Let $A$ be an Hurwitz stable matrix (i.e. the real part of the eigenvalues of $A$ lie in the left-half plane) and $Q$ be a positive semidefinite matrix ($Q\ge 0$, for short). Let $P>0$ be the solution of the following Lyapunov equation
$$
AP+PA^\top = -Q.
$$
and define the set $\mathcal{Q}:=\{Q\ge 0\,:\, P>0\}$.

>**My question.** Let $\lambda_{\max}(P)$ denote the largest eigenvalue of $P$. For any $Q\in\mathcal{Q}$, it is quite easy to see that
$$\tag{$\star$}\label{star}
-2\,\mathrm{tr}(A) = \mathrm{tr}(P^{-1}Q) \ge \frac{\mathrm{tr}(Q)}{\lambda_{\max}(P)},
$$
where $\mathrm{tr}(\cdot)$ denotes the trace operator.
However, does there *always* (i.e., for any choice of $A$ Hurwitz stable) exist a $Q=Q^\star\in \mathcal{Q}$ for which \eqref{star} is attained with equality, that is
$$
-2\,\mathrm{tr}(A) = \frac{\mathrm{tr}(Q^\star)}{\lambda_{\max}(P^\star)},
$$
with $P^\star$ being the solution of $AP^\star+P^\star A^\top = -Q^\star$?