LemmaStatement: AI-generatedProof: AI-generated

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.

Coordinate faces span their annihilator subspaces

Statement

Let $V$ be a real vector space with finite basis $(α_{s})_{s \in S}$ , let $I \subseteq S$ , and put $E_{I} = {f \in V^{*} : f (α_{i}) = 0 (i \in I)}$ , $C_{I} = {f \in E_{I} : f (α_{s}) > 0 (s \notin I)}$ . Then $C_{I}$ is nonempty, relatively open in $E_{I}$ , and spans $E_{I}$ . For any linear automorphism $x$ of $V^{*}$ , $x C_{I}$ is relatively open in $x E_{I}$ and spans $x E_{I}$ . These assertions include $I = S$ , where $C_{S} = E_{S} = {0}$ .

Statement references

Annihilators, relative hyperplanes and complexification

Linear combination of a finite list, and the span span(S) as the smallest linear subspace containing S

Invertible linear maps, linear isomorphisms, and inverse linear maps

The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement

Facts & Assumptions

[F1]

Coordinate dual functionals and the coordinate Euclidean metric on $V^{*}$ , restricted to subspaces for relative topology, are supplied by Annihilators, relative hyperplanes and complexification. Write $f_{s} = f (α_{s})$ and $∥ f ∥_{2} = (\sum_{s \in S} ∣ f_{s} ∣^{2})^{1 / 2}$ .

[F2]

The span of a subset is the smallest linear subspace containing it; a subset spans a space when that span is the whole space (Linear combination of a finite list, and the span span(S) as the smallest linear subspace containing S).

[F3]

A linear automorphism is an invertible linear self-map; its inverse is linear and satisfies both inverse equations (Invertible linear maps, linear isomorphisms, and inverse linear maps).

[F4]

A subset of a metric space is open precisely when each of its points has a positive-radius ball contained in the subset (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement). Here the metrics on $E_{I}$ and $x E_{I}$ are restrictions of the coordinate Euclidean metric in [F1].

Proof

Given: The objects and hypotheses of the statement, with the coordinate metrics in [F1]. Goal: Establish all the stated conclusions.

1.1F1F2F3F4algebra

If $I = S$ , then $C_{I} = E_{I} = {0}$ , which is nonempty, open in itself and spans itself; also $x C_{I} = x E_{I} = {0}$ by linearity. All conclusions hold in this case, including $S = \emptyset$ . Henceforth suppose $S ∖ I \neq \emptyset$ . In dual coordinates $E_{I}$ is the subspace with coordinates zero on $I$ , and $v = \sum_{s \notin I} α_{s}^{*}$ belongs to $C_{I}$ . For each $s \notin I$ , both $v$ and $v + α_{s}^{*}$ belong to $C_{I}$ , so every subspace containing $C_{I}$ contains their difference $α_{s}^{*}$ . Every $f \in E_{I}$ equals $\sum_{s \notin I} f_{s} α_{s}^{*}$ ; thus $C_{I}$ spans $E_{I}$ .

2.1F1F4step 1.1algebra

For $f \in C_{I}$ put $r_{f} = \frac{1}{2} \min_{s \notin I} f_{s} > 0$ . If $g \in E_{I}$ and $∥ g - f ∥_{2} < r_{f}$ , then $∣ g_{s} - f_{s} ∣ \leq ∥ g - f ∥_{2} < r_{f} \leq f_{s} / 2$ for each $s \notin I$ , so $g_{s} > 0$ . Thus the ball of radius $r_{f}$ about $f$ in $E_{I}$ lies in $C_{I}$ . This proves relative openness.

3.1F1F2F3F4step 1.1step 2.1algebra∎

Let $(b_{s t})_{s, t \in S}$ be the coordinate matrix of the linear inverse $x^{- 1}$ and put $L = 1 + \sum_{s, t \in S} ∣ b_{s t} ∣ > 0$ . For any $h \in V^{*}$ and $s \in S$ , linearity and $∣ h_{t} ∣ \leq ∥ h ∥_{2}$ give $∣ (x^{- 1} h)_{s} ∣ \leq \sum_{t} ∣ b_{s t} ∣ ∣ h_{t} ∣ \leq L ∥ h ∥_{2}$ . Fix $f \in C_{I}$ and $y = x f$ . If $z \in x E_{I}$ and $∥ z - y ∥_{2} < r_{f} / L$ , then $g = x^{- 1} z \in E_{I}$ and $∣ g_{s} - f_{s} ∣ = ∣ (x^{- 1} (z - y))_{s} ∣ < r_{f} \leq f_{s} / 2$ for every $s \notin I$ . Hence $g \in C_{I}$ and $z \in x C_{I}$ . This supplies a positive-radius ball in $x E_{I}$ about every $y \in x C_{I}$ , proving relative openness directly. Finally every subspace containing $x C_{I}$ contains $x (v + α_{s}^{*}) - x v = x α_{s}^{*}$ for $s \notin I$ , and these vectors span $x E_{I}$ by applying $x$ to the coordinate expansion in step 1.1. Since $x C_{I} \subseteq x E_{I}$ , it spans $x E_{I}$ . Together with step 1.1 this also covers the zero-dimensional case.

Depends on

Used by · 2

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.