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LU decomposition with 5 output arguments

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Duy Hai
Duy Hai 2024 年 3 月 22 日
編集済み: Torsten 2024 年 3 月 22 日
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

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Christine Tobler
Christine Tobler 2024 年 3 月 22 日
編集済み: Torsten 2024 年 3 月 22 日
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
[L,U,P,Q,D] = lu(S) satisfies P*(D\S)*Q = L*U
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 件のコメント
Duy Hai
Duy Hai 2024 年 3 月 22 日
When the "vector" option is used, the solving process should be
c = D\b;
x(q) = U\(L\c(p));
Torsten
Torsten 2024 年 3 月 22 日
編集済み: Torsten 2024 年 3 月 22 日
Thanks. Except that x is a row vector instead of a column vector, this seems to work.

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その他の回答 (1 件)

Matt J
Matt J 2024 年 3 月 22 日
編集済み: Matt J 2024 年 3 月 22 日
dA=decomposition(A,'ldl');
x=dA\b;
  3 件のコメント
Torsten
Torsten 2024 年 3 月 22 日
編集済み: Torsten 2024 年 3 月 22 日
[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.
outputFormShape 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
Matt J 2024 年 3 月 22 日
編集済み: Matt J 2024 年 3 月 22 日
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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