If you use the normal * operator, Matlab assumes you want to do a traditional matrix multiplication, where you take the inner products of the rows of one matrix against the columns of another. Therefore, if you multiply two vectors together using *, Matlab will report an error. For division (/), there is no "matrix division" that is defined, so any time you divide two vectors or matrices, they must be the same size, and you must use ./ . In your problem, the operations in the denominator result in a vector of results. There is no way to divide over a whole vector, but what you want is to divide over each element of the vector. None of this matters for scalars though. In your problem, mu is a scalar, so you can multiply mu by a vector or matrix using just * with no problems. You could write .* in your problem, and still get the same answer with no errors, but it would not be the correct code.

W = bsxfun(@times,V,M) If V is a row vector you will get V element-wise times each row of M. Similarly if V is a column vector.

Use the matrix multiplication operator:

sum_temp = A.' * B;

You could do something like this, Which seems to be what you want and provides the same results as what you post.

A = [2 3;
     4 5];
B = [9 8 7 6;
     8 7 6 5];
A = imresize(A,size(B),'nearest');
C = A.*B

This will only really work in the particular case you are presenting though. It also assumes you have the image processing toolbox

There are several ways to multiply each column of a matrix by the corresponding element of the vector. The first is to use the REPMAT function to expand the vector to the same size as the matrix and them perform elementwise multiplication using .* -- however, this will require a large amount of memory. For example: To multiply each column of vector b with the row a: a = [1 2 3]'; b = [1 2; 3 4; 5 6]; x = repmat(a,1,size(b,2)); x.*b % or dot(x,b) If the matrix involved is a square matrix then the vector can be converted to a diagonal matrix and then multiplied as follows: Amat = rand(3,3); b = rand(1,3); c = Amat*diag(b); %or %c = Amat*diag(1./Bvec); The second method is to use a FOR loop. For long vectors prior to MATLAB 6.5 (R13), this may take a significant amount of time; however, making use of the acceleration technology introduced in MATLAB 6.5 (R13), this time is drastically reduced. If you're using MATLAB 6.5 (R13), copy the following code into a function file for an example that does this: function B=DivideColumns(N) % Avoiding creating too large a matrix if nargin<1 || N >2500 N = 500; end tic A = 1./hilb(N); c = 1:N; B = A; % So that we don't overwrite A T1 = toc; tic for i=1:N B(:,i) = B(:,i)/c(i); end T2 = toc; S1 = sprintf('Constructing the %d by %d ... A matrix took %f seconds.', N, N, T1); S2 = sprintf('\nDividing the columns of A ... by elements of c took %f seconds', T2); disp([S1 S2]) For large values of N, you can see that generating A will take much more time than performing the division. Another option is to use a diagonal or sparse diagonal matrix form of your vector. For example, the following code will produce a B that agrees with the B returned by the code above: function B = TimesDiag(N) % Avoiding creating too large a matrix if nargin<1 || N >2500 N = 500; end tic A = 1./hilb(N); c = 1:N; T1 = toc; tic % Replace this with "B = A*spdiag(c(:),0,N,N);" % if N is very large: B = A*diag(c); T2 = toc; S1 = sprintf('Constructing the %d by %d ... A matrix took %f seconds.', N, N, T1); S2 = sprintf('\nDividing the columns of ... A by elements of c took %f seconds', T2); disp([S1 S2])

Use cellfun: A = magic(3) C = {rand(3,4); rand(3,4); rand(3,4); rand(3,4); rand(3,4)} result = cellfun(@(x)mtimes(A,x),C,'UniformOutput',false) C{1} result{1}