Is the resulting matrix from this lyapunov operator always negative definite?

Question

Let $A$ be an arbitrary $n \times n$ real matrix such that $\mathrm{Re}(\lambda_i(A)) < 0$ for all $i=1,...,n$, and let $P = P^T \succ 0$.

Is the resulting matrix $A^T P + P A$ necessarily negative definite?

Answer 1

No. Take $$ A = \begin{bmatrix} -10 & 0 \\ 0 & -1 \end{bmatrix}, \qquad P = \begin{bmatrix} 1.1 & 1 \\ 1 & 1.1 \end{bmatrix} .$$ Then the eigenvalues of $A^T P + P A$ are approximately $2.6990$ and $-26.8990$.

Answer 2

Here is another counterexample: take $$ P=I_2,\ A=\pmatrix{-1&3\\ 0&-1},\ B=A^TP+PA=\pmatrix{-2&3\\ 3&-2}. $$ $B$ has a negative determinant. Therefore it cannot possess two negative eigenvalues.

In general, since $A^TP+PA$ is congruent to $P^{-1/2}(A^TP+PA)P^{-1/2}$, it is negative definite if and only if the symmetric part of $P^{1/2}AP^{-1/2}$ is negative definite.