For the symmetric case, SVD is eigen decomposition.
On Mon, Feb 17, 2014 at 1:12 PM, peng <pc...@uowmail.edu.au> wrote: > If SSVD is not designed for such eigenvector problem. Then I would vote > for retaining the Lanczos algorithm. > However, I would like to see the opposite case, I have tested both > algorithms on symmetric case and SSVD is much faster and more accurate than > its competitor. > > Yours Peng > > On Wed 12 Feb 2014 03:25:47 PM EST, peng wrote: > >> In PageRank I'm afraid I have no other option than eigenvector >> \lambda, but not singular vector u & v:) The PageRank in Mahout was >> removed with other graph-based algorithm. >> >> On Tue 11 Feb 2014 06:34:17 PM EST, Ted Dunning wrote: >> >>> SSVD is very probably better than Lanczos for any large decomposition. >>> That said, it does SVD, not eigen decomposition which means that the >>> question of symmetrical matrices or positive definiteness doesn't much >>> matter. >>> >>> Do you really need eigen-decomposition? >>> >>> >>> >>> On Tue, Feb 11, 2014 at 2:55 PM, peng <pc...@uowmail.edu.au> wrote: >>> >>> Just asking for possible replacement of our Lanczos-based PageRank >>>> implementation. - Peng >>>> >>>> >>>
