Eigenvalues of a special block matrix associated with strongly connected graph 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 adjacency 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 
$$M = 
\begin{bmatrix}
B-A, & -A \\\\
A-B, & D
\end{bmatrix}\in \mathbb{R}^{2n\times 2n}$$ 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 all have positive real parts. Anyone can help prove or disprove the above claim? (Gershgorin Circle Theorem does not apply here because $M$ is not diagonally dominate)
Some facts: Both $(B-A)$ and $(D-A)$ have exactly one zero eigenvalue and all the rest eigenvalues lie in the open right half complex plane because the directed graph is strongly connected. In particular, $(B-A)$ is called the Laplacian matrix of the graph. 
 A: I think your claim is true. Taking your matrix $M$ and multiplying from the left by $$H=\begin{bmatrix}
I, & 0 \\\
-I, & I
\end{bmatrix},$$ and from the right by its transpose, 
$$H^t=\begin{bmatrix}
I, & -I \\\
0, & I
\end{bmatrix},$$ the eigenvalues don't change, yet the matrix you get is
$$HMH^t=\begin{bmatrix}
B-A, &B \\\
0, & D-A
\end{bmatrix}.$$
The eigen values of the above matrix are those of $B-A$ and those of $D-A$ and are all non-negative as you already explained. 
A: the answer was essentially correct but i believe H should be defined as H11=H21=H22=1 and H12=0.
the resulting matrix HMH^T is block upper triangular with B-A and D-A on the diagonal, hence
is hurwitz stable.
although the eigenvalues change, the inertia theorem guarantees that their stability type
cannot change, i.e. they cannot cross the imaginary axis.
A: I think I have a counterexample. Try this:
$A=\begin{bmatrix}1 & 1 & 0 & 1  \\\\ 0 & 1 & 1 & 0 \\\\ 1 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 1 \end{bmatrix}$.
When I construct $M$ I get a pair of conjugate complex eigenvalues in its spectrum.
