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From Sean Owen <sro...@gmail.com>
Subject Re: Understanding the SVD recommender
Date Thu, 17 Nov 2011 13:26:24 GMT
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 <jake.mannix@gmail.com> 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" <srowen@gmail.com> 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 <ted.dunning@gmail.com> wrote:
> > You are correct.  The...
>

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