LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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.

A linear functional annihilating the kernel of a surjection is a unique transpose multiple

Statement

Let $A : R^{m} \to R^{n}$ be a surjective linear map, with $m, n \geq 1$ , and let $ℓ : R^{m} \to R$ be linear. If $ℓ$ vanishes on $\ker A$ , then there is a unique $λ \in R^{n}$ such that $ℓ (v) = ⟨ λ, A v ⟩ for every v \in R^{m} .$ Equivalently, the row vector of $ℓ$ is $A^{T} λ$ .

Facts & Assumptions

Given: The surjective linear map $A$ and the linear functional $ℓ$ vanishing on $\ker A$ .

[L1]

Surjectivity provides a preimage of each standard basis vector, and the standard basis gives the coordinate expansion of every vector in $R^{n}$ (Injection, surjection, bijection, The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L2]

Kernels are the vectors mapped to zero, and linear maps preserve finite linear combinations (Kernel and image of a linear map, Linear map between vector spaces over the same field, The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial); the Euclidean inner product is the coordinate dot product (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Proof

technique · direct

1.1givenL1choose

For each standard basis vector $e_{j}$ , choose $w_{j} \in R^{m}$ with $A w_{j} = e_{j}$ , and define $λ_{j} = ℓ (w_{j})$ . These are finitely many choices.

2.1step 1.1L1L2algebra

For $v \in R^{m}$ , put $w = \sum_{j < n} (A v)_{j} w_{j}$ . Then $A w = A v$ by [L1] and linearity, so $v - w \in \ker A$ and $ℓ (v) = ℓ (w) = \sum_{j < n} (A v)_{j} λ_{j} = ⟨ λ, A v ⟩$ .

3.1step 2.1L1L2∎

If both $λ$ and $μ$ work, surjectivity gives $v$ with $A v = λ - μ$ ; then $0 = ⟨ λ - μ, A v ⟩ = ∥ λ - μ ∥_{2}^{2}$ , so $λ = μ$ . This proves existence and uniqueness.

Depends on

Used by

Lagrange multipliers for a regular vector-valued level-set constraint Theorem

Dependency tree · two levels

35 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

J. M. Lee, Introduction to Smooth Manifolds, Lagrange multipliers discussion (standard reference, not scraped)
University of Toronto MAT237 notes, Section 2.8 (standard reference, not scraped)