Is it true that $A^{T}P+PA>0$ for an unstable matrix $A$ and a positive definite $P$?

Question

For a positive definite matrix $P$ and a matrix $A$ with all positive eigenvalues, how to guarantee that the matrix $Q=A^{T}P+PA$ is positive definite?

I know if $A$ is a stable matrix (i.e. all the eigenvalues of $A$ are in open left half plane) then $Q_{1}=A^{T}P+PA$ is negative definite but I was wondering if the analogous conclusion can be made about the definiteness of $Q$ when $A$ is an unstable matrix.

Answer 1

No, this is not true. Consider the counterexample $$ A=\left(\begin{array}{rr} 2 & 2\\ 1 & 2\\ \end{array}\right),\quad P=\left(\begin{array}{rr} 2 & 1\\ 1 & 1\\ \end{array}\right). $$ The matrix $$ Q=\left(\begin{array}{rr} 10 & 9\\ 9 & 8\\ \end{array}\right) $$ is not positive definite.

Just for clarification: if $A$ is a stable matrix, then for any negative definite $Q$ there exists such positive definite $P$ that $A^TP+PA=Q$.