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 roundoff
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
<pourbaix@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 tridiagonal
> transformation. The tridiagonal maxtrix still has 3 nonzero main
> diagonal
> elements. The smallest one is about 1.e13, 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.
>
