power method with rayleigh coeff

Hello, I'm working on Rayleigh iteration and my program is as follows. I am getting wrong values. I was wondering if someone could suggest something.

function [eigval, eigvec,itern] = rayl(A, x0, tol, nmax)

x = x0;

iter = 0;

e0 = 0;

while iter < nmax

y = A * x;

en = (x' * y) / (x' * x);

if abs(en - e0) <= tol

break;

end

x = y / norm(y);

e0 = en;

iter = iter + 1;

end

eigval = en;

eigvec = x;

itern =iter;

end

.........................................................................

Calling in function:

A=[4 1 -1 0;1 3 -1 0;-1 -1 5 2;0 0 2 4]

tol=1.d-8; nmax=100; x0=[1 1 1 1]';

[eig_val_ray, xr,iter] = rayl(A, x0, nmax, tol)

................................................................

output:

eig_val_ray = 4.5000

xr = 4×1

1

1

1

1

iter = 0

Answer 1

[Alan Stevens]

Make sure you call the function with the arguments in the same order as those defined in the function!

A=[4 1 -1 0;1 3 -1 0;-1 -1 5 2;0 0 2 4];
tol=1.d-8;  nmax=100; x0=[1 1 1 1]';
[eig_val_ray, xr,iter] = rayl(A, x0, tol, nmax);  %%%%%%%%%%%%%%%%%%%

eig_val_ray

eig_val_ray = 7.0861

xr

xr = 4x1

-0.3325 -0.2671 0.7590 0.4919

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iter

iter = 24

function [eigval, eigvec,itern] = rayl(A, x0, tol, nmax)%%%%%%%%%%%%%%
    x = x0;
    iter = 0;
    e0 = 0;
    while iter < nmax
        y = A * x;
        en = (x' * y) / (x' * x); 
        if abs(en - e0) <= tol
            break;
        end
        x = y / norm(y); 
        e0 = en;
        iter = iter + 1;
    end
    eigval = en;
    eigvec = x;
    itern =iter;
end

[Sonali]

Thank you. It worked.