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 <ste...@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.bick...@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 <ste...@wienert.cc> > wrote: > > > >> After reading this thread: > >> > >> > http://mail-archives.apache.org/mod_mbox/mahout-user/201102.mbox/%3caanlktinq5k4xrm7nabwn8qobxzgvobbot2rtjzsv4...@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 <ste...@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+dbfrj8pcds2n...@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 <ste...@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.dunn...@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 <ste...@wienert.cc> > >> wrote: > >> >>> > >> >>>> compared to SVD, is the result is the "right singular value"? > >> >>>> > >> >>> > >> >> > >> >> > >> >> > >> >> -- > >> >> Stefan Wienert > >> >> > >> >> http://www.wienert.cc > >> >> ste...@wienert.cc > >> >> > >> >> Telefon: +495251-2026838 > >> >> Mobil: +49176-40170270 > >> >> > >> > > >> > > >> > > >> > -- > >> > Stefan Wienert > >> > > >> > http://www.wienert.cc > >> > ste...@wienert.cc > >> > > >> > Telefon: +495251-2026838 > >> > Mobil: +49176-40170270 > >> > > >> > >> > >> > >> -- > >> Stefan Wienert > >> > >> http://www.wienert.cc > >> ste...@wienert.cc > >> > >> Telefon: +495251-2026838 > >> Mobil: +49176-40170270 > >> > > > > > > -- > Stefan Wienert > > http://www.wienert.cc > ste...@wienert.cc > > Telefon: +495251-2026838 > Mobil: +49176-40170270 >
