I have a rectangular matrix A(m x n, m <= n) and A is full rank. All elements of A are either 1 or 0. If any two rows have a 1 in same column, then they need to be put in same row-group. As a minimal example;
A = [0 0 1 1 0 0
1 1 0 0 0 0
0 0 0 1 1 1]
Then A can be decomposed into two row groups A_1 and A_2 such that
A = [A_1
A_2]
where
A_1 = [1 1 0 0 0 0]
A_2 = [0 0 1 1 0 0
0 0 0 1 1 1]
This is because both row vectors of A_2 are orthogonal to the (only) row vector in A_1.
EDIT 1 :
I am adding More Context to hopefully making the problem complete and logical. If it isn't then please let me know.
1) No Solution is not possible.
2) If no such row groups exist, then all the row vectors form a single group.
3) A bigger example to hopefully make things clearer.
A = [1 1 0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0
0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 1 1 1 0
0 0 0 0 0 0 0 0 1 1]
then there are three row groups containing rows that are pairwise orthogonal to rows from other groups.
A_1 = [1 1 0 0 0 0 0 0 0 0]
A_2 = [0 0 1 1 0 0 0 0 0 0
0 0 0 1 1 1 0 0 0 0]
A_3 = [0 0 0 0 0 0 1 1 1 0
0 0 0 0 0 0 0 0 1 1]
Please give a link if this is a known concept from Linear Algebra.