$M$ is row strict diagonally dominant matrix and Hurwitz. What about $M+M^T$?

Question

Consider a row strict diagonally dominant matrix $M$ such that:

$$\begin{cases} m_{i,i} < 0 ~\forall i\\ m_{i,j} \geq 0 ~\forall j \neq i\\ \displaystyle \sum_{j=1, j \neq i}^N m_{i,j} < -m_{i,i} ~\forall i. \end{cases}.$$

Of course, all eigenvalues of $M$ have negative real part.

Now consider the matrix $P = M + M^T.$

Can I conclude that also all the eigenvalues of $P$ are negative?

Intuition (and optimism) says me "yes". But I need a result that assures that the claim is right.

I found very few stuff googling around about the spectrum of the sum of matrix with its transpose. I only found this on MathSE. But it refer to a generic matrix $M$.

Answer 1

This is NOT true. I will give you a counterexample: $M=\begin{bmatrix}-1.01 & 1 \\2& -2.01 \end{bmatrix}, P=M+M^T=\begin{bmatrix}-2.02 & 3 \\3 & -4.02 \end{bmatrix}$.

Using matlab command eig I found out that the eigenvalues of $M$ are $-0.01$ and $-3.01$ while those of $P$ are $-6.1823$ and $0.1423$. It is noted that $M$ is strictly diagonally dominant and Hurwitz as requried in the question. However, the second eigenvalue of $P$ is positive and therefore proves that your proposition is false. Actually, there are many counterexamples to this questions.

BTW: The definitions of diagonally dominant and strictly diagonally dominant are as follows [*]:

A matrix $A=[a_{ij}]\in M_n$ is diagonally dominant if \begin{equation} |a_{ij}| \ge \sum_{j \ne i}|a_{ij}|=R_i^{'} ~~for~all > ~i=1,..,n \end{equation} It is strictly diagonally dominant if \begin{equation} |a_{ij}| > \sum_{j \ne i}|a_{ij}|=R_i^{'} ~~for~all > ~i=1,..,n \end{equation}

Base on this, your question is not accurate enough. You should change the condition to strictly diagonally dominant instead.

[*]: Horn, Roger A., and Charles R. Johnson. Matrix analysis. Cambridge university press, 2012.