SVD/Eigenvector confusion

At 01:17 AM 29/02/2004, Prof Brian Ripley wrote:

On Sun, 29 Feb 2004, Philip Warner wrote:

My understanding of SVD is that, for A an mxn matrix, m > n:

     A = UWV*

where W is square root diagonal eigenvalues of A*A extended with zero
valued rows, and U and V are the left & right eigen vectors of A. But this
does not seem to be strictly true and seems to require specific
eigenvectors, and I am not at all sure how these are computed.

(A %*% t(A) is required, BTW.)  That is not the definition of the SVD.
It is true that U are eigenvectors of A %*% t(A) and V of t(A) %*% A, but
that does not make them left/right eigenvectors of A (unless that is your
private definition).

Sorry, that should have read 'left & right singular vectors', and I'm 
beginning to suspect that they are only the starting point for deriving the 
singular vectors (based on 
http://www.cs.utk.edu/~dongarra/etemplates/node191.html)

  Since eigenvectors are not unique, it does mean that
you cannot reverse the process, as you seem to be trying to do.

...cut...

which seems a little off the mark.

It is not expected to work.

Maybe not by you... 8-}

There is no rule: the SVD is computed by a different algorithm.

So I assume my approach will not give me the singular vectors, and I need a 
different way of deriving them, is that right?

Thanks for your help, it is much appreciated.




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