How to prove that the Laplacian for a directed graph has an eigenvalue at 0?

Question

If Laplacian L = D-A where D is the degree matrix and A is an 0-1 adjacency matrix for a directed graph. How can we prove that this Laplacian has an eigenvalue at 0?

Answer 1

Let $\textbf{1}$ be the function taking the value $1$ at each vertex. Then by explicit calculation, we see that both $D\textbf{1}$ and $A\textbf{1}$ are equal to the function taking the value $\deg(v)$ at each vertex $v$. Consequently, $$ (D-A)\textbf{1}=0, $$ which means that $\textbf{1}$ is an eigenvector of the Laplacian with eigenvalue $0$.