# [R] SVD/Eigenvector confusion

Philip Warner pjw at rhyme.com.au
Sat Feb 28 15:54:16 CET 2004

```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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```