[proofplan] The proof is a duality argument between the continuous-time Kalman filter Riccati equation and the continuous-time linear-quadratic regulator Riccati equation. We apply the stabilizing algebraic Riccati theorem for LQR to the dual system with dynamics matrix $A^\top$, input matrix $C^\top$, control weight $R$, and state cost matrix $Q = GG^\top$. Detectability of $(A,C)$ becomes stabilizability of $(A^\top,C^\top)$, while stabilizability of $(A,G)$ becomes detectability of $(A^\top,G^\top)$. The LQR theorem gives the stabilizing solution and convergence of the dual Riccati flow; transposing the closed-loop matrix converts the dual Hurwitz statement into the filter Hurwitz statement. [/proofplan]

[step:Form the dual linear-quadratic regulator data] Define the dual control matrices \begin{align*} A_d := A^\top \in \mathbb{R}^{n \times n}, \qquad B_d := C^\top \in \mathbb{R}^{n \times m}, \qquad S_d := R \in \mathbb{R}^{m \times m}. \end{align*} Define the dual state-cost factor \begin{align*} H_d := G^\top \in \mathbb{R}^{r \times n}. \end{align*} Then \begin{align*} H_d^\top H_d = G G^\top = Q. \end{align*} Since $R$ is symmetric positive definite by hypothesis, $S_d$ is an admissible positive definite control-weight matrix for the dual LQR problem. The LQR algebraic Riccati equation for the data $(A_d,B_d,H_d,S_d)$ is \begin{align*} 0 = A_d^\top X + XA_d + H_d^\top H_d - XB_dS_d^{-1}B_d^\top X. \end{align*} Substituting the definitions of $A_d$, $B_d$, $H_d$, and $S_d$ gives \begin{align*} 0 = AX + XA^\top + Q - XC^\top R^{-1}CX. \end{align*} Thus the dual LQR Riccati equation is exactly the filter algebraic Riccati equation with the unknown matrix renamed from $X$ to $P$. [/step]

[step:Translate the detectability and stabilizability hypotheses to the dual system]For $\lambda \in \mathbb{C}$, define the observation Hautus map \begin{align*} \mathcal{O}_\lambda(A,C): \mathbb{C}^n \to \mathbb{C}^n \times \mathbb{C}^m, \qquad v \mapsto ((\lambda I_n-A)v,Cv). \end{align*} The detectability of $(A,C)$ is equivalent to the stabilizability of $(A^\top,C^\top)$. Indeed, by the Hautus criterion, detectability of $(A,C)$ means that $\operatorname{rank}\mathcal{O}_\lambda(A,C)=n$ for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$. The transpose of the matrix representing $\mathcal{O}_\lambda(A,C)$ is the matrix representing the control Hautus map \begin{align*} \mathcal{K}_\lambda(A^\top,C^\top): \mathbb{C}^n \times \mathbb{C}^m \to \mathbb{C}^n, \qquad (v,w) \mapsto (\lambda I_n-A^\top)v+C^\top w. \end{align*} Since a matrix and its transpose have the same rank, $\operatorname{rank}\mathcal{K}_\lambda(A^\top,C^\top)=n$ for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$, which is the Hautus stabilizability criterion for $(A^\top,C^\top)$. Similarly, for $\lambda \in \mathbb{C}$ define \begin{align*} \mathcal{K}_\lambda(A,G): \mathbb{C}^n \times \mathbb{C}^r \to \mathbb{C}^n, \qquad (v,w) \mapsto (\lambda I_n-A)v+Gw. \end{align*} By the Hautus criterion, stabilizability of $(A,G)$ means that $\operatorname{rank}\mathcal{K}_\lambda(A,G)=n$ for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$. Transposing the representing matrix gives the observation Hautus map \begin{align*} \mathcal{O}_\lambda(A^\top,G^\top): \mathbb{C}^n \to \mathbb{C}^n \times \mathbb{C}^r, \qquad v \mapsto ((\lambda I_n-A^\top)v,G^\top v). \end{align*} Rank preservation under transpose gives $\operatorname{rank}\mathcal{O}_\lambda(A^\top,G^\top)=n$ for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$, which is the Hautus detectability criterion for $(A^\top,G^\top)$.[/step]

[guided]The duality step is not only a formal transpose of the Riccati equation; the hypotheses must transpose into exactly the hypotheses required by the LQR Riccati theorem. For $\lambda \in \mathbb{C}$, define \begin{align*} \mathcal{O}_\lambda(A,C): \mathbb{C}^n \to \mathbb{C}^n \times \mathbb{C}^m, \qquad v \mapsto ((\lambda I_n-A)v,Cv). \end{align*} Detectability of $(A,C)$ says that every unstable or marginally stable mode of $A$ is seen by $C$. The Hautus detectability criterion states this as $\operatorname{rank}\mathcal{O}_\lambda(A,C)=n$ for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$. The transpose of the matrix representing $\mathcal{O}_\lambda(A,C)$ represents the control Hautus map \begin{align*} \mathcal{K}_\lambda(A^\top,C^\top): \mathbb{C}^n \times \mathbb{C}^m \to \mathbb{C}^n, \qquad (v,w) \mapsto (\lambda I_n-A^\top)v+C^\top w. \end{align*} A matrix and its transpose have the same rank, so $\operatorname{rank}\mathcal{K}_\lambda(A^\top,C^\top)=n$ for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$. This is precisely the Hautus stabilizability criterion for $(A^\top,C^\top)$. Now define \begin{align*} \mathcal{K}_\lambda(A,G): \mathbb{C}^n \times \mathbb{C}^r \to \mathbb{C}^n, \qquad (v,w) \mapsto (\lambda I_n-A)v+Gw. \end{align*} Stabilizability of $(A,G)$ means $\operatorname{rank}\mathcal{K}_\lambda(A,G)=n$ for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$. Transposing the representing matrix gives \begin{align*} \mathcal{O}_\lambda(A^\top,G^\top): \mathbb{C}^n \to \mathbb{C}^n \times \mathbb{C}^r, \qquad v \mapsto ((\lambda I_n-A^\top)v,G^\top v). \end{align*} Again rank is preserved by transpose, so $\operatorname{rank}\mathcal{O}_\lambda(A^\top,G^\top)=n$ for every $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$. This is the Hautus detectability criterion for $(A^\top,G^\top)$. Therefore the filter assumptions are exactly the LQR assumptions after passing to the dual data $(A^\top,C^\top,G^\top,R)$.[/guided]

[step:Apply the stabilizing Riccati theorem to the dual regulator problem]We apply the Algebraic Riccati Equation Existence Theorem in its continuous-time stabilizing LQR form, including the associated autonomous Riccati-flow convergence statement. For data $(A_d,B_d,H_d,S_d)$, the theorem requires $S_d=S_d^\top>0$, stabilizability of $(A_d,B_d)$, and detectability of $(H_d,A_d)$. The first condition holds because $S_d=R$ and $R=R^\top>0$; the second and third conditions were verified in the previous step. Therefore there exists a unique symmetric positive semidefinite matrix $X_\infty \in \mathbb{R}^{n \times n}$ solving \begin{align*} 0 = A_d^\top X_\infty + X_\infty A_d + H_d^\top H_d - X_\infty B_dS_d^{-1}B_d^\top X_\infty. \end{align*} The theorem also states that the dual closed-loop matrix $A_d-B_dS_d^{-1}B_d^\top X_\infty$ is Hurwitz and that the autonomous Riccati flow converges to $X_\infty$ from every symmetric positive semidefinite initial condition. Substituting $A_d=A^\top$, $B_d=C^\top$, $S_d=R$, and $H_d^\top H_d=Q$, the equation for $X_\infty$ becomes \begin{align*} 0 = AX_\infty + X_\infty A^\top + Q - X_\infty C^\top R^{-1}CX_\infty. \end{align*} Thus $X_\infty$ is a symmetric positive semidefinite solution of the filter algebraic Riccati equation.[/step]

[guided]This is the point where the proof consumes the main external Riccati theorem. We use the Algebraic Riccati Equation Existence Theorem in its continuous-time stabilizing LQR form, including the autonomous Riccati-flow convergence statement. For the dual LQR data $(A_d,B_d,H_d,S_d)$, the theorem requires three hypotheses: $S_d=S_d^\top>0$, the pair $(A_d,B_d)$ is stabilizable, and the pair $(H_d,A_d)$ is detectable. We verify them in order. Since $S_d=R$ and the theorem statement assumes $R=R^\top>0$, the control weight condition holds. The previous step proved that detectability of $(A,C)$ is exactly stabilizability of $(A^\top,C^\top)$, which is stabilizability of $(A_d,B_d)$. The previous step also proved that stabilizability of $(A,G)$ is exactly detectability of $(A^\top,G^\top)$, which is detectability of $(A_d,H_d)$ in the notation of the LQR theorem. Therefore the LQR Riccati theorem gives a unique symmetric positive semidefinite matrix $X_\infty \in \mathbb{R}^{n \times n}$ satisfying \begin{align*} 0 = A_d^\top X_\infty + X_\infty A_d + H_d^\top H_d - X_\infty B_dS_d^{-1}B_d^\top X_\infty. \end{align*} It also gives that $A_d-B_dS_d^{-1}B_d^\top X_\infty$ is Hurwitz and that every solution of the dual autonomous Riccati flow starting from a symmetric positive semidefinite matrix converges to $X_\infty$. Now substitute the dual definitions. Since $A_d=A^\top$, $B_d=C^\top$, $S_d=R$, and $H_d^\top H_d=Q$, the algebraic equation becomes \begin{align*} 0 = AX_\infty + X_\infty A^\top + Q - X_\infty C^\top R^{-1}CX_\infty. \end{align*} This is exactly the filter algebraic Riccati equation with unknown $X_\infty$.[/guided]

[step:Transpose the dual closed-loop stability statement] Define \begin{align*} P_\infty := X_\infty. \end{align*} The dual closed-loop matrix is $A_d-B_dS_d^{-1}B_d^\top X_\infty=A^\top-C^\top R^{-1}CP_\infty$. Taking the transpose gives \begin{align*} \left(A^\top - C^\top R^{-1}CP_\infty\right)^\top = A - P_\infty C^\top R^{-1}C, \end{align*} where we used $P_\infty^\top=P_\infty$ and $R^\top=R$, hence $(R^{-1})^\top=R^{-1}$. A real square matrix and its transpose have the same characteristic polynomial, so they have the same eigenvalues with algebraic multiplicity. Therefore, since the dual closed-loop matrix is Hurwitz, the filter closed-loop matrix \begin{align*} A - P_\infty C^\top R^{-1}C \end{align*} is Hurwitz. [/step]

[step:Identify the covariance flow with the dual Riccati flow and pass to the limit]Let $P_0 \in \mathbb{R}^{n \times n}$ be symmetric positive semidefinite. The dual LQR Riccati flow with initial value $X(0)=P_0$ is \begin{align*} \frac{dX}{dt}(t) = A_d^\top X(t) + X(t)A_d + H_d^\top H_d - X(t)B_dS_d^{-1}B_d^\top X(t). \end{align*} Substituting the dual data gives \begin{align*} \frac{dX}{dt}(t) = AX(t) + X(t)A^\top + Q - X(t)C^\top R^{-1}CX(t). \end{align*} This is exactly the covariance Riccati equation with the same initial condition. Hence, by uniqueness of solutions for this finite-dimensional autonomous ordinary differential equation, $P(t)=X(t)$ for every $t$ in their common interval of existence. The convergence part of the dual stabilizing Riccati theorem gives global existence, preservation of symmetric positive semidefiniteness, and \begin{align*} \lim_{t \to \infty} X(t)=X_\infty. \end{align*} Since $P(t)=X(t)$ and $P_\infty=X_\infty$, it follows that \begin{align*} \lim_{t \to \infty} P(t)=P_\infty. \end{align*} The uniqueness of $X_\infty$ among symmetric positive semidefinite stabilizing solutions of the dual LQR algebraic Riccati equation is exactly the uniqueness of $P_\infty$ among symmetric positive semidefinite stabilizing solutions of the filter algebraic Riccati equation, because the equations coincide term by term. This proves all asserted conclusions.[/step]

[guided]We now transfer the convergence statement from the dual Riccati flow back to the covariance Riccati flow. Let $P_0 \in \mathbb{R}^{n \times n}$ be symmetric positive semidefinite, and let $X: [0,\infty) \to \mathbb{R}^{n \times n}$ denote the dual LQR Riccati solution with $X(0)=P_0$. The dual flow is \begin{align*} \frac{dX}{dt}(t) = A_d^\top X(t) + X(t)A_d + H_d^\top H_d - X(t)B_dS_d^{-1}B_d^\top X(t). \end{align*} Substituting $A_d=A^\top$, $B_d=C^\top$, $S_d=R$, and $H_d^\top H_d=Q$ gives \begin{align*} \frac{dX}{dt}(t) = AX(t) + X(t)A^\top + Q - X(t)C^\top R^{-1}CX(t). \end{align*} This is the covariance Riccati equation with the same initial value $P_0$. The right-hand side is a polynomial map in the entries of the matrix variable, hence is locally Lipschitz on the finite-dimensional [vector space](/page/Vector%20Space) $\mathbb{R}^{n \times n}$. Uniqueness for finite-dimensional autonomous ordinary differential equations therefore implies that the covariance solution $P(t)$ and the dual solution $X(t)$ agree for every $t$ on their common interval of existence. The version of the stabilizing LQR Riccati theorem invoked above gives global existence of $X(t)$, preservation of symmetric positive semidefiniteness, and convergence to $X_\infty$: \begin{align*} \lim_{t \to \infty} X(t)=X_\infty. \end{align*} Because $P(t)=X(t)$ for all $t$ and $P_\infty=X_\infty$, the same limit holds for the filter covariance: \begin{align*} \lim_{t \to \infty} P(t)=P_\infty. \end{align*} Finally, uniqueness transfers without any additional argument: the dual and filter algebraic Riccati equations are the same equation after substituting the dual data, so a second symmetric positive semidefinite stabilizing filter solution would be a second symmetric positive semidefinite stabilizing dual LQR solution. The LQR theorem rules this out.[/guided]