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

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?

Answer 1

[Bruno Luong]

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

A=rand(3,3,1e5);

tic
    n = size(A,1);
    % FEX https://fr.mathworks.com/matlabcentral/fileexchange/68976-multipleqr
    [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.

tic
    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.

[Christian Schröder]

Excellent! Thank you!

Answer 2

[Alex Mcaulley]

delta = arrayfun(@(t) det(squeeze(G(t,:,:))),1:size(G,1));

[Christian Schröder]

Thanks! This is exactly what I need.

Answer 3

[Fabio Freschi]

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));

[Christian Schröder]

Thank you! I wasn't aware of arrayfun --- this is just what I need.

Answer 4

[Jos (10584)]

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

[Christian Schröder]

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!