Eigenvalues by power method

hello everyone, is there's anyone could help me to solve this problem using matlab code

[Sam Chak]

Hi @Sofy

Can you share the knowledge of solving this problem according to the textbook or lecture notes?

It helps in planning what to write in the MATLAB code later.

[James Tursa]

What have you done so far? What specific problems are you having with your code?

[Sofy]

actually it wasn't home work I'm just try to know the way of coding this issue coz I'm beginner in matlab

Answer 1

[Hiro Yoshino]

See eig and eigs. Looks like eigs is more suitable for you though ...

I suppose u represents an eigen vector but the initial point seems to have to be u_0. In this case you may need to keep the norms of the eigen vectors <= 1.

[John D'Errico]

Note that this is clearly a homework question, where the power method was explicitly requested. So eig or eogs are both not useful, but we need to avoid writing code to do student homework assignments here.

[Hiro Yoshino]

Agreed

[Sofy]

Thx for your comments

Answer 2

[Kashish]

function [m,y_final]=power_method(A,x);
m=0;
n=length(x);
y_final=zeros(n,1);
y_final=x;
tol=1e-3;
while(1)
    mold=m;
    y_final=A*y_final;
    m=max(y_final);
    y_final=y_final/m;
    if (m-mold)<tol
        break;
    end
end
end

[John D'Errico]

I'm sorry, but this code is terribly poor as an answer to the question. It will fail far too often. (I'm not even sure it will ever work correctly. My guess is, under a specific set of circumstances, it MAY produce an estimate of an eigenvalue.) In fact, there are multiple flaws in the code I can see. If I tell you what you did wrong, you won't learn anything. You need to spend some time understanding what you did.

A = rand(3) - 3*eye(3)
A =
-2.1537      0.59001      0.50669
0.45027      -2.5355      0.19874
0.08948      0.47973      -2.5368

eig(A)
ans =
-1.6144 +          0i
-2.8058 +    0.13422i
-2.8058 -    0.13422i

I've created a simple 3x3 matrix. Surely power_method might find even one of the eigenvalues? How hard can that be? Yes, two of the eigenvalues were complex, but the other is not.

power_method(A,ones(3,1))
    ans =
    -1.057

I would suggest you consider why the code failed, even on that simple example. If it bothers you that my example matrix had complex eigenvalues, then I'll do this:

A = A + A'
A =
    -4.3073       1.0403      0.59617
    1.0403       -5.071      0.67847
    0.59617      0.67847      -5.0736

That simple operation insures all of the eigenvalues are real. We can test that claim easily enough.

eig(A)
ans =
    -5.8925
    -5.387
    -3.1725

So purely real eigenvalues. But surely your code should work now?

power_method(A,ones(3,1))
ans =
    -2.6709

So strange. It still fails miserably.

Ok, maybe your code will only work on symmetric, positive definite matrices? Anyway, that was what was requested in the homework assignment as posted. Sigh. It does not. I can swap the signs on the eigenvalues of A simply enough.

A = -A
A =
    4.3073      -1.0403     -0.59617
    -1.0403        5.071     -0.67847
    -0.59617     -0.67847       5.0736
    
eig(A)
ans =
    3.1725
    5.387
    5.8925
    

Surely with 3 positive eigenvalues on a symmetric matrix, surely it will work?

power_method(A,ones(3,1))
ans =
       5.2965

How strange. It still fails completely, even on what should be a trivially simple problem now. It came a little closer this time, but really...

Again, you should seriously look at your code, and think about what happened, why it failed. Use the debugger, running on that last example matrix and think about why your code is failing.