Re: is SVD decomposition unique?

From: Zdislav V. Kovarik (kovarik_at_mcmaster.ca)
Date: 11/21/04 

Date: Sun, 21 Nov 2004 16:50:45 -0500

On Sat, 20 Nov 2004, lucy wrote:

> Hi all,
>
> I am wondering if SVD decomposition is unique or not... given a matrix A,
> can it be composed into two different SVD forms:
>
> A=U1*S1*V1'=U2*S2*V2'
>
> ???

Re-think the question from your follow-up inquiry (about similarity): what
are your assumptions?

To the present question: Have you experimented? Look at the 1-by-1 matrix
[-2]: two of its SVD's (real scalars assumed) are

  [1]*[2]*[-1] and [-1]*[2]*[1]

And in higher dimensions, the variety will increase.

Next experiment: How many SVD's does the 2-by-2 identity matrix have?

And if you feel "cheated" by the simplicity of the previous examples,
consider all the SVD's of an orthogonal matrix.

Cheers, ZVK(Slavek).