Re: question on the svd

From: Ian Bush (I.J.Bush_at_nospam.dl.ac.uk)
Date: 12/16/04 

Date: Thu, 16 Dec 2004 09:54:22 +0000

Jentje Goslinga wrote:

> This is not really any different from the symmetric eigenvalue
> problem where eigenvectors are also determined up to a
> multiplication by -1.
>

More strictly for all normalised eigenvectors that correspond
to non-degenerate eigenvalues they are determined up to a factor
k, where k * CONJG( k ) = 1. When there are degeneracies any linear
combination of the evecs that correspond to the degenerate evals is
also an evec,

Ian