how to create a 3D diagonal matrix from a 2D matrix ?

Hi, I have a 2D matrix A=rand(3,5) and i want to create a 3D matrix B, where

B(1,:,:) = diag(A(1,:)) ; B(2,:,:) = diag(A(2,:)) ; B(3,:,:) = diag(A(3,:))

so the final size of B is expected to be 3x5x5

I am looking for a command to calculate B without passing through loops or for .

thanks

[Matt Fig]

I am confused about the point of this....

B = rand(3,5,5);
C = rand(3,5,5);
D = rand(3,5);
try
x1 = (B-C).\D  % Error
catch
disp('first one made an error!')
x2 = (B-C)\D  % Error
end

What are you trying to do again??

[yasser]

My previous problem was the following:

the dimensions of B and C are 380x380 and the dimensions of D = 380x1

A = rand(380,1); B = diag(A); C = rand(380,380); D = rand(380,1); X = inv(B-C)*D; cost = sum(X);

the following step is to vectorize my input variable A from 1x380 to 100x380, this way i compute an array of cost functions X.

any help is appreciated

[Matt Fig]

But that is not what you wrote above! You clearly have B and C as 3D and the equation you show produces an error. So which is it? 3D or 2D? How are we to help if you say 3D one time and 2D the next??

[yasser]

your comment is correct, but that's another issue to solve. I will open another thread to explain in details my problem.

Answer 1

[the cyclist]

This is a little kludgy, but it works. I deliberately did not simplify some of the operations, to keep it clearer where the 5's and 3's come in.

A = rand(3,5);
linearIndex = logical([ones(3*1,1); repmat([zeros(3*5,1); ones(3*1,1)],[5-1 1])]);
B = zeros(3,5,5);
B(linearIndex) = A(:);

[yasser]

This command is working perfectly. However is it time-cost efficient for a global optimization tool where the dimensions of B are 100x380x380. It looks a bit costly ?

Answer 2

[Sean de Wolski]

So each slice of B should contain a column vector (like you have above)? Why not just use a FOR-loop? or bsxfun? For example:

x = bsxfun(@minus,C,A).\D

Actually, bsxfun won't fill in zeros here so a for-loop is your best bet.

[yasser]

loop for wont help much, because i have simplified my problem to a 3x5x5, my real problem is much bigger 100x380x380.

[Sean de Wolski]

And how is a for-loop not perfectly applicable to that?