Consider a Lyapunov equation $$ PA + A^TP = -Q $$ where $A$ is stable and $Q = Q^T\ge0$. It is well known that if $Q>0$ (i.e., positive definite), then there are close relations between eigenvalues of $A$, $P$, and $Q$. For example, Lancaster showed the inequalities: $$ \frac{m(Q)}{|m(A+A^*)|} \le m(P) \le \frac{M(Q)}{2|m(A)|},\\ \frac{m(Q)}{2|M(A)|} \le M(P) \le \frac{M(Q)}{|M(A+A^*|} $$ where $m(A)$ and $M(A)$ are the minimum and the maximum of the real part of the eigenvalues of $A$, respectively.
Now, assume that $Q = C^TC\ge0$ and $(A,C)$ is observable. We still have a unique solution $P = P^T>0$ of the Lyapunov equation. I am wondering if we have inequalities similar to the one shown above in this case. Specifically, I am wondering if the lower bound of $m(P)$ and the upper bound of $M(P)$ can be written by using $A$ and $C$. Any help would be appreciated.
Edit
Probably, it would be better to show one of my trials. The solution $P$ can be written as $$ P = \int_0^\infty e^{tA^T}C^TCe^{tA}dt. $$ Since $P>0$, there exists $x\neq0$ such that $$ m(P) = \frac{x^TPx}{x^Tx} = \int_0^\infty\frac{\|Ce^{tA}x\|^2}{\|x\|^2}dt > 0. $$ Thus, I may need to find a lower bound of $\|Ce^{tA}x\|$ by using $A$, $C$, and $\|x\|$; this is the point I was stuck. Since $C$ is not assumed to have full rank, I may need to use the property that $(A,C)$ is observable but do not have any idea so far...