Consider a binary square matrix ${\bf B} \in \{0,1\}^{N \times N}$, with $b_{i,i} = 0$ (this can be, for example, the adjacency matrix of a directed graph).
Let's define $r_i = \sum_{j=1}^N b_{i,j}$ and $c_i = \sum_{j=1}^N b_{j,i}$.
Consider a vector ${\bf v} \in \mathbb{R}^{N}$, and let ${\bf D}({\bf v})$ be a diagonal matrix which diagonal correspond to ${\bf v}$.
Finally, consider the matrix ${\bf M} = {\bf B} - {\bf D}({\bf v}).$
We know that:
- The matrix ${\bf M}$ is strictly diagonally dominant by rows if $v_i > r_i, \forall i \in \{1, \ldots, N\}.$
- The matrix ${\bf M}$ is strictly diagonally dominant by columns if $v_i > c_i, \forall i \in \{1, \ldots, N\}.$
In both cases, all the eigenvalues of ${\bf M}$ have strictly negative real part.
I was wondering if:
- The eigenvalues of matrix ${\bf M}$ have strictly negative real part if $$v_i > \max\{r_i, c_i\}, \forall i \in \{1, \ldots, N\}.$$
- The eigenvalues of matrix ${\bf M}$ have strictly negative real part if $$v_i > \min\{r_i, c_i\}, \forall i \in \{1, \ldots, N\}.$$
For the case (3), it should work, since we can say that the matrix ${\bf M}$ is strictly diagonally dominant by rows and by columns at the same time.
Anyway, what can we say about (4)? Is it also true?
Motivation
If (4) is true, then I can "smartly" add negative terms to the diagonal in order to get a stable spectrum (contained in the left complex plane) with the "minimum effort".
I took a matrix ${\bf B}= \begin{pmatrix}0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}$. Then $\min\{r_i, c_i\}=1, \forall i \in \{1, \ldots, N\}$. So I picked $v_i$ not very bigger than $1$, namely, $v_i=\frac{11}{10}$ for each $i$.
Then $$\det ({\bf M}-\lambda{\bf I})=\det ({\bf B-D(v)-\lambda{\bf I}})=-\lambda^3-\frac {33}{10}\lambda^2- \frac {263}{100}\lambda+\frac {769}{1000}.$$
Mathcad calculated the roots of this equation and one of them is approximately $0.225>0$.