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Find smallest Eigenvalue and the corresponding eigenvector.

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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)
  2 comentarios
Torsten
Torsten el 9 de Jun. de 2022
Editada: Torsten el 9 de Jun. de 2022
Iterate for the largest eigenvalue of A^(-1) - it's the inverse of the smallest of A.
Eiman Hakimy
Eiman Hakimy el 9 de Jun. de 2022
so which part i need to changes ? sorry because i'm still confuse. can you show which line i need to changes

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Respuesta aceptada

John D'Errico
John D'Errico el 9 de Jun. de 2022
Editada: John D'Errico el 9 de Jun. de 2022
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.
  2 comentarios
Eiman Hakimy
Eiman Hakimy el 9 de Jun. de 2022
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
Torsten el 9 de Jun. de 2022
Editada: Torsten el 9 de Jun. de 2022
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.

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Más respuestas (1)

SALAH ALRABEEI
SALAH ALRABEEI el 9 de Jun. de 2022
You can find it here

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