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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

An unobserved state mode

Example

Take $d = 2$ , one input, one output, $A = [\begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array}]$ , $B = 0_{2 \times 1}$ , and $C = [\begin{array}{cc} 1 & 0 \end{array}]$ . For every finite horizon $q \geq 1$ with zero inputs, the distinct initial states $x_{0} = (0, 0)^{T}$ and $x_{0}^{'} = (0, 1)^{T}$ produce the same exact output sequence, namely $y_{k} = 0$ for all $0 \leq k < q$ . The finite observability matrix has rank $1 < 2$ .

Example references

State recurrence and output equation

Finite observability criterion

Dahleh, Dahleh, and Verghese, Lectures on Dynamic Systems and Control, §25.2

Lall, State-Space Realization, §9.2

Facts & Assumptions

Given: The displayed matrices, zero inputs, a finite horizon $q \geq 1$ , and the two stated initial states.

[F1]

Discrete linear input-state-output experiments: the state evolves by $x_{k + 1} = A x_{k} + B u_{k}$ and each exact output is $y_{k} = C x_{k}$ .

[F2]

Finite observability identifies the initial state: for known $A, C$ and inputs, the initial-state output map is injective exactly when $O_{q} = [C; C A; \dots; C A^{q - 1}]$ has rank $d$ .

Verification

1.1givenF1algebra

The first initial state remains zero because $B = 0$ , so its outputs are all zero. For the difference $e_{2} = (0, 1)^{T}$ , $A e_{2} = 2 e_{2}$ ; induction gives $A^{k} e_{2} = 2^{k} e_{2}$ for each $k \geq 0$ . Since $C e_{2} = 0$ , every output difference is $C A^{k} e_{2} = 0$ . Thus the two distinct initial states produce identical outputs at all samples.

2.1givenF2step 1.1algebra

For each $j \geq 0$ , $C A^{j} = [1, 0]$ , so $O_{q}$ consists of $q$ copies of $[1, 0]$ and has rank $1$ . Since $1 < d = 2$ , [F2] implies that the initial-state output map is noninjective for every $q \geq 1$ .

3.1step 1.1step 2.1∎

The initial states in step 1.1 are different but have the same complete observed sequence, so no data-only rule can recover both states exactly in this experiment.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Mohammed Dahleh, Munther A. Dahleh, and George Verghese, Lectures on Dynamic Systems and Control, Chapter 25: Minimal State-Space Realization (standard reference, not scraped)
Sanjay Lall, Engr207c: Operator Theory and Linear Systems, State-Space Realization lecture notes (standard reference, not scraped)