if $A+A^T$ is stable, then is $A$ also stable?

Question

Let $A \in \mathbb{R}^{n\times n}$ such that $A+A^T$ has stable eigenvalues (their real part is strictly negative). I remember seeing somewhere that this involves that $A$ is also stable but I do not know exactly where and what ... Can someone help me?

It was an inequality, I recall, saying that the greatest eigenvalue of $A+A^T$ is greater than the greatest eigenvalue of $A$ or something like that ...

Edit

use Lyapunov with $P= I$

$A+A^T$ is a symmetric real matrix and so its eigenvalues are real. — user376343, Sep 21 '18 at 16:20
At least for a $2$-by-$2$ real matrix $A$, the stability of $A+A^\top$ implies that of $A$. — Batominovski, Sep 21 '18 at 18:07

score 4 · Accepted Answer · answered Sep 21 '18 at 18:31

Proposition. For an $n$-by-$n$ real matrix $A$, if the matrix $A+A^\top$ is stable, then $A$ is stable.

Let $B:=-A$. Write $S:=B+B^\top=-(A+A^\top)$. Then, $A+A^\top$ is stable if and only if $S$ is positive-definite. We suppose that $A+A^\top$ is stable.

For a nonzero $v\in\mathbb{R}^n$, we have $$2\,\left(v^\top\,B\,v\right)=v^\top\,B\,v+v^\top\,B\,v=v^\top \,B^\top \,v+v^\top\,B\,v=v^\top\,S\,v>0\,.$$ That is, $v^\top\,B\,v>0$ for every nonzero $v\in\mathbb{R}^n$. Consequently, $B$ is a real positive-definite matrix (in a broader sense). From this link, the real part of every eigenvalue of $B$ is positive. Therefore, the real part of every eigenvalue of $A=-B$ is negative. This shows that $A$ is stable.

The converse does not hold. As in achille hui's example (see the link), the eigenvalues of $$A:=-\begin{bmatrix}3&7\\1&3\end{bmatrix}$$ are $-3+\sqrt{7}<0$ and $-3-\sqrt{7}<0$. Thus, $A$ is stable. However, the eigenvalues of $$A+A^\top=-\begin{bmatrix}6&8\\8&6\end{bmatrix}$$ are $2>0$ and $-14<0$, whence $A+A^\top$ is not stable.

if $A+A^T$ is stable, then is $A$ also stable?

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