TITLE: Invertibility of a square matrix with zero diagonal elements and positive non-diagonal elements
QUESTION [5 upvotes]: $M$ 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:

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

REPLY [7 votes]: 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.