eigen value problem

1. Shree S’ad Vidhya Mandal
Institude of Technology
Eigenvalue Problems
(Group No.11)
SUBJECT : VCLA (2110015)
Department of electrical engineering
second semester
Regards:N.M.Patel

2. NO. NAME ENROLLMENT NO.
1 PATEL SHIVAM 150450109036
2 RAJPUT DHRMENDRA 150450109043
3 DESAI VISHAL 150450109058

3. Eigenvalue Problems
A matrix eigenvalue problem considers the
vector equation
(1)
Here A is a given square matrix, an unknown
scalar, and an unknown vector
is called as the eigen value or
characteristic value or latent value or proper
roots or root of the matrix A, and is called as
eigen vector or charecteristic vector or latent
vector or real vector.
xxA 

x

x

4. What are Eigenvalues ?
If A is an n x n matrix and λ is a scalar for which
has a nontrivial solution , then is an
eigenvalue of A and x is a corresponding eigenvector
of A. is called the eigenvalue problem for A.
Note that we can rewrite the equation
as follows:
or is the trivial solution.
But our solutions must be nonzero vectors called
eigenvectors that correspond to each of the distinct
eigenvalues.
xlA nxx  
0 xn Axl
xxA 
xxA  
0.0)(  xxAxln
n
Rx

5. Eigenvalue Problems
How to Find Eigenvalues
EXAMPLE 1
Determination of Eigenvalues
We illustrate all the steps in terms of the
matrix
5 2
.
2 2
 
   
A

6. EXAMPLE 1
Solution.
(a) Eigenvalues. These must be
determined first.
Equation (1) is
in components
1 1
2 2
5 2
;
2 2
x x
x x

     
          
Ax

7. EXAMPLE 1
Solution. (continued 1)
(a)Eigenvalues. (continued 1)
Transferring the terms on the right to
the left, we get
1 2 1
1 2 2
5 2
2 2 .
x x x
x x x


  
 
1 2
1 2
( 5 ) 2 0
2 ( 2 ) 0
x x
x x


   
   

8. EXAMPLE 1
This can be written in matrix
Because (1) is n..
which gives
( ) 0 A I x
,0)(  xIAIAA xxxx 

9. EXAMPLE 1
Solution. (continued 2)
(a) Eigenvalues. (continued 2)
We see that this is a homogeneous linear
system. it has a nontrivial solution (an
eigenvector of A we are looking for) if and
only if its coefficient determinant is zero,
that is,
2
5 2
( ) det( )
2 2
( 5 )( 2 ) 4 7 6 0.
D

 

   
 
  
 
         
A I

10. EXAMPLE 1
Solution. (continued 3)
(a) Eigenvalues. (continued 3)
We call the characteristic determinant or,
if expanded, the characteristic polynomial,
and the characteristic equation of A.
The solutions of this quadratic equation are
and . These are the eigenvalues of
A.
Eigen vector corresponding to eigen value
is
)(D
0)( D
11  62 
11 
01  IA 

11. EXAMPLE 1
 
 202
1024
0
0
12
24






















yx
yx
y
x
Solving these two equations we get,
and
There for eigenvector corresponding to eigenvalue
is
Eigen vector space is
0x 0y
11     0,0,1  yxE
  .0,01 E

12. EXAMPLE 2
Example 2: Find the eigenvalues A.
Step 1: Find the eigenvalues for A.
The determinant of a triangular matrix
is the product of the elements at the
diagonal. Thus, the characteristic
equation of A is
A 
3 4 0
0 3 0
0 0 1











13. EXAMPLE 2
has algebraic multiplicity 1 and
has algebraic multiplicity 2.
p()  det(I  A)  I  A  0

  3 4 0
0   3 0
0 0  1
 (  3)2
( 1)1
 0.
11  62 

14. EXAMPLE 2
Eigen vector corresponding to is
Here,
and
There for eigen vector corrous ponding to eigen value
Eigen space is
31    01  IA 
02
04
,
100
000
040
3
2
2















x
x
Rrrx
02 x 03 x
      RrrrxxxE  ,0,0,10,0,,, 3213
  RrrE  /0,0,13

15. EXAMPLE 2
Now, eigen vector corresponding to
Here,
and
.32 
 













200
000
040
02IA 
02
04
,
3
2
1



x
x
Rrrx
02 x 03 x

16. EXAMPLE 2
Therefore eigen vector corresponding to eigen value
Eigen space is
      .,0,0,10,0,,, 3213 RrrrxxxE 
  RrrE  0,0,13
Now,eigen vector corresponding to 13 
 
0
02
042
000
020
042
0
2
2
21
3












 

x
x
xI
IA 

17. EXAMPLE 2
Here
Therefor eigen vector of is
Eigen space is
0
;
1
3


x
Rrrx
13     0,,0,, 321 rxxx 
  .0,1,01 RrrE 

18. Thank you