Sparse matrix from the columns of an initial square matrix

Hello everyone, given a matrix A of size (n,n),

I would like to construct a matrix B of size (n,m*n) the following way:

B = zeros(n,m*n);
for j = 1:m
    col_start = (j-1)*n+1;
    col_end = j*n;
    B(:,col_start:col_end) = diag(A(:,j));
end

This version uses a for loop, is there any faster way of constructing B?

The difficulty for me is to achieve the same result without a for loop. I precised "sparse" in the summary, but I do not necessarily refer to the sparse matrix data type in Matlab; B is sparse in a mathematical sense.

Thank you in advance.

PS: I asked a similar question last year, but in this question, A is assumed to be a square matrix.

[KSSV]

Did you try to initialize B? Why do you think the present loop is slow?

[Michel Pohl]

I have just edited the question to include the initialization of B in the code snipped. I think that there might be a faster way to accomplish that. For instance, if all columns of A were identical, I would use the repmat function, which is faster than a brutal implementation with a for loop.

[Matt J]

You haven't shown us what the output should be. Here is what I get when I run your code with some input data. Is this correct?

n=3;m=2;
A=rand(n);
B = zeros(n,m*n);
for j = 1:m
    col_start = (j-1)*n+1;
    col_end = j*n;
    B(:,col_start:col_end) = diag(A(:,j));
end
A,B

A = 3×3

0.7619 0.2149 0.5172 0.2228 0.1111 0.0854 0.6459 0.4688 0.1326

B = 3×6

0.7619 0 0 0.2149 0 0 0 0.2228 0 0 0.1111 0 0 0 0.6459 0 0 0.4688

[Michel Pohl]

Ah yes, you understood my question well, thank you for your contribution!

Answer 1

[Bruno Luong]

n=3;m=2;
A=randi(10,n,m) %  m columns are used; not n

A = 3×2

7 10 2 2 2 9

J=(1:m*n);
I=mod(J-1,n)+1;
B = sparse(I, J, A)

B = 
   (1,1)        7
   (2,2)        2
   (3,3)        2
   (1,4)       10
   (2,5)        2
   (3,6)        9

C = full(B) % if full is prefered

C = 3×6

7 0 0 10 0 0 0 2 0 0 2 0 0 0 2 0 0 9