Can I calculate the inverse of a matrix using arrayfun?

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Benson Gou
Benson Gou am 25 Jun. 2021
Kommentiert: Benson Gou am 29 Jun. 2021
Dear All,
I have a sparse matrix A. I want to calculate its inverse. I used the following way to calculate invA.
invA = A \ speye(size(A));
My matrix size is 3000 by 3000. I found it took me 1.3 seconds to get invA. It is longer than I expected.
I am wondering if I could use arrayfun to calculate invA. The idea is to calculate each column in invA by invA(:,i) = A \ ei, where ei is a zero column vector except its ith element is 1.0.
I tried the following code to implement the above idea:
invA = arrayfun(@(x) A\x ,speye(size(A,1)));
But I got error message.
Thanks a lot.
Benson
  1 Kommentar
Matt J
Matt J am 25 Jun. 2021
I don't see how arrayfun could help, unless this is on the GPU.

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Walter Roberson
Walter Roberson am 25 Jun. 2021
Bearbeitet: Walter Roberson am 25 Jun. 2021
invA = cell2mat(arrayfun(@(C) A\sparse(C,1,1,size(A,1),1), 1:size(A,2), 'uniform', 0))
I would be surprised if it is faster.
  8 Kommentare
Benson Gou
Benson Gou am 29 Jun. 2021
Hi, Walter,
I do not have the Image Processing Toolbox. So I cannot run decompose(A).
But the goal that I want to obtain the inverse of A is becaus I need to find out the nonzero elements in invA. So I do not know if decompose(A) is able to be useful for my case.
Thanks a lot.
Benson

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Weitere Antworten (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov am 25 Jun. 2021
Bearbeitet: Sulaymon Eshkabilov am 25 Jun. 2021
In fact, you can employ arrayfun for the matric inverse calc, e.g.:
tic;
iA =arrayfun(@inv,A);
toc;
Note that arrayfun is not the best option and does not take the sparse matrix.
Just direct inv() is the fastest so far.
In fact, for solving linear systems, to compute the inverse is not advised.
  4 Kommentare
Walter Roberson
Walter Roberson am 25 Jun. 2021
By the way, you cannot apply arrayfun() directly to a sparse matrix.
Benson Gou
Benson Gou am 29 Jun. 2021
Hi, Walter,
Yes, the decomposition is faster than \ operator. Thanks a lot.
Benson

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