Calculating dominant eigenvector for each matrix in a large array

Hi,

I am trying to write a for loop to make an array of dominant eigenvectors for each of the matrices in a 11 X 11 X 10,000 array. My problem is in decoupling the V of the [V,D] = eig(A). I hope the below code can give an idea of what I am trying to do, but please let me know if I need to be more clear about anything.

LHS_matrices;

R_EIGENVECTORS = zeros(nstages,1,nsample);
for k = 1:nsample
            [V,D] = eig(LHS_MATRICES(:,:,k));
            R_EIGENVECTORS(:,1,k) = V(:,1);
end

That code runs, but all of the eigenvector outputs are the same, and seem to be incorrect, e.g.:

val(:,:,1) =

     0
     1
     0
     0
     0
     0
     0
     0
     0
     0
     0


val(:,:,2) =

     0
     1
     0
     0
     0
     0
     0
     0
     0
     0
     0

The below code was perfect for determining the dominant eigenvalue for population growth rate (I am a conservation biologist). If anyone knows of something like this I could use to determine the dominant eigenvectors (and thus the stable population distribution), that would be amazing!

LHS_matrices;
EIGENVALUES = zeros(nsample,1);
for k = 1:nsample
    EIGENVALUES(k,1) = max(abs(eig(LHS_MATRICES(:,:,k))));
end

[the cyclist]

Can you upload your data in a mat file (using the paper clip icon), or a representative sample that illustrates the issue?

[Jenna Parker]

I think I figured out the issue, but thank you for responding! My matrices are reducible, since there is one stage class that does not have a path to the others. This is why the right eigenvectors are so weird. From what I understand a stable age distribution does not exist when it comes to reducible leslie matrices.

[David Goodmanson]

Hi Jenna,

be aware that the eigenvalues produced by eig are not necessarily sorted, so the largest one might not be first.

[Jenna Parker]

Thank you for the heads up David, I appreciate it. I knew that about the eigenvalues, but is the same true of eigenvectors? I had read online that for right eigenvectors, the first column is always dominant.

[David Goodmanson]

HI Jenna, Since the nth column belongs with the nth listed eigenvalue in the diagonal eigenvalue matrix, and those values can be in arbitrary order, the first column is not always dominant. It's not like the svd.

[Jenna Parker]

That is extremely helpful! Thanks so much!