Find smallest Eigenvalue and the corresponding eigenvector.

I need to write a program which computes the largest and the smallest (in terms of absolute value) eigenvalues using power method. I can find the largest one using the power method. But I have no idea how to find the smallest one using the power method.

How can I modify the power method so that it computes the smallest eigenvalue?

my power method algorithm :

1. Start

2. Define matrix X

3. Calculate Y = AX

4. Find the largest element in the magnitude of matrix Y and assign it to K.

5. Calculate fresh value X = (1/K) * Y

6. If [Kn – K(n-1)] > delta, go to step 3.

7. Stop.

Below is the coding :

function [ v d] = power_method( A )
% for finding the largest eigen value by power method
 
 
disp ( ' Enter the matrix whose eigen value is to be found')
 
% Calling matrix A
 
A = input ( ' Enter matrix A :   \n')
% check for matrix A
 
% it should be a square matrix
 
[na , ma ] = size (A);
if na ~= ma
    disp('ERROR:Matrix A should be a square matrix')
    return
end
 
% initial guess for X..?
 
% default guess is [ 1 1 .... 1]'
 
disp('Suppose X is an eigen vector corresponding to largest eigen value of matrix A')
r = input ( 'Any guess for initial value of X? (y/n):   ','s');
switch r 
    case 'y'
        % asking for initial guess
    X0 = input('Please enter initial guess for X :\n')
        % check for initial guess
    [nx, mx] = size(X0);
    if nx ~= na || mx ~= 1
        disp( 'ERROR: please check your input')
        return
    end
    otherwise
    X0 = ones(na,1);
end
 
%allowed error in final answer
 
t = input ( 'Enter the error allowed in final answer:  ');
tol = t*ones(na,1);
% initialing k and X
 
k= 1;
X( : , 1 ) = X0;
 %initial error assumption
err= 1000000000*rand(na,1);
% loop starts
 
while sum(abs(err) >= tol) ~= 0
    X( : ,k+ 1 ) = A*X( : ,k); %POWER METHOD formula
    % normalizing the obtained vector 
    [ v i ] = max(abs(A*X( : ,k+ 1 )));
    E = X( : ,k+ 1 );
    e = E( i,1);
    X(:,k+1) = X(:,k+1)/e;
    err = X( :,k+1) - X( :, k);% finding error
    k = k + 1;  
end
 
%display of final result
 
fprintf (' The largest eigen value obtained after %d itarations is  %7.7f  \n', k, e)
disp('and the corresponding eigen vector is ')
X( : ,k)

[Torsten]

Iterate for the largest eigenvalue of A^(-1) - it's the inverse of the smallest of A.

[Eiman Hakimy]

so which part i need to changes ? sorry because i'm still confuse. can you show which line i need to changes

Answer 1

[John D'Errico]

Just compute the matrix

Ainv = inv(A);

Now use your code on the matrix Ainv. This works because the smallest eigenvalue is now the largest.

Will this faiil for a singular matrix A? Of course. But then you cannot use the power method there anyway.

So there is absolutely no need to modify your code. Just use it on a different matrix.

[Eiman Hakimy]

okay so i need to uses the inverse method right ?

so you says that's i don't need to modify my code so which one that's i need to changes to inverse method ?

sorry if i'm asking question too much because i'm still confuse with this. @John D'Errico

[Torsten]

@Eiman Hakimy

d will be the smallest eigenvalue of A if you execute

A_input = inv(A);
[~, d] = power_method(A_input);
d = 1/d

But first try to understand the code you copied for the power method.

Answer 2

[SALAH ALRABEEI]

You can find it here