If $A+A^T$ is negative definite, then the eigenvalues of $A$ have negative real parts?

I think the following simple argument works, although I'm surprised not to have heard of it before:

If $\lambda \in \mathbb C$ is an eigenvalue of $A$ with eigenvector $x$, then $\lambda |x|^2 = \langle Ax,x \rangle = \langle x,A^T x \rangle = \overline{\langle A^T x, x\rangle}$, so $\langle A^T x, x\rangle = \bar{\lambda} |x|^2.$

Therefore $2\Re{\lambda} |x|^2 = (\lambda + \bar\lambda)|x|^2 = \langle Ax,x\rangle + \langle A^T x,x \rangle = \langle (A+A^T)x,x \rangle > 0.$

The other direction is false even for diagonalizable matrices: $\left(\begin{array}{cc} 1&100\\0&2\end{array}\right)$ has positive eigenvalues, but not when added to its transpose.

If we make the very strong assumption that $A$ is normal, then the converse holds by looking at $\exp(-At) \exp(-A^T t) = \exp(-(A+A^T)t)$.

consider the following dynamical system: $$\dot{x}=Ax$$ Lets study the stability of this system knowing that $A+A^T<0$.

Consider the following Lyapunov function $$V=x^Tx$$ Taking the derivative along the system dynamics we have $$\dot{V}=\dot{x}^Tx+x^T\dot{x}=x^T(A^T+A)x<0$$ Therefore, the system is globally (the Lyapunov function is radially unbounded) asymptotically stable (here actually it is globally exponentially stable), meaning that $A$ is Hurwitz (the real part of all eigenvalues is negative).