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. >
