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Eigenvalues and Eigenvectors
Consider multiplying nonzero vectors by a given square matrix, such as
We want to see what influence the multiplication of the given matrix has on the vectors.
In the first case, we get a totally new vector with a different direction and different length
when compared to the original vector. This is what usually happens and is of no interest
here. In the second case something interesting happens. The multiplication produces a
vector which means the new vector has the same direction as
the original vector. The scale constant, which we denote by is 10. The problem of
systematically finding such ’s and nonzero vectors for a given square matrix will be the
theme of this chapter. It is called the matrix eigenvalue problem or, more commonly, the
eigenvalue problem.
We formalize our observation. Let be a given nonzero square matrix of
dimension Consider the following vector equation:
......(1).Ax ϭ lx
n ϫ n.
A ϭ [ajk]
l
l
[30 40]T
ϭ 10 [3 4]T
,
c
6 3
4 7
d c
5
1
d ϭ c
33
27
d, c
6 3
4 7
d c
3
4
d ϭ c
30
40
d.
c08.qxd 11/9/10 3:07 PM Page 323
(2))X  0(
0


A I
AX X
To have a non-zero solution of this set of homogeneous linear equation (2) | A-λI | must
.be equal to zero i.e
  (3 )A   I  0
.The following procedure can find the eigen values & eigen vector of n order matrix A
1. to find the characteristic polynomial P(λ) = det [A−λI
2. to find the roots of the characteristic equations the roots are eigen
values that we required
P()  0
3. To solve the homogenous system
]
.To find n- eigen vectors
[Α−λΙ]Χ=0
1
wallaa alebady
wallaa.alebady@yahoo.com
Example: Determine the eigen value and corresponding eigen vector at
the matrix







23
14
A
Solution:












 X  

1


 

a




  x 
4 


x 


 

x 







 







 

 


 X 
 

 



 
 
  x  x  x  x  x 

 


 

 









 
   
  (4 









 





 0 


1/ 2
1/ 2
3/ 10
1/ 10
theeigenvectors 
1/ 2
1/ 2
2
1
5theeigen vector corresponding to
,0523
05
5
23
14
5
3/ 10
1/ 10
3
1
10
31for X

X is defindby
eigenvector maybe normalized to unit length the normalized eigenvector
3
Theeigen vector corresponding to 1
3 where a is arbitrary constant
0323
0341
23
14
To find theeigen vector for  1
51 ,&
05)(05
)(2  ) 3  00
23
14
0
0
0
23
14
22
2
2121121
21121
2
1
2
1
2
1
1
21
21221
21121
2
1
2
1
2
length  aa
a
X
is
a x  ax x  x  x  x
x  x  x  x  x
xx
AX  Xfor
a
a
length  a
length
X
X
a
a
X
is
alet x  a , x
x  xx  x  x
x
x
x
x
AX  X
A I



 1)(  6



2
wallaa alebady
wallaa.alebady@yahoo.com
Example: Determine the eigen values and corresponding eigen vectors of
the matrix
A
P
P




X
3c








2 / 3 


 X 
 9 
 6 


















 X 
  



















 c  




















 




 A 








 




2 / 31 / 3
2 / 3 1 / 32 / 3
2 / 31 / 32 / 3
0(:
0(:
1 / 3
2 / 3
2 / 3
1
2
2
3144,
1
2
2
2
2
2
042
022
0223
0
402
022
223
0)(3
)(96 ,3 ,
9  18 )  09)((01629918( )
0
702
052
226
0( )
702
052
226
333
222
1
1
213
31
21
321
3
2
1
11
321
223
I ) XAcase
I ) XAcase
c
clength  c
length
X
c
c
c
c
the eigenvecto rs is X
cx  xlet x
xx
xx
xxx
x
x
x
I XAcase
distinct
Simplifyin g we have
I




 



3
wallaa alebady
wallaa.alebady@yahoo.com
.Find the eigenvalues. Find the corresponding eigenvectors
1. 2. 3. 4. D
13 5 2
2 7 Ϫ8
5 4 7
TD
3 5 3
0 4 6
0 0 1
Tc
5 Ϫ2
9 Ϫ6
dc
3.0 0
0 Ϫ0.6
d
5. E
Ϫ3 0 4 2
0 1 Ϫ2 4
2 4 Ϫ1 Ϫ2
0 2 Ϫ2 3
U
c08.qxd 10/30/10 10:56 AM Page 329
H.W
4
wallaa alebady
wallaa.alebady@yahoo.com

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Eigenvalues and eigenvectors

  • 1. Eigenvalues and Eigenvectors Consider multiplying nonzero vectors by a given square matrix, such as We want to see what influence the multiplication of the given matrix has on the vectors. In the first case, we get a totally new vector with a different direction and different length when compared to the original vector. This is what usually happens and is of no interest here. In the second case something interesting happens. The multiplication produces a vector which means the new vector has the same direction as the original vector. The scale constant, which we denote by is 10. The problem of systematically finding such ’s and nonzero vectors for a given square matrix will be the theme of this chapter. It is called the matrix eigenvalue problem or, more commonly, the eigenvalue problem. We formalize our observation. Let be a given nonzero square matrix of dimension Consider the following vector equation: ......(1).Ax ϭ lx n ϫ n. A ϭ [ajk] l l [30 40]T ϭ 10 [3 4]T , c 6 3 4 7 d c 5 1 d ϭ c 33 27 d, c 6 3 4 7 d c 3 4 d ϭ c 30 40 d. c08.qxd 11/9/10 3:07 PM Page 323 (2))X  0( 0   A I AX X To have a non-zero solution of this set of homogeneous linear equation (2) | A-λI | must .be equal to zero i.e   (3 )A   I  0 .The following procedure can find the eigen values & eigen vector of n order matrix A 1. to find the characteristic polynomial P(λ) = det [A−λI 2. to find the roots of the characteristic equations the roots are eigen values that we required P()  0 3. To solve the homogenous system ] .To find n- eigen vectors [Α−λΙ]Χ=0 1 wallaa alebady [email protected]
  • 2. Example: Determine the eigen value and corresponding eigen vector at the matrix        23 14 A Solution:              X    1      a       x  4    x       x                          X                x  x  x  x  x                             (4                   0    1/ 2 1/ 2 3/ 10 1/ 10 theeigenvectors  1/ 2 1/ 2 2 1 5theeigen vector corresponding to ,0523 05 5 23 14 5 3/ 10 1/ 10 3 1 10 31for X  X is defindby eigenvector maybe normalized to unit length the normalized eigenvector 3 Theeigen vector corresponding to 1 3 where a is arbitrary constant 0323 0341 23 14 To find theeigen vector for  1 51 ,& 05)(05 )(2  ) 3  00 23 14 0 0 0 23 14 22 2 2121121 21121 2 1 2 1 2 1 1 21 21221 21121 2 1 2 1 2 length  aa a X is a x  ax x  x  x  x x  x  x  x  x xx AX  Xfor a a length  a length X X a a X is alet x  a , x x  xx  x  x x x x x AX  X A I     1)(  6    2 wallaa alebady [email protected]
  • 3. Example: Determine the eigen values and corresponding eigen vectors of the matrix A P P     X 3c         2 / 3     X   9   6                     X                         c                              A                2 / 31 / 3 2 / 3 1 / 32 / 3 2 / 31 / 32 / 3 0(: 0(: 1 / 3 2 / 3 2 / 3 1 2 2 3144, 1 2 2 2 2 2 042 022 0223 0 402 022 223 0)(3 )(96 ,3 , 9  18 )  09)((01629918( ) 0 702 052 226 0( ) 702 052 226 333 222 1 1 213 31 21 321 3 2 1 11 321 223 I ) XAcase I ) XAcase c clength  c length X c c c c the eigenvecto rs is X cx  xlet x xx xx xxx x x x I XAcase distinct Simplifyin g we have I          3 wallaa alebady [email protected]
  • 4. .Find the eigenvalues. Find the corresponding eigenvectors 1. 2. 3. 4. D 13 5 2 2 7 Ϫ8 5 4 7 TD 3 5 3 0 4 6 0 0 1 Tc 5 Ϫ2 9 Ϫ6 dc 3.0 0 0 Ϫ0.6 d 5. E Ϫ3 0 4 2 0 1 Ϫ2 4 2 4 Ϫ1 Ϫ2 0 2 Ϫ2 3 U c08.qxd 10/30/10 10:56 AM Page 329 H.W 4 wallaa alebady [email protected]