LDL does not support complex symmetric matrices

조회 수: 12 (최근 30일)
Nathan Zhao
Nathan Zhao 2022년 1월 6일
댓글: Yaroslav Urzhumov 2023년 1월 19일
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?

채택된 답변

Christine Tobler
Christine Tobler 2022년 1월 7일
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.
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Nathan Zhao
Nathan Zhao 2022년 1월 7일
thanks! Just to follow-up, LAPACK is meant for dense matrices so this answer is not applicable for sparse matrices, correct?
Christine Tobler
Christine Tobler 2022년 1월 10일
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.

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추가 답변 (1개)

David Goodmanson
David Goodmanson 2022년 1월 6일
편집: David Goodmanson 2022년 1월 7일
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.
  댓글 수: 2
Nathan Zhao
Nathan Zhao 2022년 1월 24일
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
Yaroslav Urzhumov 2023년 1월 19일
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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