[R] SVD/Eigenvector confusion
Prof Brian Ripley
ripley at stats.ox.ac.uk
Sat Feb 28 15:57:59 CET 2004
On Sun, 29 Feb 2004, Philip Warner wrote:
> 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
> > Since eigenvectors are not unique, it does mean that
> >you cannot reverse the process, as you seem to be trying to do.
> > >
> > > 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?
I think there are ways to derive the correct signs, but your approach is a
poor way to do the calculations as it squares the condition number of A.
There are standard algorithms for computing the SVD from A alone.
> Thanks for your help, it is much appreciated.
> Philip Warner | __---_____
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Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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