Eigenvalues of $A+A^*$

Question

Suppose that $A$ is a real square matrix with eigenvalues that have negative real part. Is it necessarily true that $A+A^*$ has negative eigenvalues?

I am not entirely sure how to go about doing this as this is the sum of two matrices but it seems like it might be true.

Answer 1

The answer is NO. Consider $$ A := \begin{pmatrix} -1 & x \\ 0 &-1\end{pmatrix}.$$ Then the eigenvalues of $A$ are $-1,-1$ no matter what $x\in\mathbb{R}$ is.

On the other hand, $$ A + A^* = \begin{pmatrix} -2 & x \\ x &-2\end{pmatrix}$$ has eigenvalues $-2-x$ and $-2+x$ the latter of which is positive for $x>2$.

Answer 2

$\newcommand{tr}{\operatorname{tr}}$The complex eigenvalues of a real $2\times 2$ matrix are $$\lambda_j=\frac{\tr A+(-1)^j\sqrt{(\tr A)^2-4\det A}}2$$So:

• if the eigenvalues are not real, then $\Re\lambda_j=\frac12\tr A$.

• if $\det A< 0$, then the eigenvalues are real with opposite signs.

Therefore, any $2\times 2$ real matrix $A$ such that $$\begin{cases}\tr A<0\\ (\tr A)^2<4\det A\\ \det(A+A^t)<0\end{cases}$$ is a counterexample.

This means a solution for the system $$\begin{cases}a_{11}+a_{22}<0\\ (a_{11}-a_{22})^2+4a_{21}a_{12}<0\\ 4a_{11}a_{22}-(a_{12}+a_{21})^2<0\end{cases}$$

So, here's what you need: fix $a_{11}>0$, pick $a_{22}<-a_{11}$ and then choose $a_{12}$ and $a_{21}$ very large in moduli with opposite signs. For instance, $a_{11}=1$, $a_{22}=-3$, $a_{21}=-100$, $a_{12}=100$ should do. Or even $a_{11}=-1$, $a_{22}=-3$, $a_{21}=-100$, $a_{12}=200$.