LDL Factorization Fails on Sparse Matrix

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Nathan Zhao
Nathan Zhao el 2 de En. de 2022
Comentada: Nathan Zhao el 4 de En. de 2022
I was playing with matlab's built in finite difference laplacian operator and noticed that the ldl factorization in matlab works when it is dense but somehow fails when it is sparse:
A = (delsq(numgrid('L', 40)));
A2 = full(delsq(numgrid('L', 40)));
bA = sum(A,2);
[LA, DA, PA] = ldl(A);
x = LA'\(DA\(LA\bA));
[L2, D2, P2] = ldl(A2);
x2 = L2'\(D2\(L2\bA));
The Laplacian should be positive definite and is real symmetric so hermitian so the LDL should be applicable but am stumped on why factorizing the sparse version fails.
  2 comentarios
Torsten
Torsten el 2 de En. de 2022
From the documentation of LDL:
L = ldl(A) returns only the permuted lower triangular matrix L as in the two-output form. The permutation information is lost, as is the block diagonal factor D. By default, ldl references only the diagonal and lower triangle of A, and assumes that the upper triangle is the complex conjugate transpose of the lower triangle. Therefore [L,D,P] = ldl(TRIL(A)) and [L,D,P] = ldl(A)both return the exact same factors. Note, this syntax is not valid for sparse A.
Nathan Zhao
Nathan Zhao el 4 de En. de 2022
ah, many thanks! In that case, I see that the appropriate syntax is the [L,D,P,S] = ldl(A). I'd like to follow up on the previous question with this: how is the ldl implemented in Matlab? Or to be more general, I want to confirm the ldl implementation is not using the same algorithm as the lu() function (which I'm aware uses umfpack). The reason I am asking is that when I compare the speed of the ldl and lu on the same laplacian but for larger matrix sizes, say 200 instead of 40, the ldl comes out a lot slower than doing [L,U,P,Q] = lu(A) of the same matrix even though we expect the ldl to be, at best, up to 50% faster. My hunch here is that the ldl is not using some multifrontal solver

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