Definition
Let $G=(V,E,A)$ be a strongly connected directed graph, where $V=\{1,2,...,n\}$ denotes the vertex set, $E$ is the edge set, and $A$ is the associated adjacent matrix with $0-1$ weighting, that is $a_{i,j}=1$ if $(j,i)\in E$, and $a_{i,j}=0$ otherwise.
$B$ and $D$ are two diagonal matrices, where $b_{ii}=\sum_{j=1}^na_{i,j}$ and $d_{ii}=\sum_{j=1}^na_{j,i}$. In other words, the diagonal entries of $B$ are the row sum of $A$, and the diagonal entries of $D$ are the column sum of $A$.
Problem
Now define a new matrix $M=\left[\begin{array}{c|c} B-A & -A \\ \hline A-B & D \end{array}\right]$. Since the column sum of $M$ are identical zeros, zero must be one of its eigenvalue. Can I claim that the rest eigenvalues all have positive real parts?
I tried many numerical examples, the rest eigenvalues have positive real parts. Anyone can help prove or disprove it? (Gershgorin Circle Theorem does not apply here because $M$ is not diagonally dominate)