Oh do you only need the top k x k bit of the tridiagonal to find the top k eigenvalues?
I really don't want to write a QR decomposer, phew. On Sun, Nov 6, 2011 at 11:26 PM, Ted Dunning <[email protected]> wrote: > The tridiagonal is much smaller than you would need if you wanted all the > eigenvalues. Since you only want a small number, you only have a > tri-diagonal matrix that is some multiple of that size. In-memory > decomposition makes total sense. >
