MATLAB Answers

Get the diagonal without calculating the explicit matrix

2 views (last 30 days)
Long Hong
Long Hong on 26 Mar 2020
Edited: Matt J on 26 Mar 2020
Dear all:
I am trying to calculate a diagonal of a matrix (denoted A), which is formed by multiplying two large-dimensional matrices (denoted as B*C).
A naive way to do it is: first, calculating explicitly A = B*C, then get diagonal out from A. However, the first step takes forever to run due to the high-dimension of B and C. But the only thing I need is the diagonal of A.
Another straightforward way in my mind is: I could create a loop by calculating each element of the diagonal of A one by one. It will surely save a lot of time, but I am not sure if this is the most efficient way.
I am wondering if anyone knows a faster/smarter way to calculate it.
Thank you very much in advance!
Best,
Long

  1 Comment

Matt J
Matt J on 26 Mar 2020
The best approach will depend on the dimensions of the matrices, and whether they are of sparse-type or not.

Sign in to comment.

Accepted Answer

Matt J
Matt J on 26 Mar 2020
Edited: Matt J on 26 Mar 2020
Assuming B*C results in a square matrix,
diagonal=sum(B.' .* C, 1);

  8 Comments

Show 5 older comments
Long Hong
Long Hong on 26 Mar 2020
Thank you Matt! I have tested it in a relatively large subset of my original data, your algorithm is indeed faster. Thank you for the valuable advice!
Long Hong
Long Hong on 26 Mar 2020
Thank you the cyclist! Do you have any insight in doing the transpose quicker? I am a bit confused here.
Matt J
Matt J on 26 Mar 2020
the cyclist means you might avoid the transpose by loading data column-wise instead of row-wise when you first build B.

Sign in to comment.

More Answers (1)

the cyclist
the cyclist on 26 Mar 2020
Here is one way:
% Make up some inputs
N = 4;
B = rand(N);
C = rand(N);
% Calculate the diagonal
A_diag = 0;
for nr = 1:N
A_diag = A_diag + B(:,nr).*C(nr,:)';
end

  1 Comment

Long Hong
Long Hong on 26 Mar 2020
Thanks the cyclist! This is a method I have applied currently.

Sign in to comment.

Sign in to answer this question.


Translated by