[proofplan]
We pass from the plant-observer variables $(x,\hat{x})$ to the state-error variables $(x,e)$, where $e = x - \hat{x}$. In these coordinates the closed-loop dynamics become triangular: the $x$-equation is driven by $x$ and $e$, while the $e$-equation evolves autonomously under $A - LC$. This triangular form makes the characteristic polynomial factor into the product of the characteristic polynomials of $A - BK$ and $A - LC$, so the closed-loop spectrum is the multiset union of the controller and observer spectra. Since both diagonal blocks are Hurwitz, every closed-loop eigenvalue has negative real part, which gives exponential stability of the origin.
[/proofplan]
[step:Write the closed-loop equations in plant and observer variables]
Let $X: \mathbb{R} \to \mathbb{R}^{n}$ denote a plant trajectory, and let $\widehat{X}: \mathbb{R} \to \mathbb{R}^{n}$ denote the corresponding observer trajectory. The controller is the map $U: \mathbb{R}^{n} \to \mathbb{R}^{m}$ defined by $U(\xi) = -K\xi$.
For a closed-loop trajectory we therefore have $u(t) = U(\widehat{X}(t)) = -K\widehat{X}(t)$ and $y(t) = CX(t)$ for each time $t$. Substituting these expressions into the plant and observer equations gives
\begin{align*}
\dot{X}(t) = AX(t) - BK\widehat{X}(t).
\end{align*}
Similarly,
\begin{align*}
\dot{\widehat{X}}(t) = A\widehat{X}(t) - BK\widehat{X}(t) + L(CX(t) - C\widehat{X}(t)).
\end{align*}
Thus the closed-loop system is a linear autonomous system on $\mathbb{R}^{n} \times \mathbb{R}^{n}$.
[/step]
[step:Change coordinates from observer state to estimation error]
Define the estimation error trajectory $E: \mathbb{R} \to \mathbb{R}^{n}$ by $E(t) = X(t) - \widehat{X}(t)$. Equivalently, $\widehat{X}(t) = X(t) - E(t)$ for every $t$. Substituting this identity into the plant equation gives
\begin{align*}
\dot{X}(t) = AX(t) - BK(X(t) - E(t)).
\end{align*}
After collecting terms,
\begin{align*}
\dot{X}(t) = (A - BK)X(t) + BKE(t).
\end{align*}
For the error equation, differentiate $E(t) = X(t) - \widehat{X}(t)$ and use the two closed-loop equations:
\begin{align*}
\dot{E}(t) = \dot{X}(t) - \dot{\widehat{X}}(t).
\end{align*}
Substituting the formulas already obtained gives
\begin{align*}
\dot{E}(t) = AX(t) - BK\widehat{X}(t) - A\widehat{X}(t) + BK\widehat{X}(t) - LCX(t) + LC\widehat{X}(t).
\end{align*}
The two terms involving $BK\widehat{X}(t)$ cancel, and the remaining terms factor through $X(t) - \widehat{X}(t) = E(t)$:
\begin{align*}
\dot{E}(t) = (A - LC)E(t).
\end{align*}
[guided]
The point of introducing $E(t) = X(t) - \widehat{X}(t)$ is that the observer correction term is designed to make the estimation error independent of the feedback gain $K$. We define the error trajectory as the map $E: \mathbb{R} \to \mathbb{R}^{n}$ given by $E(t) = X(t) - \widehat{X}(t)$. This definition gives $\widehat{X}(t) = X(t) - E(t)$. Since the controller is $u(t) = -K\widehat{X}(t)$, the plant equation becomes
\begin{align*}
\dot{X}(t) = AX(t) + B(-K\widehat{X}(t)).
\end{align*}
Using $\widehat{X}(t) = X(t) - E(t)$, we get
\begin{align*}
\dot{X}(t) = AX(t) - BK(X(t) - E(t)).
\end{align*}
Distributing the matrix product gives
\begin{align*}
\dot{X}(t) = (A - BK)X(t) + BKE(t).
\end{align*}
Now we compute the error dynamics directly. Differentiating the defining identity $E(t) = X(t) - \widehat{X}(t)$ gives
\begin{align*}
\dot{E}(t) = \dot{X}(t) - \dot{\widehat{X}}(t).
\end{align*}
The plant equation is $\dot{X}(t) = AX(t) - BK\widehat{X}(t)$, while the observer equation after substituting $y(t) = CX(t)$ and $u(t) = -K\widehat{X}(t)$ is
\begin{align*}
\dot{\widehat{X}}(t) = A\widehat{X}(t) - BK\widehat{X}(t) + LCX(t) - LC\widehat{X}(t).
\end{align*}
Subtracting these expressions gives
\begin{align*}
\dot{E}(t) = AX(t) - BK\widehat{X}(t) - A\widehat{X}(t) + BK\widehat{X}(t) - LCX(t) + LC\widehat{X}(t).
\end{align*}
The feedback terms cancel exactly:
\begin{align*}
-BK\widehat{X}(t) + BK\widehat{X}(t) = 0.
\end{align*}
The remaining terms are
\begin{align*}
\dot{E}(t) = (A - LC)X(t) - (A - LC)\widehat{X}(t).
\end{align*}
Factoring the common matrix $A - LC$ and using $E(t) = X(t) - \widehat{X}(t)$ yields
\begin{align*}
\dot{E}(t) = (A - LC)E(t).
\end{align*}
Thus the estimation error evolves autonomously, and its stability is controlled only by the observer matrix $A - LC$.
[/guided]
[/step]
[step:Represent the transformed closed-loop system as a triangular linear map]
Define the coordinate-change map $T: \mathbb{R}^{n} \times \mathbb{R}^{n} \to \mathbb{R}^{n} \times \mathbb{R}^{n}$ by $T(\xi,\widehat{\xi}) = (\xi,\xi - \widehat{\xi})$. Its inverse is the map $T^{-1}: \mathbb{R}^{n} \times \mathbb{R}^{n} \to \mathbb{R}^{n} \times \mathbb{R}^{n}$ defined by $T^{-1}(\xi,\eta) = (\xi,\xi - \eta)$. Thus $T$ is a linear isomorphism. In the coordinates $(x,e) = T(x,\hat{x})$, the closed-loop system is generated by the [linear map](/page/Linear%20Map) $M: \mathbb{R}^{n} \times \mathbb{R}^{n} \to \mathbb{R}^{n} \times \mathbb{R}^{n}$ defined by $M(\xi,\eta) = ((A - BK)\xi + BK\eta,(A - LC)\eta)$. Therefore the original closed-loop matrix in $(x,\hat{x})$-coordinates is similar to $M$, so it has the same characteristic polynomial, eigenvalues, and algebraic multiplicities as $M$.
[/step]
[step:Factor the characteristic polynomial from the triangular form]
Let $\lambda \in \mathbb{C}$. Complexify $M$ to the linear map $M_{\mathbb{C}}: \mathbb{C}^{n} \times \mathbb{C}^{n} \to \mathbb{C}^{n} \times \mathbb{C}^{n}$ defined by $M_{\mathbb{C}}(\xi,\eta) = ((A - BK)\xi + BK\eta,(A - LC)\eta)$. The map $\lambda I - M_{\mathbb{C}}$ has the block upper triangular form with diagonal maps $\lambda I_n - (A - BK)$ and $\lambda I_n - (A - LC)$. Expanding the determinant along this block triangular structure gives
\begin{align*}
\det(\lambda I_{2n} - M_{\mathbb{C}}) = \det(\lambda I_n - (A - BK))\det(\lambda I_n - (A - LC)).
\end{align*}
Hence the zeros of the closed-loop characteristic polynomial are exactly the zeros of the two diagonal characteristic polynomials, with multiplicities added. Therefore the closed-loop eigenvalues are precisely the eigenvalues of $A - BK$ together with the eigenvalues of $A - LC$, counted with algebraic multiplicity.
[/step]
[step:Use the Hurwitz hypotheses to conclude exponential stability]
By hypothesis, $A - BK$ is a Hurwitz matrix, meaning every eigenvalue of $A - BK$ has strictly negative real part. Also by hypothesis, $A - LC$ is a Hurwitz matrix, meaning every eigenvalue of $A - LC$ has strictly negative real part. The preceding factorization shows that every eigenvalue of the closed-loop linear map has strictly negative real part. Hence the closed-loop matrix is Hurwitz.
By the standard finite-dimensional linear stability criterion for autonomous systems, a Hurwitz system matrix is equivalent to exponential stability of the origin. Therefore the origin $(x,e) = (0,0)$ is exponentially stable in the transformed coordinates. Since $T$ is a linear isomorphism, exponential stability is preserved under the coordinate change between $(x,\hat{x})$ and $(x,e)$. Consequently the origin $(x,\hat{x}) = (0,0)$ of the plant-observer closed-loop system is exponentially stable. This proves that the observer-based controller stabilises the plant, and the closed-loop pole statement has already been established by the characteristic-polynomial factorization.
[/step]
