive to small perturbations (Theorem 10.5.1). 4. Applications of the SVD: The singular values and the singular vectors of a matrix A are useful and most reliable tools for determining the (numerical) rank and the rank-de ciency of A; nding the orthonormal bases for range and the null space of A, nding the distance of A from another matrix of lower rank (in particular the nearness to singularity of a nonsingular matrix); solving both full-rank and the rank-de cient least squares problems, and nding the ipseudoinverse of A, etc. These remarkable abilities and the fact that the singular values are insensitive to small perturbations, have made the SVD an indispensable tool in a wide variety of applications areas such as the control and systems theory, signal processing, statistical applications, biomedical engineering, image processing, etc. 5. Computing the SVD: The most widely used approach for computing the SVD of A is the Golub-Kahan-Reinsch algorithm. This algorithm comes into two phases. In phase 1, the matrix A is reduced to a bidiagonal matrix by orthogonal equivalence, and in phase 2, the bidiagonal matrix is further reduced to a diagonal matrix by orthogonal similarity using implicit QR iteration with the Wilkinson's shift. Unfortunately, very tiny singular values may not be computed with very high relative accuracy by this method. A modi cation of this method, known as the zero shift QR iteration or the QR iteration with a zero shift has been proposed by Demmel and Kahan in 1990. The Demmel-Kahan method computes all the singular values with high relative accuracy. 711
10.12 Some Suggestions for Further Reading As mentioned before, applications of the SVD are varied. There are books in each area of applications where the SVD plays an important role. For applications of the SVD to classical control problem, see an earlier survey of Klema and Laub (25). The SVD also plays an important role in modern control theory, especially in H -in ty and Robust Control Theory. Curious readers are referred to a growing number of interesting papers in these areas. For applications of the SVD to robust pole assignment problem, see Kautsky, Nichols and van Dooren (1985). For some applications of the SVD to system identi cation and signal processing problems, see an interesting survey paper a variety of applications of singular value decomposition in identi cation and signal processing by Joos Van Dewalle and Bart DeMoor (1988). Two important books in in the area of \SVD and Signal processing" are: 1) SVD and Signal Processing, Algorithms, Applications and Architecture, edited by Ed. F. Deprettere, Elsevier Science Publishers B.W. (North Holland), Amsterdam, 1988. 2) SVD and Signal Processing II, Algorithms, Analysis and Applications, edited by R. Vaccaro, Elsevier, Amsterdam, 1991. For applications to image processing, see the book Fundamentals of Digital Image Processing by A. K. Jain, Prentice Hall, Information and System Sciences Series, edited by Thomas Kailath, 1989 and Digital Image Processing by H. C. Andrews and B. R. Hunt, Prentice-Hall, Inc., 1977. Many matrices arising in signal processing and control, and systems theory applications are structured, such as, Hankel, Toeplitz, etc. Finding ecient and reliable algorithms for di erent types of computations that can exploit the structures of these matrices is a challenging problem to researchers. Much work has been done and it is presently an active area of research. For some interesting papers in this context, see the book Linear Algebra in Signals, Systems, and Control, edited by B. N. Datta, et al. SIAM, 1988. See also the recent paper of Van Dooren (1990). For some statistical applications, see the excellent survey papers by Golub (1969), and Hammarling (1985). Discussions on mathematical applications of the SVD, such as nding numerical rank, nearness to singularity, orthonormal bases for the range and null-space, etc. are contained in all numerical linear algebra books. The books by Golub and Van Loan (MC) and Watkins (FMC), in particular, have treated these aspects very well. For discussions on singular values sensitivity, see Stewart (1979) and (1984). 712
For computational aspects of singular values and singular vectors, the original paper by Golub and Kahan (1965), and the recent papers by Demmel and Kahan (1990) and Fernando and Parlett (1992) are valuable. Codes for the GKR method appear in Golub and Reinsch (1970), and Businger and Golub (1969). Fortran codes for the SVD computations also appear in Forsythe, Malcom and Moler (CMMC), and in Lawson and Hanson (SLS). The SVD Theorem (Theorem 10.2.1) has been generalized to a pair of matrices (A; B ) by Van Loan (1976). This is known as the generalized singular value decomposition theorem. For the statement and a proof of this theorem, see Golub and Van Loan (MC 1984, pp. 318-319). For other papers in this area see Kagstrom (1985) and Paige and Saunders (1981). For computational algorithms for the generalized SVD, see Stewart (1983), Van Loan (1985), and Paige (1986). For further generalizations of the SVD, see an excellent paper by DeMoor and Zha (1991). For the SVD of the product of two matrices, see Heath, Laub, Paige and Ward (1986).
713
Exercises on Chapter 10
PROBLEMS ON SECTIONS 10.2 AND 10.3 1. (a) Let A be m n, and U and V be orthogonal. Then from the de nitions of singular values prove that the singular values of A and V T AV are the same. (b) How are the singular vectors of A related with those of U T AV ? (c) If m = n and A is symmetric positive de nite, then prove that the eigenvalues of A are the same as its singular values. 2. An economic version of the SVD. Let A be m n (m n). Let rank(A) = r n. Then prove that A = V SU; where V is an m n matrix with orthonormal columns, S is a nonsingular r r diagonal matrix, and U is an r n matrix with orthonormal rows. Why is the version called the Economic Version? 3. Let be a singular value of A with multiplicity `; that is, i = i+1 = = i+`;1. Let U V T be the singular value decomposition of A. Then construct U~ and V~ such that U~ (V~ )T is also a singular value decomposition. 4. (a) Derive Theorem 10.3.1 from Theorem 10.3.2. (b) Given 01 21 B 3 4 CC ; A=B @ A 5 6 Find the singular values 1 and 2 of A by computing the eigenvalues of AT A. Then nd the orthogonal matrix P such that
P T SP = diag(1; 2; ;1 ; ;2 ; 0);
where S =
033
A !
AT 022
:
714
5. Using the constructive proof of Theorem 10.2.1, nd the SVD of the following matrices: 01 21 B 3 4 CC ; A = ( 1 2 3 ) A=B @ A 0 51 16 B CC A=B A = diag(1; 0; 2; 0; ;5); @1A; 0 11 1 1 B 0 CC ; where is small: A=B @ A 0
PROBLEMS ON SECTIONS 10.5, 10.6, and 10.7 6. (a) Find the rank, k k2; k kF , Cond(A), and the orthonormal bases for the null-space and the range of each of the matrices in problem #5. Find the orthogonal projections onto the range and the null-space of A and of their orthogonal complements. (b) Prove that for an mxn matrix A i. rank(AT A) = rank(AAT ) = rank(A) = rank(AT ) ii. AT A and AAT have the same nonzero eigenvalues. iii. If the eigenvectors u1 and u2 of AT A are orthogonal, then Au1 and Au2 are orthogonal. 7. (a) Let A be an invertible matrix. Then show that kAk2 = 1 if and only if A is a multiple of an orthogonal matrix. (b) Let U be an orthogonal matrix. Then prove that
kAU k2 = kAk2 and
kAU kF = kAkF
(c) Let A be an m n matrix. Using the SVD of A, prove that i. Cond2(AT A) = (Cond2 (A))2 ii. kAT Ak2 = kAk22 iii. Cond2(A) = Cond2 (U T AV ), where U and V are orthogonal. 715
(d) Let rank(Amn ) = n, and let Bmr be a matrix obtained by deleting (n ; r) columns from A, then prove that Cond2(B ) Cond2 (A): 8. Prove that if A is an m n matrix with rank r, and if B is another m n matrix satisfying kA ; Bk2 < r , then B has at least rank r. 9. (a) Give a proof of Theorem 10.6.3. (b) Given
01 21 B CC A=B @3 4A;
5 6 nd a rank-one matrix B closest to A in 2-norm. What is kB ; Ak2 ? 10. Let A and B be n n real matrices. Let Q be an orthogonal matrix such that kA ; BQkF kA ; BX kF for any orthogonal matrix X . Then prove that Q = UV T , where BT A = U V T . 11. Given
01 2 31 B 2 3 4 CC ; A=B @ A
5 6 7 and using the result of problem #10, show that out of all the orthogonal matrices X , the matrix 0 ;:2310 ;:3905 :8912 1 B ;:4824 :8414 :2436 CC Q=B @ A :8449 :3736 :3827 is such that kA ; QkF kA ; X kF : Find kA ; QkF by computing the singular values of A ; Q. 01 21 B CC 12. (a) Let A = B @ 1 3 A. Find the outer product expansion of A using the SVD of A. 1 4 (b) Compute (AT A);1 using the SVD of A.
716
PROBLEMS ON SECTION 10.8 0 01 1 BB . . . CC BB CC BB CC r D=B CC ; BB 0 C BB ... C CA @
13. Let
Then show that
0
01 BB . . . BB B Dy = B BB BB B@
i > 0; i = 1; : : :; r:
0
01
1
0
1
r
0
...
0
CC CC CC CC : CC CA
14. Verify that the matrix Ay = V y U T where y = diag( 1 ) (with the convention that if j = 0, j 1 we use = 0), is the pseudoinverse of A. (Check all four conditions for the de nition of the
j
pseudoinverse.)
15. For any nonzero matrix A, show that (a) (b) (c) (d)
AAyv = v for any vector v in R(A). Ayx = 0 for any x in N (AT ). (AT )y = (Ay )T (Ay )y = A:
16. Let A be an m n matrix. Show that (a) If A has full column rank, then (b) If A has full row rank, then
Ay = (AT A);1AT Ay = AT (AAT );1:
17. >From the SVD of A, compute the singular value decompositions of the projection matrices: P1 = AyA; P2 = I ; Ay A; P3 = AAy and P4 = I ; AAy . Also verify that each of these is a projection matrix. 717
18. Let
01 21 B 2 4 CC ; A=B @ A
031 B 6 CC : b=B @ A
3 6 9 Find the minimum 2-norm solution x to the least squares problem min kAx ; bk2: x What is kxk2? Obtain x and kxk2 using both the SVD of A and the pseudoinverse of A. 011 011 B 2 CC ; v = BB 1 CC ; nd Ay where 19. Given u = B @ A @ A 3 1
A = uvT : 20. Let
01 1 1 01 B C A=B @ 0 1 1 1 CA ;
031 B CC b=B @ 3 A:
2 3 4 5 14 (a) Find a least squares solution x : min kAx ; bk2. What is kxk2? x (b) Find the minimum-norm solution x, and kxk2.
PROBLEMS ON SECTION 10.9 21. Let B be an upper bidiagonal matrix having a multiple singular value. Then prove that B must have a zero either on its diagonal or superdiagonal. 22. Consider the family of bidiagonal matrices 0 1 ; (1 + ) 1 BB CC ... ... BB CC ... ... CC ; B ( ) = B BB . . . (1 + ) C B@ CA 1; 1: It can be shown (Demmel and Kahan 1990) that the smallest singular value of B( ) is approximately 1;n(1 ; (2n ; 1) ). Taking = 106; and using = 0, verify the above result. 718
23. Let
Find the SVD of A
0 1 :9999 1 B 2 1:9999 CC A=B @ A 3 2:9999
(a) using the Golub-Kahan-Reinsch algorithm. (b) using the Demmel-Kahan algorithm. (c) Compare your results of (a) and (b). 24. Show that an approximate op-count for the Golub-Kahan-Reinsch and the Chan SVD are, respectively, 2m2n + 4mn2 + 2 n3 and 2m2n + 11n3 for an m n matrix A. Compute also the 9
op-count for the Demmel-Kahan algorithm.
719
MATLAB PROGRAMS AND PROBLEMS ON CHAPTER 10 You will need housqr, parpiv, reganl from MATCOM. 1. (The purpose of this exercise is to study the sensitivities (insensitivities) of the singular values of a matrix). Using MATLAB commands svd and norm, verify the inequalities in Theorems 10.5.1 and 10.5.2, respectively.
Test-Data (i)
0 1 1 1 1 1 BB C BB 0 0:99 1 1 CCC A=B ; B@ 0 0 0:99 1 CCA 0
0
0
0:99
(ii) A = The Wilkinson Bidiagonal matrix of order 20. In each case, construct a suitable E so that (A + E ) di ers from A in the (n; 1)th element only by an = 10;5. (Note that the eigenvalues of both matrices are ill-conditioned). 2. Generate randomly a 15 5 matrix using MATLAB command: A = rand (15,5). Find s = svd (A). Compute now jjAjj2; jjAjjF ; the condition number of A with respect to 2norm using the entries of s and then compare your results with those obtained by MATLAB commands norm(a), norm(a, 'fro'), cond(a), respectively. 3. (a) Compute the orthonormal bases for the range and the null space of a matrix A as follows: (i) Use [U; S; V ] = svd (A) from MATLAB. (ii) Use housqr or housqrp from Chapter 5, as appropriate. (iii) Use orth(A) and null(A) from MATLAB. (b) Compare the results of (i) and (ii) and (iii) and op-count for each algorithm.
Test-Data:
720
(i)
0 1 0 0 1 1 BB C BB 0 0 0 0 CCC A=B ; B@ 1 1 1 0 CCA
1 1 1 0 (ii) A = A randomly generated matrix of order 10.
4. Compute the rank of each of the following matrices using (i) MATLAB command rank (that uses the singular values of A) and (ii) the program housqrp (Householder QR factorization with pivoting) and parpiv from Chapter 5, compare the results.
Test-Data and the Experiment (a) (Kahan Matrix)
0 1 1 ; c ; c ; c BB C BB 0 1 ;c ;c CCC B .. .. C ... A = diag (1; s; ; sn;1) B . C BB . CC ; . . . ;c C BB ... CA @ 0 0 0 1
where, c2 + s2 = 1; c; s > 0; c = 0:2; s = 0:6; n = 100. (b) A 15 10 matrix A created as follows: A = xy T ; where x = round (10 rand (15,1)) y = round (10 rand (10,1)). 5. (a) Generate randomly a matrix A of order 6 4 by using MATLAB command rand (6,4). Now run the following MATLAB command: [U; S; V ] = svd (A): Set S (4; 4) = 0; compute B = U S V 0. What is rank (B )? (b) Construct a matrix C of rank 3 as follows:
C = qr (rand (3)): Verify that jjC ; Ajj2F jjB ; Ajj2F using MATLAB command norm for computing the Frobenius norm. (c) What is the distance of B from A? 721
(d) Find a matrix D of rank 2 that is closest to A. (This exercise is based on Theorems 10.6.2 and 10.6.3). 6. Let
0 1 1 1 1 1 BB CC B C 0 : 0001 1 1 A=B BB 0 0 0:0001 1 CCC @ A
0 0 0 1 Find the distance of A from the nearest singular matrix. Find a perturbation which will make A singular. Compare the size of this perturbation with jr j. 7. Let A = U V T be the singular value decomposition (SV D) of a randomly generated 15 10 matrix A = rand (15,10), obtained by using MATLAB command [U; S; V ] = svd(A): Set S (8; 8) = S (9; 9) = S (10; 10) = 0 all equal to zero. Compute B = U S V 0. Find the best approximation of the matrix B in the form
B
r X i=1
xi yiT
P such that jjB ; ri=1 xi yiT jj2 = minimum, where xi and yi are vectors, and r is the rank of B . 8. For matrices A and B in problem #7, nd an unitary matrix Q such that jjA ; BQjj is minimized. (Hint: Q = V U T , where AT B = U V T ). (Use MATLAB command svd to solve this problem). 9. (a) Write a MATLAB program, called covsvd to compute (AT A);1 using the singular value decomposition. Use it to nd (AT A);1 for the 8 8 Hilbert matrix and compare your results and
op-count with those obtained by running reganl from MATCOM. (b) Using covsvd nd the Pseudoinverse of A and compare your result with that obtained by running MATLAB command pinv. (A is same as in part (a)). 10. Let A be a 10 10 Hilbert matrix and b be a vector generated such that all entries of the vector x of Ax = b are equal to 1. Solve Ax = b using the SVD of A, and compare the accuracy, op-count and elapsed time with those obtained by linsyspp from Chapter 6. 722
11. Let A = rand (10,3), and
X = pinv(A):
Verify that X satis es all the four conditions of the Pseudoinverse using MATLAB: AXA = X; XAX = X; (AX )T = AX; (XA)T = XA. 12. Write a MATLAB program, called chansvd to implement the improved SVD algorithm of Chan given in Section 10.9.3, using MATLAB commands qr and svd. [U; S; V ] = chansvd(A): Run your program with a randomly generated 30 5 matrix A = rand (30,5) and compare the op-count and elapsed time with those obtained by using svd(A). 13. Write a MATLAB program, called bidiag to bidiagonalize a matrix A (Section 10.9.2) : [B ] = bidiag(A; tol) where B is a bidiagonal matrix and tol is the tolerance. Test your program using A = rand(15,10). 14. (The purpose of this exercise is to compare the three related approaches for nding the small singular values of a bidiagonal matrix.) Write a MATLAB program to implement the Demmel-Kahan algorithm for computing the singular values of a bidiagonal matrix: [s] = dksvd(A) where s is the vector containing the singular values of A.
Experiment: Set A = rand (15,10). Compute [U; S; V ] = svd(A). Set S (10; 10) = S (9; 9) = S (8; 8) = 10;5. Compute B = U S V 0 . Run the program bidiag on B :
C = bidiag(B):
Compute now the singular values of C using (i) svd, (ii) dksvd, and (iii) chansvd and compare the results with respect to accuracy (especially for the smallest singular values),
op-count and elapsed time.
723
11.A TASTE OF ROUND-OFF ERROR ANALYSIS 11.1 11.2 11.3 11.4 11.5 11.6
Basic Laws of Floating Point Arithmetic : : : : : : : : : : : : : : : : : : : Backward Error Analysis for Forward Elimination and Back Substitutions Backward Error Analysis for Triangularization by Gaussian Elimination : Backward Error Analysis for Solving Ax = b : : : : : : : : : : : : : : : : : Review and Summary : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Suggestions for Further Reading : : : : : : : : : : : : : : : : : : : : : : :
: : : : : :
: : : : : :
: : : : : :
: : : : : :
: : : : : :
: 724 : 724 : 727 : 734 : 736 : 737
CHAPTER 11 A TASTE OF ROUND-OFF ERROR ANALYSIS
11. A TASTE OF ROUND-OFF ERROR ANALYSIS Here we give the readers a taste of round-o error analysis in matrix computations by giving backward analysis of some basic computations such as solutions of triangular systems, LU factorization using Gaussian elimination and solution of a linear system. Let's recall that by backward error analysis we mean an analysis that shows that the computed solution by the algorithm is an exact solution of a perturbed problem. When the perturbed problem is close to the original problem, we say that the algorithm is backward stable.
11.1 Basic Laws of Floating Point Arithmetic We rst remind the readers of the basic laws of oating point arithmetic which will be used in the sequel. These laws were obtained in Chapter 2. Let j j < ; where is the machine precision. Then 1: (x y ) = (x y )(1 + ) 2: (xy ) = xy (1 + ) 3: If y 6= 0, then ( xy ) = ( xy )(1 + ): Occasionally, we will also use 4: (x y ) = 1x + y ;
(11.1.1) (11.1.2) (11.1.3) (11.1.4)
where `*' denotes any of the arithmetic operations +, ;, , /.
11.2 Backward Error Analysis for Forward Elimination and Back Substitutions Case 1. Lower Triangular System Consider solving the lower triangular system
Ly = b where
L = (l ); b = (b ; : : :; b ) ; 1
ij
and
(11.2.1)
n
y = (y ; : : :; y ) ; 1
using the forward elimination Scheme. We will use s^ to denote a computed quantity of s. 724
T
n
T
(11.2.2)
Step 1.
y^ = ( lb ) = l (1b+ ) ; 1
1
where
1
11
11
1
j j 1
(using law 4). This gives
l (1 + )^y = b 11
that is,
1
1
1
^l11y^1 = b1;
where
^l11 = l11(1 + 1 ):
This shows that y^1 is the exact solution of an equation line whose coecient is a number close to l11.
Step 2.
Similarly,
l y^ ) y^ = b ;l l y^ = b ; ( l ( b ; l y (1 + ))(1 + ) = l (1 + ) 2
2
2
21 1
21 1
22
2
22
21 1
11
22
(11.2.3) (11.2.4)
22
2
(using both (11.1.2), and (11.1.4)),
where j11j; j21j and j2j are all less than or equal to . Equation (11.2.3) can be rewritten as
l (1 + )(1 + )^y + l (1 + )^y = b (1 + ) 21
That is,
11
1
22
2
2
2
22
l (1 + )^y + l (1 + )^y = b ; 21
where
22
21
1
= 1 + ; 21
11
22
22
22
2
2
; = 1+ 22
2
22
22
(neglecting 11 22, which is small). Thus, we can say that y^1 and y^2 satisfy ^l21y^1 + ^l22y^2 = b2 where and
^l21 = l21(1 + 21) ^l22 = l22(1 + 22): 725
(11.2.5)
Step k.
The above can be easily generalized and we can say that at the kth step, the unknowns y1 through y satisfy ^l 1y^1 + ^l 2y^2 + + ^l y^ = b (11.2.6) where ^l = l (1 + ); j = 1; : : :; k. The process can be continued until k = n. Thus, we see that the computed y^1 through y^ satisfy the following perturbed triangular system: ^l11y^1 = b1 ^l21y^1 + ^l22y^2 = b2 .. . ^l 1y^1 + ^l 2y^2 + + ^l y^ = b k
k
kj
kj
k
kk
k
k
nn
n
n
kj
n
n
n
where ^l = l (1 + ); k = 1; : : :; n; j = 1; : : :; k. Note that 11 = 1 . These equations can be written in matrix form L^ y = (L + L)y = b kj
kj
kj
(11.2.7)
where L is a lower triangular matrix whose (i; j )th element (L) = l . Knowing the bounds for , the bounds for can be easily computed. For example, if n is small enough so that n < 1 , then j j 1:06(k ; j + 2) (see Chapter 2, Section 2.3). Then 100 j(L) j 1:06(i ; j + 2)jl j: (11.2.8) ij
ij
ij
ij
ij
kj
ij
ij
The above can be summarized in the following theorem.
Theorem 11.2.1 The computed solution y^ to the n n lower triangular system Ly = b obtained by forward elimination, satis es a perturbed triangular system: (L + L)^y = b; 1 . where the entries of L are bounded by (11.2.8) assuming that n < 100
Case 2. Upper Triangular System The round-o error analysis of solving an upper triangular system
Ux = c using back substitution is similar to case 1. We can state 726
Theorem 11.2.2 Let U be an n n upper triangular matrix and c be a vector.
Then the computed solution x^ to the system
Ux = c using back substitution satis es (U + U )^x = c where 1 . assuming n < 100
(11.2.9)
j(U ) j 1:06(i ; j + 2)ju j; ij
(11.2.10)
ij
11.3 Backward Error Analysis for Triangularization by Gaussian Elimination The treatment here follows very closely to the one given in Ortega (Numerical Analysis: A Second Course), and in Forsythe, Malcom and Moler (CSLAS). Recall that the process of triangularization using Gaussian elimination consists of (n ; 1) steps. At step k, the matrix A( ) is constructed which is triangular in the rst k columns; that is k
0a B B B B A =B B B B @
(k ) 11
...
a
(k ) ln
(k )
(k )
.. .
kk
a
a 1 .. C . C CC a C C: . . . ... C CA a
(k ) nk
(11.3.1)
(k ) kn
(k ) nn
The nal matrix A( ;1) is triangular. We shall assume that the quantities a( ) are the computed numbers. First, let's analyze the computations of the entries of A(1) from A in the rst step. Let the computed multipliers be m^ 1 ; i = 2; 3; : : :; n: k
n
ij
i
Then
m^ = ( aa ) = aa (1 + ); i1
where
i1
i1
11
11
j j : i1
727
i1
(11.3.2)
Thus, the error e(0)1 in setting a(1)1 to zero is given by i
i
e
= a(1)1 ; a 1 + m^ 1a11 = 0 ; a 1 + ( a 1 )(1 + 1)a11 a11 = 1a 1
(0) i1
i
i
i
i
i
i
i
(11.3.3)
i
Let us now compute the errors in computing the other elements a(1) of A(1). The computed elements a(1) ; i; j = 2; : : :; n are given by ij
ij
a
(1) ij
= (a ; (m ^ 1a )) = (a ; (m ^ 1a1 ))(1 + (1) ) = [a ; m ^ 1 a1 (1 + (1))](1 + (1)); i; j = 2; : : :; n; ij
i
ij
ij
i
ij
j
j
where
ij
j
ij
ij
j j ; j j : (1)
(1)
ij
ij
(11.3.3) can be rewritten as
a = (a ; m^ a ) + e ; i; j = 2; : : :; n: (1)
where
i1
ij
ij
ij
e = a ; m^ a ; i; j = 2; : : :; n: 1+ (0)
(1)
(1)
ij
ij
ij
i1
(1)
1j
(11.3.5)
(1)
ij
ij
From (11.3.4) and (11.3.5), we have
A = A;L A+E ; (1)
where
(11.3.4)
(0)
1j
0 0 B B m^ L =B B . B @ ..
(0)
1
(11.3.6)
(0)
0 0 0 0 1 BB C e e C B CC : E =B ... B@ ... CA e e
0 0 C 0 0C 21 C; (0) ... C CA m^ 1 0 0 Analogously, at the end of step 2, we will have
(0)
n
(0) 21
(0) 2n
(0) n1
(0)
A =A ;L A +E ; (2)
(1)
(1)
where L(1) and E (1) are similarly de ned. 728
(1)
(1)
nn
(11.3.7)
Substituting (11.3.6) in (11.3.7), we have
A = A ;L A +E = A;L A+E ;L A +E (2)
(1)
(1)
(1)
(1)
(0)
(0)
(1)
(1)
(11.3.8)
(1)
Continuing in this way, we can write
A
(n
;1) + L(0)A + L(1)A(1) + + L( ;2)A( ;2) = A + E (0) + E (1) + + E ( ;2) n
Since
L
(k
;1)
(k )
(k )
01 CC CC .. C; . C .. C .C A 0
k +1;k
L A =L A (k )
n
00 0 BB BB . 0 =B BB ... m^ . .. B@ .. 0
we have
n
(n
m^
n;k
;1);
k = 0; 1; 2; : : :; n ; 2:
(11.3.9)
Thus from (11.3.8) and (11.3.9), we obtain
A
;1) + L(0) A( ;1) + L(1)A( ;1) + + L( ;2)A( ;1)
(n
n
n
n
= A + E (0) + E (1) + + E ( ;2)
n
(11.3.10)
n
That is,
(I + L(0) + L(1) + + L( ;2))A( ;1) = A + E (0) + E (1) + + E ( ;2) n
n
n
729
(11.3.11) (11.3.12)
Noting now
I + L + L + +L (0)
(1)
0 1 BB m^ BB =B BB m^. B@ ..
0 1 m^ 32 ... m^ 2
21
;
(n 2)
31
n1
;1)
(n
0 C 0C CC 0C C = L^ (Computed L);
CC A
...
m^ ; 1 = U^ (computed U ), and denoting E + E + + E ; by E , we have from (11.3.11): m^
A
1
0 0 1 ...
n
(0)
n;n
(1)
1
(n
A + E = L^ U^
2)
(11.3.13)
where the entries of the matrices E (0); : : :; E ( ;2) are given by: 00 0 0 0 1 BB 0 0 0 0 CC BB . . . ( ;1) C .. .. .. e +1 e( +1;1) C CC ; k = 1; 2; : : :; n ; 1; E ( ;1) = B BB . . . . . . B@ .. .. .. .. .. .. C CA 0 0 e( ;1) e( ;1) n
k
k
k
k
;k
k
k
(k
;1) = a( ;1) k
i;k
e
(k ij
;1) =
(k ) ij
1+
(k )
i;k
a ; m^ a (k ) ij
(11.3.14)
k n;n
n;k
e
;n
(k
ik
i;k
;1) (
kj
k)
ij
;
; i; j = k + 1; : : :; n;
(11.3.15) (11.3.16)
ij
and
j j ;
j j ; ij
(11.3.17)
j j :
(11.3.18)
(k )
ik
and
(k )
ij
We formalize the above discussions in the following theorem:
730
Theorem 11.3.1 The computed upper and lower triangular matrices L^ and U^ produced by Gaussian elimination satisfy
A + E = L^ U^ where U^ = A( ) and L^ is lower triangular matrix of computed multipliers: 0 1 0 01 BB m^ C BB .21 .1 .0 . 0. CCC L=B BB ... . . . . . . . . . ... CCC .. .. . . .. C B@ .. A m^ 1 m^ 2 m^ ;1 1 n
n
Example 11.3.1 Let t = 2.
n
n;n
0 :21 :35 :11 1 B C A=B @ :11 :81 :22 CA :33 :22 :39
Step 1. m^ = ::11 21 = :52 m^ = ::33 21 = 1:57 21
31
a = :81 ; :52 :35 = :63 a = :22 ; :52 :11 = :16 (1) 22 (1) 23
a = :22 ; 1:57 :35 = ;:33 a = :39 ; 1:57 :11 = :22 (1) 32 (1) 33
0 :21 :35 :11 1 B 0 :63 :16 CC A =B @ A 0 ;:33 :22 (1)
e
(0) 21
= 0 ; [:11 ; :52 :21] = ;:0008
e
(0) 22
= :63 ; [:81 ; :52 :35] = :0020
e
(0) 23
= :10 ; [:22 ; :52 :11] = ;:0020
e
(0) 31
= 0 ; [:33 ; 1:57 :21] = ;:0003 731
e
= ;:33 ; [:22 ; 1:57 :35] = ;:0005
e
(0) 32 (0) 33
= :22 ; [:39 ; 1:57 :11] = :0027
0 0 1 0 0 B C E =B @ ;:0008 :0020 ;:0028 CA ;:0003 ;:0005 :0027 (0)
Step 2.
m = ; ::33 a = :22 + :52 :16 = :30 63 = ;:52; 0 :21 :35 :11 1 B 0 :63 :16 CC = U: A =B @ A ^ 0 0 :30 (2) 33
32
(2)
e
= 0 ; [;:33 + :52 :63] = :0024
e
= :30 ; [:22 + :52 :16] = ;:0032 00 0 1 0 B0 0 C = B 0 C @ A 0 :0024 ;:0032
(1) 32 (1) 33
E Thus
(1)
0 0 1 0 0 B C E=E +E =B @ ;:0008 :0020 ;:0028 CA ;:003 :0019 ;:0005 0 1 1 0 0 B C L^ = B @ :52 1 0 CA 1:57 ;:52 1 (0)
Since
(1)
we can easily verify that L^ U^ = A + E .
Bounds for the elements of E We now assess how large the entries of the error matrix E can be. For this purpose we assume that pivoting has been used in Gaussian elimination so that jm ^ j 1. Recall that the growth factor is de ned by max ja( )j = max ja j : ik
k
i;j;k
i;j
732
ij
ij
Let a = max ja j. Then from (11.3.14) and (11.3.15), we have i;j
ij
je
(k ik
;1)j a;
k = 1; 2; : : :; n ; 1; i = k + 1; : : :; n
and, for i; j = k + 1; : : :; n;
je j 1 ; ja j + ja (k )
(k )
(k
ij
ij
ij
;1)j
2 1 ; a:
(since m^ 1) ik
Denote 1 ; by .
Then
jE j = jE + + E ; j jE j + + jE ; j 0 20 0 0 1 B 0 66BB 1 2 2 CC BB 0 66BB CC BB 0 1 a 66B C +B B B C 66BB CC BB 4@ A BB @ 1 2 2 00 1 0 BB 0 BB . ++ B BB .. B@ 0 0 0 0 0 10 BB 1 2 2 2 CC BB CC B a B 1 3 4 4 C CC BB .. CA . @ 1 3 5 2n ; 2 (0)
(0)
(n
2)
(n 2)
2 2 0 0
01 C 0C CC 2C C
C C CC C A
2 13 0 C77 0C C7
CC77 CC77 0 0C A75 0 1 2 (11.3.19)
Remark: The inequalities (11.3.18) hold element-wise. We can immediately obtain a bound in terms of norms. Thus,
kE k1 a(1 + 3 + + (2n ; 1)) = an : 2
733
(11.3.20)
11.4 Backward Error Analysis for Solving Ax = b We are now ready to give a backward round-o error analysis for solving Ax = b using triangularization by Gaussian elimination, followed by forward elimination and back substitution. First, from Theorem 11.3.1, we know that triangularization of A using Gaussian elimination yields L^ and U^ such that A + E = L^ U^ . These L^ and U^ will then be used to solve: L^ y = b ^ = y: Ux From Theorem 11.2.1 and Theorem 11.2.2, we know that computed solution y^ and x^ of the above two triangular systems satisfy: (L^ + L)^y = b; and (U^ + U )^x = y^: From these equations, we have or
(U + U )^x = (L + L);1 b (L^ + L)(U^ + U )^x = b
or where (Note that A + E = L^ U^ ).
(A + F )^x = b
(11.4.1)
F = E + (L)U^ + L^ (U ) + (L)(U );
(11.4.2)
Bounds for F From (11.4.2) we have
kF k1 kE k1 + kLk1 kU k1 + kLk1kU k1 + kLk1kU k1: We now obtain expressions for kLk1; kU k1 ; kLk1 and kU k1 . Since, 1 0 1 CC BB m^ . . . 0 CC ; B L^ = B ... CA B@ ... m^ m^ ; 1 21
n1
n;n
734
1
from (11.2.8), we obtain
0 1 2 B CC B 3jm ^ j 2 B CC jLj 1:06 B .. ... B CA . @ (n + 1)jm ^ j 3jm ^ ;j 2 Assuming partial pivoting, i.e., jm ^ j 1; k = 1; 2; : : :; n ; 1; i = k + 1; : : :; n, we have 21
21
n;n
(11.4.3)
1
ik
kL^ k1 n kLk1 n(n2+ 1) (1:06)
(11.4.4) (11.4.5)
kU^ k1 na;
(11.4.6)
Similarly,
(note that U = A( ;1) ) n
and
kU k1 n(n2+ 1) 1:06a:
(11.4.7)
(note that max ju j a). ij
Also recall that
kE k1 n a ; 1 :
(11.4.8)
kLk1 kU k1 n a
(11.4.9)
2
Assume that n2 1 (which is a very reasonable assumption in practice), then 2
Using (11.4.4){(11.4.8) in (11.4.2), we have
kF k1 kE k1 + kLk1kU^ k1 + kL^ k1kU k1 + kLk1 kU k1 n a ; 1 + 1:06n (n + 1)a + n a 2
2
2
(11.4.10)
l and a kAk , from (11.4.10) we can write Since ; 1 1
kF k1 1:06(n + 3n )kAk1: 3
The above discussion leads to the following theorem. 735
2
(11.4.11)
Theorem 11.4.1 The computed solution x^ to the linear system Ax = b using Gaussian elimination with partial pivoting satis es a perturbed system (A + F )^x = b where F is de ned by (11.4.2) and kF k1 1:06(n3 + 3n2 )kAk1.
Remarks: 1. The above bound for F is grossly overestimated. In practice, this bound for F is very rarely attained.
Wilkinson (AEP, pp ) states that in practice kF k1 is always less than or equal to nkAk1 . 2. Making use of (11.2.8), (11.2.10), and (11.3.17), we can also obtain an element-wise bound for F . (Exercise)
11.5 Review and Summary In this chapter, we have presented backward error analyses for: 1. Lower and upper triangular systems using forward elimination and back substitution (Theorems 11.2.1 and 11.2.2). 2. LU factorization using Gaussian elimination with partial pivoting and (Theorem 11.3.1). 3. Linear systems problem Ax = b using Gaussian elimination with partial pivoting followed by forward elimination and back substitution (Theorem 11.4.1). Bounds for the error matrix E in each case has been derived (11.2.8, 11.2.10, 11.3.19, 11.4.11). We have merely attempted here to give the readers a taste of round-o error analysis, as the title of the chapter suggests. The results of this chapter are already known to the readers. They have been stated earlier in the book without proofs. We have tried to give formal proofs here. 736
To repeat, these results say that the forward elimination and back substitution methods for triangular systems are backward stable, whereas, the stability of the Gaussian elimination process for LU factorization and therefore, for the linear system problem Ax = b using the process, depend upon the size of the growth factor.
11.6 Suggestions for Further Reading The most authoritative book for learning details of backward error analyses of algorithms for matrix computations is the Wilkinson's Algebraic Eigenvalue Problem (AEP). Wilkinson's other book in this area, Rounding Errors in Algebraic Process, is also a good source of knowledge. The books A Second Course in Numerical Analysis, by James Ortega, and Computer Solutions of Linear Algebraic Systems, by Forsythe and Moler, have also given in-depth treatment of the material of this chapter.
737
Exercises on Chapter 11 1. Using = 10, and t = 2; compute the LU decomposition using Gaussian elimination (without pivoting) for the following matrices, and nd the error matrix E in each case such that
! 3 4 (a) A = ; 5 6 :25 :79 !
(b)
:01 :12 ! 10 9 (c) ;
A + E = LU:
;
8 5 ! 4 1 (d) ; 1 2 ! :01 :05 (e) : :03 :01 2. Suppose now that partial pivoting has been used in computing the LU factorization of each of the above matrices of problem #1. Find again the error matrix E in each case, and compare the bounds of the entries in E predicted by (11.3.12) with the actual errors. 3. Consider the problems of solving linear systems:
Ax = b using Gaussian elimination with partial pivoting with each of the matrices from problem #1 ! 1 and taking b = in each case. Find F in each case such that the computed solution x 1 satis es (A + F )x = b: Compare the bounds predicated by (11.4.11) with actual errors. 4. Making use of (11.2.8), (11.2.10), and (11.3.13{(11.3.17), nd an element-wise bound for F in Theorem 11.4.1. (A + F )x = b: 5. From Theorems 11.2.1 and 11.2.2, show that the process of forward elimination and back substitution for lower and upper triangular systems, respectively, are backward stable. 6. From (11.3.18), conclude that the backward stability of Gaussian elimination is essentially determined by the size of the growth factor . 738
7. Consider the problem of evaluating the polynomial
p( ) = a + a ; n
n
n
1
;1 + + a
n
0
by synthetic division:
p = a n
n
i = n; n ; 1; : : :; 1:
p ; = ( p + a ; ); i
1
i
i
1
Then p0 = p( ): Show that
p = a (1 + ) + a ; (1 + ) 0
n
n
n
n
1
n+1
;1 + + a0 (1 + 0 ):
n
Find a bound for each ; i = 0; 1; : : :; n: What can you say about the backward stability of the algorithm from your bounds? i
739
A. A BRIEF INTRODUCTION TO MATLAB A.1 Some Basic Information on MATLAB : : : : : : : : : : : : : : : : : : : A.1.1 What is MATLAB ? : : : : : : : : : : : : : : : : : : : : : : : : : : A.1.2 Entering and Exiting MATLAB : : : : : : : : : : : : : : : : : : : A.1.3 Two most important commands: HELP and DEMO : : : : : : A.1.4 Most frequently used MATLAB operations and functions : : A.1.5 Numerical Examples : : : : : : : : : : : : : : : : : : : : : : : : : : A.2 Writing Your Own Programs Using MATLAB Commands : : : : : : A.2.1 Some Relational Operators in MATLAB : : : : : : : : : : : : : A.2.2 Some Matrix Building Functions : : : : : : : : : : : : : : : : : : A.2.3 Colon Notation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : A.2.4 for, while, if commands : : : : : : : : : : : : : : : : : : : : : : : : A.2.5 Computing Flop-Count and Elapsed Time of An Algorithm. : A.2.6 Saving a MATLAB Program : : : : : : : : : : : : : : : : : : : : : A.2.7 Getting a Hard Copy : : : : : : : : : : : : : : : : : : : : : : : : : : A.2.8 Examples of Some Simple MATLAB Programs : : : : : : : : : A.2.9 Use of `diary' Command and Printing The Output : : : : : : :
: : : : : : : : : : : : : : : :
: : : : : : : : : : : : : : : :
: : : : : : : : : : : : : : : :
740 740 740 740 741 743 746 746 747 747 748 748 748 748 749 750
APPENDIX A A BRIEF INTRODUCTION TO MATLAB
A. A BRIEF INTRODUCTION TO MATLAB A.1 Some Basic Information on MATLAB A.1.1 What is MATLAB ? MATLAB stands for MATrix LABoratory. It is an interactive software package for solving problems arising in scienti c and engineering computations. MATLAB contains programs for all fundmental matrix computations such as solutions of linear systems, various matrix factorizations, solutions of least squares problems , eigenvalues and eigenvector computations, singular values and singular vectors computations, etc. It was developed by Cleve Moler. The most current version contains programs for many other types of computations including 2-D and 3-D graphic capabilitiies.
A.1.2 Entering and Exiting MATLAB On most systems the command matlab will let you enter into MATLAB. Give the command exit to quit MATLAB, when you are all done.
A.1.3 Two most important commands: HELP and DEMO The two most important commands are help and demo. Typing 'help' you will get the listing of all the matlab functions and other matlab capabilities. Typing 'help' followed by a matlab function name from the list will give you more speci c information about that particular function.
Example A.1.1 >> help norm
NORM Matrix or vector norm. For matrices.. NORM(X) is the largest singular value of X, max(svd(X)). NORM(X,2) is the same as NORM(X). NORM(X,1) is the 1-norm of X, the largest column sum, = max(sum(abs((X)))). NORM(X,inf) is the infinity norm of X, the largest row sum, = max(sum(abs((X')))). NORM(X,'inf') is same as NORM(X,inf). NORM(X,'fro') is the F-norm, sqrt(sum(diag(X'*X))).
NORM(X,P) is available for matrix X only if P is 1, 2, inf or 'fro'.
For vectors.. NORM(V,P) = sum(abs(V)^P)^(1/P). NORM(V) = norm(V,2). NORM(V,inf) = max(abs(V)). NORM(V,-inf) = min(abs(V)). In MATLAB 4.0, if X has complex components, z, then abs(z) = sqrt(z*conj(z)), not abs(real(z)) + abs(imag(z)), which was used in some earlier versions of MATLAB.
See also COND, RCOND, CONDEST, NORMEST.
Demo teaches you how to use the matlab functions such as how to enter matrix values into a
matrix, how to nd its transpose, how to nd the rank of a matrix, etc.
A.1.4 Most frequently used MATLAB operations and functions Some Basic Matrix Operations +
Plus
-
Minus
*
Matrix multiplication
.*
Array multiplication
^
power
\
Backslash or left division
/
Slash or right division
&
Logical AND
|
Logical OR
~
Logical NOT
Matrix functions - Numerical linear algebra. Matrix analysis. cond
- Matrix condition number.
norm
- Matrix or vector norm.
741
rcond
- LINPACK reciprocal condition estimator.
rank
- Number of linearly independent rows or columns.
det
- Determinant.
trace
- Sum of diagonal elements.
null
- Null space.
orth
- Orthogonalization.
rref
- Reduced row echelon form.
Linear equations. \ and /
- Linear equation solution; use "help slash".
chol
- Cholesky factorization.
lu
- Factors from Gaussian elimination.
inv
- Matrix inverse.
qr
- Orthogonal-triangular decomposition.
qrdelete
- Delete a column from the QR factorization.
qrinsert
- Insert a column in the QR factorization.
pinv
- Pseudoinverse.
lscov
- Least squares in the presence of known covariance.
Eigenvalues and singular values. eig
- Eigenvalues and eigenvectors.
poly
- Characteristic polynomial.
hess
- Hessenberg form.
svd
- Singular value decomposition.
qz
- Generalized eigenvalues.
rsf2csf
- Real block diagonal form to complex diagonal form.
cdf2rdf
- Complex diagonal form to real block diagonal form.
schur
- Schur decomposition.
balance
- Diagonal scaling to improve eigenvalue accuracy.
Using help you can get information on any of the above routines. Here is an exapmle.
Example A.1.2 >> help lu
742
LU
Factors from Gaussian elimination.
[L,U] = LU(X) stores a upper triangular matrix in U and a "psychologically lower triangular matrix", i.e. a product of lower triangular and permutation matrices, in L , so that X = L*U.
[L,U,P] = LU(X) returns lower triangular matrix L, upper triangular matrix U, and permutation matrix P so that P*X = L*U. By itself, LU(X) returns the output from LINPACK'S ZGEFA routine.
A.1.5 Numerical Examples To enter a matrix and a vector in MATLAB >> A=[1 3 5; 2 4 6; 1 3 9] A = 1
3
5
2
4
6
1
3
9
>> b=[1 1 1]'
b = 1 1 1
To solve the Linear system Ax = b >> x=A\b
x =
743
-0.5000 0.5000 0
To compute the inverse of A >> inv(A)
ans =
-2.2500
1.5000
0.2500
1.5000
-0.5000
-0.5000
-0.2500
0
0.2500
To compute the eigenvalues of A >> eig(A)
ans = -0.3246 12.3246 2.0000
To reduce A to an upper hessenberg matrix >> hess(A) ans =
1.0000
-4.9193
3.1305
-2.2361
8.6000
-6.2000
0
-3.2000
4.4000
To nd the rank of A 744
>> rank(A)
ans =
3
To compute the 2-norm of A >> norm(A)
ans =
13.3538
To compute the condition number of A (with respect to 2-norm) >>cond(A)
ans = 42.1539
To compute the QR factorization of A using Householder matrices. >> [q,r] = qr(A) will give the qr factorization of A q =
-0.4082
0.5774
-0.7071
-0.8165
-0.5774
-0.0000
-0.4082
0.5774
0.7071
-2.4495
-5.7155
-10.6145
0
1.1547
4.6188
0
0
2.8284
r =
745
To compute the lu factorization of A using partial pivoting. >> [l,u] = lu(A)
l = 0.5000
1.0000
0
1.0000
0
0
0.5000
1.0000
1.0000
u = 2
4
6
0
1
2
0
0
4
A.2 Writing Your Own Programs Using MATLAB Commands It is easy to write your own programs using MATLAB commands. We list rst some most common uses and then give examples. We urge students to get hold of a copy of Matlab Primer written by Professor Kermit Sigmon of University of Florida. This is available via anonymous ftp from Address: math.u .edu Deirectory: pub/matlab Files: primer.tex, primer.ps or Send e-mail to
[email protected] containing the single line send matlab/primer.tex
A.2.1 Some Relational Operators in MATLAB <
less than
>
greater than
<=
less than or equal to
746
>=
greater than or equal to
==
equal
A.2.2 Some Matrix Building Functions eye
identity matrix
zeros
matrix of zeros
rand
randomly generated matrix with entries between zero and one.
max
max entry
diag
create or extract a diagonal
triu
upper triangular part of a matrix.
tril
lower triangular trianluar part of a matrix
hilb
Hilbert matrix
toeplitz
Toeplitz matrix
size
gives the dimension of a matrix
abs
absolute value of the elements of ammatrix or a vector.
Examples
1. rand(5,3) will create a 5 x 3 randomly generated matrix 2. If x is a vector diag(x) is the diagonal matrix with entries of x on the diagonal. diag(A) is the vector consisting ot the diagonal of A 3. hilb(5) will create the 5 x 5 Hilbert matrix. 4. max(A(:,2)) will give the maximum value of the second column of A. max(max(A)) will give the maximum entry of the whole matrix A.
A.2.3 Colon Notation The 'colon' notation (:) is used to conveniently indicate a whole row or column
Examples
A(:,2) denotes the entire 2nd column of A. A(1:3,:) denotes the rst 3 rows of A. A(:,[2,5,7]) denotes the second, fth and seventh columns of A The statement A(:,[2,5,7]) = B(:,1:3) will replace the columns 2,5,7 of A with the rst three columns of B
747
A.2.4 for, while, if commands These commands are most useful in writing MATLAB programs for matrix algorithms. The uses of these commands will be illustrated in the following examples.
A.2.5 Computing Flop-Count and Elapsed Time of An Algorithm. To compute the op count of an algorithm set ops(0) immediately before executing the algorithm and type ops immediately after completion.
Example
flops(0) \\ x
= a\b
flops
will give the total ops to solve a linear system with a given matrix a and a vector b. The function cputime returns the CPU time in seconds that has been used by the MATLAB process since the MATLAB process started. For example : t = cputime; your operation; cputime ; t returns the cputime to run your operation. Since the PC version of MATLAB does not have a cputime function, MATCOM contains a program cputime.m. If the version of matlab that you use contains the cputime function then you can delete the cputime.m program from MATCOM.
A.2.6 Saving a MATLAB Program The command save < filename > will store all variables in the lename ' lename.mat'. The command load < filename > will restore all the variables from the le named ' lename.mat'.
A.2.7 Getting a Hard Copy diary< filename > will cause a copy of all subsequent terminal input and most of the resulting output to be writen on the named le. diary o suspends it. 748
A.2.8 Examples of Some Simple MATLAB Programs Example A.2.1 The following code will make the elements below the diagonal of the 4 x 4 matrix A = a(i,j) zero. a = rand(4,4) for i = 1:4 for j = 1:4 if i > j a(i,j) = 0 end; end; end;
Example A.2.2 The following code will create a matrix A such that the (i; j )-th entry of the matrix A = a(i; j ) is (i + j ). a= zeros(4,4) for i = 1:4 for j = 1:4 a(i,j) = i+j end; end;
Example A.2.3 The following MATLAB program computes the matrix-matrix product of two upper triangular matrices. This program will be called matmat.m % Matrix-Matrix product with upper triangular matrices % input U and V two upper triangular matrices of order n % output C = U * V % function
C = matmat(U,V)
749
function
C = matmat(U,V)
[n,m] = size(U) C = zeros(n,n) for i = 1:n for j = i:n for k = i:j C(i,j) = C(i,j) + U(i,k) * V(k,j) end; end; end;
end;
Example A.2.4 (Matlab Implementation of Algorithm 3.1.1) % Computing
the two norm of a vector x
% input x % output
nrm, the two norm of the vector x
% function
nrm = twonorm(x)
function
nrm = twonorm(x)
[n,m] = size(x) r = max(abs(x)) y=x/r s = 0 for i=1:n s = s + y(i)^2 end; nrm = r * s^0.5 end;
A.2.9 Use of `diary' Command and Printing The Output .
Sometimes you may want a listing of your matlab program step by step as it was executed. The 750
diary command can be used to create a listing.
Example: >>diary B:diary8
<----- sets the diary on
>> C = matmat(U,V)
<----- execute your program
>> diary off
<----- set the diary off
The above command will store all the commands that are executed by the program matmat.m in the le diary8. Only that data which is printed on the screen is stored in diary8. This le can then be printed. In case you do not want the output to be printed on the screen place a semi colon at the end of the line. To write a comment-line put the % sign in the rst column. Examples of some M les are given in appendix B.
751
B. MATLAB AND SELECTED MATLAB PROGRAMS B.1 MATCOM and Some Selected Programs From MATCOM B.1.1 What is MATCOM ? : : : : : : : : : : : : : : : : : : : : B.2 How To Use MATCOM ? : : : : : : : : : : : : : : : : : : : : : B.3 Chapter-wise Listing Of MATCOM Programs : : : : : : : : B.4 Some Selected MATCOM Programs : : : : : : : : : : : : : :
: : : : :
: : : : :
: : : : :
: : : : :
: : : : :
: : : : :
: : : : :
: : : : :
: : : : :
752 752 752 755 759
APPENDIX B MATCOM AND SELECTED MATLAB PROGRAMS FROM MATCOM
B. MATLAB AND SELECTED MATLAB PROGRAMS B.1 MATCOM and Some Selected Programs From MATCOM B.1.1 What is MATCOM ? MATCOM is a MATLAB based interactive software package containing the implementation of all the major algorithms of chapter 3 through chapter 8 of the book ' Numerical Linear Algebra and Applications ' by Prof. Biswa Nath Datta. For each problem considered in this book, there are more than one (in some cases several) algorithms. By using the programs in MATCOM the students will be able to compare di erent alogrithms for the same problem with respect to accuracy , op-count, and elapsed time , etc. The students will be able to verify the statements about goodness and badness of di erent algorithms. In particular, they will be able to distinguish between a good and a bad algorithm. MATCOM has been written by Conrad Fernandes, a graduate student of Professor Datta.
B.2 How To Use MATCOM ? .
In section B3 you have a chapter wise listing of the programs in MATCOM
Example
create a householder matrix
housmat.m 5.3.1 / /
The program name is /
\ \ \ is the algorithm or the section number
To use this program in matlab, you need to know the input and output variables. To nd out these variables type >> help housmat create a householder matrix 5.3.1 housmat.m H = I - 2 *u*u'/(u'*u) input
vector x
output [ u,H]
function [u,H] = housmat(x)
The input is a vector x and the output are a matrix H and a vector u. To execute the program you have to do the following 1. create the input vector x 2. then type [u,H] = housmat(x)
As output you will get the Householder matrix H such that Hx is a multiple of e1 and the vector u out of which the Householder matrix H has been formed. >> x = rand(4,1)
x =
0.2190 0.0470 0.6789 0.6793
>> [u,H] = housmat(x) u =
1.7740 0.0693 0.9994 1.0000 H = -0.2220
-0.0477
-0.6884
-0.6888
-0.0477
0.9981
-0.0269
-0.0269
753
-0.6884
-0.0269
0.6122
-0.3880
-0.6888
-0.0269
-0.3880
0.6117
754
B.3 Chapter-wise Listing Of MATCOM Programs CHAPTER 3 Title Program Name Number Back substitution backsub.m 3.1.3 The Inverse of an Upper Triangular Matrix invuptr.m 3.1.4 Basic Gaussian Elimination gauss.m 3.1.5 CHAPTER 4 Title Program Name Number T T Computing (I ; 2(u u )=(u u)) A housmul.m 4.2.1 Computing A (I ; 2(u uT )=(uT u)) houspostmul.m 4.2.1 (sec. num.) CHAPTER 5 Title Program Name Triangularization Using Gaussian elimination lugsel.m
Without Pivoting Triangularization Using Gaussian Elimination With Partial Pivoting Triagularization Using Gaussian elimination With Complete Pivoting Creating Zeros in a Vector With a Householder Matrix Householder QR Factorization Householder QR Factorization for a Nonsquare Matrix Householder Hessenberg Reduction Creating Zeros in a Vector Using Givens Rotations Creating Zeros in a Speci ed Position of a Matrix Using Givens Rotations QR factorization Using Givens Rotations Givens Hessenberg Reduction 755
Number
5.2.2
parpiv.m
5.2.3
compiv.m
5.2.4
housmat.m housqr.m
5.4.1 5.4.2
housqrnon.m houshess.m
5.4.2 5.4.3
givcs.m and givrot.m 5.5.1 (sec. num.) givrota.m givqr.m givhs.m
5.5.1 5.5.2 5.5.3
CHAPTER 6 Forward Elimination forelm.m Solving Ax = b with Partial Pivoting gausswf
without Explicit Factorization Cholesky Algorithm Sherman Morrison Formula Inverse by LU Factorization without Pivoting Inverse by Partial Pivoting Inverse by Complete Pivoting Hager's norm-1 condition number estimator Iterative Re nement The Jacobi Method The Gauss-Seidel Method The Succesvie Overrelaxation Method The Basic Conjugate Gradient Algorithm Incomplete Cholesky Factorization No-Fill Incomplete LDLT
756
6.4.1 6.4.3
choles.m shermor.m inlu.m
6.4.4 6.5.2 (sec. num.) 6.5.3 (sec. num.)
inparpiv.m incompiv.m hagnormin1.m iterref.m jacobi.m gaused.m sucov.m congrd.m icholes.m nichol.m
6.5.3 (sec. num.) 6.5.3 (sec. num.) 6.7.1 6.9.1 6.10.1 6.10.2 6.10.3 6.10.4 6.10.6 6.10.7
CHAPTER 7 Title Program Name Number Least Squares Solution Using Normal Equations lsfrnme.m 7.8.1 The Householder-Golub Method For the Full-Rank lsfrqrh.m 7.8.2
Least Squares Problem Classical Gram-Schmidt for QR factorization Modi ed Gram-Schmidt for QR factorization Least Squares Solution by MGS Least Squares Solution for the Rank-De cient Problem Using QR Minimum Norm Solution for the Full-Rank Underdetermined Problem Using Normal Equations Minimum Norm solution for the Full-Rank Underdetermined Problem Using QR Linear Systems Analog Least Squares Iterative Re nement Iterative Re nement for Least Squares Solution Computing the Variance-Covariance Matrix
Title
clgrsch.m mdgrsch.m lsfrmgs.m
7.8.3 7.8.4 7.8.5
lsrdqrh.m
7.8.6
lsudnme.m
7.9.1
lsudqrh.m
7.9.2
lsitrn1.m lsitrn2.m reganal.m
7.10.1 7 .10.2 7.11.1
CHAPTER 8
Program Name power.m invitr.m rayqot.m senseig.m qritrb qritrh.m
Power Method Inverse Iteration Rayleigh-Quotient Iteration Sensitivities of Eigenvalues The Basic QR Iteration The Hessenberg QR Iteration The Explicit Single Shift QR Iteration qritrsse.m The Explicit Double Shift QR Iteration qritrdse One Iteration-Step of the Implicit Double Shift QR qritrdsi.m
NO CHAPTER 757
Number
8.5.1 8.5.2 8.5.3 8.7.2 (sec. num.) 8.9.1 (sec. num.) 8.9.2 (sec. num.) 8.9.4 (sec. num.) 8.9.5 (sec. num.) 8.9.1 (section 8.9.6)
Title
Program Name Number The Absolute Maximum of a Vector absmax.m Interchange Two Vectors inter.m Computing the CPU Time cputime.m
758
B.4 Some Selected MATCOM Programs % Back substitution 3.1.3
backsub.m
% input upper triangular T and vector b % output [ y] the solution to Ty = b by back subst % function [y] = backsub(T,b); function [y] = backsub(T,b); %
!rm diary8
%
diary diary8
[m,n] = size(T); if m~=n disp('matrix T
is not square')
return end; y = zeros(n,1) for i = n:-1:1 sum = 0 for j = i+1:n sum = sum + T(i,j)*y(j) end; y(i) = (b(i) - sum ) / T(i,i) end;
end; % inverse of an upper triangular matrix
3.1.4 invuptr.m
% the matrix T is overwritten by its inverse % function [T] = invuptr(T); function [T] = invuptr(T); %
!rm diary8
%
diary diary8
[m,n] = size(T); if m~=n disp('matrix T
is not square')
return end;
759
s = eye(n,n) for k = n:-1:1 T(k,k) = 1/T(k,k) for i = k-1 :-1 :1 sum = 0 for j = i+1:k sum = sum + T(i,j)*T(j,k) end; T(i,k) = -sum/T(i,i) end; end;
end; % pre multiply a
matrix by a Housholder
matrix
% compute [I - 2uu'/u'*u]*A % input
A and vector u
% output [A] % function [A] = housmul(A,u) function [A] = housmul(A,u) %
!rm diary8
%
diary diary8
[m1,n] = size(A);
beta = 2/(u'*u) for j = 1 : n alpha = 0 for i = 1 : m1 alpha =
alpha + u(i) * A(i,j)
end; alpha =
beta * alpha
for i = 1:m1 A(i,j) = A(i,j) - alpha * u(i) end; end;
760
housmul.m
4.2.1
end; % LU factorization using Gaussian Elimination Without Pivoting 5.2.2 % input
A
% output [l, u] % function [l,u] = lugsel(A) function [l,u] = lugsel(A) %
!rm diary8
%
diary diary8
[m1,n] = size(A); for k = 1 : n-1 for i = k+1:n A(i,k) =
A(i,k)/ A(k,k)
end; for i = k+1:n for j = k+1 :n A(i,j)
= A(i,j)
- A(i,k) * A(k,j)
end; end; end; u = triu(A) l =
tril(A,-1)
for i = 1:n l(i,i) = 1 end; end;
% Creating Zeros in a Specified Posn. of a % matrix A using Givens Rotations 5.5.5 givrota.m % this program calls the MATCOM program givcs.m % input
i j
and matrix
A
% output [A] , A = J * A % function [A] = givrota(i,j,A)
761
lugsel.m
function [A] = givrota(i,j,A) [m,n] = size(A) %
!rm diary89
%
diary diary89 x = zeros(2,1) x(1) = A(i,i) x(2) = A(j,i) [c,s] = givcs(x) J = eye(n,n) J(i,i) = c J(i,j) = s J(j,i) =-s J(j,j) = c
A = J*A
end;
%Solving Ax = b with partial pivoting without Explicit Factorization % gausswf.m 6.4.2 % input
matrix A and vector b
% output [A, b]
matrix A is overwritten with the upper triangular
% part A^(n-1) and the multipliers are stored in the lower % triangular part of A. The vector b is overwritten by b^(n-1). % function [A,b] = gausswf(A,b) function [A,b] = gausswf(A,b) %
!rm diary8
%
diary diary8
[m1,n] = size(A); for k = 1 : n-1 for i = k+1:n A(i,k) =
A(i,k)/ A(k,k)
end;
762
for i = k+1:n b(i) = b(i) + A(i,k) * b(k) for j = k+1 :n A(i,j)
= A(i,j)
- A(i,k) * A(k,j)
end; end; end; u = triu(A) l =
tril(A,-1)
for i = 1:n l(i,i) = 1 end; end; % Jacobi method 6.10.1 jacobi.m % input
matrix A x b and numitr
% output [xold] % function [xold] = jacobi(A,x,b,numitr); function [xold] = jacobi(A,x,b,numitr); %
!rm diary8
%
diary diary8
[m,n] = size(A); if m~=n disp('matrix a
is not square')
return end; xold = x Bsub = zeros(n,n) for
i = 1 : n for j = 1 : n if i ~= j Bsub(i,j) = -A(i,j) / A(i,i)
end; end;
763
end;
bsub = zeros(n,1) for i = 1 : n bsub(i) = b(i) / A(i,i) end;
for i = 1 : numitr disp('the iteration number') i xnew = Bsub * xold
+ bsub
xold = xnew; end end; % Householder-Golub method for the least squares problem 7.8.2 % input matrix A and vector b %
output [ x]
% function [x] = lsfsqrh(A,b); function [x] = lsfrqrh(A,b); %
!rm diary8
%
diary diary8
[m,n] = size(A); y = zeros(n,1) [q,r1] = qr(A) c = q'*b ran = rank(r1) r = r1(1:ran,:) [r,c,x] = backsub(r,c)
[m1,n] = size(A); if m1~=n disp('matrix A
is not square')
764
lsfrqrh.m
return end;
[q,r] = qr(A) for k = 1 : numitr disp('the iteration number') k Anew = r*q [q,r] = qr(Anew) end;
xold = diag(Anew) end;
765
