用Python / Scipy对角大型稀疏矩阵

Question

I am working with a large (complex) Hermitian matrix and I am trying to diagonalize it efficiently using Python/Scipy.

Using the eigh function from scipy.linalg it takes about 3s to generate and diagonalize a roughly 800x800 matrix and compute all the eigenvalues and eigenvectors.

The eigenvalues in my problem are symmetrically distributed around 0 and range from roughly -4 to 4. I only need the eigenvectors corresponding to the negative eigenvalues, though, which turns the range I am looking to calculate into [-4,0).

My matrix is sparse, so it's natural to use the scipy.sparse package and its functions to calculate the eigenvectors via eigsh , since it uses much less memory to store the matrix.

Also I can tell the program to only calculate the negative eigenvalues via which='SA' . The problem with this method is, that it takes now roughly 40s to compute half the eigenvalues/eigenvectors. I know, that the ARPACK algorithm is very inefficient when computing small eigenvalues, but I can't think of any other way to compute all the eigenvectors that I need.

Is there any way, to speed up the calculation? Maybe with using the shift-invert mode? I will have to do many, many diagonalizations and eventually increase the size of the matrix as well, so I am a bit lost at the moment.

I would really appreciate any help!

Answer 1

This question is probably better to ask on http://scicomp.stackexchange.com as it's more of a general math question, rather than specific to Scipy or related to programming.

If you need all eigenvectors, it does not make very much sense to use ARPACK. Since you need N/2 eigenvectors, your memory requirement is at least N*N/2 floats; and probably in practice more. Using eigh requires N*N+3*N floats. eigh is then within a factor of 2 from the minimum requirement, so the easiest solution is to stick with it.

If you can process the eigenvectors "on-line" so that you can throw the previous one away before processing the next, there are other approaches; look at the answers to similar questions on scicomp.