Rank of a sum of matrices

[Pere Callahan]

HI,

I came across the following question, which I could only solve for one trivial special case. I'm hoping for help from your side on how to deal with the general case.

Assume we are in the situation that we have a decomposition of a full-rank d x d matrix, M, into a sum of N rank-1 matrices, N>d, in formulas,

M = m_1+\ldots+m_N.


I'm interested in whether or not one can in general conclude that there exists a subset \{m_{k_1},\ldots,m_{k_d}\} whose sum is a full-rank (that is rank d) matrix.

The special case I mentioned is the case d=1, in which case there is nothing to prove:smile:
What I tried is writing the rank 1 matrices m_n as an outer product of vectors, that is m_n=b_n\otimes a_n. Then the assumption that the sum of the m_n have full rank certainly implies that the b_n span all of \mathbb{R}^d, so in looking for a subset whose sum is rank d I started with choosing a basis from among the b_n; i did not succeed, however, in showing that the sum of the corresponding m_n is a full rank matrix.


I would appreciate any tips from you,

Thanks,
Pere

[fantispug]

Yes, it is true. I'll outline the proof, and let you fill in the details.

For each column there exists a matrix in the decomposition with non-trivial elements in that column.

So if we choose the kth element of our subset as the one with non-zero entries in the kth column the sum of this subset must be full rank.