Path: news.mathworks.com!not-for-mail
From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: SVD which one is which
Date: Fri, 26 Dec 2008 23:21:02 +0000 (UTC)
Organization: The MathWorks, Inc.
Lines: 14
Message-ID: <gj3osu$8gs$1@fred.mathworks.com>
References: <gj3n2r$akd$1@fred.mathworks.com>
Reply-To: <HIDDEN>
NNTP-Posting-Host: webapp-02-blr.mathworks.com
Content-Type: text/plain; charset="ISO-8859-1"
Content-Transfer-Encoding: 8bit
X-Trace: fred.mathworks.com 1230333662 8732 172.30.248.37 (26 Dec 2008 23:21:02 GMT)
X-Complaints-To: news@mathworks.com
NNTP-Posting-Date: Fri, 26 Dec 2008 23:21:02 +0000 (UTC)
X-Newsreader: MATLAB Central Newsreader 1187260
Xref: news.mathworks.com comp.soft-sys.matlab:508855

"Ali Saleemi" <naumansaleemi@hotmail.com> wrote in message <gj3n2r$akd$1@fred.mathworks.com>...
> SVD gives eigenvectors in descending order, is there anyway to know which eigen vector belongs to which variable in the original matrix?  I dont have a square matrix in my original data so I cannot use eig() which according to my knowledge does not sort the eigenvectors.
> .......

  I refer you to the Wikipedia website for a discussion on singular value decompositions.  In particular read the section entitled "Relation to eigenvalue decomposition."

http://en.wikipedia.org/wiki/Singular_value_decomposition#Relation_to_eigenvalue_decomposition

  In answer to your question "which eigen vector belongs to which variable in the original matrix", there is no such relationship.  There is a relationship in the pairings between eigenvalues and eigenvectors, which is a different concept.  When matlab's 'svd' does give an eigenvalue decomposition, the pairing is between the eigenvalues of S and the corresponding columns of V in the decomposition

 M = U*S*V';

Roger Stafford