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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 $\mathcal{Q}:=\{Q\ge 0\,:\, P>0\}$. Moreover, for $Q\in\mathcal{Q}$, define $f(Q) := \mathrm{tr}(P^{-1}Q)$, where $\mathrm{tr}(\cdot)$ denotes the trace operator.

My question. Let $\lambda_{\max}(P)$ denote the largest eigenvalue of $P$. It is quite easy to see that $$\tag{$\star$}\label{star} f(Q) \ge \frac{\mathrm{tr}(Q)}{\lambda_{\max}(P)}. $$ However, does there always (i.e., for any choice of $A$ Hurwitz stable) exist a $Q\in \mathcal{Q}$, for which \eqref{star} is attained with equality?

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 $AP^\star+P^\star A^\top = -Q^\star$?

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 $f(Q) := \mathrm{tr}(P^{-1}Q)$, where $\mathrm{tr}(\cdot)$ denotes the trace operator.

My question. Let $\lambda_{\max}(P)$ denote the largest eigenvalue of $P$. It is quite easy to see that $$\tag{$\star$}\label{star} f(Q) \ge \frac{\mathrm{tr}(Q)}{\lambda_{\max}(P)}. $$ However, does there always (i.e., for any choice of $A$ Hurwitz stable) exist a $Q\ge 0$, for which \eqref{star} is attained with equality?

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 $\mathcal{Q}:=\{Q\ge 0\,:\, P>0\}$. Moreover, for $Q\in\mathcal{Q}$, define $f(Q) := \mathrm{tr}(P^{-1}Q)$, where $\mathrm{tr}(\cdot)$ denotes the trace operator.

My question. Let $\lambda_{\max}(P)$ denote the largest eigenvalue of $P$. It is quite easy to see that $$\tag{$\star$}\label{star} f(Q) \ge \frac{\mathrm{tr}(Q)}{\lambda_{\max}(P)}. $$ However, does there always (i.e., for any choice of $A$ Hurwitz stable) exist a $Q\in \mathcal{Q}$, for which \eqref{star} is attained with equality?

Let $A$ be an Hurwitz stable matrix (i.e. the real part of the eigenvalues of $A$ lie in the left-half plane) and $Q\ne 0$ be a positive semidefinite matrix ($Q\ge 0$, for short). Let $P\ge 0$ be the solution of the following Lyapunov equation $$ AP+PA^\top = -Q, $$ and define $f(Q) := \mathrm{tr}(PQ)$, where $\mathrm{tr}(\cdot)$ denotes the trace operator.

My question. Let $\lambda_{\max}(P)$ denote the largest eigenvalue of $P\ge 0$. It is quite easy to see that $$\tag{$\star$}\label{star} f(Q) \ge \frac{\mathrm{tr}(Q)}{\lambda_{\max}(P)}. $$ However, does there always (i.e., for any choice of $A$ Hurwitz stable) exist a $Q\ge 0$, $Q\ne 0$, for which \eqref{star} is attained with equality?

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 $f(Q) := \mathrm{tr}(P^{-1}Q)$, where $\mathrm{tr}(\cdot)$ denotes the trace operator.

My question. Let $\lambda_{\max}(P)$ denote the largest eigenvalue of $P$. It is quite easy to see that $$\tag{$\star$}\label{star} f(Q) \ge \frac{\mathrm{tr}(Q)}{\lambda_{\max}(P)}. $$ However, does there always (i.e., for any choice of $A$ Hurwitz stable) exist a $Q\ge 0$, for which \eqref{star} is attained with equality?