Given matrices $Q>0,F,A$ and a number $\alpha\in(0,1)$, find some $P>0,X,\Psi$ such that $$ \begin{aligned} \Psi^T P \Psi\leq \alpha P\\ \begin{bmatrix} A^TP+PA-Q & F^TX^T\\ XF & -\Psi^TP\Psi \end{bmatrix}&\leq 0 \end{aligned} $$ In particular, about $\Psi$ I only know that it is nonsingular.
Intuitively, we may let $\Psi=\gamma I$, and this problem is reduced to a LMI problem. However, I would not like to solve it that way. The following are what I want to do:
If possible, add some more constraints (possibly LMI ) or some objective function to avoid $\Psi$ being identity matrices like $\gamma I$.
So far I am unable to turn this problem into a SDP, nor LMI, BMI, nor cone complementary linearization problem.
I have tried the approach: If I can found some $\beta>0$ with $\Psi^TP\Psi\geq \beta P$, the problem can also be solved by LMI, which is why I asked the question(bounds of Lyapunov operator) with the hope that some constraints about the singular value or eigenvalue of $\Psi$ can be obtained. However, it is not easy to found some $\beta=\beta(\Psi)$ such that $\Psi^TP\Psi\geq \beta P$.
I am stuck here, hoping for some help.
What about letting $Z:=\Psi^TP\Psi\ge0$? In this case, the problem is linear in the decision variables $P,X,Z$ (assuming $\alpha$ is fixed).
Once the problem is solved, we need to solve for $\Psi$ in $\Psi^TP\Psi=Z$. This can be rewritten as $\Psi^TQ^TQ\Psi=R^TR$ where we have used a Cholesky decomposition; i.e. $P=Q^TQ$ and $Z=R^TR$. Therefore, a possible solution for $\Psi$ is given by $\Psi=Q^{-1}R$.