ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (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.

Dual lattice and covolume for a diagonal scaling

Example

Let $a_{1}, \dots, a_{n} \in R ∖ {0}$ and let $A = diag (a_{1}, \dots, a_{n})$ . For $Λ = A Z^{n}$ the covolume is $covol (Λ) = \prod_{i = 1}^{n} ∣ a_{i} ∣$ and the dual lattice is $Λ^{*} = diag (a_{1}^{- 1}, \dots, a_{n}^{- 1}) Z^{n} = {(\frac{m_{1}}{a_{1}}, \dots, \frac{m_{n}}{a_{n}}) : m \in Z^{n}},$ since $A^{- T} = A^{- 1}$ for a diagonal matrix. In one dimension, $covol (a Z) = ∣ a ∣$ and $(a Z)^{*} = a^{- 1} Z$ ; for the sampling spacing $a = h > 0$ this is the pair $h Z$ and $h^{- 1} Z$ of the sampling results on this pair.

Facts & Assumptions

Given: Reals $a_{1}, \dots, a_{n} \neq 0$ , the diagonal matrix $A = diag (a_{1}, \dots, a_{n})$ with entries $A_{i j} = a_{i}$ when $i = j$ and $A_{i j} = 0$ otherwise, and the lattice $Λ = A Z^{n}$ with the covolume and dual lattice of Full-rank lattices, covolume, and the dual lattice.

[F1]

For a diagonal matrix the only nonzero term of the Leibniz sum $\det A = \sum_{σ \in S_{n}} sgn (σ) \prod_{i = 1}^{n} A_{σ (i), i}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix) is the identity permutation, and $A^{- 1} = diag (a_{i}^{- 1})$ because $(A A^{- 1})_{i j} = \sum_{k} A_{i k} a_{k}^{- 1} δ_{k j} = δ_{i j}$ (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, Invertible matrices and the general linear group ${GL}_{n} (F)$ ).

[F2]

$covol (Λ) = ∣ \det A ∣$ and the dual lattice is $Λ^{*} = A^{- T} Z^{n}$ with $A^{- T} = (A^{- 1})^{T}$ (Full-rank lattices, covolume, and the dual lattice); the transpose of a diagonal matrix is itself.

Verification

1.1F1F2givenalgebra

Each off-diagonal entry of $A$ vanishes, so for every permutation $σ \neq id$ some factor $A_{σ (i), i}$ with $σ (i) \neq i$ is zero and only $σ = id$ contributes: $\det A = \prod_{i = 1}^{n} a_{i} \neq 0$ by [F1]. Hence $A$ is invertible with inverse $diag (a_{1}^{- 1}, \dots, a_{n}^{- 1})$ , as the product computation $(A A^{- 1})_{i j} = δ_{i j}$ shows [F1]. Therefore $covol (Λ) = ∣ \det A ∣ = \prod_{i = 1}^{n} ∣ a_{i} ∣$ , and since a diagonal matrix equals its transpose, $A^{- T} = A^{- 1} = diag (a_{i}^{- 1})$ , so $Λ^{*} = diag (a_{1}^{- 1}, \dots, a_{n}^{- 1}) Z^{n}$ , which is the displayed set of tuples $(m_{1} / a_{1}, \dots, m_{n} / a_{n})$ with $m \in Z^{n}$ by [F2] and the definition of $A Z^{n}$ .

2.1F1F2givenalgebra∎

For $n = 1$ the matrix is $(a_{1})$ with $a_{1} = a \neq 0$ , giving $covol (a Z) = ∣ a ∣$ and $(a Z)^{*} = a^{- 1} Z$ ; at the sampling spacing $a = h > 0$ this is exactly the pair $h Z$ , $h^{- 1} Z$ used by the sampling results, whose dual lattice is again full-rank and satisfies $covol (Λ^{*}) = covol (Λ)^{- 1} = h^{- 1}$

Depends on

Used by

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Sources

Lior Silberman, Fourier series and the Poisson summation formula (Math 604/613 notes, UBC) (standard reference, not scraped)
Michael E. Taylor, Fourier Analysis, Distributions, and Constant-Coefficient Linear PDE (author PDF) (standard reference, not scraped)