Question

6. (20") Given the 3 x 3 matrix A- 20 00 (a) compute A'A. (b) find all eigenvalues of AA and their associated eigenwectors (c) write down all singular values of A in descending order (d) find the singular-value decomposition(SVD) A-UEV"

Answer 1

Solution: Given

A= 0 0 1 0 2 0 4 0 0

a).

ATA 0 0 4 0 2 2 0 0 0 0 0 1 0 2 0 4 0 0

ATA 16 0 0 0 4 4 0 0 0 1

b). Now eigen values of $A^TA$ are given by

ATA-XI = 0

16 - X 0 0 0 0 4- 0 0 1-1 = 0

(16 - 1) (4 - ) (1 - 1) = 0

>= 16, 4, 1

Now, eigen vector $v_1$ corresponding to
- 16
is given by

Α' Α. 161]υι = 0

T 0 0 0 0 -12 0 0 0 -15 y 2 0

> y=0,2 = 0

by choosing
=
, we get

1

Now, eigen vector $v_2$ corresponding to
12 = 4
is given by

Α' Α - 4I]x2 = 0

12 O 0 0 0 0 0 о —3 у

I= 0,2 0

by choosing
hi
, we get

2 - 1 0

Now, eigen vector $v_3$ corresponding to $\lambda_3=1$ is given by

Α' Α - Ι]ug = 0

(150 0 0 0 30 0 00 之

> I = 0, y = 0

by choosing $z=1$ , we get

U3 1

c). Now, singular values of are given below

01 = V16, 02= V4, 03 = V1

но1 = 4, O2 = 2, 03

Which are the required singular values.

d). Now the matrix $\sum$ in a singular value decomposition of is given by

Σ- 4 Ο Ο Ο 2 Ο Ο Ο 1

Now, a matrix of orthonormal basis of eigen vector is given by

V = 1 0 0 0 0 1 0 0 0 1

we now find the matrix .

the first column of is given by

1 Loi Avi 0 0 1 0 2 0 4 0 0 90

oil Avi 4 4

oil Avi 0 1

the second column of is given by

1 02' Av2 0 0 1 0 2 0 4 0 0 90

> O2 AU2 2 0

> 02 Av2 1 0

the third column of is given by

1 -1 03 Auz 0 0 1 0 2 0 4 0 0 90

-1 03 Auz o

Thus

0 0 1 U = 0 1 0 1 0 0

Hence, the singular value decomposition is given by

Α = UΣvΤ

T 0 0 1 0 20 4 0 0 0 0 1 0 1 0 1 0 0 4 0 02 2 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1

Which is the required solution.

This complete the solution.