Efficient Matrix Multiplication

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Sam Da
Sam Da 2011-2-26
I have A(2000x5000). I need to perform the following:
P1 = A(:,1)*A(:,1)';
for i=2:5000
P1 = P1 + AA(:,i)*A(:,i)'
end
What is the most efficient way to do above? It takes so much time to do it right now due to size of the arrays.
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the cyclist
the cyclist 2011-2-26
From his initialization step, I would infer that "AA" is just a typo of "A."
Jan
Jan 2011-2-27
Just an actually too obvious comment: If AA is not typo, A*A' is not a matching solution. So, Sam Da, we need your help.

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the cyclist
the cyclist 2011-2-26
P1 = A * A';
On my machine, that cut the execution time from 330 seconds to 1.5. :-)
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Oleg Komarov
Oleg Komarov 2011-2-26
A = rand(10);
P1 = A(:,1)*A(:,1)';
for i=2:10
P1 = P1 + A(:,i)*A(:,i)';
end
P2 = A * A';
abs(P1-P2) < eps*3
Cyclist's method is essentially the same.
James Tursa
James Tursa 2011-2-26
Yep. I went back & checked the code I used to double check the result & saw my mistake. Thanks.

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