ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-23 (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.

One sensor cannot identify two independent coefficients

Example

Let $V = span {ϕ_{1}, ϕ_{2}}$ have the ordered basis $(ϕ_{1}, ϕ_{2})$ , and represent a candidate field by the coefficient vector $a = (a_{1}, a_{2})^{T} \in R^{2}$ . A single sensor has the linear observation matrix $G = [1 0] \in R^{1 \times 2}$ , the weighting covariance is $R = [1]$ , and a datum is $y = G a + e$ for a fixed realized error $e \in R$ . The sensor depends on $a_{1}$ alone, so it cannot determine $a_{2}$ on the full two-dimensional coefficient space.

Example references

Whitened observation identifiability and stability

Axler, Linear Algebra Done Right, §7E, pp. 270–278

Law, Stuart, and Zygalakis, Chapter 2 §2.1, pp. 25–26

Finite coefficient observation and innovation interface

Facts & Assumptions

Given: The two-dimensional coefficient space, the matrix $G = [1 0]$ , the scalar weighting covariance $R = [1]$ , and fixed realized error $e$ in the Example.

[F1]

Identifiability is injectivity of the observation map on the chosen coefficient subspace; a nonzero kernel vector makes distinct coefficients observationally indistinguishable. Whitened observation identifiability and stability

Verification

Given: The hypotheses and notation in the Example.

1.1givenalgebra

Direct multiplication gives $G (a_{1}, a_{2})^{T} = a_{1}$ and $G (0, 1)^{T} = 0$ , while $(0, 1)^{T} \neq 0$ . Thus $v = (0, 1)^{T}$ is a nonzero vector in the kernel of $G$ .

2.1F1step 1.1algebra∎

For every $a \in R^{2}$ and every $t \in R$ , linearity gives $G (a + t v) = G a + t G v = G a$ . Taking any $t \neq 0$ produces a distinct coefficient vector with the same predicted observation and the same realized error $e$ ; by [F1], the coefficient vector is not identifiable.

Depends on

Whitened observation identifiability and stability