TITLE: Finding the smallest eigenvalues of a large, but structured, matrix
QUESTION [8 upvotes]: I'm trying to find the eigenvector corresponding to the second smallest eigenvalue of a large $(4,000,000 \times 4,000,000)$ matrix $M$. $M$ is a Laplacian matrix, and it has the following structure: $M = D - AA'$, where $D$ is a diagonal matrix and $A$ is a large sparse matrix. $A$ has dimension $4,000,000 \times 10,000,000$, but only about $40,000,000$ non-zero entries. So I can rapidly perform matrix-vector multiplication: $Mv = Dv - A(A'v)$.
Currently I'm using Scipy (which calls ARPACK) to find the smallest eigenvalues and corresponding eigenvectors of $M$. The implementation is a variant of Lanczos method, as far as I can tell. Unfortunately the implementation fails to converge with some frequency. Even when it does converge, since I need fairly high accuracy, it takes many iterations to converge. Any suggestions?
Lanczos method is much more reliable when finding the largest eigenvalues and the corresponding eigenvectors, as I understand it. So I was thinking that I could transform the problem, perhaps instead finding the largest eigenvalues of M's left inverse. But, even if I somehow manage to find a good approximation of M's left inverse (most methods don't converge), the largest eigenvalue is infinite (the smallest eigenvalue of $M$ -- a Laplacian matrix -- is $0$), so it seems there would still be convergence issues, since I can't solve for the second largest eigenvalue without also solving for the largest, as I understand it. Is there an easier way?

REPLY [3 votes]: If it is a Laplacian then you not only know the smallest eigenvalue is zero, but you also know its corresponding eigenvector.  You can use this information by essentially adding a single step to your procedure, it's something like adding back the mean of v to each element of Mv.  Then use arnoldi to find the smallest (instead of second smallest) eigenpair.