True though I thought it was intended to run through all m steps - is it simply that you stop after some number k and that still leads you to compute the k largest eigenvectors of the original? On Nov 6, 2011 11:36 PM, "Danny Bickson" <[email protected]> wrote:
> To be exact, the size of the tridiagonal matrix > is number of iterations + 1. See description of the matrix T_mm > in Wikipedia: http://en.wikipedia.org/wiki/Lanczos_algorithm > > Best, > > DB > > On Sun, Nov 6, 2011 at 6:30 PM, Sean Owen <[email protected]> wrote: > > > 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. > > > > > >
