Sum each element in a matrix with the previous elements on the same diagonal
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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;]
回答（3 个）
Andrei Bobrov
2016-2-3
B = fliplr(spdiags(cumsum(spdiags(A))));
B = B(:,all(B));
Walter Roberson
2016-2-3
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.
Ajay Goyal
2016-2-3
编辑：Ajay Goyal
2016-2-3
0 个投票
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
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