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Dynamical observations and finite system realization

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Discrete linear input-state-output experiments

Definition

Let $d \geq 0$ , $m, p \geq 1$ , and $q \geq 1$ be integers, and let $h > 0$ be a fixed sampling interval. A real discrete linear input-state-output experiment of state dimension $d$ , input dimension $m$ , output dimension $p$ , and horizon $q$ consists of matrices $A \in M_{d \times d} (R)$ , $B \in M_{d \times m} (R)$ , and $C \in M_{p \times d} (R)$ , an initial state $x_{0} \in R^{d}$ , and known inputs $(u_{k})_{0 \leq k < q - 1}$ in $R^{m}$ . The states are defined by $x_{k + 1} = A x_{k} + B u_{k} (0 \leq k < q - 1),$ and the exact sampled outputs at times $k h$ are $y_{k} = C x_{k}$ for $0 \leq k < q$ . When observations are perturbed, the observed outputs are ${\tilde{y}}_{k} = y_{k} + η_{k}$ for deterministic error vectors $η_{k} \in R^{p}$ ; the exact-data convention is $η_{k} = 0$ for all $k$ . No probability law is assigned to the errors. A numerical or discretization error that changes the forward map is distinct from this additive observation error.

For the fixed matrices $A, B, C$ , the Markov parameters are $M_{k} : = C A^{k} B \in M_{p \times m} (R) (k \geq 0),$ interpreted using the infinite-horizon impulse experiment: extend the same state and output equations to all $k \geq 0$ , with zero initial state, $u_{0} = v \in R^{m}$ , and $u_{k} = 0$ for $k \geq 1$ . Its output at sample $k + 1$ is $y_{k + 1} = M_{k} v$ for every $k \geq 0$ . The horizon- $q$ restriction records this identity only for $0 \leq k < q - 1$ ; when $q = 1$ , it records only $y_{0} = 0$ and has no input samples. Matrix powers use $A^{0} = I_{d}$ and $A^{k + 1} = A^{k} A$ . If $d = 0$ , the state space consists of its single zero vector and the matrices have the corresponding empty shapes.

Definition references

Finite real matrix spaces

Matrix products, powers, and identities

Standard Euclidean inner products

Dahleh, Dahleh, and Verghese, Lectures on Dynamic Systems and Control, Chapter 25

Lall, State-Space Realization, Engr207c lecture notes

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Exact block-Hankel realization

Definition

Fix integers $d \geq 0$ , $m, p \geq 1$ , and $q \geq 1$ . The input is a list of exact Markov matrices $M_{0}, \dots, M_{2 q - 1} \in M_{p \times m} (R) .$ Use the exact-real linear-algebra model: an exact rank test returns the rank of a finite real matrix; a thin rank- $d$ singular-value decomposition of a rank- $d$ matrix $H_{0}$ returns $H_{0} = U_{d} Σ_{d} V_{d}^{T}$ with $Σ_{d}$ diagonal and strictly positive, and an exact Moore--Penrose routine returns a matrix satisfying the four Penrose equations. A routine failure or a returned object that fails its stated exact identities is reported explicitly. This specification makes no finite-bit computability or complexity claim for arbitrary real inputs.

Form the $q$ -by- $q$ block Hankel matrices $H_{0} = [M_{i + j}]_{0 \leq i, j < q}, H_{1} = [M_{i + j + 1}]_{0 \leq i, j < q},$ each of shape $q p$ by $q m$ . Execute the following steps.

Apply the exact rank test to $H_{0}$ . If it fails, return failed-solve; if its value is not $d$ , return inconsistent-rank-data.
If $d = 0$ , use the empty factorization $H_{0} = \hat{O} \hat{R}$ with $\hat{O}$ of shape $q p$ by $0$ and $\hat{R}$ of shape $0$ by $q m$ . If $d > 0$ , obtain the thin rank- $d$ SVD in the declared exact model and set $\hat{O} : = U_{d} Σ_{d}^{1 / 2}, \hat{R} : = Σ_{d}^{1 / 2} V_{d}^{T} .$ Verify $H_{0} = \hat{O} \hat{R}$ , $rank \hat{O} = d$ , and $rank \hat{R} = d$ ; if any operation or verification fails, return failed-solve.
Compute the exact Moore--Penrose inverses ${\hat{O}}^{+}$ and ${\hat{R}}^{+}$ and verify their four Penrose equations. If a computation or verification fails, return failed-solve.
Set $\hat{A} : = {\hat{O}}^{+} H_{1} {\hat{R}}^{+}$ , let $\hat{B}$ be the first $m$ columns of $\hat{R}$ , and let $\hat{C}$ be the first $p$ rows of $\hat{O}$ . Return success with $(\hat{A}, \hat{B}, \hat{C})$ . For $d = 0$ , these are the corresponding empty matrices.

The routine can report failed-solve for a failed or invalid exact-operation backend. The existence theorems for finite-matrix SVDs and Moore--Penrose inverses ensure that the exact mathematical operations requested in steps 2–3 exist; this definition alone asserts no noisy rank selection or stable parameter recovery.

Definition references

Markov parameter convention $M_{k} = C A^{k} B$

Rank of a finite-dimensional linear map

Finite-dimensional singular-value decomposition

Four Penrose equations

Existence and uniqueness of the finite-matrix pseudoinverse

Finite matrix product laws

Dahleh, Dahleh, and Verghese, Lectures on Dynamic Systems and Control, Chapter 25

Lall, State-Space Realization, Engr207c lecture notes

TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite observability identifies the initial state

Statement

Let $d \geq 0$ , $m, p \geq 1$ , and $q \geq 1$ be integers, and consider a real discrete input-state-output experiment with known matrices $A, B, C$ , known inputs $u_{0}, \dots, u_{q - 2}$ , and unknown initial state $x_{0}$ . For $0 \leq k < q$ , set $f_{0} : = 0, f_{k} : = \sum_{j = 0}^{k - 1} C A^{k - 1 - j} B u_{j} (k \geq 1),$ and define the finite observability matrix $O_{q} : = [\begin{array}{c} C \\ C A \\ ⋮ \\ C A^{q - 1} \end{array}] .$ The corrected noiseless output stack is $z : = [\begin{array}{c} y_{0} - f_{0} \\ y_{1} - f_{1} \\ ⋮ \\ y_{q - 1} - f_{q - 1} \end{array}] = O_{q} x_{0} .$ The map $x_{0} \mapsto z$ is injective if and only if $rank O_{q} = d$ . If $d = 0$ , both injectivity and the rank condition hold for the singleton state space. If $d \geq 1$ and $O_{q}$ has full column rank, let $σ_{\min} (O_{q}) > 0$ be its least singular value. For any additive stacked observation error $e \in R^{q p}$ , the minimum-norm least-squares estimate of $x_{0}$ from $z + e$ is ${\hat{x}}_{0} = O_{q}^{+} (z + e)$ , and it satisfies $∥ {\hat{x}}_{0} - x_{0} ∥_{2} \leq \frac{∥ e ∥_{2}}{σ_{\min} (O_{q})} .$

Statement references

Finite-horizon recurrence and exact observation convention

Rank as the dimension of the image

Rank-nullity for finite-dimensional domains

Finite-dimensional singular-value decomposition

Rank equals the number of positive singular values

Finite-matrix Moore--Penrose pseudoinverse

Existence and uniqueness of the pseudoinverse

Full-column-rank pseudoinverse identity

The product $O O^{+}$ is an orthogonal projection

The pseudoinverse gives the minimum-norm least-squares solution

Pythagoras for orthogonal vectors

Dahleh, Dahleh, and Verghese, Lectures on Dynamic Systems and Control, Chapter 25

Lall, State-Space Realization, Engr207c lecture notes

Facts & Assumptions

Given: A real discrete experiment of horizon $q$ with known matrices $A, B, C$ , known inputs $u_{0}, \dots, u_{q - 2}$ , unknown initial state $x_{0}$ , and a stacked additive error $e \in R^{q p}$ .

[F1]

Discrete linear input-state-output experiments: states obey $x_{k + 1} = A x_{k} + B u_{k}$ , outputs obey $y_{k} = C x_{k}$ , and observation errors are additive deterministic vectors.

[F2]

Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication: compatible matrix products are associative, identity matrices act on either side, and multiplication distributes over finite sums.

[F3]

Rank and nullity of a linear map with finite-dimensional domain: $rank T = \dim (im T)$ and $nullity T = \dim (\ker T)$ for finite-dimensional domains.

[F4]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a linear map with finite-dimensional domain, $\dim V = nullity T + rank T$ .

[F5]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: a linear map between finite-dimensional real or complex inner-product spaces has orthonormal left and right singular bases, with its positive singular values ordered nonincreasingly.

[F6]

The rank of a linear map is the number of its nonzero singular values: the rank equals the number of positive singular values.

[F7]

The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations: $O^{+}$ satisfies $O O^{+} O = O$ , $O^{+} O O^{+} = O^{+}$ , and $O O^{+}$ and $O^{+} O$ are self-adjoint.

[F8]

Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse: every finite real matrix has a unique Moore--Penrose pseudoinverse.

[F9]

If $A$ has full column rank, then $A^{+} = (A^{*} A)^{- 1} A^{*}$ : if $O$ has full column rank, then $O^{+} O = I_{d}$ .

[F10]

$A A^{+}$ and $A^{+} A$ are the orthogonal projections onto $im A$ and $im A^{*}$ : $O O^{+}$ is the orthogonal projection onto $im O$ .

[F11]

For every right-hand side $b$ , $A^{+} b$ is the unique least-squares solution of minimum Euclidean norm: for a finite real matrix $O$ and vector $b$ , $O^{+} b$ is the unique least-squares solution of minimum Euclidean norm.

[F12]

Pythagoras and finite orthogonal sums: if $a$ and $b$ are orthogonal, then $∥ a + b ∥_{2}^{2} = ∥ a ∥_{2}^{2} + ∥ b ∥_{2}^{2}$ .

Proof

technique · direct

1.1givenF1F2algebra

For each $0 \leq k < q$ , the recurrence gives $x_{k} = A^{k} x_{0} + \sum_{j = 0}^{k - 1} A^{k - 1 - j} B u_{j}$ , with the sum empty at $k = 0$ . At $k = 0$ this is $x_{0} = A^{0} x_{0}$ . If it holds at $k < q - 1$ , substitution into $x_{k + 1} = A x_{k} + B u_{k}$ and the power recursion give the formula at $k + 1$ , using associativity and distributivity from [F2]. Multiplying by $C$ and subtracting $f_{k}$ gives $y_{k} - f_{k} = C A^{k} x_{0}$ , so stacking yields $z = O_{q} x_{0}$ .

1.2givenF5F6algebra

Suppose $d \geq 1$ and $O_{q}$ has full column rank. By [F5] and [F6], an SVD has exactly $d$ positive singular values $s_{1} \geq \dots \geq s_{d} = σ_{\min} (O_{q}) > 0$ . Writing $w$ in the right singular basis gives $∥ O_{q} w ∥_{2}^{2} = \sum_{j = 1}^{d} s_{j}^{2} ∣ w_{j} ∣^{2} \geq σ_{\min} (O_{q})^{2} ∥ w ∥_{2}^{2}$ .

2.1givenF3F4step 1.1algebra

Two states $x, x^{'} \in R^{d}$ produce the same corrected stack exactly when $O_{q} (x - x^{'}) = 0$ . Thus injectivity is equivalent to $\ker O_{q} = {0}$ . By [F3] and [F4], this is equivalent to $rank O_{q} = d$ . When $d = 0$ , the map has a singleton domain and the empty matrix has rank zero, so the same equivalence holds.

2.2givenF7F8F9F10F11F12step 1.1algebra

By [F7] and [F8], $O_{q}^{+}$ is the unique Moore--Penrose inverse. Set ${\hat{x}}_{0} = O_{q}^{+} (z + e)$ ; [F11] identifies it as the minimum-norm least-squares estimate, and [F9] with $z = O_{q} x_{0}$ gives $w : = {\hat{x}}_{0} - x_{0} = O_{q}^{+} e$ . Hence $O_{q} w = O_{q} O_{q}^{+} e$ . By [F10], this is an orthogonal projection of $e$ ; the decomposition $e = O_{q} O_{q}^{+} e + (e - O_{q} O_{q}^{+} e)$ has orthogonal summands, so [F12] gives $∥ O_{q} w ∥_{2} \leq ∥ e ∥_{2}$ .

3.1step 1.2step 2.2step 2.1algebra∎

Combining steps 1.2 and 2.2 gives $σ_{\min} (O_{q}) ∥ {\hat{x}}_{0} - x_{0} ∥_{2} \leq ∥ O_{q} ({\hat{x}}_{0} - x_{0}) ∥_{2} \leq ∥ e ∥_{2}$ . Division by the positive singular value proves the claimed sensitivity bound; step 2.1 proves the injectivity assertion.

TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

State similarity preserves input-output data

Statement

Let $d \geq 0$ , $m, p \geq 1$ , and $q \geq 1$ be integers. Suppose a noiseless experiment has matrices $A \in M_{d \times d} (R)$ , $B \in M_{d \times m} (R)$ , and $C \in M_{p \times d} (R)$ , initial state $x_{0} \in R^{d}$ , and common inputs $(u_{k})_{0 \leq k < q - 1}$ . For any invertible $T \in M_{d} (R)$ define $\hat{A} = T^{- 1} A T, \hat{B} = T^{- 1} B, \hat{C} = C T, {\hat{x}}_{0} = T^{- 1} x_{0} .$ The experiment with $(\hat{A}, \hat{B}, \hat{C}, {\hat{x}}_{0})$ and the same inputs has the same output $y_{k}$ at every sample $0 \leq k < q$ . The equality holds for every horizon and every common input sequence.

Statement references

Discrete input-state-output recurrence

Invertible matrices and their inverses

Associativity and identity laws for matrix products

Dahleh, Dahleh, and Verghese, Lectures on Dynamic Systems and Control, §25.3

Lall, State-Space Realization, §9.2

Facts & Assumptions

Given: A noiseless real discrete experiment of horizon $q$ , matrices $A, B, C$ , initial state $x_{0}$ , common inputs $u_{0}, \dots, u_{q - 2}$ , and an invertible matrix $T \in M_{d} (R)$ .

[F1]

Discrete linear input-state-output experiments: the state and output equations are $x_{k + 1} = A x_{k} + B u_{k}$ and $y_{k} = C x_{k}$ at the stated samples.

[F2]

Invertible matrices and the general linear group ${GL}_{n} (F)$ : an invertible $T$ has a two-sided inverse $T^{- 1}$ , so $T T^{- 1} = T^{- 1} T = I_{d}$ .

[F3]

Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication: conformable matrix products are associative and identity matrices act on either side.

Proof

1.1givenF1F2F3algebra

At $k = 0$ , the transformed initial state is ${\hat{x}}_{0} = T^{- 1} x_{0}$ . If ${\hat{x}}_{k} = T^{- 1} x_{k}$ for $k < q - 1$ , then [F1] and the definitions give ${\hat{x}}_{k + 1} = \hat{A} {\hat{x}}_{k} + \hat{B} u_{k} = T^{- 1} A T T^{- 1} x_{k} + T^{- 1} B u_{k} = T^{- 1} (A x_{k} + B u_{k}) = T^{- 1} x_{k + 1}$ ; [F2] cancels $T T^{- 1}$ and [F3] permits regrouping. Induction proves ${\hat{x}}_{k} = T^{- 1} x_{k}$ for $0 \leq k < q$ .

2.1givenF1F2step 1.1algebra∎

For every such $k$ , the transformed output is ${\hat{y}}_{k} = \hat{C} {\hat{x}}_{k} = C T T^{- 1} x_{k} = C x_{k} = y_{k}$ by [F1] and [F2]. Since the horizon, initial state, and inputs were arbitrary, this holds for every common finite input experiment.

Coordinate diagram

The square records the state-coordinate identity $T^{- 1} A = \hat{A} T^{- 1}$ : both paths send $x$ to $T^{- 1} A x$ . The proof above establishes the output identity by induction and does not rely on the diagram.

TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Hankel realization recovers a controllable and observable system

Statement

Let $d \geq 0$ , $m, p \geq 1$ , and $q \geq 1$ be integers, and let $A \in M_{d \times d} (R)$ , $B \in M_{d \times m} (R)$ , and $C \in M_{p \times d} (R)$ define a discrete linear input-state-output experiment. Let $M_{k} : = C A^{k} B (0 \leq k \leq 2 q - 1),$ $O_{q} : = [\begin{array}{c} C \\ C A \\ ⋮ \\ C A^{q - 1} \end{array}], R_{q} : = [\begin{array}{cccc} B & A B & \dots & A^{q - 1} B \end{array}] .$ Assume $rank O_{q} = rank R_{q} = d$ , the finite-horizon observability and controllability conditions. Assume that every exact rank, SVD, and Moore--Penrose call made by the block-Hankel realization algorithm completes and passes its stated exact checks. Then the algorithm returns success with matrices $(\hat{A}, \hat{B}, \hat{C})$ , and there is an invertible $T \in M_{d} (R)$ such that $\hat{A} = T^{- 1} A T, \hat{B} = T^{- 1} B, \hat{C} = C T .$ For $d = 0$ , this means the unique zero-dimensional realization and the unique invertible map on the zero space.

Statement references

Markov parameters and discrete realization convention

Exact block-Hankel realization algorithm

Rank of finite-dimensional maps

Rank-nullity

Matrix multiplication identities

Invertible matrices

Finite-dimensional singular-value decomposition

Moore--Penrose pseudoinverse equations

Exact finite-matrix pseudoinverse

Input-output invariance under state similarity

Dahleh, Dahleh, and Verghese, Lectures on Dynamic Systems and Control, §25.4

Lall, State-Space Realization, Engr207c lecture notes, §9.2

Facts & Assumptions

Given: The real matrices $A, B, C$ , a horizon $q \geq 1$ , Markov parameters $M_{k} = C A^{k} B$ for $0 \leq k \leq 2 q - 1$ , full-rank matrices $O_{q}, R_{q}$ as in the statement, and exact rank, SVD, and Moore--Penrose calls that complete and pass the algorithm's stated checks.

[F1]

Discrete linear input-state-output experiments: the Markov parameter with index $k$ is $M_{k} = C A^{k} B$ , and the impulse output at sample $k + 1$ is $M_{k} v$ .

[F2]

Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication: matrix multiplication is associative and unital, including conformable rectangular and zero-sized shapes.

[F3]

Rank and nullity of a linear map with finite-dimensional domain: rank is the dimension of the image and nullity is the dimension of the kernel for a finite-dimensional domain.

[F4]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a finite-dimensional domain, rank plus nullity equals the domain dimension.

[F5]

Exact block-Hankel realization: the algorithm forms $H_{0} = [M_{i + j}]$ and $H_{1} = [M_{i + j + 1}]$ , uses a rank- $d$ factorization $H_{0} = \hat{O} \hat{R}$ with both factors of rank $d$ , and returns $\hat{A} = {\hat{O}}^{+} H_{1} {\hat{R}}^{+}$ , the first input block of $\hat{R}$ , and the first output block of $\hat{O}$ .

[F6]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition and Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse: every finite real matrix has an exact finite SVD and a unique Moore--Penrose pseudoinverse.

[F7]

The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations: for a matrix $X$ , $X X^{+} X = X$ and $X^{+} X X^{+} = X^{+}$ ; the two square products are self-adjoint.

[F8]

Invertible matrices and the general linear group ${GL}_{n} (F)$ : an invertible square matrix has a two-sided inverse; $I_{0}$ is the inverse of the unique zero-sized matrix.

[F9]

State similarity preserves input-output data: if $(\hat{A}, \hat{B}, \hat{C}) = (T^{- 1} A T, T^{- 1} B, C T)$ and initial states are related by ${\hat{x}}_{0} = T^{- 1} x_{0}$ , common inputs give identical outputs at every sample.

Proof

technique · direct

1.1givenF1F5F6F8algebra

If $d = 0$ , then $B$ has no rows and $C$ has no columns, so $M_{k} = 0$ for every $k$ and $H_{0} = H_{1} = 0$ . Its rank is zero, and the algorithm's exact empty factorization and empty pseudoinverses return the unique empty matrices. Taking $T = I_{0}$ gives the asserted similarity. Henceforth suppose $d \geq 1$ .

1.2givenF1F2F3algebra

For $0 \leq i, j < q$ , the $(i, j)$ block of $H_{0}$ is $M_{i + j} = C A^{i + j} B = (C A^{i}) (A^{j} B)$ by [F1] and [F2], and the corresponding block of $H_{1}$ is $M_{i + j + 1} = (C A^{i}) A (A^{j} B)$ . Thus $H_{0} = O_{q} R_{q}$ and $H_{1} = O_{q} A R_{q}$ . Because $R_{q}$ has rank $d$ and codomain $R^{d}$ , it is surjective; because $O_{q}$ has rank $d$ and domain $R^{d}$ , it is injective. Therefore $im H_{0} = im O_{q}$ and $rank H_{0} = d$ .

2.1givenF5F6step 1.2

The exact rank test in the algorithm therefore returns $d$ . The SVD factorization and pseudoinverses exist by [F6], and the stated successful-call hypothesis ensures that the algorithm returns factors $H_{0} = \hat{O} \hat{R}$ with each factor of rank $d$ and returns success. In particular $\hat{R}$ is surjective, so $im H_{0} = im \hat{O}$ . Combining this with step 1.2 gives $im \hat{O} = im O_{q}$ .

3.1givenF3F4F5F8step 2.1algebra

The columns of $O_{q}$ and $\hat{O}$ are bases of the same $d$ -dimensional image by step 2.1. Thus there is a unique $d \times d$ matrix $T$ with $\hat{O} = O_{q} T$ . If $T v = 0$ , then $\hat{O} v = O_{q} T v = 0$ ; since $\hat{O}$ has rank $d$ , its kernel is zero, so $v = 0$ . Rank-nullity [F3] and [F4] makes the square matrix $T$ surjective as well as injective; hence $T$ is invertible by [F8]. From $O_{q} R_{q} = H_{0} = O_{q} T \hat{R}$ and injectivity of $O_{q}$ , $R_{q} = T \hat{R}$ , so $\hat{R} = T^{- 1} R_{q}$ .

4.1F2F5F7step 1.2step 3.1algebra

Substituting $O_{q} = \hat{O} T^{- 1}$ and $R_{q} = T \hat{R}$ into the identity in step 1.2 gives $H_{1} = \hat{O} (T^{- 1} A T) \hat{R}$ . Since $\hat{O}$ has full column rank, ${\hat{O}}^{+} \hat{O} = I_{d}$ : the equation $\hat{O} {\hat{O}}^{+} \hat{O} = \hat{O}$ from [F7] and injectivity imply this identity. Since $\hat{R}$ has full row rank, $\hat{R} {\hat{R}}^{+} = I_{d}$ : [F7] gives $\hat{R} {\hat{R}}^{+} \hat{R} = \hat{R}$ , and surjectivity implies the identity. The algorithm's formula then yields $\hat{A} = {\hat{O}}^{+} H_{1} {\hat{R}}^{+} = T^{- 1} A T$ .

5.1F5step 3.1step 4.1algebra

The first $m$ columns of $R_{q}$ form $B$ , and the first $p$ rows of $O_{q}$ form $C$ . Taking those blocks in $\hat{R} = T^{- 1} R_{q}$ and $\hat{O} = O_{q} T$ gives $\hat{B} = T^{- 1} B$ and $\hat{C} = C T$ . Hence the triple returned by the algorithm satisfies all three similarity identities.

6.1F1F9step 5.1algebra∎

By [F9], the similarity identities imply identical outputs for every finite common-input experiment when initial states are related by $T^{- 1}$ . With zero initial state and an impulse input, this gives $\hat{C} {\hat{A}}^{k} \hat{B} = C A^{k} B$ for every $k \geq 0$ . This is exact realization modulo state similarity; it asserts no stability under perturbed Markov data.

4 · Examples, counterexamples and false statements

None yet.

Dynamical observations and finite system realization

2 · Logical flowchart

3 · Definitions, theorems and proofs

Discrete linear input-state-output experiments

Definition

Definition references

Exact block-Hankel realization

Definition

Definition references

Finite observability identifies the initial state

Statement

Statement references

Facts & Assumptions

Proof

State similarity preserves input-output data

Statement

Statement references

Facts & Assumptions

Proof

Coordinate diagram

Hankel realization recovers a controllable and observable system

Statement

Statement references

Facts & Assumptions

Proof

4 · Examples, counterexamples and false statements

Sources