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From Stefan Wienert <ste...@wienert.cc>
Subject Re: Need a little help with SVD / Dimensional Reduction
Date Mon, 06 Jun 2011 17:30:27 GMT
Hi Danny!

I understand that for M*M' (and for M'*M) the left and right
eigenvectors are identical. But that is not exactly what I want. The
lanczos solver from mahout gives me the eigenvectors of M*M', which
are the left singular vectors of M. But I need the right singular
vectors of M (and not M*M'). How do I get them?

Sorry, my matrix math is not as good as it should be, but I hope you
can help me!

Thanks,
Stefan

2011/6/6 Danny Bickson <danny.bickson@gmail.com>:
> 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
>>
>



-- 
Stefan Wienert

http://www.wienert.cc
stefan@wienert.cc

Telefon: +495251-2026838
Mobil: +49176-40170270

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