LU decomposition with 5 output arguments
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Dear all,
I am trying to solve a huge linear system Ax = b using LU decomposition with five output arguments:
[L, U, p, q, D] = lu(A, 'vector');
My matrix A is sparse with nnz(A) =1e+9 and size(A) = (1e+7, 1e+7)
However, I cannot find any official help to use these 5 arguments to solve the system.
Does anyone have any idea how to ultilize these arguments?
Best,
Doan
0 Comments
Accepted Answer
Christine Tobler
on 22 Mar 2024
Edited: Torsten
on 22 Mar 2024
As Matt says, the simplest is likely to just use decomposition object.
In terms of how to solve a linear system, the doc page gives the formula
from this we can get a formula for the matrix S by just reformulating:
S = D*P'*L*U*Q'
So you can write the initial formula to solve
S*x = b
and insert that formula
(D*P'*L*U*Q') * x = b
now applying each factor's inverse from the left results in
x = Q*(U\(L\(P*(D\x))));
To use the 'vector' syntax instead, each of P*... and Q*... will become an indexing operation instead.
4 Comments
More Answers (1)
Matt J
on 22 Mar 2024
Edited: Matt J
on 22 Mar 2024
dA=decomposition(A,'ldl');
x=dA\b;
3 Comments
Torsten
on 22 Mar 2024
Edited: Torsten
on 22 Mar 2024
[L,U,P,Q,D] = lu(S) also returns a diagonal scaling matrix D such that P*(D\S)*Q = L*U. Typically, the row-scaling leads to a sparser and more stable factorization.
outputForm — Shape of permutation outputs
'matrix' (default) | 'vector'
Shape of permutation outputs, specified as 'matrix' or 'vector'. This flag controls whether lu returns the row permutations P and column permutations Q as permutation matrices or permutation vectors.
As matrices, the outputs P and Q satisfy these identities:
- Three outputs — P satisfies P*A = L*U.
- Four outputs — P and Q satisfy P*S*Q = L*U.
- Five outputs — P, Q, and D satisfy P*(D\S)*Q = L*U.
As vectors, the outputs P and Q satisfy these identities:
- Three outputs — P satisfies A(P,:) = L*U
- Four outputs — P and Q satisfy S(P,Q) = L*U
- Five outputs — P, Q, and D satisfy D(:,P)\S(:,Q) = L*U.
Example: [L,U,P] = lu(A,'vector')
Matt J
on 22 Mar 2024
Edited: Matt J
on 22 Mar 2024
Thanks, but my matrix is not Hermitian so that LDL doesn't apply.
Well, you can just do,
dA=decomposition(A)
and let it choose what is best for thte matrix
I have no idea on which option that decomposition uses: 'vector' format or the default format 'matrix'.
I don't think you need to know. You can just check how much memory the dA object consumes.
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