Brute force method:
D3 = length(C(1,1,:));
D2 = length(C(1,:,1));
D1 = length(C(:,1,1));
for i = 2:D1
for j = 1:i-1
for k = 1:i-1
C(i,j,k) = 0;
end
end
end
Above provides [1 2 3; 0 4 6; 0 0 9] ; [2 4 6; 4 8 12; 0 0 18]; [3 6 9; 6 12 18; 9 18 27], which I believe would be the 'true' lower triangle. Don't quote me on that. To get what you requested, via 'brute force':
b = zeros([1 length(C(1,:,1))]);
for k = 2:D3
for i = 1:k-1
C(i,:,k) = b;
end
end
