Rectangular matrix manipulation with triu and tril

Hey guys,

I am currently working with matrices that have several different sizes such as 256x153, 303x106, 127x89, and many more. Whenever I use a perfect square matrix, the cut made by the triu and tril commands are from the top left corner of the matrix to the bottom right corner; however, when I use a rectangular matrix this is not the case. Is there a line of code that I can implement so that the cut made by tril and triu ALWAYS cuts a matrix of ANY SIZE through the bottom right corner?

Thanks,

Pranav

Answer 1

[John D'Errico]

tril and triu are set so they work from the top left corner. You don't want them to work that way. So you could write your own version of the codes, to just then use tril and triu with a tweak to the diagonal chosen.

The fix is pretty simple.

function Ahat = trilb(A,K)
  % like tril, but it acts relative to the lower right corner of the matrix
  if (nargin <2) || isempty(K)
      K = 0;
  end
  da = diff(size(A));
  Ahat = tril(A,K + da);
end

function Ahat = triub(A,K)
  % like triu, but it acts relative to the lower right corner of the matrix
  if (nargin <2) || isempty(K)
      K = 0;
  end
  da = diff(size(A));
  Ahat = triu(A,K + da);
end

Do they work? So it would seem.

triu(ones(3,5))
ans =
     1     1     1     1     1
     0     1     1     1     1
     0     0     1     1     1
>> triub(ones(3,5))
ans =
     0     0     1     1     1
     0     0     0     1     1
     0     0     0     0     1
>> tril(ones(3,5))
ans =
     1     0     0     0     0
     1     1     0     0     0
     1     1     1     0     0
>> trilb(ones(3,5))
ans =
     1     1     1     0     0
     1     1     1     1     0
     1     1     1     1     1

Effectively, it could be thought of as a one line fix, but easier if you just write the functions to do it for you. Otherwise, you would need to remember whether to add or subtract the difference in dimension from K. And clearly, the result is no different from tril and triu when the matrix is square.

Answer 2

[the cyclist]

This will use the largest square matrix from the bottom right corner, as input to triu:

d = min(size(A))-1;
triu(A(end-d:end,end-d:end))