It is relatively easy to prove that if there exists a positive definite matrix $Q$ such that $Q - A^{H}QA$ is positive definite, where $A^{H}$ means the conjugate transpose of $A$, then the spectral radius of $A$ is less than $1$. Just look at every eigenpair $(v,\lambda)$. But as for the reverse problem, I am wondering how to construct a $Q$ such that $Q - A^{H}QA$ is positive definite, when the spectral radius of $A$ is less than $1$?



If the spectral radius of matrix $A$ is less than $1$, how to construct a positive definite $Q$ such that $Q - A^{H}QA$ is also positive definite?
I'll follow the suggestions in the comments and post my comment as an answer:


Take your favourite positive definite matrix $P$ and define $Q := \sum\_{k=0}^\infty (A^{\operatorname{H}})^k P A^k$; note that this series converges since, as the spectral radius of $A$ is $<1$, there exist numbers $\delta \in [0,1)$ and $M \ge 1$ such that $\|A^k\| \le M \delta^k$ for all integers $k \ge 0$.
The matrix $Q$ is positive definite and satisfies $Q−A^{\operatorname{H}}QA=P$.


*Remarks:*


* Both the result and the proof above are standard in systems and control theory; see for instance Section 3.3.5 in the book *Mathematical systems theory. I. Modelling, state space analysis, stability and robustness* (2005) by Hinrichsen and Pritchard ([link to zbMATH](https://zbmath.org/1074.93003)).
* The same argument also works for bounded linear operators on infinite dimensional Hilbert spaces.


