To get an SVD isn't it more common to compute the eigen decomposition of

[ 0  A' ]
[ A  0  ]

Rather than A' A?

I have heard that this is supposed to avoid some problems with round-off
such as  you are seeing.  Moreover, many algorithms can be restated so that
this matrix never needs to be constructed explicitly.

On Wed, Mar 17, 2010 at 4:23 AM, Dimitri Pourbaix
<pourb...@astro.ulb.ac.be>wrote:

> I just checked the code and notice the problem is already present at the
> exit of TriDiagonalTransform.  In order to compute SVD, one computes the
> eigen decomposition of A^tA which relies upon the Householder tri-diagonal
> transformation.  The tri-diagonal maxtrix still has 3 non-zero main
> diagonal
> elements.  The smallest one is about 1.e-13, i.e. slightly too large to be
> considered null against the largest, about 900.0
>
> Right now, I see no option but avoiding the computation of A^tA.  It is
> fair to ask for 2.2.
>

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