# LDL does not support complex symmetric matrices

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Nathan Zhao on 6 Jan 2022
Commented: Yaroslav Urzhumov on 19 Jan 2023
Given that you cannot factorize a complex symmetric matrix with LDL in MatLab, I was wondering if there was a particular reason? Is there some package which can perform this factorization?

Christine Tobler on 7 Jan 2022
Largely the reason LDL doesn't support this is that it's less commonly requested for complex symmetric matrices than for complex Hermitian ones, so LDL only supports the more common variant. We would also need some option to indicate which of the two modes is meant to be used.
There is a LAPACK function ZSYTRF that you can call from a mex-file to get the complex symmetric LDL decomposition.
Nathan Zhao on 7 Jan 2022
thanks! Just to follow-up, LAPACK is meant for dense matrices so this answer is not applicable for sparse matrices, correct?
Christine Tobler on 10 Jan 2022
Yes, LDL for sparse matrices doesn't support complex at all, neither the Hermitian nor the complex symmetric case. This is again not based on mathematics, just on practical considerations of resources and commonality of the use case.

David Goodmanson on 6 Jan 2022
Edited: David Goodmanson on 7 Jan 2022
Hi Nathan,
the LDL decomposition works for hermitian matrices. In LDL the D matrix is hermitian, so
if A = L*D*L', then A' = L*D' *L' = L*D*L' = A, and A has to be hermitian.
But a symmetric complex matrix is not hermitian, so LDL won't work.
Nathan Zhao on 24 Jan 2022
Hi David Goodmanson,
thanks for your response! For complex symmetric matrices, I'm aware a naive LDL implementation does not work, but if you use a bunch-kauffman factorization (which uses 2x2 pivoting and makes D block-diagonal rather than diagonal, but the blocks are either 1x1 or 2x2), then the factorization should work?
Best,
Nathan
Yaroslav Urzhumov on 19 Jan 2023
David,
LDL for a symmetric matrix implies A=L*D*L.', where .' is the usual (rather than complex conjugate) transpose. You can easily see that this decomposition is a symmetric matrix - even if D is complex-valued. It's a different kind of decomposition.

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