I have a non-symmetric matrix, where the non-symmetry only appears at a subset of points. This arises due to the particular manner on which boundary conditions are applied in a Cartesian grid method. I'm looking for a way to factor the matrix into the product of a symmetric sparse matrix and a diagonal matrix with a low rank correction so I can symmetrize the system as suggested here:
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There is no way you really mean product since the product of a low rank matrix with anything is also low rank. Given non-symmetric $A$, perhaps you are looking for
$$S = \frac 1 2 (A + A^T)$$ $$N = \frac 1 2 (A - A^T).$$
If $A = S + N$ is symmetric in most of the domain, then the non-symmetric part $N$ will be low rank and sparse.
Of course a sum is much less useful than a product if you were interested in solving this system. In particular, if $N$ is nonzero at boundaries, the rank is not small enough for specialized tricks to apply. What was really intended in the question you reference is to reformulate your problem, not to brute-force decompose or symmetricze.
answered Aug 9, 2012 at 14:15
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$\begingroup$ I suppose I should have said an identity matrix plus a low rank correction. $\endgroup$ Commented Aug 9, 2012 at 14:29
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$\begingroup$ @JohnMousel Please fix the title and body of your question. $\endgroup$ Commented Aug 9, 2012 at 17:19