LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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.

One-step subspace test: a nonempty $W \subseteq V$ is a linear subspace if and only if $λ u + v \in W$ for all $λ \in F$ and $u, v \in W$

Statement

Let $V$ be a vector space over a field $F$ (Vector space over a field) and let $W \subseteq V$ be nonempty. Then $W$ is a linear subspace of $V$ (Linear subspace of a vector space) if and only if

$λ u + v \in W for all λ \in F and all u, v \in W .$

Nonemptiness cannot be dropped: the empty set satisfies the displayed condition vacuously and is not a linear subspace, since it does not contain $0_{V}$ .

Facts & Assumptions

Given: A field $F$ , a vector space $V$ over $F$ , and a nonempty subset $W \subseteq V$ .

[L1]

A linear subspace of $V$ is a subset satisfying (W1) $0_{V} \in W$ , (W2) closure under $+$ , and (W3) closure under scalar multiplication (Linear subspace of a vector space).

[L2]

The vector space axioms, in particular (V5) $1_{F} v = v$ , and that $(V, +, 0_{V})$ is an abelian group with $0_{V}$ a two-sided identity (Vector space over a field).

[L3]

$(- 1_{F}) v = - v$ for every $v \in V$ , and $v + (- v) = 0_{V}$ (In any vector space $0_{F} v = 0_{V}$ , $λ 0_{V} = 0_{V}$ , $(- λ) v = - (λ v)$ , $(- 1_{F}) v = - v$ , and $λ v = 0_{V}$ forces $λ = 0_{F}$ or $v = 0_{V}$ , Vector space over a field).

[L4]

$F$ has elements $0_{F}$ and $1_{F}$ , and every $μ \in F$ has an additive inverse $- μ$ ; in particular $1_{F} \in F$ and $- 1_{F} \in F$ (Field).

Proof

technique · direct

1.1

Necessity. Suppose $W$ is a linear subspace, and let $λ \in F$ and $u, v \in W$ . Then $λ u \in W$ by (W3), and hence $λ u + v \in W$ by (W2).

1.2

Sufficiency, the zero vector. Suppose $λ u + v \in W$ for all $λ \in F$ and $u, v \in W$ . Since $W$ is nonempty, choose $w \in W$ ; applying the condition with $λ = - 1_{F}$ and $u = v = w$ gives $(- 1_{F}) w + w \in W$ , and $(- 1_{F}) w + w = - w + w = 0_{V}$ , so $0_{V} \in W$ , which is (W1).

givenL3L4choose

1.3

Sufficiency, closure under addition. Let $u, v \in W$ . Applying the condition with $λ = 1_{F}$ gives $1_{F} u + v \in W$ , and $1_{F} u = u$ by (V5), so $u + v \in W$ , which is (W2).

givenL2L4

2.1

Sufficiency, closure under scalars. Let $λ \in F$ and $u \in W$ . By step 1.2 we have $0_{V} \in W$ , so the condition applies to $λ$ , $u$ and $0_{V}$ and gives $λ u + 0_{V} \in W$ ; since $0_{V}$ is a two-sided identity, $λ u + 0_{V} = λ u$ , so $λ u \in W$ , which is (W3).

step 1.2givenL2

3.1

Steps 1.2, 1.3 and 2.1 verify (W1), (W2) and (W3), so a nonempty $W$ satisfying the displayed condition is a linear subspace; with step 1.1 this proves the equivalence.

step 1.1step 1.2step 1.3step 2.1L1∎

Remarks

The order of the two verifications matters. Closure under scalar multiplication is deduced after $0_{V}$ is known to lie in $W$ , because it is obtained by applying the test to the pair $u$ , $0_{V}$ . Running the argument in the other order would use $0_{V} \in W$ before it had been established.
One test, three conditions. The single condition is exactly as strong as the three of Linear subspace of a vector space, and it is the form used in practice: to check that a set is a linear subspace one shows it is nonempty and closes under a single mixed expression. It is the linear analogue of the one-step subgroup test $g h^{- 1} \in H$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ), and, exactly as there, the nonemptiness hypothesis is what rules out the empty set.
Neither closure condition implies the other, so a test combining them is not extravagant. A subset of a vector space can be closed under addition and not under scalar multiplication, and another can be closed under scalar multiplication and not under addition; the companion examples page records a witness of each kind.

Depends on

Used by

Dependency tree · two levels

13 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

Linear subspace (Wikipedia) (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed., Ch. 1 (standard reference, not scraped)

One-step subspace test: a nonempty W⊆VW⊆V is a linear subspace if and only if λu+v∈Wλu+v∈W for all λ∈Fλ∈F and u,v∈Wu,v∈W