What can I say about the eigenvalues of $A+A^T$?

Question

So I have as hypothesis that $A\in M_{d\times d}(\mathbb{R})$ is exponentially stable. This means, usually, that the the real part of its eigenvalues is always negative. Now, I was hopping that if $A$ is also normal, then I could say something about this characteristic. Ideally, show that $B=A+A^T$ is also exponentially stable.

So, I am reaching to the community in case anyone knows what can I say about $B$, besides that it is symmetric, and specially related to the positivity/negativity of its eigenvalues. Also, any sources where I can look into are appreciated. Maybe start even by assuming that the eigenvalues are real or something like that.

Answer 1

If $A$ is normal (i.e. $AA^T = A^TA$) and the real part of the eigenvalues of $A$ are positive, then it follows that the eigenvalues of $B = A + A^T$ are positive and real. In particular, the eigenvalues of $B$ are necessarily equal to $2 \operatorname{Re}[\lambda]$ for all eigenvalues $\lambda$ of $A$.

We can see this as a consequence of the spectral theorem. There exists a unitary matrix $U$ and a complex diagonal matrix $D$ such that $A = UDU^*$, where $U^*$ denotes the conjugate-transpose of $U$. From there, we have $$ A + A^T = A + A^* = [UDU^*] + [UDU^*]^* = UDU^* + UD^*U^* \\= U[D + D^*]U^* = 2U \operatorname{Re}[D]U^*. $$

Answer 2

$B$ is symmetric, and so all of its eigenvalues are real.

Answer 3

Here is another less satisfying take:

$A$ is exponentially stable iff $\lim_{t \to \infty} e^{At} = 0$.

Since $e^{A^*t} = (e^{At})^*$, we see that $A^*$ is also exponentially stable.

If $A,A^*$ commute (that is, $A$ is normal) we see that $e^{(A+A^*)t} = e^{At} (e^{At})^*$ from which it follows that $A+A^*$ is exponentially stable.