How can I prove this statement regarding a discrete-time algebraic Riccati inequality?

Question

Consider the square matrix $X \in \mathbb R^{n \times n}$. In a paper that I'm currently reading, the authors state that, if the following two conditions are met: $$ \begin{align} &(1) \quad AXA^T - X + Q - AXC^T(CXC^T + R)^{-1}CXA^T \preceq 0 \\ &(2) \quad \textrm{$(A,Q^{1/2})$ is controllable} \end{align} $$ where

• $A \in \mathbb R^{n \times n},C \in \mathbb R^{m \times n}, Q \succeq 0,R \succ 0$, and
• $(A,Q^{1/2})$ being controllable means that there exists a $P_1 \succ 0$ such that $A^TP_1A - P_1 \prec 0$, or equivalently that there exists a $P_2 \succ 0$ such that $AP_2A^T - P_2 + Q = 0$,

then $X \succ 0$. I'm not entirely sure how to check whether this statement is true or false, so I would appreciate any hints.

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For reference, the statement given in the paper is actually more general than the one I mentioned above. That is, let $\lambda \in [0,1]$ and define the function $$ g_\lambda(X) = AXA^T + Q - \lambda AXC^T(CXC^T + R)^{-1}CXA^T $$ Then, in lemma 1(g) in Appendix A in the paper (page 1460, right column), the authors state that, for every $\lambda \in [0,1]$, if $(A,Q^{1/2})$ is controllable and $X \succeq g_\lambda(X)$, then $X \succ 0$.

To prove this, the authors first show that, for every $\lambda \in [0,1]$, $g_\lambda(X) \succeq (1-\lambda)AXA^T + Q$ (this is lemma 1(f)). The proof of this is straightforward. Finally, the authors give the following proof for lemma 1(g) (emphasis on the word "detectable" is mine):

From fact f), it follows that $X \succeq g_\lambda(X) \succeq (1-\lambda)AXA^T + Q$. Let $\hat X$ such that $\hat X = (1-\lambda)A\hat X A^T + Q$. Such $\hat X$ must clearly exist. Therefore, $X - \hat X \succeq (1-\lambda)A(X - \hat X)A^T \succeq 0$. Moreover, the matrix $\hat X$ solves the Lyapunov equation $\hat X = \tilde A\hat X \tilde A^T + Q$ where $\tilde A = \sqrt{1-\lambda} A$. Since $(\tilde A,Q^{1/2})$ is detectable, it follows that $\hat X \succ 0$ and so $X \succ 0$, which proves the fact.

I'm not sure if it is possible to use this proof for my case (when $\lambda = 1$). Also, I think the authors made a mistake when they wrote "Since $(\tilde A,Q^{1/2})$ is detectable..." and I think it should instead be "Since $(\tilde A,Q^{1/2})$ is controllable...", since they never assumed that $(A,Q^{1/2})$ is detectable anywhere in the paper, only controllable.

Answer 1

I was able to find the answer in section 13.1 in the book

P. Lancaster and L. Rodman, Algebraic Riccati equations. in Oxford science publications. Oxford : New York: Clarendon Press ; Oxford University Press, 1995.

More specifically, using theorem 13.1.1, corollary 13.1.2, and theorem 13.1.3 in section 13.1 in this book, we can show that, if $(A,Q)$ is controllable and $(C,A)$ is detectable, and if there exists a matrix $X \succeq 0$ such that $$ X \preceq AXA^T + Q - AXC^T(CXC^T + R)^{-1}CXA^T $$ then $X \succ 0$.