The document discusses eigenvalues and eigenvectors. It defines an eigenvalue problem as finding scale constants (λ) and nonzero vectors (X) such that when a square matrix (A) multiplies a vector (X), it produces a vector in the same direction but scaled by λ. The characteristic polynomial is used to find the eigenvalues by setting its determinant equal to 0. Once the eigenvalues are obtained, the corresponding eigenvectors can be found by solving the homogeneous system (A - λI)X = 0. Examples are provided to demonstrate finding the eigenvalues and eigenvectors of different matrices.

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
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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
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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
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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
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