LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

Finite Hermite data at distinct eigenvalues have a unique polynomial representative

Statement

Let $s \geq 0$ , let $λ_{1}, \dots, λ_{s} \in C$ be distinct, let $m_{1}, \dots, m_{s}$ be positive integers, and put $N = \sum_{i = 1}^{s} m_{i}$ . Let $V_{N} = span {1, z, \dots, z^{N - 1}}$ , with the empty span $V_{0} = {0}$ . For arbitrary data $c_{i, k} \in C$ with $1 \leq i \leq s$ and $0 \leq k < m_{i}$ , there is exactly one polynomial $p \in V_{N}$ satisfying $p^{(k)} (λ_{i}) = c_{i, k} (1 \leq i \leq s, 0 \leq k < m_{i}) .$

Facts & Assumptions

Given: A finite list of distinct nodes (possibly empty), positive multiplicities when nodes are present, and the derivative data $c_{i, k}$ .

[F1]

The space $V_{N}$ and the coordinate space $C^{N}$ both have dimension $N$ ; the monomial basis is $1, z, \dots, z^{N - 1}$ , empty when $N = 0$ .

[F2]

The map that sends a polynomial to the finite list of its derivatives at the fixed nodes is linear.

Proof

1.1givenF1F2construct

Let $T : V_{N} \to C^{N}$ be given by $T (p) = (p^{(k)} (λ_{i}))_{i, k}$ . By [F2], $T$ is linear, and by [F1] its domain and codomain both have dimension $N$ .

2.1givenstep 1.1algebra

If $N = 0$ , then $V_{N} = {0}$ , so $T$ is injective. If $N > 0$ and $T (p) = 0$ , the exact Taylor expansion of the polynomial at $λ_{i}$ shows $(z - λ_{i})^{m_{i}}$ divides $p$ for every $i$ . The factors for distinct $λ_{i}$ are pairwise coprime, so their product of degree $N$ divides $p$ . Since $p \in V_{N}$ has degree less than $N$ , this forces $p = 0$ . Thus $T$ is injective in either case.

3.1step 1.1step 2.1algebra∎

An injective linear map between finite-dimensional spaces of the same dimension is surjective, so the prescribed vector $(c_{i, k})_{i, k}$ has a preimage under $T$ and that preimage is unique.

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Jan Zur, Matrix Functions, TU Berlin full lecture notes (standard reference, not scraped)
Nicholas J. Higham, Functions of Matrices, Handbook of Linear Algebra, Chapter 17 (standard reference, not scraped)