Is this matrix of full rank ? Matrix given by a directed graph I constructed this matrix $M$ given a directed graph $G = (E,V)$ :
$$M \in \mathcal M^{|E| \times |V|}$$
with this definition.
On every line it corresponds to an edge. If the arc is going from a vertex $v$, then in the corresponding column $v$ there is a $1$. If it goes to $v$, there is a $-1$.
I think this matrix is of full rank. Do you have any counterexample ?
Maybe if there is a cycle?
 A: The matrix will always have rank less than $|V|$, because the sum of all the columns (that is, over all vertices) gives the $0$ vector. So if $|V|\le|E|$, $M$ cannot possibly have full rank (which is full column rank in this case). 
More generally, adding up all the columns in a connected component (connected ignoring the edge orientation) gives the $0$ vector. So a graph with $k$ connected components gives a matrix $M$ with column rank at most $|V|-k$.
In fact, the rank of $M$ is exactly $|V|-k$, because $M^{\mathsf T}M$ is the Laplacian matrix of the underlying undirected graph. The rank of $M^{\mathsf T}M$ is $|V|-k$, because the multiplicity of the $0$ eigenvalue is always the number of connected components. Therefore the rank of $M$ is $|V|-k$.
$M$ can still have full row rank if $|E|=|V|-k$. There is only one way for this to happen. If $V_1, V_2, \dots, V_k$ are the connected components, then the graph must have at least $|V_i|-1$ edges spanning $V_i$, for at least $|V|-k$ edges total; equality holds just if each connected component is a tree (ignoring edge orientation).
So $M$ has full row rank exactly when the underlying undirected graph is a forest.
