[proofplan] We prove both directions by comparing transfer functions through their Markov parameters $CA^kB$. If the realization is not reachable, we restrict the dynamics to the reachable subspace; if it is not observable, we quotient by the unobservable subspace. Conversely, for a reachable and observable realization, the infinite Hankel map formed from the Markov parameters factors as an observable map after a reachable map, so its rank is exactly the state dimension. Any competing realization factors the same Hankel map through its own state space, forcing its dimension to be at least $n$. [/proofplan]

[step:Show that equality of transfer functions is equality of Markov parameters] Let $X := \mathbb{R}^n$, $U := \mathbb{R}^m$, and $Y := \mathbb{R}^p$. We regard $A: X \to X$, $B: U \to X$, $C: X \to Y$, and $D: U \to Y$ as linear maps represented by the given matrices. When evaluating the transfer function at a complex parameter $\lambda \in \mathbb{C}$, we extend these real linear maps complex-linearly to $X_{\mathbb{C}} := X \otimes_{\mathbb{R}} \mathbb{C}$, $U_{\mathbb{C}} := U \otimes_{\mathbb{R}} \mathbb{C}$, and $Y_{\mathbb{C}} := Y \otimes_{\mathbb{R}} \mathbb{C}$. Define the transfer function as the map \begin{align*} G_{A,B,C,D}: \{\lambda \in \mathbb{C} : \lambda I_n-A \text{ is invertible}\} \to \mathcal{L}(U_{\mathbb{C}},Y_{\mathbb{C}}). \end{align*} For each $\lambda$ in its domain, this map is given by \begin{align*} G_{A,B,C,D}(\lambda)=D+C(\lambda I_n-A)^{-1}B. \end{align*} The Markov parameters of the realization are the linear maps $CA^kB: U \to Y$ for $k \in \mathbb{N}\cup\{0\}$. For $|\lambda|$ sufficiently large, define $w := \lambda^{-1}$. Then $|w|$ is sufficiently small that the spectral radius of $wA$ is strictly less than $1$. Hence $I_n-wA$ is invertible and the Neumann series converges in the finite-dimensional operator norm to its inverse: \begin{align*} (I_n-wA)^{-1}=\sum_{k=0}^{\infty} w^k A^k. \end{align*} Thus \begin{align*} (\lambda I_n-A)^{-1}=w(I_n-wA)^{-1} \end{align*} and hence \begin{align*} G_{A,B,C,D}(\lambda)=D+\sum_{k=0}^{\infty} CA^kB\,\lambda^{-k-1}. \end{align*} By [uniqueness of Laurent series coefficients](/page/Laurent%20Series) for matrix-valued holomorphic functions on a punctured neighbourhood of $\lambda=\infty$, applied entrywise to the Laurent expansions above, if two finite-dimensional realizations have the same transfer function, then their Laurent expansions at $\lambda=\infty$ have the same coefficients. Conversely, if two realizations have equal direct terms and equal Markov parameters $CA^kB$ for every $k \in \mathbb{N}\cup\{0\}$, then the displayed Laurent expansions agree for all sufficiently large $|\lambda|$. Each transfer function entry is a rational function of $\lambda$, hence holomorphic away from the finite set of poles determined by the corresponding characteristic polynomial. Equality on the large-$|\lambda|$ pole-free region therefore implies equality of the corresponding rational matrix entries on the common pole-free domain, by the identity theorem applied entrywise on every connected component of that domain. In particular, if \begin{align*} (\hat A,\hat B,\hat C,\hat D) \in \mathbb{R}^{\hat n \times \hat n} \times \mathbb{R}^{\hat n \times m} \times \mathbb{R}^{p \times \hat n} \times \mathbb{R}^{p \times m} \end{align*} has the same transfer function, then \begin{align*} D=\hat D \end{align*} and, for every $k \in \mathbb{N}\cup\{0\}$, \begin{align*} CA^kB=\hat C\hat A^k\hat B. \end{align*} [/step]

[step:Reduce a nonreachable realization to its reachable subspace] Define the reachable subspace \begin{align*} \mathcal{R}(A,B) := \operatorname{span}\{A^kBu : k \in \mathbb{N}\cup\{0\},\ u \in U\} \subset X. \end{align*} Assume $\mathcal{R}(A,B) \neq X$. Let $r := \dim \mathcal{R}(A,B)$, so $r<n$. The subspace $\mathcal{R}(A,B)$ is $A$-invariant because, for every generator $A^kBu$ with $k \in \mathbb{N}\cup\{0\}$ and $u \in U$, \begin{align*} A(A^kBu)=A^{k+1}Bu \in \mathcal{R}(A,B). \end{align*} It also contains $B(U)$ by taking $k=0$. Define the restricted linear maps \begin{align*} A_{\mathcal{R}}: \mathcal{R}(A,B) \to \mathcal{R}(A,B), \qquad A_{\mathcal{R}}x := Ax, \end{align*} \begin{align*} B_{\mathcal{R}}: U \to \mathcal{R}(A,B), \qquad B_{\mathcal{R}}u := Bu, \end{align*} and \begin{align*} C_{\mathcal{R}}: \mathcal{R}(A,B) \to Y, \qquad C_{\mathcal{R}}x := Cx. \end{align*} For every $k \in \mathbb{N}\cup\{0\}$ and $u \in U$, induction on $k$ gives \begin{align*} A_{\mathcal{R}}^kB_{\mathcal{R}}u=A^kBu. \end{align*} Applying $C_{\mathcal{R}}$ gives \begin{align*} C_{\mathcal{R}}A_{\mathcal{R}}^kB_{\mathcal{R}}u=CA^kBu. \end{align*} Choose a basis of the finite-dimensional space $\mathcal{R}(A,B)$ and represent $A_{\mathcal{R}}$, $B_{\mathcal{R}}$, and $C_{\mathcal{R}}$ by matrices in that basis and the fixed bases of $U$ and $Y$. Thus the reduced realization $(A_{\mathcal{R}},B_{\mathcal{R}},C_{\mathcal{R}},D)$ has the same Markov parameters and the same direct term $D$ as $(A,B,C,D)$. By the Laurent-expansion converse proved in the first step, the two transfer functions agree. Its state dimension is $r<n$, so the original realization is not minimal. [/step]

[step:Reduce a nonobservable realization by quotienting the unobservable subspace] Define the unobservable subspace \begin{align*} \mathcal{N}(C,A) := \{x \in X : CA^k x=0 \text{ for every } k \in \mathbb{N}\cup\{0\}\} \subset X. \end{align*} Assume $\mathcal{N}(C,A) \neq \{0\}$. Let $q: X \to X/\mathcal{N}(C,A)$ be the quotient map. The subspace $\mathcal{N}(C,A)$ is $A$-invariant: if $x \in \mathcal{N}(C,A)$, then for every $k \in \mathbb{N}\cup\{0\}$, \begin{align*} CA^k(Ax)=CA^{k+1}x=0. \end{align*} Therefore the induced [linear map](/page/Linear%20Map) \begin{align*} A_q: X/\mathcal{N}(C,A) \to X/\mathcal{N}(C,A), \qquad A_q(qx) := q(Ax) \end{align*} is well-defined. Define \begin{align*} B_q: U \to X/\mathcal{N}(C,A), \qquad B_q u := q(Bu). \end{align*} Since $\mathcal{N}(C,A) \subseteq \ker C$, the map \begin{align*} C_q: X/\mathcal{N}(C,A) \to Y, \qquad C_q(qx) := Cx \end{align*} is well-defined: if $qx=qx'$, then $x-x' \in \mathcal{N}(C,A) \subseteq \ker C$, so $Cx=Cx'$. For every $k \in \mathbb{N}\cup\{0\}$ and $u \in U$, induction gives \begin{align*} A_q^kB_q u=q(A^kBu). \end{align*} Applying $C_q$ gives \begin{align*} C_qA_q^kB_q u=CA^kBu. \end{align*} Choose a basis of the finite-dimensional quotient space $X/\mathcal{N}(C,A)$ and represent $A_q$, $B_q$, and $C_q$ by matrices in that basis and the fixed bases of $U$ and $Y$. Thus $(A_q,B_q,C_q,D)$ has the same Markov parameters and the same direct term $D$ as $(A,B,C,D)$. By the Laurent-expansion converse proved in the first step, the two transfer functions agree. Its state dimension is \begin{align*} \dim(X/\mathcal{N}(C,A))=n-\dim\mathcal{N}(C,A)<n. \end{align*} So the original realization is not minimal. [/step]

[step:Factor the Hankel map through the state space]Assume now that $\mathcal{R}(A,B)=X$ and $\mathcal{N}(C,A)=\{0\}$. Let \begin{align*} E := \{(u_j)_{j=0}^{\infty} : u_j \in U \text{ for every } j,\ u_j=0 \text{ for all but finitely many } j\} \end{align*} be the [vector space](/page/Vector%20Space) of finitely supported input sequences. Let \begin{align*} S := \{(y_i)_{i=0}^{\infty} : y_i \in Y \text{ for every } i\} \end{align*} be the vector space of output sequences. Define the reachability map \begin{align*} \mathscr{R}: E \to X, \qquad \mathscr{R}((u_j)_{j=0}^{\infty}) := \sum_{j=0}^{\infty} A^jBu_j. \end{align*} The sum is finite because $(u_j)_{j=0}^{\infty}$ has finite support. Since $\mathcal{R}(A,B)=X$, the map $\mathscr{R}$ is surjective. Define the observability map \begin{align*} \mathscr{O}: X \to S, \qquad \mathscr{O}(x) := (CA^i x)_{i=0}^{\infty}. \end{align*} Since $\mathcal{N}(C,A)=\{0\}$, the map $\mathscr{O}$ is injective. Define the Hankel map \begin{align*} \mathscr{H}: E \to S, \qquad \mathscr{H} := \mathscr{O}\mathscr{R}. \end{align*} For $(u_j)_{j=0}^{\infty} \in E$, its $i$th output component is \begin{align*} (\mathscr{H}(u))_i=\sum_{j=0}^{\infty} CA^{i+j}Bu_j. \end{align*} Thus $\mathscr{H}$ is exactly the block Hankel map with block entries $CA^{i+j}B$. Since $\mathscr{R}$ is surjective and $\mathscr{O}$ is injective, \begin{align*} \operatorname{rank}\mathscr{H}=\dim \mathscr{O}(X)=\dim X=n. \end{align*}[/step]

[guided]The purpose of this step is to turn reachability and observability into one rank statement. We define the input-sequence space \begin{align*} E := \{(u_j)_{j=0}^{\infty} : u_j \in U \text{ for every } j,\ u_j=0 \text{ for all but finitely many } j\} \end{align*} and the output-sequence space \begin{align*} S := \{(y_i)_{i=0}^{\infty} : y_i \in Y \text{ for every } i\}. \end{align*} The finite-support condition on $E$ ensures that all sums below are finite algebraic sums, so no convergence issue is involved. Define \begin{align*} \mathscr{R}: E \to X, \qquad \mathscr{R}((u_j)_{j=0}^{\infty}) := \sum_{j=0}^{\infty} A^jBu_j. \end{align*} This map records all state directions produced by finitely many input impulses. Its range is precisely \begin{align*} \operatorname{span}\{A^jBu : j \in \mathbb{N}\cup\{0\},\ u \in U\}=\mathcal{R}(A,B). \end{align*} Since the realization is reachable, $\mathcal{R}(A,B)=X$, so $\mathscr{R}$ is surjective. Next define \begin{align*} \mathscr{O}: X \to S, \qquad \mathscr{O}(x) := (CA^i x)_{i=0}^{\infty}. \end{align*} This map sends an initial state $x$ to its entire future output history under zero input. Its kernel is \begin{align*} \ker \mathscr{O}=\{x \in X : CA^i x=0 \text{ for every } i \in \mathbb{N}\cup\{0\}\}=\mathcal{N}(C,A). \end{align*} Since the realization is observable, $\mathcal{N}(C,A)=\{0\}$, so $\mathscr{O}$ is injective. Now define \begin{align*} \mathscr{H}: E \to S, \qquad \mathscr{H}:=\mathscr{O}\mathscr{R}. \end{align*} For $u=(u_j)_{j=0}^{\infty}\in E$, the $i$th component of $\mathscr{H}(u)$ is \begin{align*} (\mathscr{H}(u))_i=C A^i\left(\sum_{j=0}^{\infty} A^jBu_j\right). \end{align*} By linearity of $A^i$ and $C$, this becomes \begin{align*} (\mathscr{H}(u))_i=\sum_{j=0}^{\infty} CA^{i+j}Bu_j. \end{align*} Thus the matrix of $\mathscr{H}$ is the infinite block Hankel matrix whose $(i,j)$ block is $CA^{i+j}B$. Because $\mathscr{R}$ is surjective, the image of $\mathscr{H}=\mathscr{O}\mathscr{R}$ is exactly $\mathscr{O}(X)$. Because $\mathscr{O}$ is injective, $\dim \mathscr{O}(X)=\dim X=n$. Therefore \begin{align*} \operatorname{rank}\mathscr{H}=n. \end{align*} This is the core minimality mechanism: reachable states supply $n$ independent state directions, and observability prevents two distinct state directions from producing the same output history.[/guided]

[step:Compare every competing realization through the same Hankel map] Let \begin{align*} (\hat A,\hat B,\hat C,\hat D) \in \mathbb{R}^{\hat n \times \hat n} \times \mathbb{R}^{\hat n \times m} \times \mathbb{R}^{p \times \hat n} \times \mathbb{R}^{p \times m} \end{align*} be any realization with the same transfer function as $(A,B,C,D)$. Let $\hat X := \mathbb{R}^{\hat n}$. Define \begin{align*} \widehat{\mathscr{R}}: E \to \hat X, \qquad \widehat{\mathscr{R}}((u_j)_{j=0}^{\infty}) := \sum_{j=0}^{\infty} \hat A^j\hat B u_j \end{align*} and \begin{align*} \widehat{\mathscr{O}}: \hat X \to S, \qquad \widehat{\mathscr{O}}(\hat x) := (\hat C\hat A^i\hat x)_{i=0}^{\infty}. \end{align*} The first step gives \begin{align*} CA^kB=\hat C\hat A^k\hat B \end{align*} for every $k \in \mathbb{N}\cup\{0\}$. Therefore, for every $u=(u_j)_{j=0}^{\infty}\in E$ and every $i \in \mathbb{N}\cup\{0\}$, \begin{align*} (\mathscr{H}(u))_i=\sum_{j=0}^{\infty} CA^{i+j}Bu_j=\sum_{j=0}^{\infty} \hat C\hat A^{i+j}\hat B u_j=(\widehat{\mathscr{O}}\widehat{\mathscr{R}}(u))_i. \end{align*} Hence \begin{align*} \mathscr{H}=\widehat{\mathscr{O}}\widehat{\mathscr{R}}. \end{align*} Since this factorization passes through the vector space $\hat X$, its rank is at most $\dim \hat X=\hat n$: \begin{align*} \operatorname{rank}\mathscr{H}\leq \hat n. \end{align*} From the previous step, $\operatorname{rank}\mathscr{H}=n$. Therefore \begin{align*} n\leq \hat n. \end{align*} Thus no realization with the same transfer function has state dimension smaller than $n$. [/step]

[step:Conclude the equivalence] If $(A,B,C,D)$ is not reachable or not observable, the preceding reduction steps construct a realization of the same transfer function with strictly smaller state dimension, so $(A,B,C,D)$ is not minimal. Conversely, if $\mathcal{R}(A,B)=\mathbb{R}^n$ and $\mathcal{N}(C,A)=\{0\}$, then every realization of the same transfer function has state dimension at least $n$. Hence $(A,B,C,D)$ is minimal. This proves the equivalence. [/step]