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Multiplication of very large matrix

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MA
MA 2020-12-13
评论： MA 2020-12-14
Hello everyone,
Could anyone help me in re-writing the following equation to make the processing faster. N here is the number of points and it could be 50,000. D is a diagonal matrix where only the diagonal contains values and the other elements are zeros.
M=eye(N)-((1./max((ones(N)'*D*ones(N)),eps))*(ones(N)*ones(N)'*D));
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MA
MA 2020-12-14
Thanks very much for your help guys. I now know how to simplify it. So, if anyone is interesting. Here is how to simplify it:
M=eye(N)-repmat(((N.*N.*diag(D))./sum(diag(D)))',[N,1]);
MA
MA 2020-12-14
You are right David Goodmanson. It is a vector of size N by 1. So, multiplying by its transpose make sense. Thanks alot for the help.

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Jan
Jan 2020-12-14
编辑：Jan 2020-12-14
How strange: The comments above have not been displayed on my other computer. So this answer was written 1 hour after MA's comment.
In ones(N)*ones(N)' the transposition is completely meaningless. The result can be obtained much cheaper by: repmat(N, [N, N]). The matrox multiplication with this matrix can be formulated much cheaper:
The multiplication by () is an extremly expensive way to calculate:
A = ones(N) * ones(N)' * D
B = sum(D, 1) * N
Now B is a row vector only, but A is only a blownup version with N repetitions.
Because D is a diagonal matrix, you can omit the sum also.
Equivalently for ones(N)'*D*ones(N) : Of course you cannot simply omit the multiplication, but you can express it much cheaper by: sum(D(:)) or cheaper: sum(diag(D)) .
DD = diag(D);
M = eye(N) - N^2 * DD.' ./ max(sum(DD), eps);
Note, that this matrix is extremely redundant: All columns contain the same value except for the diagonal, which is 1 smaller. An efficient implementation would exploit this instead of creating a large matrix.

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