|Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices|
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- To: eigen@xxxxxxxxxxxxxxxxxxx
- Subject: Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices
- From: Ben Goodrich <bgokgm@xxxxxxxxxxxxxx>
- Date: Wed, 24 Feb 2010 09:45:20 -0500
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On Wed, Feb 24, 2010 at 5:53 AM, Jitse Niesen <jitse@xxxxxxxxxxxxxxxxx> wrote:
> On Wed, 24 Feb 2010, Gael Guennebaud wrote:
>> note that the LDLT decomp does pivoting, so it is normal that you don't
>> get the same diagonal matrix. If you reconstruct the matrix from the decomp
>> you will see that in this case the decomp is correct.
> To further clarify, because this also threw me off for a while: I think the
> important thing here is that the matrix A is singular. The diagonal D is
> unique (up to reordering) for positive-definite matrices. However, there is
> no such uniqueness for singular matrices. Simple example
> A = [ 0 0; 0 1 ] (these are 2-by-2 matrices in Matlab notation)
> L = [ 0 0; sqrt(3) 1 ]
> D = [ 1/4 0; 0 1/4 ]
> The matrix A has two LDL^T decompositions, A = A * I * A^T = L * D * L^T,
> but the diagonal parts are not the same.
On pages 12 -- 13 (section 5.3) of this paper by Benson and Vanderbei
they claim (without proof) that D is unique in the positive
semi-definite case (although part of L is not when A is singular).
Maybe they are incorrect, but I don't think your example shows that
L = [ 0 0; sqrt(3) 1 ]
is not *unit* lower triangular. If we restrict L to be unit lower
triangular, then in your example the (a?) LDL' decomposition of A is
L = I and A = D.
So, I am still confused :) but thanks to everyone for their comments
and to Gael for implementing reconstructedMatrix() .