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From Danny Bickson <danny.bick...@gmail.com>
Subject Re: Need a little help with SVD / Dimensional Reduction
Date Mon, 06 Jun 2011 17:08:46 GMT
Hi Stefan!
For a positive semidefinite matrix, the lest and right eigenvectors are
identical.
See SVD wikipeida text: When *M* is also positive
semi-definite<http://en.wikipedia.org/wiki/Positive-definite_matrix>,
the decomposition *M* = *U**D**U* * is also a singular value decomposition.
So you don't need to be worried about the other singular vectors.

Hope this helps!

On Mon, Jun 6, 2011 at 12:57 PM, Stefan Wienert <stefan@wienert.cc> wrote:

> Hi.
>
> Thanks for the help.
>
> The important points from wikipedia are:
> - The left singular vectors of M are eigenvectors of M*M' .
> - The right singular vectors of M are eigenvectors of M'*M.
>
> as you describe, the mahout lanczos solver calculate A=M'*M (I think
> it does A=M*M', but it is not a problem). Therefore it does already
> calculate the right (or left) singular vector of M.
>
> But my question is, how can I get the other singular vector? I can
> transpose M, but then I have to calculated two SVDs, one for the right
> and one for the left singular value... I think there is a better way
> :)
>
> Hope you can help me with this...
> Thanks
> Stefan
>
>
> 2011/6/6 Danny Bickson <danny.bickson@gmail.com>:
> > Hi
> > Mahout SVD implementation computes the Lanzcos iteration:
> > http://en.wikipedia.org/wiki/Lanczos_algorithm
> > Denote the non-square input matrix as M. First a symmetric matrix A is
> > computed by A=M'*M
> > Then an approximating tridiagonal matrix T and a vector matrix V are
> > computed such that A =~ V*T*V'
> > (this process is done in a distributed way).
> >
> > Next the matrix T is next decomposed into eigenvectors and eignevalues.
> > Which is the returned result. (This process
> > is serial).
> >
> > The third step makes the returned eigenvectors orthogonal to each other
> > (which is optional IMHO).
> >
> > The heart of the code is found at:
> >
> ./math/src/main/java/org/apache/mahout/math/decomposer/lanczos/LanczosSolver.java
> > At least that is where it was in version 0.4 I am not sure if there are
> > changes in version 0.5
> >
> > Anyway, Mahout does not compute directly SVD. If you are interested in
> > learning more about the relation to SVD
> > look at: http://en.wikipedia.org/wiki/Singular_value_decomposition,
> > subsection: relation to eigenvalue decomposition.
> >
> > Hope this helps,
> >
> > Danny Bickson
> >
> > On Mon, Jun 6, 2011 at 9:35 AM, Stefan Wienert <stefan@wienert.cc>
> wrote:
> >
> >> After reading this thread:
> >>
> >>
> http://mail-archives.apache.org/mod_mbox/mahout-user/201102.mbox/%3CAANLkTinQ5K4XrM7naBWn8qoBXZGVobBot2RtjZSV4yOd@mail.gmail.com%3E
> >>
> >> Wiki-SVD: M = U S V* (* = transposed)
> >>
> >> The output of Mahout-SVD is (U S) right?
> >>
> >> So... How do I get V from (U S)  and M?
> >>
> >> Is V = M (U S)* (because this is, what the calculation in the example
> is)?
> >>
> >> Thanks
> >> Stefan
> >>
> >> 2011/6/6 Stefan Wienert <stefan@wienert.cc>:
> >> >
> https://cwiki.apache.org/confluence/display/MAHOUT/Dimensional+Reduction
> >> >
> >> > What is done:
> >> >
> >> > Input:
> >> > tf-idf-matrix (docs x terms) 6076937 x 20444
> >> >
> >> > "SVD" of tf-idf-matrix (rank 100) produces the eigenvector (and
> >> > eigenvalues) of tf-idf-matrix, called:
> >> > svd (concepts x terms) 87 x 20444
> >> >
> >> > transpose tf-idf-matrix:
> >> > tf-idf-matrix-transpose (terms x docs) 20444 x 6076937
> >> >
> >> > transpose svd:
> >> > svd-transpose (terms x concepts) 20444 x 87
> >> >
> >> > matrix multiply:
> >> > tf-idf-matrix-transpose x svd-transpose = result
> >> > (terms x docs) x (terms x concepts) = (docs x concepts)
> >> >
> >> > so... I do understand, that the "svd" here is not SVD from wikipedia.
> >> > It only does the Lanczos algorithm and some magic which produces the
> >> >> Instead either the left or right (but usually the right) eigenvectors
> >> premultiplied by the diagonal or the square root of the
> >> >> diagonal element.
> >> > from
> >>
> http://mail-archives.apache.org/mod_mbox/mahout-user/201102.mbox/%3CAANLkTi=Rta7tfRm8Zi60VcFya5xF+dbFrJ8pcds2N0-V@mail.gmail.com%3E
> >> >
> >> > so my question: what is the output of the SVD in mahout. And what do I
> >> > have to calculate to get the "right singular value" from svd?
> >> >
> >> > Thanks,
> >> > Stefan
> >> >
> >> > 2011/6/6 Stefan Wienert <stefan@wienert.cc>:
> >> >>
> >>
> https://cwiki.apache.org/confluence/display/MAHOUT/Dimensional+Reduction
> >> >>
> >> >> the last step is the matrix multiplication:
> >> >>  --arg --numRowsA --arg 20444 \
> >> >>  --arg --numColsA --arg 6076937 \
> >> >>  --arg --numRowsB --arg 20444 \
> >> >>  --arg --numColsB --arg 87 \
> >> >> so the result is a 6,076,937 x 87 matrix
> >> >>
> >> >> the input has 6,076,937 (each with 20,444 terms). so the result of
> >> >> matrix multiplication has to be the right singular value regarding
to
> >> >> the dimensions.
> >> >>
> >> >> so the result is the "concept-document vector matrix" (as I think,
> >> >> these is also called "document vectors" ?)
> >> >>
> >> >> 2011/6/6 Ted Dunning <ted.dunning@gmail.com>:
> >> >>> Yes.  These are term vectors, not document vectors.
> >> >>>
> >> >>> There is an additional step that can be run to produce document
> >> vectors.
> >> >>>
> >> >>> On Sun, Jun 5, 2011 at 1:16 PM, Stefan Wienert <stefan@wienert.cc>
> >> wrote:
> >> >>>
> >> >>>> compared to SVD, is the result is the "right singular value"?
> >> >>>>
> >> >>>
> >> >>
> >> >>
> >> >>
> >> >> --
> >> >> Stefan Wienert
> >> >>
> >> >> http://www.wienert.cc
> >> >> stefan@wienert.cc
> >> >>
> >> >> Telefon: +495251-2026838
> >> >> Mobil: +49176-40170270
> >> >>
> >> >
> >> >
> >> >
> >> > --
> >> > Stefan Wienert
> >> >
> >> > http://www.wienert.cc
> >> > stefan@wienert.cc
> >> >
> >> > Telefon: +495251-2026838
> >> > Mobil: +49176-40170270
> >> >
> >>
> >>
> >>
> >> --
> >> Stefan Wienert
> >>
> >> http://www.wienert.cc
> >> stefan@wienert.cc
> >>
> >> Telefon: +495251-2026838
> >> Mobil: +49176-40170270
> >>
> >
>
>
>
> --
> Stefan Wienert
>
> http://www.wienert.cc
> stefan@wienert.cc
>
> Telefon: +495251-2026838
> Mobil: +49176-40170270
>

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