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In the so-called incremental SVD used for collaborative filtering:

http://www.machinelearning.org/proceedings/icml2007/papers/407.pdf

http://www2.research.att.com/~volinsky/papers/ieeecomputer.pdf

http://www.quuxlabs.com/blog/2010/09/matrix-factorization-a-simple-tutorial-and-implementation-in-python/

The user x item matrix R is factored as QP using gradient descent. In the classical SVD there is the diagonal matrix S which holds the singular values. What happens(ed) to that matrix in this formulation? Is it just omitted and they still call it SVD or is it implicitly part of Q and/or P?

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  • $\begingroup$ There are many, many ways to factorize matrices. There need not be any relationship at all between two factorizations (apart from the defining fact that they represent--or in this case approximate--the same matrix). $\endgroup$ – whuber Oct 29 '13 at 13:50
  • $\begingroup$ But doesn't an "SVD" require three matrices instead of the two? My confusion is the terminology : The result being two matrices being called an SVD. It does not appear that this "SVD" even exposes the singular values. $\endgroup$ – B_Miner Oct 29 '13 at 15:57
  • $\begingroup$ I looked at only one of your references in order to find out what you mean by "QP." Where exactly is this decomposition called an "SVD"? $\endgroup$ – whuber Oct 29 '13 at 15:59
  • $\begingroup$ Page 6 of the first link. Pretty much every source related to the Netflix prize calls it SVD or incremental SVD. I think from another source: files.grouplens.org/papers/webKDD00.pdf that this technique creates the two matrices with sqrt(S) in each. $\endgroup$ – B_Miner Oct 29 '13 at 16:44
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    $\begingroup$ It's the same thing.. the diagonal is rolled into the other two matrices when they talk about two matrices. $\endgroup$ – ignorant Jan 15 '14 at 5:49

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