Mathematical Foundations - Control of Color Imaging Systems: Analysis and Design - page 144

" # " #

U 1 AV 1 U 1 AV 2
U 2 AV 1 U 2 AV 2

L 0

00

H AV 1 V 2 U H AV ¼ U 1 U 2 ½ ½ ¼ ¼ ¼ S ( 3 : 140 )

Pre- and post-multiplying both sides of Equation 3.140 by U and V H , we have

A ¼ U S V H

( 3 : 141 )

The nonnegative numbers 0 are

called singular values of A and they are square roots of eigenvalues of A H A or AA H .

s 1 s 2 s r > s r þ 1 ¼ s r þ 2 ¼¼s n ¼

Example 3.38

Find the SVD of matrix A given by

246

A ¼

26 4

S OLUTION
We first form S ¼ A T A

2
4

3
5

8 420

S ¼ A T A ¼ 4 20

20 0 52

2
1

2
2

2
3

The eigenvalues of S are s ¼ 60, s ¼ 52, and s ¼ 0. The corresponding

eigenvectors are

2
4

3
5 ,

2
4

3
5 ,

2
4

3
5

0 : 3651 0 0 : 9309

v 1 ¼ 0 : 1826 v 2 ¼ 0 : 9806 and v 3 ¼ 0 : 0716

0 : 9129 0 : 1961 0 : 3581

Therefore,

2
4

3
5

2
4

3
5

0 : 3651 0 0 : 9309

V 1 ¼ 0 : 1826 0 : 9806 and V 2 ¼ 0 : 0716

0 : 9129 0 : 1961 0 : 3581

and

2
4

3
5

" # 1

0 : 3651 0 p

60

246

0

U 1 ¼ AV 1 L 1 ¼ 0 : 1826 0 : 9806 p

52 26 4 0

0 : 9129 0 : 1961

0 : 707 0 : 707

¼

0 : 707 0 : 707