ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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A small Hankel singular value amplifies a realization perturbation

Example

Use the exact block-Hankel realization procedure with $q = d = m = p = 1$ , and assume that every exact rank, SVD, and Moore--Penrose call made in both runs completes and passes its stated exact checks. For any $0 < ε \leq 1$ , the scalar system $A = 0$ , $B = ε$ , $C = 1$ has Markov data $M_{0} = ε$ and $M_{1} = 0$ , and $H_{0} = [ε]$ has singular value $ε$ . The exact algorithm returns $\hat{A} = 0$ . If only $M_{1}$ is perturbed to ${\tilde{M}}_{1} = \sqrt{ε}$ , the pair $(M_{0}, {\tilde{M}}_{1})$ is itself generated by the scalar system $\tilde{A} = 1 / \sqrt{ε}$ , $\tilde{B} = ε$ , $\tilde{C} = 1$ , so the exact algorithm returns $\tilde{A}$ . Thus an absolute Markov-data perturbation $\sqrt{ε}$ can change the recovered transition value by $1 / \sqrt{ε}$ .

Example references

Markov parameter convention

Exact block-Hankel procedure

Exact recovery under full finite ranks

Singular values as eigenvalues of the positive square root of $H_{0}^{T} H_{0}$

Facts & Assumptions

Given: $0 < ε \leq 1$ , the two scalar Markov-data pairs, the exact block-Hankel procedure with $q = d = m = p = 1$ , and exact rank, SVD, and Moore--Penrose calls in both runs that complete and pass the stated checks.

[F1]

Discrete linear input-state-output experiments: $M_{k} = C A^{k} B$ for scalar real state-space systems.

[F2]

The singular values of a linear map as the eigenvalues of the positive square root of T^*T: singular values of a finite-dimensional endomorphism are the eigenvalues of $∣ H_{0} ∣ = \sqrt{H_{0}^{T} H_{0}}$ .

[F3]

Exact block-Hankel realization: for $q = 1$ , $H_{0} = [M_{0}]$ , $H_{1} = [M_{1}]$ , and the returned transition is $\hat{A} = {\hat{O}}^{+} H_{1} {\hat{R}}^{+}$ for the exact rank-one factorization $H_{0} = \hat{O} \hat{R}$ .

[F4]

Hankel realization recovers a controllable and observable system: when $O_{1}$ and $R_{1}$ have rank $1$ and the algorithm's exact calls complete and pass their checks, the data yield a successful realization similar to the generating scalar system.

Verification

1.1givenF1F2F3F4algebra

For the first system, $M_{0} = C B = ε$ and $M_{1} = C A B = 0$ by [F1]. Thus the $q = 1$ blocks are $H_{0} = [ε]$ and $H_{1} = [0]$ by [F3], and [F2] gives the singular value of $H_{0}$ as $\sqrt{ε^{2}} = ε$ . Here $O_{1} = [1]$ and $R_{1} = [ε]$ both have rank $1$ . By [F4] and the given successful-call hypothesis, the exact algorithm succeeds and its scalar transition is similar to $A = 0$ ; scalar similarity leaves that transition equal to $0$ .

1.2givenF1F3F4algebra

For the perturbed data, the scalar system with $\tilde{A} = 1 / \sqrt{ε}$ , $\tilde{B} = ε$ , and $\tilde{C} = 1$ has ${\tilde{M}}_{0} = ε$ and ${\tilde{M}}_{1} = \sqrt{ε}$ by [F1]; [F3] identifies these as its $q = 1$ Hankel blocks. Its $O_{1} = [1]$ and $R_{1} = [ε]$ again have rank $1$ , so [F4] and the given successful-call hypothesis imply that the exact algorithm returns a scalar transition similar to $\tilde{A}$ . A one-dimensional similarity does not change a scalar, hence $\tilde{A}$ is returned.

2.1givenstep 1.1step 1.2algebra∎

Take $ε_{n} = 1 / n^{2}$ for positive integers $n$ . The perturbation size is $\sqrt{ε_{n}} = 1 / n \to 0$ , while the change in transition value is $1 / \sqrt{ε_{n}} = n \to \infty$ and the singular value of $H_{0}$ is $1 / n^{2} \to 0$ . This is an exact finite example of unbounded sensitivity as the Hankel singular value approaches zero; it makes no claim of noisy rank-selection correctness.

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