Assume that we have generalized eigenvalue problem:
$B^HB\textbf{x} = \lambda A\textbf{x}$
where $A$ is an nxn Hermitian sparse matrix (n is very large, so we do not have $A^{-1}$ but can solve using iterative methods) and full-rank, and $B$ is a 2xn matrix such that $B^HB$ is also nxn but only rank 2. Thus, we know that this problem can only have 2 non-zero eigenvalues. Is there any simple way for finding the two eigenpairs corresponding to nonzero eigenvalues by taking advantage of the very low rank of $B^HB$? Assume that we have the two eigenvectors of $B$.
If I am only interested in the eigenvector corresponding to the largest eigenvalue, is there a faster way of finding it than using simple power iteration on the transformed standard eigenvalue problem: $A^{-1}B^HB\textbf{x} = \lambda\textbf{x}$?
Thanks!