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Eigen Value Problems Let [ A ] be an n x n square matrix. Suppose, there exists a scalar and a vector
X = ( x 1 (^) x 2 … xn ) T such that
A = λ( X )
d (^) ( eax ) a e ( ax ) dx
(^22) 2 (sin^ )^ (sin^ )
d (^) ax a ax dx
Then λ is the eigen value and X is the corresponding eigenvector of the matrix [ A ].
We can also write it as A − λ I =( O )
This represents a set of n homogeneous equations possessing non-trivial solution, provided
A − λ I = 0
This determinant, on expansion, gives an n-th degree polynomial which is called characteristic polynomial of [ A ], which has n roots. Corresponding to each root, we can solve these equations in principle, and determine a vector called eigenvector. Finding the roots of the characteristic equation is laborious. Hence, we look for better methods suitable from the point of view of computation. Depending upon the type of matrix [ A ] and on what one is looking for, various numerical methods are available.
Power Method and Jacobi’s Method
Note! We shall consider only real and real-symmetric matrices and discuss power and Jacobi’s methods
Power Method
To compute the largest eigen value and the corresponding eigenvector of the system
A = λ( X )
where [ A ] is a real, symmetric or un-symmetric matrix, the power method is widely used in practice.
Procedure Step 1: Choose the initial vector such that the largest element is unity.
Step 2 : The normalized vector v (0) is pre-multiplied by the matrix [ A ].
Step 3: The resultant vector is again normalized.
Step 4: This process of iteration is continued and the new normalized vector is repeatedly pre-multiplied by the matrix [ A ] until the required accuracy is obtained. At this point, the result looks like ( k ) (^) [ ] ( k 1) ( k ) u A v q vk = − = Here, qk is the desired largest eigen value and v (^ k ) is the corresponding eigenvector.
Example Find the eigen value of largest modulus, and the associated eigenvector of the matrix by power method 2 3 2 [ ] 4 3 5 3 2 9
A
= ^
Solution
We choose an initial vector υ(0)
as (1,1,1). T Then, compute first iteration
(1) (0)
[ ] 4 3 5 1 12
u A v
= ^ ^ =^
^ ^
^ ^
Now we normalize the resultant vector to get (^12) (1) 6 (1) (^1471) 1
u q v
= ^ =
The second iteration gives, (^12 ) (2) (1) (^6 ) 7 7 (^17114)
2 (2)
[ ] 4 3 5
u A v
q v
= ^ ^ = ^
= ^ =
Continuing this procedure, the third and subsequent iterations are given in the following slides
(3) (2)
[ ] 4 3 5 0.
u A v
= = ^ ^
^
^
[ ]
(0)
(1) (0)
we choose finitial vector as v^ t first iteration
u A v
by diagon
−^ +^ −
= = ^ − − ^ = ^ − − + ^ = ^ −
[ ]
(1) (1) 1
(2) (1)
sin 8 10 0. 2 0. sec 7 6 3 1 7 4.8 0.6 2. 12 20 24 0.8 12 16 4.8 0. 6 12 16 0.2 6 9.6 3.2 0.
ali g u q v
ond iteration
u A v
−^ −^ +
= = ^ − − ^ − ^ = ^ − + − ^ = ^ −
(1) 2 (2)
sin 0.8 2.8 0. 0.4 0.
by diagonali g u q v
(3) [ ] (2)
sin
third iteration
u A v
now daigonali g
−^ −^ −
= = ^ − − ^ − ^ = ^ − + + ^ = ^ −
8 sin 4.8574 0. 0.2868 0.
now normali g
Example Find the first three iteration of the power method applied on the following matrices
0
use x^ t
Solution
[ ]
(0)
(1) (0)
(
tan
USE x^ t
st iterations
u x
now we normalize the resul t vector to get
u
−^ −^ − −^ +^ −
= Α = ^ − − ^ ^ = ^ + − ^ =^
) 1 (1)
q x
− ^
= ^ = ^ =
^
^
[ ]
[ ]
(2) (1)
(2)
(3) (2)
u x
u
u x
−^ ^ − −^ +
^
−^ −
= Α = ^ − − ^ ^ = ^ + + =^
^
− ^ −
−^ −
= ^ ^ = ^
^ − −^ −
Exercise Find the largest eigen value and the corresponding eigen vector by power method after
fourth iteration starting with the initial vector υ (0)^ =(0, 0,1) T
1 1 1 1 1
r^ r p r (^) r r p
A v Lt p n A v λ λ λ
= = (^) →∞ = …
Here, the index p stands for the p-th component in the corresponding vector Sometimes, we may be interested in finding the least eigen value and the corresponding eigenvector. In that case, we proceed as follows.
We note that A = λ( X ).
Pre-multiplying by [ A −^1 ], we get
[ A −^1 ] A = [ A −^1 ] ( λ X ) =λ A −^1
Which can be rewritten as
A^1 1 ( X )
which shows that the inverse matrix has a set of eigen values which are the reciprocals of the eigen values of [ A ]. Thus, for finding the eigen value of the least magnitude of the matrix [ A ], we have to apply power method to the inverse of [ A ].
