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From Ted Dunning <>
Subject Re: Using SVD-conditioned matrix
Date Sun, 16 Sep 2012 18:13:17 GMT
A is in there implicitly.

U_k provides a basis of the row space and V_k provides a basis of the
column space of A.  A itself has a representation in either of these
depending on whether you think of it as rows or columns.

The original question (possibly misphrased and not in so many words) asked
for a way to project any vector into the space spanned by A.  What space
this is depends on whether we have a new column vector (with n rows x 1
column) or a new row vector (with 1 row x m columns).  In either case, the
way to compute this is to project the vector onto the basis of interest
(U_k for column vectors, V_k for row vectors) and then reconstruct that
projection in the original space.

This is not usually a practical operation when we are working with sparse
vectors because the projected vector is not usually sparse.  Thus it is
usually better to stop before projecting back into the original space.

On Sun, Sep 16, 2012 at 10:26 AM, Sean Owen <> wrote:

> I don't quite get these formulations -- shouldn't Ak be in there
> somewhere? you have a new row of that (well, some piece of some new
> row of Ak), and need a new row of Uk. Or: surely the expression
> depends on V?
> On Sun, Sep 16, 2012 at 5:33 PM, Ted Dunning <>
> wrote:
> > And if you want the reduced rank representation of A, you have it already
> > with
> >
> >     A_k = U_k S_k V_k'
> >
> > Assume that A is n x m in size.  This means that U_k is n x k and V_k is
> m
> > x k
> >
> > The rank reduced projection of an n x 1 column vector is
> >
> >     u_k = U_k U_k' u
> >
> > Beware that v_k is probably not sparse even if v is sparse.
> >
> > Similarly, the rank reduced projection of a 1 x m row vector is
> >
> >     v_k = v V_k V_k'
> >
> > A similar sparsity warning applies to v_k.  This is why it is usually
> > preferable to just work in the reduced space directly.

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