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 $1$?
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1$\begingroup$ Welcome to MathOverflow! Please note that it is possible - and strongly encouraged - to use LaTeX typesetting in your posts to enhance readability. I've edited your question accordingly. $\endgroup$– Jochen GlueckCommented Jan 23, 2023 at 4:32
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4$\begingroup$ Regarding your question: Just take your favourite positive definite matrix $P$ and define $Q := \sum_{k=0}^\infty (A^H)^k P A^k$. Then $Q$ is positive definite and satisfies $Q - A^H Q A = P$. This is a standard argument in systems and control theory; see for instance Section 3.3.5 in this book. $\endgroup$– Jochen GlueckCommented Jan 23, 2023 at 4:41
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1$\begingroup$ @FedericoPoloni: Yes, I usually don't do this - but the question is clearly not research level, so I'm a bit reluctant to post an answer to it. (The proper process would, of course, have been to vote to close, and suggest to the OP to post the question on Math StackExchange. But since the answer is just two lines and I knew it off my head, it felt somehow unfair towards the OP to not even mention the solution.) $\endgroup$– Jochen GlueckCommented Jan 23, 2023 at 7:39
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2$\begingroup$ @JochenGlueck This question has presently no close votes, 3 upvotes and no downvotes. Therefore, I don't see a reason not to post a proper answer. $\endgroup$– Stefan Kohl ♦Commented Jan 23, 2023 at 11:23
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1$\begingroup$ @StefanKohl: Ok, I did as Federico Poloni and you suggest. $\endgroup$– Jochen GlueckCommented Jan 23, 2023 at 18:01
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1 Answer
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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).
The same argument also works for bounded linear operators on infinite dimensional Hilbert spaces.
answered Jan 23, 2023 at 18:00