Efficient Matrix Multiplication

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.

[Walter Roberson]

What is AA in this?

[the cyclist]

From his initialization step, I would infer that "AA" is just a typo of "A."

[Jan]

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.

Answer 1

[the cyclist]

P1 = A * A';

On my machine, that cut the execution time from 330 seconds to 1.5. :-)

[James Tursa]

Except that does not do the same calculation. The loop in OP's post above only does the outer product of each column with itself, not the entire A*A' product.

[Oleg Komarov]

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]

Yep. I went back & checked the code I used to double check the result & saw my mistake. Thanks.