LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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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.

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Functionals vanishing on a common kernel are combinations of an independent family

Statement

Let $X$ be a real vector space (Vector space over a field), let $m \geq 1$ , and let $ψ_{1}, \dots, ψ_{m} \in X^{*}$ be linearly independent linear functionals on $X$ (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ , Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent). Let $φ \in X^{*}$ satisfy $⋂_{i = 1}^{m} \ker ψ_{i} \subseteq \ker φ$ (Kernel and image of a linear map). Then there is a unique $λ \in R^{m}$ with $φ = \sum_{i = 1}^{m} λ_{i} ψ_{i}$ . In particular, for $m = 1$ : if $ψ \neq 0$ and $\ker ψ \subseteq \ker φ$ , then $φ = λ ψ$ for a unique $λ \in R$ .

Facts & Assumptions

Given: A real vector space $X$ , an integer $m \geq 1$ , linearly independent linear functionals $ψ_{1}, \dots, ψ_{m} ⁣ : X \to R$ , and a linear functional $φ ⁣ : X \to R$ with $⋂_{i = 1}^{m} \ker ψ_{i} \subseteq \ker φ$ .

[F1]

Linear functionals and the algebraic dual $V^{*} = L (V, F)$ , The space $L (V, W)$ of linear maps with pointwise addition and scalar multiplication: $X^{*} = L (X, R)$ is the vector space of linear functionals on $X$ with pointwise operations, so a linear combination $x \mapsto \sum_{i} c_{i} ψ_{i} (x)$ of elements of $X^{*}$ is again an element of $X^{*}$ , and the zero of $X^{*}$ is the functional vanishing identically on $X$ .

[F2]

Kernel and image of a linear map, The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial: for a linear map $T$ one has $\ker T = {x : T (x) = 0}$ , and $\ker T$ is a linear subspace of the domain of $T$ .

[F3]

Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent: the list $ψ_{1}, \dots, ψ_{m}$ is linearly independent exactly when $\sum_{i = 1}^{m} c_{i} ψ_{i} = 0$ in $X^{*}$ forces $c_{1} = \dots = c_{m} = 0$ ; in particular every $ψ_{i}$ is nonzero, since otherwise the list would carry the nontrivial relation with coefficient $1$ on the zero term.

[F4]

Linear map between vector spaces over the same field: each $ψ_{i}$ and $φ$ satisfies $ψ_{i} (a x + b y) = a ψ_{i} (x) + b ψ_{i} (y)$ and $φ (a x + b y) = a φ (x) + b φ (y)$ for all $x, y \in X$ and $a, b \in R$ .

Proof

technique · induction on $m$

Given: A real vector space $X$ , an integer $m \geq 1$ , linearly independent functionals $ψ_{1}, \dots, ψ_{m}$ on $X$ , and a functional $φ$ on $X$ with $⋂_{i = 1}^{m} \ker ψ_{i} \subseteq \ker φ$ .

1.1basegivenF1F3F4algebra

Base case $m = 1$ . Let $ψ \neq 0$ and $\ker ψ \subseteq \ker φ$ . As $ψ \neq 0$ there is $y \in X$ with $ψ (y) \neq 0$ , and $x_{1} : = y / ψ (y)$ satisfies $ψ (x_{1}) = 1$ by homogeneity [F4]; for arbitrary $x \in X$ the vector $x - ψ (x) x_{1}$ lies in $\ker ψ \subseteq \ker φ$ , so $0 = φ (x) - ψ (x) φ (x_{1})$ by linearity of $φ$ [F4], that is, $φ = φ (x_{1}) ψ$ ; conversely if $λ ψ = λ^{'} ψ$ with $ψ \neq 0$ , then $(λ - λ^{'}) ψ = 0$ is the zero functional, so evaluating at $y$ gives $λ = λ^{'}$ .

1.2ihassume-hyp

Induction step setup. Let $P (m)$ denote the assertion of the statement for the fixed integer $m$ and an arbitrary real vector space, and suppose $m \geq 2$ while $P (m - 1)$ is known as the induction hypothesis; the goal is to prove $P (m)$ .

2.1step 1.1F1F2F3algebra

Put $V : = \ker ψ_{m}$ , a linear subspace of $X$ [F2], and let $ρ_{i} : = ψ_{i} ∣_{V}$ for $1 \leq i \leq m - 1$ ; each $ρ_{i}$ is a linear functional on $V$ [F1, F2]. The list $ρ_{1}, \dots, ρ_{m - 1}$ is linearly independent: if $\sum_{i < m} c_{i} ρ_{i} = 0$ , then the functional $σ : = \sum_{i < m} c_{i} ψ_{i} \in X^{*}$ vanishes on $V = \ker ψ_{m}$ , so $\ker ψ_{m} \subseteq \ker σ$ and the base case of step 1.1 gives $σ = λ ψ_{m}$ for some $λ \in R$ ; subtracting yields $\sum_{i < m} c_{i} ψ_{i} - λ ψ_{m} = 0$ in $X^{*}$ , whence $c_{1} = \dots = c_{m - 1} = 0$ and $λ = 0$ by independence of $ψ_{1}, \dots, ψ_{m}$ [F1, F3].

3.1step 1.2step 2.1F1F2algebra

The inclusion hypothesis transfers: if $v \in ⋂_{i < m} \ker ρ_{i}$ , then $v \in V = \ker ψ_{m}$ and $ψ_{i} (v) = 0$ for all $i < m$ , so $v \in ⋂_{i \leq m} \ker ψ_{i} \subseteq \ker φ$ ; hence $⋂_{i < m} \ker ρ_{i} \subseteq \ker (φ ∣_{V})$ , and $φ ∣_{V}$ is a linear functional on $V$ [F1, F4]. The induction hypothesis $P (m - 1)$ applied to the real vector space $V$ and the linearly independent list $ρ_{1}, \dots, ρ_{m - 1}$ of step 2.1 therefore provides $μ_{1}, \dots, μ_{m - 1} \in R$ with $φ ∣_{V} = \sum_{i < m} μ_{i} ρ_{i}$ , that is, with $φ - \sum_{i < m} μ_{i} ψ_{i}$ vanishing on $V$ .

4.1step 1.1step 3.1F1F3algebra

Let $τ : = φ - \sum_{i < m} μ_{i} ψ_{i} \in X^{*}$ , a functional vanishing on $V = \ker ψ_{m}$ by step 3.1, so that $\ker ψ_{m} \subseteq \ker τ$ ; since $ψ_{m} \neq 0$ [F3], the base case of step 1.1 yields $τ = μ_{m} ψ_{m}$ for some $μ_{m} \in R$ ; setting $λ_{i} : = μ_{i}$ for $i < m$ gives $φ = \sum_{i = 1}^{m} λ_{i} ψ_{i}$ by [F1].

5.1step 4.1F1F3discharge-induction∎

Uniqueness and discharge of the induction. If $\sum_{i = 1}^{m} λ_{i} ψ_{i} = \sum_{i = 1}^{m} λ_{i}^{'} ψ_{i}$ , then $\sum_{i = 1}^{m} (λ_{i} - λ_{i}^{'}) ψ_{i}$ is the zero element of $X^{*}$ [F1], so $λ_{i} = λ_{i}^{'}$ for every $i$ by linear independence [F3]. Thus $P (m - 1)$ implies $P (m)$ , and with the base case of step 1.1 the principle of induction gives $P (m)$ for every $m \geq 1$ , which is the assertion, including the stated uniqueness.

Depends on

Used by

Dependency tree · two levels

22 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

Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)