computation time-eigen vectors multiplication

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Hello,
I've written a simple for loop to compute the Q matrix shown in the formula attached. For every k(from k=2 to k=N), we multiply the eigen vector z_k with its transpose z_k^T and divide the result by the correspding eigen value mu_k. Here, N is the number of nodes in the network and q is a NxN symmetric matrix.
[x,y] =size(q);
N =x;
[V,W] =eig(q);
Q =zeros(N);
for i=2:1:N;Q=Q+(((V(:,i))*(V(:,i).'))/(W(i,i)));end
It works fine for small networks(takes roughly 20 secs for N=1200). However, I intend on using it for networks with as many as N=12000 nodes. This seems to take forever.
I'm using R2018a. My license does not include the parallel computing toolbox so I cannot use the parfor function. Is there any other way I can reduce the computation time? Perhaps to a few minutes?
Thank you
Kind Regards,
Sharadhi
  2 Commenti
John D'Errico
John D'Errico il 24 Set 2018
You are computing a rank N-1 variation of a quasi pseudo-inverse? Not sure why. And why assume the first eigenvalue is always a specific one? That is a REALLY BAD idea, because eig does not insure the order of the eigenvalues.
sharadhi
sharadhi il 24 Set 2018
As q here is the Laplacian matrix of a network, by considering only k= 2 to N, I was discarding the zero eigenvalue and corresponding vector.

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John D'Errico
John D'Errico il 24 Set 2018
Modificato: John D'Errico il 24 Set 2018
Well, disregarding my questions about why in the name of god and little green apples you want to do this, the trivial answer is to use linear algebra.
[x,y] =size(q);
N =x;
[V,W] =eig(q);
w = diag(W);
Q = V(:,2:end)*diag(1./w(2:end))*V(:,2:end).';
We should be able to do it a little faster for large arrays, if you create a sparse diagonal matrix from w. That makes the matrix multiply more efficient.
q = rand(1200,1200); q = q + q';
[V,W] = eig(q);
tic
w = diag(W);
Q = V(:,2:end)*diag(1./w(2:end))*V(:,2:end).';
toc
Elapsed time is 0.204658 seconds.
So instead of 20 seconds, only .2 seconds.
tic
w = diag(W);
Q = V(:,2:end)*spdiags(1./w(2:end),0,N-1,N-1)*V(:,2:end).';
toc
Elapsed time is 0.150572 seconds.
So slightly faster for this relatively small matrix. But when N is much larger, we gain much more, because now those matrix multiplies take a significant amount of time.
N = 12000;
q = rand(N,N); q = q + q';
[V,W] = eig(q);
w = diag(W);
tic
Q = V(:,2:end)*diag(1./w(2:end))*V(:,2:end).';
toc
tic
Q = V(:,2:end)*spdiags(1./w(2:end),0,N-1,N-1)*V(:,2:end).';
toc
  1 Commento
sharadhi
sharadhi il 24 Set 2018
It does work well. My overall code only takes 17 minutes now. I can work on the other parts and get it to be lower hopefully. Thanks, John.

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