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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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

1.1

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.

givenL1choose

2.1

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 $ℓ (v) = ℓ (w) = \sum_{j < n} (A v)_{j} λ_{j} = ⟨ λ, A v ⟩$ .

step 1.1L1L2algebra

3.1

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.

step 2.1L1L2∎

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)