Sum each element in a matrix with the previous elements on the same diagonal

I'm looking for an "efficient way", that will work on larger matrices like 2000 by 2000 in reasonable time, to do this operation on a given square symmetric and anti-symmetric matrix (Hankel matrix)

    A =[1 2 3;
        2 3 4; 
        3 4 5;];

Then the output should be

B =   [1  2  3;
       2  4  6;
       3 6  9;]

Answer 1

[Andrei Bobrov]

B = fliplr(spdiags(cumsum(spdiags(A))));
B = B(:,all(B));

Answer 2

[Walter Roberson]

No. Your B(2,1) is 4 even then there is nothing proceeding it on the same diagonal from upper left to lower right. But 4 is the cumulative sum from the diagonals from the upper right to the lower left, the anti-diagonals. This establishes that you want both diagonal and anti-diagonal cumulative sums to be taken. But your B(3,1) is 4, which is not the cumulative sum along the anti-diagonal

    x x 3
    x 3 x
    4 x x

Therefore there is no rule for what values should appear along the left column of B, and therefore the result cannot be computed.

[Mohamed Abdalmoaty]

Hi, Thanks for your answer. I had a typo in my question. B should be correct now.

Answer 3

[Ajay Goyal]

Simple Brother, Use a(i,j)=a(i,j)+a(i-1,j-1) in your double loop (for i=1:2000(for rows); for j=1:2000(for columns)) Thanks for accepting the answer formally

[Mohamed Abdalmoaty]

Thanks for your answer, the suggestion is not helpful when you have large matrices. Thanks