LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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

$\lambda u + v \in W \qquad \text{for all } \lambda \in F \text{ and all } u, v \in W .$

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 $0_F v = 0_V$ , $\lambda 0_V = 0_V$ , $(-\lambda)v = -(\lambda v)$ , $(-1_F)v = -v$ , and $\lambda v = 0_V$ forces $\lambda = 0_F$ or $v = 0_V$ , Vector space over a field).

[L4]

$F$ has elements $0_F$ and $1_F$ , and every $\mu \in F$ has an additive inverse $-\mu$ ; 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 $\lambda \in F$ and $u, v \in W$ . Then $\lambda u \in W$ by (W3), and hence $\lambda u + v \in W$ by (W2).

1.2

Sufficiency, the zero vector. Suppose $\lambda u + v \in W$ for all $\lambda \in F$ and $u, v \in W$ . Since $W$ is nonempty, choose $w \in W$ ; applying the condition with $\lambda = -1_F$ 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 $\lambda = 1_F$ 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 $\lambda \in F$ and $u \in W$ . By step 1.2 we have $0_V \in W$ , so the condition applies to $\lambda$ , $u$ and $0_V$ and gives $\lambda u + 0_V \in W$ ; since $0_V$ is a two-sided identity, $\lambda u + 0_V = \lambda u$ , so $\lambda u \in W$ , 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 $gh^{-1} \in H$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $gh^{-1} \in H$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 18 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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 \subseteq VW⊆V is a linear subspace if and only if λu+v∈W\lambda u + v \in Wλu+v∈W for all λ∈F\lambda \in Fλ∈F and u,v∈Wu, v \in Wu,v∈W