Mathematical Foundations - Control of Color Imaging Systems: Analysis and Design - page 134

S OLUTION
First we compute the two eigenvalues l 1 and l 2 . The characteristic polynomial of

matrix A is

¼ l

¼jlI Aj¼ l þ

2 2

2

P( l ) 10 l þ 24 ¼ ( l 4)( l 6) ¼ 0

24 l 12

Therefore, 6. Now we compute the eigenvectors corresponding

to the two eigenvalues. Eigenvector corresponding to

l 1 ¼ 4 and l 2 ¼

l 1 ¼ 4 is computed as

Ax 1 ¼ l 1 x 1

<
:

a
b

¼ !

a þ b ¼ 2a

22

4 a
b

! b ¼ 3a

24 12

12a þ 6b ¼ 2b

Let a ¼ 1, then b ¼ 3 and

¼

3

a
b

1

x 1 ¼

Eigenvector corresponding to l 2 ¼ 6 is given by

Ax 2 ¼ l 2 x 2

or

<
:

a
b

¼ !

a þ b ¼ 3a

11

3 a
b

! b ¼ 4a

12 6

12a þ 6b ¼ 3b

Let a ¼ 1, then b ¼ 4 and

¼

4

a
b

1

x 2 ¼

Example 3.31

Find the eigenvalues and eigenvectors of the 3 3 matrix A:

2
4

3
5

2 : 61 : 3 2 : 5

A ¼ 0 : 85 : 4 5

0 : 81 : 4 1

S OLUTION
The characteristic polynomial of matrix A is

3 2

P( l ) ¼jlI Aj¼l 7 l þ 14 l 8 ¼ ( l 1)( l 2)( l 4) ¼ 0