[proofplan] The proof is a finite-dimensional subspace decomposition argument. We first record the invariance properties of the reachable and unobservable subspaces, then choose complements so that the state space splits into reachable-observable, reachable-unobservable, unreachable-observable, and unreachable-unobservable parts. The block zeros follow directly from the two invariant subspaces $\mathcal R(A,B)$ and $\mathcal N(A,C)$ and from the containments $\operatorname{im} B \subset \mathcal R(A,B)$ and $\mathcal N(A,C) \subset \ker C$. Finally, the reachable and observable properties of the first diagonal subsystem are proved on the quotient $\mathcal R(A,B)/(\mathcal R(A,B)\cap\mathcal N(A,C))$. [/proofplan]

[step:Verify the invariant subspaces that force the Kalman block structure]Let \begin{align*} R := \mathcal R(A,B) \subset X, \qquad N := \mathcal N(A,C) \subset X. \end{align*} Here $m,p\in\mathbb N$ are the input and output dimensions, $B:\mathbb R^m\to X$ is the input map, and $C:X\to \mathbb R^p$ is the output map from the realization. By the definition of the reachable subspace, \begin{align*} R = \operatorname{span}\{A^kBu : k \geq 0,\ u \in \mathbb R^m\}. \end{align*} Thus $\operatorname{im} B \subset R$. Also $AR \subset R$, because each spanning vector $A^kBu$ satisfies $A(A^kBu)=A^{k+1}Bu\in R$. Similarly, by the definition of the unobservable subspace, \begin{align*} N = \bigcap_{k=0}^{\infty}\ker(CA^k). \end{align*} Since the term $k=0$ occurs in the intersection, $N \subset \ker C$. We also have $AN \subset N$. If $x \in N$, then for every integer $k\geq 0$, \begin{align*} CA^k(Ax)=CA^{k+1}x=0. \end{align*} Therefore $Ax \in N$.[/step]

[guided]We need only two structural facts about the realization: the reachable subspace is invariant under the state map $A$, and the unobservable subspace is invariant under $A$ and killed by $C$. Define $R := \mathcal R(A,B)$ and $N := \mathcal N(A,C)$. The input and output dimensions are $m,p\in\mathbb N$, and the input and output maps in the realization are $B:\mathbb R^m\to X$ and $C:X\to \mathbb R^p$. The reachable subspace is \begin{align*} R = \operatorname{span}\{A^kBu : k \geq 0,\ u \in \mathbb R^m\}. \end{align*} Taking $k=0$ gives $Bu \in R$ for every $u \in \mathbb R^m$, so $\operatorname{im} B \subset R$. Next we verify $AR \subset R$. It is enough to check the spanning vectors $A^kBu$ with $k\geq 0$. For each such vector, \begin{align*} A(A^kBu)=A^{k+1}Bu \in R, \end{align*} because $k+1\geq 0$ and the vector $A^{k+1}Bu$ is again one of the generators used to define $R$. This proves $AR \subset R$ without needing any finite truncation of the spanning set. For the unobservable subspace, \begin{align*} N = \bigcap_{k=0}^{\infty}\ker(CA^k). \end{align*} Because $k=0$ is included, every $x \in N$ satisfies $Cx=0$, so $N \subset \ker C$. It remains to prove $AN \subset N$. Let $x \in N$. To show $Ax \in N$, we must prove $CA^k(Ax)=0$ for every integer $k\geq 0$. For every such $k$, \begin{align*} CA^k(Ax)=CA^{k+1}x=0, \end{align*} because $k+1\geq 0$ and $x\in N$ means $CA^j x=0$ for every integer $j\geq 0$. Thus $Ax \in N$, proving $AN \subset N$.[/guided]

[step:Choose complements adapted to reachability and observability] Let $X_{r\bar o} := R \cap N$. Choose a subspace $X_{ro} \subset R$ such that \begin{align*} R = X_{ro} \oplus X_{r\bar o}. \end{align*} Choose a subspace $X_{\bar r\bar o} \subset N$ such that \begin{align*} N = X_{r\bar o} \oplus X_{\bar r\bar o}. \end{align*} Since $X_{ro} \cap N=\{0\}$ and $X_{\bar r\bar o} \cap R=\{0\}$, the sum \begin{align*} X_{ro} \oplus X_{r\bar o} \oplus X_{\bar r\bar o} \end{align*} is direct. Choose a complement $X_{\bar r o}$ of this direct sum in $X$, so that \begin{align*} X = X_{ro} \oplus X_{r\bar o} \oplus X_{\bar r o} \oplus X_{\bar r\bar o}. \end{align*} This decomposition satisfies \begin{align*} R = X_{ro} \oplus X_{r\bar o}, \qquad N = X_{r\bar o} \oplus X_{\bar r\bar o}. \end{align*} [/step]

[step:Write the transformed matrices in the adapted basis] Choose an ordered basis of $X$ by first taking a basis of $X_{ro}$, then a basis of $X_{r\bar o}$, then a basis of $X_{\bar r o}$, and finally a basis of $X_{\bar r\bar o}$. Let $T \in \mathbb R^{n \times n}$ be the matrix whose columns are these basis vectors. Since the chosen vectors form a basis of $X=\mathbb R^n$, the matrix $T$ is invertible. For $z \in \mathbb R^n$, write $x=Tz$. Then define the transformed realization by $\tilde A := T^{-1}AT$, $\tilde B := T^{-1}B$, $\tilde C := CT$, and $\tilde D := D$. The coordinate groups of $z$ correspond respectively to $X_{ro}$, $X_{r\bar o}$, $X_{\bar r o}$, and $X_{\bar r\bar o}$. [/step]

[step:Use reachability invariance to obtain the zero input blocks and lower-left zero blocks of $\tilde A$] Since $\operatorname{im} B \subset R=X_{ro}\oplus X_{r\bar o}$, the components of $Bu$ in $X_{\bar r o}$ and $X_{\bar r\bar o}$ are zero for every $u \in \mathbb R^m$. Therefore the third and fourth block rows of $\tilde B=T^{-1}B$ vanish. Equivalently, \begin{align*} \tilde B_1=B_{ro}, \quad \tilde B_2=B_{r\bar o}, \quad \tilde B_3=0, \quad \tilde B_4=0. \end{align*} Since $AR \subset R$, the image under $A$ of every vector in $X_{ro}\oplus X_{r\bar o}$ has no component in $X_{\bar r o}$ or $X_{\bar r\bar o}$. Equivalently, in the first two block columns of $\tilde A$, the third and fourth block rows vanish: \begin{align*} \tilde A_{31}=0, \qquad \tilde A_{32}=0, \qquad \tilde A_{41}=0, \qquad \tilde A_{42}=0. \end{align*} [/step]

[step:Use unobservability invariance to obtain the zero output columns and forbidden observable couplings] Since $N=X_{r\bar o}\oplus X_{\bar r\bar o}$ and $N \subset \ker C$, the second and fourth block columns of $\tilde C=CT$ vanish: \begin{align*} \tilde C = \begin{pmatrix} C_{ro} & 0 & C_{\bar r o} & 0 \end{pmatrix}. \end{align*} Since $AN \subset N$, the image under $A$ of every vector in $X_{r\bar o}\oplus X_{\bar r\bar o}$ has no component in $X_{ro}$ or $X_{\bar r o}$. Equivalently, in the second and fourth block columns of $\tilde A$, the first and third block rows vanish: \begin{align*} \tilde A_{12}=0, \qquad \tilde A_{32}=0, \qquad \tilde A_{14}=0, \qquad \tilde A_{34}=0. \end{align*} Combining these zero constraints with the unrestricted blocks gives the block identities \begin{align*} \tilde A_{11}=A_{ro}, \quad \tilde A_{12}=0, \quad \tilde A_{13}=A_{13}, \quad \tilde A_{14}=0, \quad \tilde A_{21}=A_{21}, \quad \tilde A_{22}=A_{r\bar o}, \quad \tilde A_{23}=A_{23}, \quad \tilde A_{24}=A_{24}, \quad \tilde A_{31}=0, \quad \tilde A_{32}=0, \quad \tilde A_{33}=A_{\bar r o}, \quad \tilde A_{34}=0, \quad \tilde A_{41}=0, \quad \tilde A_{42}=0, \quad \tilde A_{43}=A_{43}, \quad \tilde A_{44}=A_{\bar r\bar o}. \end{align*} [/step]

[step:Identify the reachable-observable quotient subsystem]Let \begin{align*} Q := R/(R \cap N) \end{align*} and let \begin{align*} \pi: R \to Q \end{align*} be the quotient map. Since $AR \subset R$ and $A(R\cap N)\subset R\cap N$, the [linear map](/page/Linear%20Map) $A|_R:R\to R$ induces a linear map \begin{align*} A_Q: Q \to Q, \qquad A_Q(\pi x) := \pi(Ax). \end{align*} This definition is well-defined because if $\pi x=\pi y$, then $x-y\in R\cap N$, hence $A(x-y)\in R\cap N$, so $\pi(Ax)=\pi(Ay)$. The input map $B:\mathbb R^m\to X$ has image in $R$, so it induces \begin{align*} B_Q: \mathbb R^m \to Q, \qquad B_Q u := \pi(Bu). \end{align*} The output map on the quotient is \begin{align*} C_Q: Q \to \mathbb R^p, \qquad C_Q(\pi x):=Cx. \end{align*} This is well-defined because if $\pi x=\pi y$, then $x-y\in R\cap N\subset N\subset\ker C$, so $Cx=Cy$. Under the identification $Q \cong X_{ro}$ induced by the direct sum $R=X_{ro}\oplus X_{r\bar o}$, the maps $A_Q$, $B_Q$, and $C_Q$ are represented respectively by the blocks $A_{ro}$, $B_{ro}$, and $C_{ro}$. More explicitly, the chosen complement gives a section $s:Q\to R$ whose image is $X_{ro}$, and the matrix $A_{ro}$ represents the map $q\mapsto \pi(A s(q))$ on $Q$. Thus $A_{ro}$ is the quotient-induced state map in these coordinates; it need not be the restriction of $A$ to an $A$-invariant subspace $X_{ro}$.[/step]

[guided]The delicate point is that $X_{ro}$ is a chosen complement, not necessarily an invariant subspace for $A$. The quotient is the invariant object. Define \begin{align*} Q := R/(R \cap N) \end{align*} and let \begin{align*} \pi: R \to Q \end{align*} be the quotient map. Since $AR\subset R$ and $A(R\cap N)\subset R\cap N$, the formula \begin{align*} A_Q(\pi x) := \pi(Ax) \end{align*} defines a linear map $A_Q:Q\to Q$. This is well-defined because replacing $x$ by $y$ with $\pi x=\pi y$ changes $x$ by an element of $R\cap N$, and applying $A$ keeps that change inside $R\cap N$. The input map descends because $\operatorname{im}B\subset R$: define \begin{align*} B_Q:\mathbb R^m \to Q, \qquad B_Q u := \pi(Bu). \end{align*} The output map descends because $R\cap N\subset N\subset \ker C$: define \begin{align*} C_Q:Q\to \mathbb R^p, \qquad C_Q(\pi x):=Cx. \end{align*} If $\pi x=\pi y$, then $x-y\in R\cap N\subset\ker C$, so $Cx=Cy$. Now the complement $R=X_{ro}\oplus X_{r\bar o}$ gives a section $s:Q\to R$ by selecting the unique representative in $X_{ro}$. With respect to the basis of $X_{ro}$, the state map of the quotient is represented by \begin{align*} q \mapsto \pi(A s(q)), \end{align*} not by the unrestricted action $x\mapsto Ax$ on $X_{ro}$. This distinction matters because $A(X_{ro})$ may have an $X_{r\bar o}$ component. That component vanishes after applying $\pi$, and the remaining $X_{ro}$ component is precisely the block denoted $A_{ro}$. The same chosen section identifies $B_Q$ and $C_Q$ with the coordinate blocks $B_{ro}$ and $C_{ro}$.[/guided]

[step:Prove the quotient subsystem is reachable and observable] The quotient subsystem $(A_Q,B_Q,C_Q,D)$ is reachable. Let \begin{align*} d := \dim Q. \end{align*} First, using the all-time reachable subspace, the definitions of $A_Q$ and $B_Q$ give \begin{align*} A_Q^kB_Qu = \pi(A^kBu) \end{align*} for every integer $k\geq 0$ and every $u\in\mathbb R^m$. Since $R$ is spanned by the vectors $A^kBu$ with $k\geq 0$, their images span $\pi(R)=Q$. It remains to reduce this to the standard $d$-dimensional reachability test. If $d=0$, then $Q=\{0\}$ and reachability is vacuous after deleting the zero-dimensional block. Assume now that $d\geq 1$. Since $Q$ is finite-dimensional with $d=\dim Q$, define the characteristic polynomial map $p_Q:\mathbb R\to\mathbb R$ of the endomorphism $A_Q:Q\to Q$ by \begin{align*} p_Q(\lambda):=\det(\lambda I_Q-A_Q), \end{align*} where $I_Q:Q\to Q$ is the identity map. Write \begin{align*} p_Q(\lambda)=\lambda^d+a_{d-1}\lambda^{d-1}+\cdots+a_1\lambda+a_0 \end{align*} for real coefficients $a_0,\dots,a_{d-1}\in\mathbb R$. By the [Cayley-Hamilton theorem](/theorems/865) for the finite-dimensional endomorphism $A_Q:Q\to Q$, we have \begin{align*} A_Q^d+a_{d-1}A_Q^{d-1}+\cdots+a_1A_Q+a_0I_Q=0. \end{align*} Multiplying this operator identity on the left by $A_Q^\ell$ for any integer $\ell\geq 0$ gives the explicit recurrence \begin{align*} A_Q^{d+\ell}=-a_{d-1}A_Q^{d-1+\ell}-\cdots-a_1A_Q^{1+\ell}-a_0A_Q^\ell. \end{align*} Induction on $k\geq d$ therefore expresses every $A_Q^k$ as a linear combination of $I_Q,A_Q,\dots,A_Q^{d-1}$. Hence the span of all vectors $A_Q^kB_Qu$ with $k\geq 0$ is already equal to \begin{align*} \operatorname{span}\{A_Q^kB_Qu : 0 \leq k \leq d-1,\ u \in \mathbb R^m\}. \end{align*} Therefore this finite span is $Q$. By definition, the quotient subsystem is reachable. The quotient subsystem is observable. If $d=0$, then $Q=\{0\}$ and observability is vacuous after deleting the zero-dimensional block. Assume now that $d\geq 1$. Suppose $q \in Q$ satisfies \begin{align*} C_QA_Q^k q = 0 \end{align*} for every integer $k$ with $0 \leq k \leq d-1$. Choose $x \in R$ with $q=\pi x$. Since $A_Q^k\pi x=\pi(A^kx)$ and $C_Q\pi(A^kx)=CA^kx$, the displayed equalities give \begin{align*} CA^k x=0 \end{align*} for $0 \leq k \leq \dim Q-1$. By the explicit Cayley-Hamilton recurrence just obtained from $p_Q(A_Q)=0$, each $A_Q^k$ with $k\geq d$ is a linear combination of $I_Q,A_Q,\dots,A_Q^{d-1}$. Therefore the vanishing of $C_QA_Q^kq$ for $0 \leq k \leq d-1$ implies the same vanishing for every $k \geq 0$. Thus $CA^kx=0$ for every $k \geq 0$, and hence $x \in N$. Since also $x \in R$, we have $x \in R\cap N$, so $q=\pi x=0$. Thus the only state $q\in Q$ whose output sequence $C_QA_Q^kq$ vanishes for every integer $k\geq 0$ is $0$. By definition, this proves observability. Under the chosen complement $Q\cong X_{ro}$, this reachable and observable quotient subsystem is represented by $(A_{ro},B_{ro},C_{ro},D)$ in the quotient sense described above: $A_{ro}$ is the coordinate matrix of $q\mapsto \pi(A s(q))$, where $s:Q\to R$ is the section with image $X_{ro}$. Omitting any zero-dimensional blocks gives exactly the asserted Kalman decomposition. [/step]