Almost exactly. If you stop at k iterations, you don't get the top k eigenvalues precisely. You need some extra slop to make sure your top eigenvalues are accurate.
On Sun, Nov 6, 2011 at 3:46 PM, Danny Bickson <[email protected]>wrote: > Exactly, in each iteration you extract one more eigenvalue from largest to > smallest. If you run more iterations then the matrix rank then you get > again all eigenvalues. In that case you may encounter each eigenvalue a few > times. > > On Sun, Nov 6, 2011 at 6:42 PM, Sean Owen <[email protected]> wrote: > > > 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. > > > > > > > > > > > > > > >
