One more question. OK, so I use Lanczos to find V_k by finding the top k eigenvectors of AT * A. A is sparse. But isn't AT * A dense, then? Is that just how it is?
This will be my last basic question for the week: I understand that A ~= U_k * S_k * V_kT. Let's call the product on the right A_k. A_k = A * V_k * V_kT right? And then A_k is just my complete prediction matrix right? It's dense so it's not formed all at once. But all I need are V_k and its transpose to do the work. I somehow thought it was more complicated than this -- having come back to this I keep wondering if I forget something here. On Sun, Nov 6, 2011 at 6:08 PM, Jake Mannix <[email protected]> wrote: > Re: Lanczos, yes, it operates by finding V as you describe. The user is > required to do more work to recapture U. Practical reason is that the > assumption is numCols(A) = numFeatures which is much less than numRows(A) = > numTrainingSamples > > On Nov 6, 2011 9:52 AM, "Sean Owen" <[email protected]> wrote: > > Following up on this very old thread. > > I understood all this bit, thanks, that greatly clarified. > > You multiply a new user vector by V to project it into the new > "pseudo-item", reduced-dimension space. > And to get back to real items, you multiply by V's inverse, which is > its transpose. > And so you are really multiplying the user vector by V VT, which is > not a no-op, since those are truncated matrices and aren't actually > exact inverses (?) > > The original paper talks about cosine similarities between users or > items in the reduced-dimension space, but, can anyone shine light on > the point of that? From the paper also, it seems like they say the > predictions are just computed as vector products as above. > > > Finally, separately, I'm trying to understand the Lanczos method as > part of computing an SVD. Lanczos operates on a real symmetric matrix > right? And am I right that it comes into play when you are computing > and SVD... > > A = U * S * VT > > ... because U is actually the eigenvectors of (symmetric) A*AT and V > is the eigenvectors of AT*A? And so Lanczos is used to answer those > questions to complete the SVD? > > On Fri, Jun 4, 2010 at 6:48 AM, Ted Dunning <[email protected]> wrote: > > You are correct. The... >
