Solve Lyapunov equation for $Q$

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Given Lyapunov equation: $AP+PA^T+Q=0$ and the linear system $\dot{x}=Ax$ is globally asymptotically stable i.e. the real part of all eignvalues of $A$ is strictly negative . The theorem says that choosing $Q>0 \implies P>0 $ and is unique . However , Does choosing $P>0$ guarantee that $Q>0$ ?

Edit: Is there a sufficient condition less strict than $A$ being symmetric such as $A^T+A<0$ ?

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No, $Q$ may be indefinite. Consider

$$ A = \begin{bmatrix} 1 & 4 \\ -1 & -3 \end{bmatrix}, \quad P = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$

But what you guarantee is that $Q \not <0$, since otherwise you would prove instability.