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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 $\mathcal{Q}:=\{Q\ge 0\,:\, P>0\}$.

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

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    $\begingroup$ I am surely missing something: why is $f(Q)$ not constant? Pre-multiplying both sides of Lyapunov equation with $P^{-1}$ and taking trace gives $f(Q) = -2\text{tr}(A)$? $\endgroup$
    – Abhishek Halder
    Commented Sep 11, 2018 at 7:49
  • $\begingroup$ @AbhishekHalder: Yes, $f(Q)$ is indeed constant; however the RHS of ($\star$) depends on $Q$. I will edit the OP in order to clarify this fact. $\endgroup$
    – Ludwig
    Commented Sep 11, 2018 at 13:44
  • $\begingroup$ @Mahdi: could you please elaborate a little more your comment? $\endgroup$
    – Ludwig
    Commented Sep 11, 2018 at 14:28
  • $\begingroup$ For every Hermitian matrices $A$ and $B$, we have $\langle \lambda^{\downarrow}(A),\lambda^{\uparrow}(B) \rangle \leq tr(AB)$. So, if there exists $Q, P$ such that $tr(P^{-1}Q) = tr(Q) / \lambda_{max}(P)$, then all eigenvalues of $P$ are equal. am I missing something? $\endgroup$
    – Mahdi - Free Palestine
    Commented Sep 11, 2018 at 14:44
  • $\begingroup$ @Mahdi: It seems to me that, for two matrices $X>0$, $Y\ge 0$, we have $\mathrm{tr}(X^{-1}Y)=\mathrm{tr}(Y)/\lambda_{\max}(X)$, for any choice of $X$ when $Y=v_{\min}v_{\min}^\top$, with $v_{\min}$ being the eigenvector corresponding to the smallest eigenvalue of $X^{-1}$. $\endgroup$
    – Ludwig
    Commented Sep 11, 2018 at 15:31

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