Lyapunov's criterion for Hurwitz matrices

($\Leftarrow$). Let $Av=\lambda v$, $v\in\mathbb{C}^n\setminus\{0\}$. Since $A$ is real, $v^*A^T=(Av)^*=\bar\lambda v^*$. Therefore
$$v^*(A^TP+PA)v=\bar\lambda\,v^*Pv+\lambda\,v^*Pv=2\operatorname{Re}\lambda\ v^*Pv.$$
The left side is $<0$ (a negative definite real symmetric matrix is negative definite on $\mathbb{C}^n$ too), and $v^*Pv>0$. Hence $\operatorname{Re}\lambda<0$.

($\Rightarrow$). Suppose $A$ is Hurwitz. Then there are $C,c>0$ with $\|e^{At}\|\le Ce^{-ct}$ for $t\ge0$ (Jordan form: entries of $e^{At}$ are $t^je^{\lambda t}$ with $\operatorname{Re}\lambda<0$). So the integral
$$P=\int_0^\infty e^{A^Tt}e^{At}\,dt$$
converges absolutely (integrand $\le C^2e^{-2ct}$). It is symmetric, and for $x\ne0$,
$$x^TPx=\int_0^\infty|e^{At}x|^2dt>0,$$
because $e^{At}$ is invertible so $e^{At}x\ne0$ for all $t$ and the integrand is continuous and positive. So $P>0$. Moreover
$$A^TP+PA=\int_0^\infty\frac{d}{dt}\Big(e^{A^Tt}e^{At}\Big)dt=\lim_{T\to\infty}e^{A^TT}e^{AT}-I=-I,$$
which is negative definite.

Example. $A=\begin{pmatrix}-1&10\\0&-1\end{pmatrix}$ is Hurwitz but $A^T+A$ is indefinite: $P=I$ does not work, yet a suitable $P=\begin{pmatrix}1&5\\5&51\end{pmatrix}$-like matrix does. The Lyapunov function $V(x)=x^TPx$ decreases along the flow $\dot x=Ax$.

Remark. In topological terms, the set of Hurwitz matrices is open (eigenvalues are continuous), and it is the union over $P>0$ of the open convex cones $\{A:A^TP+PA<0\}$; each cone is convex but the union is not (the set of Hurwitz matrices is *not* convex).