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Fri, 26 Dec 2008 22:50:03 +0000
SVD which one is which
http://nl.mathworks.com/matlabcentral/newsreader/view_thread/241607#618909
Ali Saleemi
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.<br>
<br>
Regards<br>
<br>
Saleemi

Fri, 26 Dec 2008 23:05:03 +0000
Re: SVD which one is which
http://nl.mathworks.com/matlabcentral/newsreader/view_thread/241607#618910
Chris Conner
the eigenvectors/values are listed in order of descending power. there is a 11 correspondence, so the first eigenvector matches with the first eigenvalue.

Fri, 26 Dec 2008 23:21:02 +0000
Re: SVD which one is which
http://nl.mathworks.com/matlabcentral/newsreader/view_thread/241607#618911
Roger Stafford
"Ali Saleemi" <naumansaleemi@hotmail.com> wrote in message <gj3n2r$akd$1@fred.mathworks.com>...<br>
> 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.<br>
> .......<br>
<br>
I refer you to the Wikipedia website for a discussion on singular value decompositions. In particular read the section entitled "Relation to eigenvalue decomposition."<br>
<br>
<a href="http://en.wikipedia.org/wiki/Singular_value_decomposition#Relation_to_eigenvalue_decomposition">http://en.wikipedia.org/wiki/Singular_value_decomposition#Relation_to_eigenvalue_decomposition</a><br>
<br>
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<br>
<br>
M = U*S*V';<br>
<br>
Roger Stafford

Sat, 27 Dec 2008 10:48:02 +0000
Re: SVD which one is which
http://nl.mathworks.com/matlabcentral/newsreader/view_thread/241607#618932
Johan Carlson
"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <gj3osu$8gs$1@fred.mathworks.com>...<br>
> "Ali Saleemi" <naumansaleemi@hotmail.com> wrote in message <gj3n2r$akd$1@fred.mathworks.com>...<br>
> > 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.<br>
> > .......<br>
> <br>
> I refer you to the Wikipedia website for a discussion on singular value decompositions. In particular read the section entitled "Relation to eigenvalue decomposition."<br>
> <br>
> <a href="http://en.wikipedia.org/wiki/Singular_value_decomposition#Relation_to_eigenvalue_decomposition">http://en.wikipedia.org/wiki/Singular_value_decomposition#Relation_to_eigenvalue_decomposition</a><br>
> <br>
> 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<br>
> <br>
> M = U*S*V';<br>
> <br>
> Roger Stafford<br>
<br>
<br>
As Roger wrote, there is no direct link between any ONE variable in X and the eigenvectors.<br>
<br>
The columns of U form an orthonormal basis of the columns space of X, while the columns of V form an orthonormal basis of the row space of X. As such, the basis vectors will most likely be linear combinations of the variables in X, which is actually often the point.<br>
<br>
There are some more interesting properties as well, namely that the eigenvector corresponding to the largest eigenvalue represents the direction in which the variance of the data is the largest, and then similarly for the remaining vectors.<br>
<br>
Check the Wiki for more details, or the excellent book "Linear Algebra and Its Applications" by Gilbert Strang. There's a good appendix explaining all this.<br>
<br>
/JC

Sat, 27 Dec 2008 17:43:47 +0000
Re: SVD which one is which
http://nl.mathworks.com/matlabcentral/newsreader/view_thread/241607#618964
Greg Heath
On Dec 26, 5:50 pm, "Ali Saleemi" <naumansale...@hotmail.com> wrote:<br>
> SVD gives eigenvectors in descending order, is there<br>
> anyway to know which eigen vector belongs to which<br>
> variable in the original matrix?<br>
<br>
In general, SVD and SVDS yields singular vectors of A,<br>
not eigenvectors of A. The nonnegative real singular<br>
values and corresponding singular vectors are<br>
ordered w.r.t ascending singular values in SVD.<br>
I am not sure about the ordering in SVDS..<br>
<br>
doc SVD<br>
dov SVDS<br>
<br>
In general, each singular vector is a linear combination of<br>
all the variables. In realworldproblems, it is rare that<br>
singular vectors are coordinate aligned.<br>
<br>
The singular vectors of A are eigenvectors of<br>
A*A^T and A^T*A.<br>
<br>
> I dont have a square matrix<br>
> in my original data so I cannot use eig() which according<br>
> to my knowledge does not sort the eigenvectors.<br>
<br>
If A is square, both EIG and EIGS yield eigenvectors<br>
of A. If A is symmetric, the eigenvalues are real and the<br>
corresponding eigenvectors are sorted (ascending for EIG<br>
and descending for EIGS).<br>
<br>
doc EIG<br>
doc EIGS<br>
<br>
Hope this helps.<br>
<br>
Greg