equivalent statement for positive definitiveness

Question

In this lecture (https://engineering.purdue.edu/~ragu/confpapers/Bal302-talk.pdf) page $14$, it is said that for a real matrix $A$ and a symmetric positive definite matrix $P$ the statement $$ A^T P + P A < 0 $$ is equivalent with $$ Q A^T + A Q < 0 $$ with $Q = P^{-1}$. I am unable to prove this. Can someone give me a hint?

Answer 1

You are given that $$A^T P + P A < 0$$ which means that for any $z \in \mathbb{R}^n$, you have $$z^T( A^T P + P A) z < 0$$ Since $Q$ is full rank, this means that if $y$ spans $\mathbb{R}^n$, then so does $z = Qy$. So, for any $y$ the following is true $$y^TQ^T( A^T P + P A) Qy < 0$$ Since $Q$ is symmetric (because its inverse $P$ is symmetric) $$y^TQ( A^T P + P A) Qy < 0$$ or $$y^T( QA^T PQ + QP AQ) y < 0$$ But $PQ = QP = I$ so $$y^T( QA^T + AQ) y < 0$$ which is true for all $y$, hence $$ QA^T + AQ < 0$$