Loops and Complicated Matrix Multiplication

Hello,

This may be simple to figure out but for some reason I can't seem to get the results I want.

I'm trying to multiply two matrices together using a loop, in which the matrix dimensions do not agree. I know this impossible to do using the standard matrix arithmetic. What I would like to do is multiply each column from matrix (Txcouple 41x40x40) against a page in the other matrix (dpTxcouple1 41x17x1600). There should be 1600 pages in the resulting answer but I am unsure how to do this. Here is what I have so far. I can manage to populate the first page with the correct result, but the other pages are filled with zeros.

for k=1:40;
    for m=1:17
       for g=1:40;
           for i=(f-1) * 40 + g;
   Tx1Final(:,m,i)=Txcouple(:,k,g).*dpTxcouple1(:,m,i);
           end 
       end
    end
end

Any help would be greatly appreciated.

Thanks!

Josh

[Matt Fig]

What is f?

[Joshua]

Sorry, f should have been replaced by k.

Thanks a lot for the help though. Working through a simpler version of the problem definitely helped!

[Matt Fig]

You're welcome!

Answer 1

[Matt Fig]

Often times it is easier to figure out such a problem by looking at a simpler version where we can actually keep track manually. Here I compare three methods that yield the same result. You have not answered the question I posted above, but from your description I tried this. Look and see if it works by comparing T,H, and C. If these do what you want, then adapt for your use.

T = randi(10,3,2,2);  % Data we can inspect without getting lost!
H = randi(10,3,5,4);  % Use these to develop a general approach.

% Method 1, the obvious approach.
C = zeros(size(H));
for ii = 1:size(H,3)
    for jj = 1:size(H,2)
        C(:,jj,ii) = T(:,ii).*H(:,jj,ii);
    end
end

% Method 2.  Vectorize inner FOR loop.  Faster.
C2 = zeros(size(H));
for ii = 1:size(H,3)
    C2(:,:,ii) = bsxfun(@times,H(:,:,ii),T(:,ii));
end

% Method 3, definitely more advanced (and faster still!)
[r,c] = size(T);
C3 = bsxfun(@times,H,reshape(T,r,1,c));

isequal(C,C2,C3)  % Yes...

Answer 2

[Azzi Abdelmalek]

 [n,m,l]=size(A);[n1,m1,l1]=size(B);v=[];
for i1=1:l
   for k=1:m
    v=[v  bsxfun(@times, A(:,i1,k),B)];
   end
end