"A matrix A is stable iff there is a positive definite matrix E such that EA is negative definite" Mas-Collel, Whinston and Green, in Microeconomic Theory (3rd Edition), define stable matrices as it follows:

Also, the following theorem is asserted in Theorem M.D.1: 

What would be its proof?
For reference, this is the definition of a negative definite matrix:

 A: Suppose that $A$ is diagonalisable over $\mathbb C$ first. Then there exists a real invertible matrix $P$ such that $D=PAP^{-1}$ is a block-diagonal matrix, where each of its diagonal sub-block $B$ is either a real diagonal matrix (corresponding to the real eigenvalues of $A$) or a scalar multiple of a $2\times2$ rotation matrix of the form $\pmatrix{a&-b\\ b&a}$ (corresponding to a non-real conjugate pair of eigenvalues $a\pm ib$). As $A$ is stable, $D$ is stable too. Therefore, in each diagonal sub-block $B$ of $D$, the diagonal of $B$ is negative. Thus $B+B^T$ and in turn $D+D^T$ are symmetric negative definite. Now pick $E=P^TP$. Then
$$
EA=(P^TP)(P^{-1}DP)=P^TDP.
$$
Thus $(EA)+(EA)^T=P^T(D+D^T)P$ is symmetric negative definite and $EA$ is negative definite.
The argument in the above can be modified to apply to the case where $A$ is not diagonalisable over $\mathbb C$. The basic idea is that we can always pick a real invertible matrix $S$ such that $J=SAS^{-1}$ is arbitrarily close to a block diagonal matrix $D$ of the aforementioned form, which is the block-diagonal part of the real Jordan form of $A$. This is done by suppressing the block super-diagonal part of the real Jordan form of $A$ via similarity transform by a diagonal matrix. The eigenvalues of $J+J^T$ (which are real because $J+J^T$ is real symmetric) therefore can be made arbitrarily close to the eigenvalues of $D+D^T$, and hence they can be made negative. It follows that $EA$ is negative definite if we put $E=S^TS$.
Alternatively, let $E=\int_0^{\infty} e^{tA^T}e^{tA}dt$. Then
$$
EA+(EA)^T
=\int_0^{\infty} (e^{tA^T}e^{tA}A + A^Te^{tA^T}e^{tA})dt
=\int_0^{\infty} \frac{d}{dt}(e^{tA^T}e^{tA})dt
=-I.
$$
But one needs to fill in the details to explain why $E$ exists and why it is symmetric positive definite (despite it is obviously symmetric positive semidefinite).
A: If $A$ is stable then $A$ and $-A^T$ have no common eigenvalues, therefore the Lyapunov equation $EA+A^TE=-Q$ has a solution for any $Q$ (see Sylvester equation). The solution can be written in the integral form as
$$
E=\int_0^\infty e^{At}Qe^{A^Tt}\,dt,
$$
which makes it obvious that $E$ is positive definite if $Q$ is.
Conversely, it is the second Lyapunov theorem on stability: take $V(x)=x^TEx$ as a Lyapunov function candidate, hence
$$
\dot V(x)=x^T(EA+A^TE)x=-x^TQx.
$$
