Linear Algebra - Solved Assignments - Fall 2006 Semester

Solution File Assignment #1 of Linear Algebra (Fall 2006) Total marks: 30

Question # 1

1  0  5  2        Let a 1   2  , a 2  1  , a 3   6  , b   1 0   2  8   6  Determine whether b can be written as a linear combination of a1, a 2, a 3 ?

Solution: If b can be written as linear combination of Then

a ,a ,a 1

x1a1  x2 a2  x3a3  b 1  0  5   2   2  1    6   1 x1   x 2   x 3     0   2 8   6  Here we have to find the value of

x1

x1 , x2 , x3 5 x3  2

2 x1  x2  6 x3  1 2 x2  8 x3  6 The argument matrix of this system is as follow 1 0 5 2  2 1 6 1    0 2 8 6  Now doing row operation on this matrix 1 0 5 2  0 1 4 3  2 R  R 2   1 0 2 8 6  And on the last row we perform the operation 2R2  R3 we get the final matrix as

2

3

1 0 5 2  0 1 4 3    0 0 0 0  Now we have solution from this matrix x1  5 x3  2 i.e x1  2  5 x3 x2  4 x3  3 i.e x2  3  4 x3 x3 Is free (we can take any value) There exist a solution of these equation therefore we can write b as linear combination for example if we take x3  2 then it will give x1  8 and x2  5 . So we can write b as linear combination as follow 1  0  5   2      x1  2   x2 1   x3  6    1 0   2  8  6  Putting the values of x1  8 , x2  5 and x3  2 we have

1  0  5   2      8  2   5 1   2  6    1 0   2  8  6  Thus b can be written as a linear combination of

a ,a a 1

2

3

Question # 2 Describe all the solutions of the following system in parametric vector form x1  3x2  5x3  4

x  4 x  8x  7 3x  7 x  9 x  6 1

2

1

3

2

3

Solution: The argument matrix of this system is  1 3 5 4   1 4 8 7     3 7 9 6  After doing row operation 1R1  R2 and 3R1  R3 on row 2 and row 3. We get

1 3 5 4   0 1 3 3     0 2 6 6  After doing row operation 2R2  R3 on row 3 we get 1 0   0 After doing row operation

3 5 4  1 3 3  0 0 0  3R2  R1 on row 1 we get the final matrix

 1 0 4 5   0 1 3 3     0 0 0 0  From matrix we have

x1  4 x3  5 i.e x1  5  4 x3 x2  3x3  3 i.e x2  3  3x3 x3 is free

 x1   5  4 x3   5  4   x    3  3x    3   x  3  3   2    3   x3   x3   0   1  here  5  v   3   0   4  p   3   1  so x  v  x3 p

Which is the required parametric vector form.

Question # 3

0  0  4      Let v1  0  , v 2   3 , v 3   1 Does {  2  8   5 Why or why not?

v ,v ,v 1

2

3

3

Solution 0  0  4        v1  0  , v 2  3 , v 3  1  2  8   5 We have to determine whether arbitrary vector b=(b1 ,b 2 ,b3 ) in R 3 can be expressed as a linear combination b=k1v1 +k 2 v 2 +k 3 v3 of a vectors v1 , v 2 , v3 Expressing this terms of components gives 0  0  4      (b1 ,b 2 ,b3 )=K1 0   K 2  3  K3  1  2  8   5 0K1  0 K 2   4 K 3  (b1,b2,b3)= 0K1    3K 2    1K 3   2K1  8 K 2   5 K 3  b1  0 K1  0 K 2  4 K 3 b2  0 K1  3K 2  K 3 b3  2 K1  8 K 2  5 K 3 0 0 4   0 3 1  2 8 5 It has non zero determinant So we have det  A   24  0, its maens that system is consis tan t Therefore { v1 , v2 , v3} span R 3

} span R ?

Solution File Of Assignment #2 of Linear Algebra (Fall 2006) Question # 1 Determine whether the columns of the Independent or not. 8 0 3 7   1 5  3 1

following matrix are linear 5 4   4  2 

Solution: We have to see that the equation Ax=0 has trival solution or non trival solution. Take the matrix  0 8 5 0   3 7 4 0      1 5 4 0     1 3 2 0  Interchange row 4 and row 1. We have  1 3 2 0   3 7 4 0      1 5 4 0     0 8 5 0  After doing row operation 3R1  R2 on row 2 we get  1 3 2 0   0 2 2 0     1 5 4 0     0 8 5 0  After doing row operation R1  R3 on row 3 we get  1 3 2 0   0 2 2 0     0 2 2 0     0 8 5 0  After doing row operation 1R2  R3 on row 3 we get

 1 3 2  0 2 2  0 0 0   0 8 5

0 0  0  0

 1 3 2  0 2 2   0 8 5  0 0 0

0 0  0  0

Interchange row 3 ad row 4

1 R2 and 8R2  R3 respectively we get following matrix 2 1 3 2 0  1 3 2 0  0 1 1 0  0 1 1 0     0 8 5 0  0 0 3 0      0 0 0 0  0 0 0 0  As we can see from last matrix that there is no free variable and there are three basic variables x1 , x2 , x3 . So the equation Ax = 0 has only the trivial solution and column are linearly independent.

After doing row operation

Question # 2 A linear transformation T is defined as T(x)=Ax. Find a vector x whose Image under T is ‘b’ and determine whether x is unique. Where 0  2 1  1  2  1 6  , b = 7  A=  3  3 2  5 

Solution: For this we have to solve Ax = b or  1 0 2   x1   1  2 1 6   x   7    2    3 2 5  x3   3 The augmented matrix is  1 0 2 1  2 1 6 7     3 2 5 3 We have to do row operations. First 2R1  R2 on row2 and then 3R1  R3 on row 3 we get following matrix respectively.

1 0 2 1 1 0 2 1 0 1 2 5   0 1 2 5    3 2 5 3 0 2 1 0  1 After doing row operations 2R2  R3 and R3 respectively on row3 we get following two 5 matrixes respectively 1 0 2 1 1 0 2 1 0 1 2 5   0 1 2 5      0 0 5 10  0 0 1 2  At last we do the row operation 2R3  R1 on row 1 and 2R3  R2 on row2 we get the following matrix 1 0 0 3  0 1 0 1    0 0 1 2  Here we have

x1  3 x2  1 x3  2 3   Vector x is  1  whose Image under T is ‘b’  2  Yes x is unique because there exist a unique solution. Question # 3

R  R be a linear transformation such that x , x ) = ( x  x , 4x  5x ) .Find x such that T(x)=(3,8) 2

Let T : T(

1

2

2

1

2

Solution:  x x  T ( x)   1 2   4 x1  5 x2  3 Here T ( x)    after putting we get 8 3  x1  x2   8      4 x1  5 x2  After equating both sides we get x1  x2  3 4 x1  5 x2  8 The augmented matrix of this system is

1

2

1 1 4 5 

3 8

First applying 4R1  R2 on row 2 and then 1R2  R1 on row 1 we get the following two matrixes respectively. 3  1 0 7 1 1 0 1 4  0 1 4     From last matrix we have x1  7

x2  4 So x is

x   7  x   1     x2   4

Solution File Of Assignment # 3 of Linear Algebra (Fall 2006) Total marks: 25

Question # 1 Solve the equation Ax=b by using LU-Decomposition for the given matrix 3 4  2 6    5 10  , b = 16  A = 4  4  2  8 2  Solution: Marks 10

2 U   4  4

3 5 8

4  10  2 

1 U   4  4

3/ 2 5 8

2 1 10  R1 2 2 

1 U  0 0

3/ 2

1 U  0 0

3/ 2 1 2

1 U  0 0

3/ 2

2 2  2   4 R1  R2 and  4 R1  R3 , L   4  4  6  2 0 0 2    2   1R2 , L  4 1 0   4  6  * 1 2 0 0 2    2   2 R2  R3 , L  4 1 0   4  2  2 1

1 2

1 0

,

,

1 L  * *

2 L  * *

0 1 *

0 1 * 0 0  1 

U 1 0  0

3/ 2 1 0

2 1  2   R3 2 1 

,

we see that A  LU , let Ly  b : 2 4   4

0 1 2

0   y1  6  0   y2   16   2   y3   2 

y1  3 y2  4 y3  1 and Ux  y 1 0  0 x1  4 x2  2 x3  1

3/ 2 1 0

2   x1  3   2   x2    4  1   x3  1 

2 L   4  4

0 1 2

0 0  1

0 0   2 

0 1 *

0 0  1

Question # 2 Solve the equation Ax=b by taking inverse of the matrix of the following system of equations. x1  2 x 3  5 3 x1  x 2  4 x 3  2 2 x1  3 x 2  4 x 3  1

Solution:

Marks 10

 2 4  4 

0 1  A=  3 1  2  3

 x1  , x   x2   x3 

,

5  b   2  1 

by [A I3 ] finding A -1 : 0 1  3 1   2  3 1  0  2 1  0 0

2

1

0

4

0

1

4

0

0

0 0  1 0 0  3R1  R2 1 

0 1

2 2

1 3

0 1

3

4

0

0

0 1

2 2

1 3

0 1

3

8

2

0

1  0 0

0 1

2 2

1 3

0 1

0

2

7

3

1  0 0

0

0

8

1 0

0 2

10 7

1  0 0

0

0

8

1 0

0 1

10 7/2

0  0   2 R1  R3 1 0 0  3R2  R3 1 

1 4 1 R3  R2 and R3  R1 3 1  3 1 4 1  R3 / 2 3/ 2 1/ 2 

3

Henc 1

we can easily check that A A  8 A  10  7 / 2 1

8 x  A b  10  7 / 2 1

 x1  35   x    43  2    x3  15 

A

1

A I

1  1  1/ 2 

3 4 3/ 2 3 4 3/ 2

1  5  1   2  1/ 2  1 

Question # 3

1 1  2 Let A =  0 5 5 the other columns.

1 3  .Find the third column of 1

1

A

without computing Marks 5

Solution

After performing hte row operations we get, 1 1 1  0 2 3 5 5 1

* * 0  * * 0 * * 1 

1  0 0 1  0 0

1 1 2 3 0 4 1 1 2 3 0 1

* * 0  * * 0 * * 1  * * 0   * * 0  * * 1/ 4 

1  0 0 1  0 0 1  0 0

1 1 2 0 0 1

* * 0   * * 3/ 4 * * 1/ 4 

1 1

* *

1 0 0 1

* * * *

1 0 1 0 0 1

* * * * * *

0   3/8 1/ 4  1/ 4   3/8 1/ 4 

R 3   5R 1 

 1    R3  4  R 2   3R 3 

1   R2 2 R 1   R 3 

R 1   R 2 

1 0 0 * *  1/ 8    * * 3/8  0 1 0 0 0 1 * *  1/ 4  Third column of without computing the other columns is given by,

A

1

* *  * * * *

 1/ 8  3 / 8   1/ 4 

Solution File OF Assignment no.4 Fall 2006 (Linear Algebra)

Question # 1

0 Let A=  2 5 space of A.

1 3 8

 4 8   2  and b= 1  .Determine whether b is in the column 1  7  Marks 10

Solution: To determine the b is in the column space of A, we see that the aug. matrix is consist ant or not. Now Row reducing the augmented matrix [A b]

0 ~  2 5 2 ~ 0 5

1 3

4 2

8

7

3 1

2 4

8

7

2 ~ 0  0 2 ~ 0  0

3 1

2 4

 1/ 2 3

8 1  1  1 8 1 

2 2

1

4

0

0

1  8   3 / 2   8  5 / 2  1

We conclude that Ax = b is inconsistent and So b is not in the Col of A.

Question # 2 Find the rank of the following matrix,

1 5  4   2

2

4

3

10

9

7

8

9

2

5

0

4

3 8  7   6

Marks 10 2 4 3 3 1 5 10  9  7 8   Solution: A= 4 8 9 2 7   4 5 0  6  2 We have to find the reduced row echelon form Applying -5R1 +R 2 , -4R1 +R 3 , 2R1 +R 4

1 0  0  0

2 4 3 0 11  22 0 7  14 0 3 6

1 0  0   0

2

4

0

11

 22

0 0

0 0

0 0

3  7  5   0  Applying 3R 3 +7R 4 ,3R 2+11R 4 3

3   7   15    21

Applying -21R 3 +15R 4 1 0  0   0

2 0 0 0

4 3 11  22 0 0 0 0

3  7   15  0 

Since The Matrix has 3 pivot columns ,so the rank A = 3

Rank(A) = 3

Question # 3

By using Cramer’s Rule, solve the following system of equations, 2 x1  x 2  x 3  4  x1  2 x 3  2 3 x1  x 2  3 x 3  2 2 x1  x 2  x 3  4  x1  2 x 3  2

Solution:

3 x1  x 2  3 x 3  2

Take determinant,

2

1 1

D  1 0 2  2 3

1 3

0 2  1 1 1 3 3

= 2(-2) - 1(-3-6) + 1(-1 ) = 2(-2) - 1(-9) - 1 = -4 +9 - 1 =4 D=4

2   1 0  1 3  3 1 

Marks 10

4 D1  2 2

1 1 0 2  16 1 3

D1 16  4 D 4  D2   2  52   x1 

x2 

D2 52   13 D 4  

D3      x3 

D3 4   1 D 4

So x1 =

-4

x2 = 13 x3 = -1

   4  

Solution File of Assignment No. 5 LINEAR ALGEBRA (Fall 2006)

Question # 1 Find the dimension of Null Space and Column Space for the matrix 4 A= 3 1

1

2

2

0

1

2

2 1  1 

SOLUTION: In order to find the dimension of the column Space we have to Row Reduced the given matrix 4 A= 3 1

1

2

2

0

1

2

2 1  1 

To Echelon Form: 4  A   3  1

2

1

2

2

0

1

2

 1   1 

1  ~  3   4 1  ~  0   0

1

2

1

2

0

1

2

1

2

5

6

5

6

1  ~  0   0

1

2

5

6

0

0

 1  change R with R 1 3  2  1   2   3R  R  ,  4 R  R  1 2  1 3    2 

1

  2    R  R  2 3   0 

1  ~  0   0

1

2

1  6/5 0

0

1

  2/5  1 R 5 2  0 

Thus A has two pivot column so the dimension of ColA=2. For Null space, we need the reduced echelon form. Further row operations on A yields 1  ~  0   0

0

4/5

1

 6/5

0

0

3/5 

  2/5   0 

A has two free variables x3 and x4.

NullA=2 So dim of ColA + dim of NullA=n(the no. of Columns of A) Which is true. Hence Dimension Theorem is true. Question # 2 Dimensions of Null Space of A,

4 Let A=  1  2

2 1 4

3  3  . An Eigenvalue of A is 3.Find a basis for 9 

the corresponding eigenspace.

Solution: The scalar 3 is an Eigenvalue of A if and only if the equation AX  3X Has the nontrivial solution. AX  3 X

 A  3I  X  0 Now we solve A-3I

 4    1   2  1    1   2

3   A  3I 1  3    0   4 9   0 2 3  2  3   4 6  Row reduced the augmented matrix for  A  3I  X  0 :  1   1    2 1  ~  0   0

2

3

2

3

4

6

2

3

0

0

0

0

A has the free variables The general solution is:

2

3

0  0   0 

0  0   0 

 R1  R2 , 2R1  R3 

x

2

and

x

3

so 3 in Eigenvalue of A.

 x1   2   3  x   x  1  x  0 2   2  3   x3   0   1    2   3     The basis is for the eigenspace is   1 .  0   .  0  1       

0 3 0

0

 0   3

Question # 3 3 Is  =4 an Eigenvalue of  2  3

0 3 4

 1 1  ? If so find the 5

corresponding eigenvector.

Solution: 3 Suppose A=  2  3

0 3 4

 1 1  5

The scalar 4 is an Eigenvalue of A if and only if the equation AX  4 X

Has the nontrivial solution. AX  4 X

 A  4I  X  0 Solve the A  4I .  3 A  4 I   2  3  1   2  3

 1  4 1   0 5 0

0 3 4 0 1 4

0 4 0

0 0  4 

 1 1  1

Row reduced the augmented matrix for A  4I :  1  2   3  1 ~  0  0

0 1 4

1 1 1

0 0  0 

0 1

1 1

4

4

0 0   2 R1  R2  ,  3R1  R3  0 

 1 ~  0  0 1 ~ 0  0

0 1 0

1 1 0

0 1

1 1

0

0

0 0  4 R2  R3 0 

0 0   R1 ,  R2 0 

A has free variable

x

3

so 4 is Eigenvalue of the matrix A.

Let

x =1 x  x  0  x 1  0 x  x  0  x 1  0 So x  1 and x  1 3

1

3

1

2

3

2

1

2

 1 1  Hence the correspondence eigenvector is  1 or 1  1   1

Solution File Assignment # 6 ( Linear Algebra) (Fall 2006) Total marks: 20

Question # 1 Diagonal the following matrix, if possible 4 1 A=   2 1 Solution: Step 1: Find the Eigen values of A. The characteristic equation is as follow: Det |A-  I |=0 1  4 0 1 2   (1   )(2   )  4(1)  0 2    2   2  4  0

2    6  0  2  3  2  6  0  (  3)  2(  3)  0 (  3)(  2)  0   3, 2 Step 2: Find two linearly independent vectors of A. Solve the characteristic equation

( A  I )x  0 For  =-3 4   x1  0  1-(-3)   1 -2-(-3)   x2  0    4 4 0  1 1 0  1 1 0  1 1 0  ~ 1 1 0  1/ 4 R1 ~ 0 0 0  R2  R1       x1  x2  0 Take x 2  t as x 2 is a free variable. x1  x2  0 x1  t  0 x1  t  x1   t   1  x   t   t 1   2      1 V1    1  For  =2 1-2 4   x1  0   1 -2-2   x   0    2    1 4 0  1 4 0  1 4 0  1 4 0  ~ 1 4 0   R1 ~ 0 0 0  R2  R1       x1  4 x2  0 Take x 2  s as x 2 is a free variable. x1  4 x2  0 x1  4 s  0 x1  4 s  x1   4 s   4  x    s   s 1     2   4 V2    1  V1 and V2 are linearly independent. Step 3: Construct p from the vectors V1 and V 2 .

p  V1 V2   1 4  p   1 1 Step 4: Construct D from the corresponding eigen values according to the arrangement in the above step. -3 0  D=    0 2 We will varify AP=PD which is the condition for a diagonalizable matrix. 1 4   1 AP=   1 -2   1  1 4  -3 PD     1 1  0

4  3  1   3 0  3  2   3

8 2  8 2 

Hence A is diagonaslizable. Question # 2 (a) Classify the following matrices as an attractor,repellor or a saddle Point of the Dynamical System x K 1  Ax K

1.7 A=   1.2 .5 B=   .3 Solution:  .3 1.7 a) A=  .8   1.2 The characteristic equation is as follow:

 .3 .8  .6 1.4 

| A   I | 0 1.7  

.3

1.2

.8  

0

(1.7   )(.8   )  (.3)(1.2)  0 1.36  1.7  .8   2  0.36  0 1  2.5   2  0

 2  2.5  1  0 Applying the quardetic equation to find the value of .

=

-b  b 2  4ac (2.5)  (2.5) 2  4(1)(1) 2.5  6.25  4   2a 2(1) 2

2.5  2.25 2.5  1.5  2 2 2.5  1.5 2.5  1.5 4 1  ,  , 2 2 2 2   2, 0.5 

As  =2>1 and  =0.5<1, hence it a saddle point of the dynamic system

.6 .5 b) B=  1.4   .3 Solution: The characteristic equation is as follow:

x

K 1

 Ax K

| B   I | 0 .5   .6 0 .3 1.4   (.5   )(1.4   )  (.6)(.3)  0 0.7  .5  1.4   2  .18  0

 2  1.9  0.88  0 Applying the quardetic equation,

=

-b  b 2  4ac (1.9)  (1.9) 2  4(1)(0.88) 1.9  3.61  3.52   2a 2(1) 2

1.9  0.09 1.9  0.3  2 2 1.9  0.3 1.9  0.3  , 2 2 2.2 1.6  , 2 2   1.1, 0.8 

As  =1.1>1 and  =0.8<1. Hence it is a saddle point for the dynamic System

(b) Solve the initial value problem

1 and A =  3 Solution:

2  4

x

x (t )  Ax(t ) for t  0 with x /

0

K 1

 A x K.

3     2

As |A- I|=0 1-

-2

3

-4-

0

(1   )(4   )  (2)(3)  0 4    4   2  6  0

 2  3  2  0  2  2    2  0  (  2)  1(  2)  0 (  2)(  1)  0   2, 1 The eigen vectors corresponding to   2 1  (2)  3  3 2 3 2 

2   x1  0   4  (2)   x2  0  0  3 2 0  ~ R2  R1 0   0 0 0 

3 x1  2 x2  0 Take x 2  p as x 2 is a free variable. 3 x1  2 p  0 3 x1  2 p 2 p 3 2  2  x1   p  2    x    3   p  3   p 3     2 p  1  Eigenvectors corresponding to the eigen value  =-1 x1 

1-(-1) -2   x1   3 -3  x2  

 2 2 0  1 1 0  1 1 3 3 0  ~ 1 1 0  2 R1 , 3 R2     1 1 0  ~  R2  R1 0 0 0  x1  x2  0 Take x 2  t as x 2 is a free variable. x1  x2  0 x1  t  0 x1  t  x1  t  1  x   t   t 1  2     So, x(t )  a1e t x1  a2 e t x2 Initially t=0 3  2 1  2   a1 3   a2 1      3   2a1  a2   2   3a  a     1 2 2a1  a2  3......(1) 3a1  a2  2......(2) By subtrating 2 from 1, we get a1  1 Put this value of a1 in equation 1. 2(-1)+a 2  3 a2  3  2  5 2 1 x(t )  e 2t    5e  t   3  1

Solution File of Assignment No. 7 (Linear Algebra) FALL SEMESTER 2006 Question # 1

1 2 Apply the Power Method to A =  with 5 4 the dominant eigenvalue.Stop when K=5

x

1   0 to estimate 0  

Solution: First Compute Ax 0  2 1 1   2  Ax0         , 0  4  4 5 0   4  1 1  2  .5 x1  Ax0       0 4  4  1   2 1  .5 1  1   2  Ax1           , 1  7  4 5 1   2  5 7  1 1  2   0.2857  x2  Ax1       1 7 7  1   2 1  0.2857  0.5714  1 1.5714  Ax2      1.1428  5  6.1428 ,  4 5 1      1 1 1.5714   0.2558 x3  Ax2    2 6.1428 6.1428 1   2 1 0.2558 0.5116  1 1.5116  Ax3      1.0232  5   6.0232  ,  4 5 1      1 1 1.5116   0.2509  x4  Ax3    3 6.0232 6.0232  1   2 1  0.2509  0.5018  1 1.5018  Ax4      1.0036  5  6.0036  ,  4 5 1      1 1 1.5018  0.2501 x5  Ax4    4 6.0036 6.0036  1   2 1 0.2501 0.5002  1 1.5002  Ax5      1.0004  5  6.0004  ,  4 5 1     

2  6.1428

3  6.0232

4  6.0036

5  6.0004

Question # 2 (a) Determine whether the set S= { u1, u 2, u 3 } is an orthogonal set?

1  Where u 1   2  , 1 

0   5     u 2  1  , u 3  2  2  1 

Solution: 1    u 1  2 , 1 

0   5     u 2  1  , u 3  2  2  1 

we calculate the products pairs of distinct vectors to check whether the set is orthagonal or not. 1  0  u1.u2   2  . 1   0  2  2  0 1   2  1   5  u1.u3   2  .  2   5  4  1  0 1  1  0   5  u2 .u3  1  .  2   0  2  2  0  2  1  So, the given set is orthagonal. 2

(b) Show that the set S= { u1, u 2 } is an orthogonal basis for R ?

6 Express the vector y=   as a linear combination of the vectors in S 3  3  2 Where u1    and u 2    1  6 

Solution:

The set s will be orthagonal if u1.u 2  0. 3  2  u1.u2    .    6  6  0 1  6  So, it is orthagonal set. Now,  6  3 y.u1    .    18  3  15 3  1   6   2  y.u2    .    12  18  30 3   6   3   3 u1.u1    .    9  1  10 1  1   2   2  u2 .u2    .    4  36  40. 6  6  Now, y

y.u1 y.u2 15 30 .u1  .u2   u1  u2 u1.u1 u2 .u2 10 40

3 3 y   u1  u2 2 4 Question # 3

5   3  3  9  ,   5 ,   Let y=   u 1   u 2   2  .Find the distance from y to the plane 5  1  1  in

Solution :

R

3

spanned by u1 and u 2 .

5   3  3  9  ,   5 ,     u1   u 2  2  5  1  1  y.u1 y.u2 yˆ  .u1  .u2 u1.u1 u2 .u2  5   3  y.u1   9  .  5  15  45  5  35 5  1   3  3 u1.u1   5 .  5  9  25  1  35 1  1  5   3 y.u2   9  .  2   15  18  5  28 5  1   3  3 u2 .u2   2  .  2   9  4  1  14 1  1  Now putting these values in the formula below: yˆ 

y.u1 y.u2 .u1  .u2 u1.u1 u2 .u2

 3  3  3  3  3 6  3  35   28     yˆ   5  2    5  (2)  2    5   4    9   35 14 1  1  1  1  1   2   1  5  3   2  y  yˆ   9    9   0  5   1  6  y  yˆ  4  0  36  40 2

y  yˆ  40  2 x 2 x10  2 10 Hence the distance of y from W is 2 10.

Solution File Assignment No.8 Linear Algebra (Fall Semester 2006) Question # 1 Consider the basis S= { u1, u 2, u 3 } for

R

3

.Where

1 0  1        u 1  1 , u 2  1  , u 3  2 1 1  3  Use the Gram-Schmidt Process to transfer S to an orthonormal basis for

R

3

Solution: Step 1

1 v1  u1  1 1 Step 2 v2  u2  projw1 u2  u2 

u2 .v1 v1

2

v1

0 1 u2 .v1  1  . 1  0(1)  1(1)  1(1)  2 1  1 v1

2

 111  3

0 1 0  2 / 3 2       v2  1   1  1    2 / 3 3 1  1 1   2 / 3

 2 / 3  v2  1/ 3  1/ 3  Step 3

v3  u3  projw2 u3  u3 

u3 .v1 v1

2

v1 

u3 .v2 v2

2

v2

1  1 u3 .v1   2  . 1  1(1)  2(1)  3(1)  6 3  1 1   2 / 3 2 1 1 u3 .v2   2  . 1/ 3   1(  )  2( )  3( ) 3 3 3 3  1/ 3  2 2 3 2  2  3 u3 .v2      1 3 3 3 3 4 1 1 4 11 6 2 2 v2       9 9 9 9 9 3 1  1  2 / 3 6  1    v3   2   1  1/ 3   3 2/3 3  1 1/ 3  1  1  2 / 3  3     v3   2   2 1  1/ 3  2 3  1 1/ 3  1   2   6 / 6  v3   2    2   3 / 6  3   2  3 / 6   1  6 / 6  v3  0   3 / 6  1  3 / 6   1  6 / 6  v3  0   3 / 6  1  3 / 6  0   v3   1/ 2  1/ 2  Thus 1  2 / 3    v1  1 , v2 = 1/ 3  , v3 = 1 1/ 3 

0   1/ 2    1/ 2  These are the orthogonal basis for R 3 . Now we have to find the orthonormal basis R 3

v1  (1) 2  (1)2  (1)2  1  1  1  3 6  2   1   1  v2           3  3  3 3 2

2

2

2

2

2  1 1 v3  0         2  2 2 1/ 3    v q1  1  1/ 3  v1   1/ 3   2 / 6    v q2  2  1/ 6  v2   1/ 6 

q3 

v3 v3

  0     1      2  1     2 

Question # 2 The orthogonal basis for the columns space of the matrix 5 1 3 1 1 1   A=  1 5 2   7 8 3 3  1  1  3  is given by   ,    1 3      3   1 A.

Solution: Let

 3 1  ,   .Find a QR- factorization of the above matrix 1    3 

3 1 v1    ,  1   3

 3 1 v3    1   3

1 3 v2    , 3    1

Normalize the three vectors to obtain u 1 , u2 , u3 .

3  20   3    1   1 1 1   20   u1  v1   v1 20  1  1     20  3    3  20  1  20   1    3   3 1 1   20   u2  v2   v2 20 3   3     20   1   1  20   3  20    3   1   1 1 1   20   u3  v3   v3 20 1   1     20  3    3  20  QR Factorization:

 3   1  Q  1  3 

3

 20   3 1 20 20   3 1  20 20   3 1 20 20  1

20 20 20 20

20

As

R=

So take the transport of the above matrix,

Q

T

A

 3   QT   1   3   3   R 1   3 

1

1 20

20

20

3 20

20

1

20 1

1 20 3 20

3 20 1

3 20

1 20

3

1 20

20

1

3 20

20

3

20 3

1 20

20

 20    20   20   3 20    1  20   1  3  20  5 1 5 7

5 1 5 7

1 1  2  8

1 3  3 1 1 3   1 1  1    R 1 3 3 1  1 2 20   3 1 1 3    8 3  9+1+1+9 15  1  5  21 3  1  2  24  1   3333 -5+3+5+7 1  3  6  8   20  9  1  1  9 15+1+5-21 3  1  2  24   20 40 26  1  R 0 20 2   20  0 0 24 

Question # 3 Find the Least Square solution and its error, where 1 2  5  2  0  and b= 8  A=  2 1  3  Solution:

2 A A 1

-2 2 0 3

2 AT b   1

-2 2 0 3

T

Then the equation

AT A  AT b becomes

1 2   4  4  4 2  0  6  12 8  0       2 2  0  6 1  0  9   8 10   2  3   5     10  16  2   26  2   24   8    5  0  3    5  3    2   1         

12 8   x1   24  8 10  x    2    2   10 8 1  AT A  561 8 12    xˆ   AT A  AT b 1

1 10 8  24  56  8 12   2  1  240  16  =  56  192  24 





1  224   4   56  168   3 

 4  xˆ    3 Now

2 Axˆ   2  2

1  8  3  5  4   0      8  0    8  3 3    8  9   1 

Hence,

 5  5  5  5 0 b  Axˆ   8    8   8  8   0  1   1  1  1  0 And

b  Axˆ  0 The least squar e error is

0