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 <srowen@gmail.com> 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" <danny.bickson@gmail.com> 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 <srowen@gmail.com> 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 <ted.dunning@gmail.com>
> > > 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
> > > > tridiagonal matrix that is some multiple of that size. Inmemory
> > > > decomposition makes total sense.
> > > >
> > >
> >
>
