Efficiently calculating the trace of a matrix product

13 次查看(过去 30 天)
I have two NxN square matrices, A and B, and I would like to calculate the trace of AB. Since the trace of AB only depends on its diagonal elements, it should hypothetically not be necessary to compute all of AB, thereby reducing the amount of operations from N^3 to N^2. My question is twofold:
  • Does calling tr(AB) in MATLAB automatically exploit this fact?
  • If not, is there an efficient way of doing this that doesn't involve calling for loops?
Thanks!

采纳的回答

Matt J
Matt J 2019-5-9
编辑:Matt J 2019-5-9
Bt=B.';
traceProduct = A(:).'*Bt(:);
  4 个评论
Matt J
Matt J 2019-5-10
Another way to test if the trace product is JIT optimized is to compare both implementations on a large matrix,
A=rand(3000); B=A;
tic;
version1 =trace(A*B);
toc;
%Elapsed time is 0.956904 seconds.
tic;
version2 = A(:).'*reshape(B.',[],1);
toc;
%Elapsed time is 0.068032 seconds.
It is pretty clear that the direct implementation is not optimized.
James Tursa
James Tursa 2019-5-10
I get the same results as Matt on various versions. And even if there is some version (maybe future) of MATLAB that does this optimization, it is still only the physical transpose part that could beat the direct implementation above ... i.e., compiled code that avoids physically forming the transpose.

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Logical 的更多信息

产品


版本

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by