Consider a real matrix $A = [a_{ij}]_{1 \leq i,j \leq n} \in \mathbb{R}^{n \times n}$. If
- for any $i \neq j$, $a_{ij} \leq 0$, and
- for any $1 \leq i \leq n$, $a_{ii} = -\sum_{j\neq i} a_{ij} \geq 0$,
is it true that $A$ can be diagonalized and has $n$ real eigenvalues $0 = \lambda_1 \leq \cdots \leq \lambda_n$?
If matrix $A$ is symmetric, I know $A$ is the Laplacian matrix of a weighted undirected graph, and the above holds. I wonder if it is still true in general? Or can we say the real parts of the eigenvalues are all non-negative? I just feel this should be a well-studied problem...
(feel free to edit the title and the body of the question)
As @AsAnExerciseProve has pointed, that all eigenvalues have non-negative real parts follows directly from the Gershgorin circle theorem.