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Re: Woodbury formula in practice
Posted:
Sep 27, 2012 12:57 PM


srobinson3130@gmail.com writes: >Hello, > >I am trying to use the Woodbury formula [1] in order to apply corrections o= >n the inverse of a matrix A I already calculated once as X (rather big: n= >=3D6000+). Once in a while I would like to compute the inverse of A with a = >"k" changes ranging from 23 to 20. > >Applying the woodbury formula should be of big help there, and it works per= >fectly when I only apply 2 or 3 changes. But when I apply 20, one of the ma= >trix I need to invert in the calculation (kind of known as the capacitance = >matrix [1]) becomes singular (determinant gets way too small). I am thus un= >able to apply the formula. Note that the corrections applied are valid, i.e= >. if I create the new matrix from scratch, it is nonsingular. > >My questions are the following:=20 > is it something I should expect? > are there anyway to solve it? > >I have been looking everywhere for practical information regarding woodbury= >'s usage in a numerical analysis context but can not find much information.= > I would be very grateful for any guidance on that topic. > > >SR > >[1] http://en.wikipedia.org/wiki/Woodbury_matrix_identity
There is one paper by Yip in SIAM J Sci Stat Comp 7, 1986, which states that ''if both A and B are well conditioned and A=B UV^T , then the application of the ShermanMorrisonWoodbury formula is stable'' but nothing more . But if you look at that formula then it is obvious that there is a danger of loss of precision if B is already illconditioned. I don't know why you are using the inverse explicitly, I personally would prefer to use a LU or QR decomposition of the original matrix , from which any operation with the inverse can easily be obtained. These decompositions can be updated in a stable manner (and there exists ready to use software for this in netlib). hth peter



