# [ Determinant of Adjacency Matrix is null ]

Let be $D=(V,E)$ . Prove then the determinant of its adjacency matrix , $det(A) = 0$ $\iff$ $\exists S \subseteq V$ such that $|v_{ext} \cap S|$ is an even number , $\forall$ v $\in V$ . $v_{ext}$ denotes the nodes $u$ such that $vu \in E$. Also , D may have loops (i.e $\exists$ $A_{uu} =1$).

I have tried the direct implication in this way : if $det(A) = 0$ , it means that the vectors of $A$ are not linearly independents. But what`s the next step ?

Thanks!