# Invertibility of a square matrix with zero diagonal elements and positive non-diagonal elements

$M$ is square and $$M(i,j)=0, i=j$$ $$M(i,j)>0, i\ne j$$ Is $M$ full-rank or invertible?

Actually the $M$ I am studying has much stronger properties but I guess the simple conditions above might be enough to make $M$ non-singular. The stronger properties of $M$ are:

1. All elements in $M$ are non-negative integers between $0$ and $N$;

2. The sum of each row is equal to $N$.

### Edit

$N$ is not the dimension of $M$. It's just a constant positive integer.

I did search for this problem, but it seemed there was no much work on such matrices. It looks easy, but I don't know how to prove it and I couldn't find a counter-example either.

The matrix $M$ doesn't have to invertible. I give a simple counterexample here. $$M=\begin{pmatrix} 0&2&1&1\\ 2&0&1&1\\ 1&1&0&2\\ 1&1&2&0 \end{pmatrix}$$ You will see $r_1+r_2-r_3-r_4=0$ which indicates they are linear dependent.
• Thanks a lot Shuchang. What if I have even stronger condition on $M$ -- the subdiagonal entries, which are directly below and to the left of the main diagonal, are also all zeros. Other properties of $M$ remain the same. Must $M$ be nonsingular now? ($Dim(M) \ge 3$) – D. Chen Oct 6 '13 at 2:44
• Thanks. I meant only the entries on the diagnoal and subdiagonal of $M$ were zeros so it was not a triangular matrix. An example could be changing your matrix above to: $$M= \begin{pmatrix} 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 1 \\ 1 & 0 & 0 & 2 \\ 1 & 2 & 0 & 0 \\ \end{pmatrix}$$ Do you think $M$ with such constraints must be nonsingular? – D. Chen Oct 6 '13 at 2:55