Best way to calculate the determinants of a series of matrices?

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I've got a series of time-dependent matrices which I'd like to calculate the determinants of. These matrices are stored as a three-dimensional array, where the first dimension indicates the period; in other words, the matrix at time t is given by
Gt = squeeze(G(t, :, :))
I'd now like a snippet of code which, being run, will ensure that
Delta(t) = det(squeeze(G(t, :, :)))
holds for all t. Of course I could do this with a loop, but I feel that there must be a more succinct, vectorized way of doing it. Sadly, MATLAB's det function itself is of no help. Is there something else I could use, or will I have to bite the proverbial bullet and use a loop after all?


Bruno Luong
Bruno Luong 2019 年 9 月 5 日
編集済み: Bruno Luong 2019 年 9 月 5 日
I reverse the order and put the page in third dimension (avoid to use squeeze).
For small size, you can save CPU time by 4 fold using MultipleQR available on FEX
n = size(A,1);
[Q,R] = MultipleQR(A);
R = reshape(R,n*n,[]);
d1 = (-1)^n * prod(R(1:n+1:end,:),1);
toc % Elapsed time is 0.087167 seconds.
d2 = arrayfun(@(k) det(A(:,:,k)), 1:size(A,3));
toc % Elapsed time is 0.376470 seconds.
% Check correctness
norm(d1-d2)/norm(d2) % 4.2026e-16
MultipleQR will be less efficient for large n.
  1 件のコメント
Christian Schröder
Christian Schröder 2019 年 9 月 5 日
Excellent! Thank you!


その他の回答 (3 件)

Alex Mcaulley
Alex Mcaulley 2019 年 9 月 5 日
delta = arrayfun(@(t) det(squeeze(G(t,:,:))),1:size(G,1));
  1 件のコメント
Christian Schröder
Christian Schröder 2019 年 9 月 5 日
Thanks! This is exactly what I need.


Fabio Freschi
Fabio Freschi 2019 年 9 月 5 日
Not sure if it you can speedup your code, but a single line code to do the job is
Delta = arrayfun(@(i)det(squeeze(G(i,:,:))),1:size(G,1));
  1 件のコメント
Christian Schröder
Christian Schröder 2019 年 9 月 5 日
Thank you! I wasn't aware of arrayfun --- this is just what I need.


Jos (10584)
Jos (10584) 2019 年 9 月 5 日
Elaborating on the answers using arrayfun, you can avoid the multiple squeeze operations by permuting the dimension order first:
G = permute(G,[2 3 1]) ; % last dimension is running fastest
D = arrayfun(@(k) det(G(:,:,k)), 1:size(G,3)) % per Fabio and Alex
  1 件のコメント
Christian Schröder
Christian Schröder 2019 年 9 月 5 日
Good to know, I hadn't previously been aware that squeeze() could be avoided if the singleton dimension was in fact last --- or that you could use permute to rearrange the dimensions of an array. Still very much a MATLAB newbie, so thanks to you and everyone who pointed this out!





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