# power method with rayleigh coeff

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Sonali 2024 年 4 月 12 日
コメント済み: Sonali 2024 年 4 月 14 日
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

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### 採用された回答

Alan Stevens 2024 年 4 月 13 日
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
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Sonali 2024 年 4 月 14 日
Thank you. It worked.

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