[proofplan] The proof first computes the differential equation for the estimation error by subtracting the observer dynamics from the plant dynamics and using the output identity $y = Cx$. This produces a homogeneous autonomous linear system whose coefficient matrix is $A - LC$, independent of the input $u$. The convergence assertion is then reduced to the standard finite-dimensional fact that every solution of $\dot{e} = Me$ converges to $0$ for every initial state exactly when every eigenvalue of $M$ has negative real part. [/proofplan]

[step:Subtract the observer equation from the plant equation]Define \begin{align*} M := A - LC \in \mathbb{R}^{n \times n}. \end{align*} Since $x$ and $\tilde{x}$ are $C^1$ maps from $[0,\infty)$ to $\mathbb{R}^n$, the error map \begin{align*} e: [0,\infty) \to \mathbb{R}^n, \qquad e(t) = x(t) - \tilde{x}(t) \end{align*} is also $C^1$. For each $t \geq 0$, differentiating the definition of $e$ gives \begin{align*} \dot{e}(t) = \dot{x}(t) - \dot{\tilde{x}}(t). \end{align*} Substitute the plant and observer equations: \begin{align*} \dot{e}(t) = \bigl(Ax(t) + Bu(t)\bigr) - \bigl(A\tilde{x}(t) + Bu(t) + L(y(t) - C\tilde{x}(t))\bigr). \end{align*} The two input terms $Bu(t)$ cancel. Since $y(t) = Cx(t)$, this becomes \begin{align*} \dot{e}(t) = A\bigl(x(t) - \tilde{x}(t)\bigr) - L\bigl(Cx(t) - C\tilde{x}(t)\bigr). \end{align*} By linearity of $C$, \begin{align*} Cx(t) - C\tilde{x}(t) = C\bigl(x(t) - \tilde{x}(t)\bigr) = Ce(t). \end{align*} Therefore \begin{align*} \dot{e}(t) = Ae(t) - LCe(t) = (A - LC)e(t) = Me(t). \end{align*}[/step]

[guided]The point of defining the error as $e(t) = x(t) - \tilde{x}(t)$ is that the plant and observer are driven by the same input $u(t)$, so subtracting the equations should remove the forcing term. Because $x$ and $\tilde{x}$ are $C^1$, the map $e$ is $C^1$, and its derivative is computed directly: \begin{align*} \dot{e}(t) = \dot{x}(t) - \dot{\tilde{x}}(t). \end{align*} Now insert the two differential equations. The plant gives $\dot{x}(t) = Ax(t) + Bu(t)$, while the observer gives \begin{align*} \dot{\tilde{x}}(t) = A\tilde{x}(t) + Bu(t) + L(y(t) - C\tilde{x}(t)). \end{align*} Hence \begin{align*} \dot{e}(t) = \bigl(Ax(t) + Bu(t)\bigr) - \bigl(A\tilde{x}(t) + Bu(t) + L(y(t) - C\tilde{x}(t))\bigr). \end{align*} The cancellation of $Bu(t)$ is exact because the observer uses the same input as the plant. Using the output equation $y(t) = Cx(t)$, we get \begin{align*} \dot{e}(t) = A\bigl(x(t) - \tilde{x}(t)\bigr) - L\bigl(Cx(t) - C\tilde{x}(t)\bigr). \end{align*} Since $C$ is a linear matrix map from $\mathbb{R}^n$ to $\mathbb{R}^p$, \begin{align*} Cx(t) - C\tilde{x}(t) = C\bigl(x(t) - \tilde{x}(t)\bigr). \end{align*} Using $e(t) = x(t) - \tilde{x}(t)$, the equation becomes \begin{align*} \dot{e}(t) = Ae(t) - LCe(t) = (A - LC)e(t). \end{align*} Thus the error dynamics are homogeneous and autonomous. In particular, the convergence of the observer error depends only on the matrix $A - LC$, not on the particular input $u$.[/guided]

[step:Express every error trajectory using the matrix exponential] Let $I_n \in \mathbb{R}^{n \times n}$ denote the identity matrix, and define \begin{align*} \Phi: [0,\infty) \to \mathbb{R}^{n \times n}, \qquad \Phi(t) = e^{Mt}. \end{align*} By the defining [power series](/page/Power%20Series) for the matrix exponential, $\Phi$ is $C^1$, satisfies $\Phi(0) = I_n$, and satisfies \begin{align*} \dot{\Phi}(t) = M\Phi(t). \end{align*} Therefore, for the initial error $e(0) = x(0) - \tilde{x}(0)$, the map \begin{align*} e_0: [0,\infty) \to \mathbb{R}^n, \qquad e_0(t) = e^{Mt}e(0) \end{align*} satisfies $e_0(0) = e(0)$ and $\dot{e}_0(t) = Me_0(t)$. Define the linear vector field \begin{align*} F: \mathbb{R}^n \to \mathbb{R}^n, \qquad F(\xi) = M\xi. \end{align*} Since $F$ is linear, it is continuous and globally Lipschitz with Lipschitz constant $\|M\|_{\mathrm{op}}$. The [Picard-Lindelöf existence and uniqueness theorem](/theorems/2774) applies locally to the initial value problem $\dot{z}=F(z)$, $z(0)=e(0)$, and gives uniqueness on every time interval on which two $C^1$ solutions are defined. The explicit map $e_0$ is defined on all of $[0,\infty)$, and the error trajectory $e$ is also defined on $[0,\infty)$ by the plant-observer hypotheses. Therefore, for each $T>0$, both $e|_{[0,T]}$ and $e_0|_{[0,T]}$ solve the same initial value problem on $[0,T]$, so local uniqueness gives $e(t)=e_0(t)$ for $0 \leq t \leq T$. Since $T>0$ is arbitrary, \begin{align*} e(t) = e^{Mt}e(0) \end{align*} for all $t \geq 0$. [/step]

[step:Show Hurwitz matrices force every error trajectory to vanish]Assume $M = A - LC$ is Hurwitz. View $M$ as a complex square matrix in $\mathbb{C}^{n \times n}$. By the [Jordan normal form theorem](/theorems/412), which applies to every complex square matrix, there exist an invertible matrix $S \in \mathbb{C}^{n \times n}$ and Jordan blocks $J_1,\dots,J_q$ such that \begin{align*} S^{-1}MS = J := \operatorname{diag}(J_1,\dots,J_q). \end{align*} For each block index $k \in \{1,\dots,q\}$, there are an eigenvalue $\mu_k \in \mathbb{C}$ of $M$, a block size $d_k \in \mathbb{N}$, the identity matrix $I_{d_k} \in \mathbb{C}^{d_k \times d_k}$, and a nilpotent matrix $N_k \in \mathbb{C}^{d_k \times d_k}$ such that \begin{align*} J_k = \mu_k I_{d_k} + N_k. \end{align*} For each block, \begin{align*} e^{J_k t} = e^{\mu_k t}e^{N_k t}. \end{align*} Since $N_k^{d_k}=0$, the exponential $e^{N_k t}$ is a matrix whose entries are polynomials in $t$ of degree at most $d_k-1$. Because $M$ is Hurwitz, $\operatorname{Re}\mu_k < 0$ for every block index $k$. Thus each polynomial-exponential factor $t^a e^{\mu_k t}$ with $0 \leq a \leq d_k-1$ tends to $0$ as $t \to \infty$. Hence \begin{align*} \lim_{t \to \infty} e^{Jt} = 0 \end{align*} entrywise, and therefore \begin{align*} \lim_{t \to \infty} e^{Mt} = \lim_{t \to \infty} S e^{Jt} S^{-1} = 0. \end{align*} Consequently, for every initial error $e(0) \in \mathbb{R}^n$, \begin{align*} \lim_{t \to \infty} |e(t)| = \lim_{t \to \infty} |e^{Mt}e(0)| = 0. \end{align*} Since $e(t) = x(t) - \tilde{x}(t)$, this is exactly \begin{align*} \lim_{t \to \infty} |\tilde{x}(t) - x(t)| = 0. \end{align*}[/step]

[guided]We now prove the stable direction of the equivalence: assume $M=A-LC$ is Hurwitz, meaning every complex eigenvalue of $M$ has strictly negative real part. The useful representation is [Jordan normal form](/theorems/864), because it separates each mode into an exponential factor and a finite polynomial factor. Viewing $M$ as an element of $\mathbb{C}^{n \times n}$, the Jordan normal form theorem applies to this complex square matrix and gives an invertible matrix $S \in \mathbb{C}^{n \times n}$ and Jordan blocks $J_1,\dots,J_q$ such that \begin{align*} S^{-1}MS = J := \operatorname{diag}(J_1,\dots,J_q). \end{align*} For each block index $k \in \{1,\dots,q\}$, there are an eigenvalue $\mu_k \in \mathbb{C}$ of $M$, a block size $d_k \in \mathbb{N}$, the identity matrix $I_{d_k} \in \mathbb{C}^{d_k \times d_k}$, and a nilpotent matrix $N_k \in \mathbb{C}^{d_k \times d_k}$ such that \begin{align*} J_k = \mu_k I_{d_k} + N_k. \end{align*} The matrices $\mu_k I_{d_k}$ and $N_k$ commute, so the exponential of the block factors as \begin{align*} e^{J_k t} = e^{\mu_k t}e^{N_k t}. \end{align*} Because $N_k^{d_k}=0$, the series for $e^{N_k t}$ terminates after the power $d_k-1$. Thus every entry of $e^{N_k t}$ is a polynomial in $t$ of degree at most $d_k-1$. Since $M$ is Hurwitz, $\operatorname{Re}\mu_k<0$ for every $k$, and therefore every scalar term $t^a e^{\mu_k t}$ with $0\leq a\leq d_k-1$ tends to $0$ as $t\to\infty$. Hence \begin{align*} \lim_{t \to \infty} e^{Jt} = 0 \end{align*} entrywise. Similarity preserves this limit after multiplying by the fixed matrices $S$ and $S^{-1}$, so \begin{align*} \lim_{t \to \infty} e^{Mt} = \lim_{t \to \infty} S e^{Jt} S^{-1} = 0. \end{align*} Using the representation of error trajectories from the previous step, every initial error $e(0)\in\mathbb{R}^n$ satisfies \begin{align*} \lim_{t \to \infty} |e(t)| = \lim_{t \to \infty} |e^{Mt}e(0)| = 0. \end{align*} Since $e(t)=x(t)-\tilde{x}(t)$, this is the desired convergence \begin{align*} \lim_{t \to \infty} |\tilde{x}(t)-x(t)| = 0. \end{align*}[/guided]

[step:Show convergence for every initial error forces the matrix to be Hurwitz]Assume that for every initial error $e(0) \in \mathbb{R}^n$, the corresponding solution of $\dot{e} = Me$ satisfies \begin{align*} \lim_{t \to \infty} |e(t)| = 0. \end{align*} We prove that $M$ is Hurwitz. Suppose, toward a contradiction, that $M$ has a complex eigenvalue $\lambda \in \mathbb{C}$ with $\operatorname{Re}\lambda \geq 0$. Let $v \in \mathbb{C}^n \setminus \{0\}$ be an eigenvector, so $Mv = \lambda v$. The complex-valued solution \begin{align*} z: [0,\infty) \to \mathbb{C}^n, \qquad z(t) = e^{\lambda t}v \end{align*} satisfies $\dot{z}(t)=Mz(t)$. If $\lambda \in \mathbb{R}$, then $v$ may be chosen real, and the real solution $e(t)=e^{\lambda t}v$ does not converge to $0$ when $\operatorname{Re}\lambda=\lambda \geq 0$, contradicting the hypothesis. If $\lambda \notin \mathbb{R}$, write \begin{align*} v = a + ib \end{align*} with $a,b \in \mathbb{R}^n$. Since $v \neq 0$, at least one of $a$ and $b$ is nonzero. The real and imaginary parts of $z(t)$ are real solutions of $\dot{e}=Me$, because $M$ has real entries. The complex solution satisfies \begin{align*} |z(t)| = |e^{\lambda t}v| = e^{\operatorname{Re}\lambda\, t}|v|. \end{align*} If both real solutions $\operatorname{Re} z(t)$ and $\operatorname{Im} z(t)$ converged to $0$, then $z(t)$ would converge to $0$ in $\mathbb{C}^n$, contradicting $|z(t)| = e^{\operatorname{Re}\lambda\, t}|v| \geq |v| > 0$. Hence at least one real solution fails to converge to $0$, again contradicting the hypothesis. Therefore no eigenvalue of $M$ has nonnegative real part, so every eigenvalue of $M$ has strictly negative real part. Thus $M=A-LC$ is Hurwitz.[/step]

[guided]We prove the converse by contradiction. Assume every real solution of $\dot{e}=Me$ converges to $0$, but suppose that $M$ has a complex eigenvalue $\lambda\in\mathbb{C}$ with $\operatorname{Re}\lambda\geq 0$. Choose an eigenvector $v\in\mathbb{C}^n\setminus\{0\}$ with $Mv=\lambda v$. Then the complex-valued map \begin{align*} z: [0,\infty) \to \mathbb{C}^n, \qquad z(t)=e^{\lambda t}v \end{align*} satisfies $\dot{z}(t)=Mz(t)$, because differentiating the scalar exponential gives $\lambda e^{\lambda t}v=e^{\lambda t}Mv$. If $\lambda\in\mathbb{R}$, then, since $M$ has real entries, the eigenvector may be chosen in $\mathbb{R}^n$. The real solution $e(t)=e^{\lambda t}v$ does not converge to $0$ when $\lambda\geq 0$: if $\lambda=0$, it is constant and nonzero, and if $\lambda>0$, its Euclidean norm grows like $e^{\lambda t}|v|$. This contradicts the assumed convergence of every real solution. It remains to handle the case $\lambda\notin\mathbb{R}$. Write \begin{align*} v=a+ib \end{align*} with $a,b\in\mathbb{R}^n$. Since $v\neq0$, at least one of $a$ and $b$ is nonzero. Because $M$ has real entries, taking real and imaginary parts commutes with applying $M$, so $\operatorname{Re}z(t)$ and $\operatorname{Im}z(t)$ are real solutions of $\dot{e}=Me$. The norm of the complex solution is \begin{align*} |z(t)|=|e^{\lambda t}v|=e^{\operatorname{Re}\lambda\,t}|v|. \end{align*} Since $\operatorname{Re}\lambda\geq0$, this quantity is at least $|v|>0$ for all $t\geq0$. If both real solutions $\operatorname{Re}z(t)$ and $\operatorname{Im}z(t)$ converged to $0$, then $z(t)$ would converge to $0$ in $\mathbb{C}^n$, contradicting the displayed lower bound. Hence at least one real solution fails to converge to $0$, again contradicting the hypothesis. Therefore no eigenvalue of $M$ has nonnegative real part, and $M=A-LC$ is Hurwitz.[/guided]

[step:Conclude the observer convergence criterion] The first step proved that the observer error satisfies \begin{align*} \dot{e}(t) = (A - LC)e(t). \end{align*} The preceding two steps show that every solution of this error equation converges to $0$ for every initial error if and only if $A - LC$ is Hurwitz. Since \begin{align*} e(0) = x(0) - \tilde{x}(0) \end{align*} can be any vector in $\mathbb{R}^n$ as the pair $x(0),\tilde{x}(0)$ varies over $\mathbb{R}^n \times \mathbb{R}^n$, this is equivalent to \begin{align*} \lim_{t \to \infty} |\tilde{x}(t) - x(t)| = 0 \end{align*} for every pair of initial states $x(0),\tilde{x}(0) \in \mathbb{R}^n$. This proves the stated equivalence. [/step]