PropositionStatement: 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.

Annihilators, relative hyperplanes and complexification

Statement

Let $V$ be a real vector space with finite basis $(e_{s})_{s \in S}$ , let $J \subseteq S$ , and let $E = {f \in V^{*} : f (e_{j}) = 0 for every j \in J}$ . For $α \in V$ , write $r_{α} : E \to R$ for the restricted evaluation $r_{α} (f) = f (α)$ . Then $r_{α}$ is zero exactly when $α \in span {e_{j} : j \in J}$ ; otherwise its kernel is a linear hyperplane of $E$ , meaning that its quotient in $E$ has dimension one. Two nonzero restricted evaluations have the same kernel exactly when one is a nonzero real scalar multiple of the other.

Identify the complexification $E \otimes_{R} C$ with $E \oplus i E$ by $f \otimes (a + b i) \mapsto a f + i b f$ . For a real hyperplane $H \subseteq E$ , its complexification, understood as the image of $H \otimes_{R} C$ under the inclusion into $E \otimes_{R} C$ , is $H \oplus i H$ .

For subsets of $E$ , relative interior and closure mean interior and closure computed in $E$ with the metric induced by the coordinate Euclidean metric on $V^{*}$ associated to the basis $(e_{s})$ . Thus balls used in these operations are balls in $E$ .

Statement references

The dual family (b^)_b∈ B associated to a Hamel basis B, defined by b^(c)=δ_bc

Linear functionals and the algebraic dual V^*=L(V,F)

Linear hyperplane

Kernel and image of a linear map

The tensor and direct-sum models of complexification are canonically complex-linearly isomorphic

Interior, closure, boundary, limit point, isolated point and dense subset of a metric space

Symmetry and associativity isomorphisms for tensor products over a commutative ring

Module homomorphisms induce tensor-product homomorphisms functorially

Isometry, isometric embedding, and the subspace metric on a subset

ℝⁿ as the set of functions n → ℝ, and d₁, d₂, d_∞ are metrics on it

Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis

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

span(S) is exactly the set of linear combinations of finite lists of elements of S, and span(∅) = {0_V}

The quotient vector space V/W and its canonical projection

Coset equality, well-defined quotient operations, and the canonical projection with kernel W

Finite-dimensional vector space, and its dimension dim_F V; infinite-dimensional means having no finite basis

Linear subspace of a vector space

Linear map between vector spaces over the same field

Facts & Assumptions

[F1]

The coordinate functional $e_{s}^{*}$ has $e_{s}^{*} (e_{t}) = δ_{s t}$ ; a functional is determined by its values on the basis, and $V^{*}$ has pointwise linear operations (The dual family (b^)_b∈ B associated to a Hamel basis B, defined by b^(c)=δ_bc, Linear functionals and the algebraic dual V^*=L(V,F)).

[F2]

The kernel of a linear map is its zero fibre. A linear hyperplane is a subspace whose quotient has dimension one (Kernel and image of a linear map, Linear hyperplane).

[F3]

For a real vector space $X$ , the tensor and direct-sum models of complexification are canonically isomorphic by $Φ_{X} : C \otimes_{R} X \to X \oplus i X$ , with $z \otimes x$ corresponding to $z (x + i 0)$ (The tensor and direct-sum models of complexification are canonically complex-linearly isomorphic).

[F4]

In a metric space, the interior consists of points having a ball contained in the set, and the closure consists of points whose every ball meets the set (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[F5]

Over the commutative ring $R$ , tensor symmetry is an isomorphism $σ_{X, Y} : X \otimes_{R} Y \to Y \otimes_{R} X$ sending $x \otimes y$ to $y \otimes x$ (Symmetry and associativity isomorphisms for tensor products over a commutative ring). Real-linear maps $u : X \to X^{'}$ and $v : Y \to Y^{'}$ induce $u \otimes v$ , sending $x \otimes y$ to $u (x) \otimes v (y)$ (Module homomorphisms induce tensor-product homomorphisms functorially).

[F6]

If $(X, d_{X})$ is a metric space and $A \subseteq X$ , then $d_{A} = d_{X} ∣_{A \times A}$ is a metric on $A$ and $B_{A} (a, r) = B_{X} (a, r) \cap A$ for $a \in A$ and $r > 0$ (Isometry, isometric embedding, and the subspace metric on a subset).

[F7]

For every integer $n \geq 1$ , the function $d_{2} (x, y) = (\sum_{k < n} (x_{k} - y_{k})^{2})^{1 / 2}$ is a metric on $R^{n}$ (ℝⁿ as the set of functions n → ℝ, and d₁, d₂, d_∞ are metrics on it).

[F8]

A basis is a linearly independent spanning subset; the empty basis is a basis exactly of the zero space (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). The span is the smallest linear subspace containing the given subset, and consists exactly of its finite linear combinations, including the empty combination (Linear combination of a finite list, and the span span(S) as the smallest linear subspace containing S, span(S) is exactly the set of linear combinations of finite lists of elements of S, and span(∅) = {0_V}).

[F9]

For a vector space $X$ and a linear subspace $K$ , the quotient $X / K$ has classes $x + K$ and operations $(x + K) + (y + K) = (x + y) + K$ and $a (x + K) = a x + K$ (The quotient vector space V/W and its canonical projection). These operations are well-defined and make $X / K$ a vector space; the projection $x \mapsto x + K$ is linear and surjective with kernel $K$ (Coset equality, well-defined quotient operations, and the canonical projection with kernel W).

[F10]

A vector space with a finite basis has dimension equal to the number of elements of that basis (Finite-dimensional vector space, and its dimension dim_F V; infinite-dimensional means having no finite basis).

[F11]

A linear subspace contains zero and is closed under addition and scalar multiplication; with the restricted operations it is a vector space (Linear subspace of a vector space). A map between vector spaces over the same field is linear when $T (a u + b v) = a T (u) + b T (v)$ for all vectors $u, v$ and scalars $a, b$ (Linear map between vector spaces over the same field).

Proof

Given: The finite real basis, subset $J$ , set $E$ , and restricted evaluations in the statement. Define $d (f, g) = (\sum_{s \in S} ∣ f (e_{s}) - g (e_{s}) ∣^{2})^{1 / 2}$ on $V^{*}$ .

Goal: Prove the assertions about restricted evaluations, their kernels, and complexification, and specify the relative operations.

1.1givenF1F2F11algebra

The zero functional belongs to $E$ . For $f, g \in E$ and $a, b \in R$ , pointwise operations give $(a f + b g) (e_{j}) = a f (e_{j}) + b g (e_{j}) = 0$ for every $j \in J$ . Thus $E$ is a real linear subspace of $V^{*}$ and inherits its vector-space operations. For every $α \in V$ , those same operations give $r_{α} (a f + b g) = a r_{α} (f) + b r_{α} (g)$ , so $r_{α}$ is real-linear and $r_{α} (0) = 0$ . Put $K_{α} = \ker r_{α}$ . It contains zero, and if $f, g \in K_{α}$ then $r_{α} (a f + b g) = 0$ for all $a, b \in R$ . Hence $K_{α}$ is a real linear subspace of $E$ . These checks apply also when $J$ is empty or $E = {0}$ .

1.2givenF1F8algebra

Write $α = \sum_{s \in S} a_{s} e_{s}$ . For $f \in E$ , linearity gives $r_{α} (f) = \sum_{s \notin J} a_{s} f (e_{s})$ . If $α$ is in the indicated span this is zero for all $f \in E$ . Conversely, if $r_{α} = 0$ , then $e_{s}^{*} \in E$ for each $s \notin J$ , so evaluation at $e_{s}^{*}$ gives $a_{s} = 0$ for every such $s$ . This proves the equivalence.

1.3givenF1F7F8choosealgebra

Put $n = ∣ S ∣$ . If $n \geq 1$ , enumerate $S = {s_{0}, \dots, s_{n - 1}}$ and define $C : V^{*} \to R^{n}$ by $C (f)_{k} = f (e_{s_{k}})$ . By [F1], $C$ is injective because a functional is determined by its basis values, and it is surjective because $x \in R^{n}$ is the image of $\sum_{k < n} x_{k} e_{s_{k}}^{*}$ . The coordinate formula gives $d (f, g) = d_{2} (C (f), C (g))$ . By [F7], $d$ is nonnegative, symmetric and satisfies the triangle inequality; also $d (f, g) = 0$ exactly when $C (f) = C (g)$ , which by injectivity is exactly when $f = g$ . If $n = 0$ , then $V = {0}$ and $V^{*} = {0}$ , and the empty-sum formula gives the unique distance $d (0, 0) = 0$ ; separation and symmetry hold, and the only triangle inequality is $0 \leq 0 + 0$ . Thus $d$ is a metric in every case.

2.1givenF2F8F9F10step 1.1choosealgebra

Suppose $r_{α}$ is nonzero. Choose $u \in E$ with $r_{α} (u) = 1$ , by rescaling a vector on which it is nonzero. Every $f \in E$ has the unique decomposition $f = (f - r_{α} (f) u) + r_{α} (f) u$ , with the first term in $\ker r_{α}$ . Hence the quotient $E / \ker r_{α}$ is spanned by the nonzero class of $u$ and has dimension one; its kernel is therefore a hyperplane.

2.2givenF2F3F5F11step 1.1algebra

A real hyperplane $H \subseteq E$ is a linear subspace by [F2], hence a real vector space by [F11]. For $X = E$ and $X = H$ , compose the tensor symmetry in [F5] with $Φ_{X}$ in [F3] to obtain the isomorphism $T_{X} = Φ_{X} \circ σ_{X, C} : X \otimes_{R} C \to X \oplus i X$ . It sends $x \otimes (a + b i)$ to $a x + i b x$ and has inverse $x + i y \mapsto x \otimes 1 + y \otimes i$ . Thus $T_{E}$ gives the stated identification.

3.1step 2.1step 1.1F2choosealgebra

Let $r$ and $q$ be nonzero restricted evaluations with the same kernel, and choose $u$ with $r (u) = 1$ . The decomposition in step 2.1 gives $q (f) = r (f) q (u)$ for every $f \in E$ . Since $q$ is nonzero, $q (u) \neq 0$ , so $q$ is a nonzero scalar multiple of $r$ . Conversely, multiplying a functional by a nonzero scalar does not change its zero fibre.

3.2givenF5F11step 2.2algebra

Let $j : H \to E$ be the inclusion. Since $H$ has the restricted operations, $j (a h + b k) = a j (h) + b j (k)$ for $h, k \in H$ and $a, b \in R$ , so $j$ is real-linear; the identity on $C$ is real-linear by the same criterion. By [F5] it induces $j_{C} = j \otimes {id}_{C} : H \otimes_{R} C \to E \otimes_{R} C$ . For $h, k \in H$ , the inverse formula in step 2.2 gives $T_{E} (j_{C} (T_{H}^{- 1} (h + i k))) = T_{E} (j (h) \otimes 1 + j (k) \otimes i) = j (h) + i j (k)$ . Hence, under these isomorphisms, $j_{C}$ is the componentwise inclusion $H \oplus i H \to E \oplus i E$ . This inclusion is injective because its value is zero only when both $h = 0$ and $k = 0$ , and its image is exactly $H \oplus i H$ . This proves the complexification assertion.

4.1givenF4F6step 1.3step 1.2step 2.1step 3.1step 2.2step 3.2∎

Apply [F6] with $X = V^{*}$ , its coordinate Euclidean metric $d$ established in step 1.3, and $A = E$ . The restriction $d_{E} = d ∣_{E \times E}$ is a metric on $E$ , and its balls are $B_{E} (f, r) = B_{V^{*}} (f, r) \cap E$ for $f \in E$ and $r > 0$ . Applying [F4] to $(E, d_{E})$ gives exactly the stated relative interior and closure, with all balls taken in $E$ . Together with steps 1.2, 2.1, 3.1, 2.2 and 3.2, this proves every assertion.

Depends on

Used by · 9

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.