LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The additive group of a vector space is an abelian group and every linear subspace is a subgroup of it; conversely a subgroup closed under scalar multiplication is a linear subspace

Statement

Let $V$ be a vector space over a field $F$ (Vector space over a field).

$(V, +, 0_{V})$ is an abelian group (Group and abelian group), called the additive group of $V$ .
Every linear subspace $W$ of $V$ (Linear subspace of a vector space) is a subgroup of $(V, +, 0_{V})$ (Subgroup). Consequently $W$ with the restricted addition is itself a group, whose identity is $0_{V}$ and whose inverses are those of $V$ .
Conversely, if $W \subseteq V$ is a subgroup of $(V, +, 0_{V})$ and $λ v \in W$ for all $λ \in F$ and $v \in W$ , then $W$ is a linear subspace of $V$ .

So the linear subspaces of $V$ are exactly the subgroups of its additive group that are closed under scalar multiplication.

Facts & Assumptions

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

[L1]

Axiom (V1): $(V, +, 0_{V})$ is an abelian group (Vector space over a field, Group and abelian group).

[L2]

A subgroup of a group $G$ with identity $e$ is a subset $H$ satisfying (S1) $e \in H$ , (S2) $x, y \in H$ implies $x y \in H$ , and (S3) $x \in H$ implies $x^{- 1} \in H$ ; such an $H$ , with the restricted operation, is itself a group whose identity and whose inverses are those of $G$ (Subgroup).

[L3]

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).

[L4]

$(- 1_{F}) v = - v$ for every $v \in V$ (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 $v = 0_{V}$ ).

Proof

technique · direct

1.1

Claim 1 is axiom (V1) of a vector space, which asserts in as many words that $(V, +, 0_{V})$ is an abelian group.

1.2

Let $W$ be a linear subspace of $V$ . Condition (W1) says $0_{V} \in W$ , which is condition (S1) for the group $(V, +, 0_{V})$ , whose identity is $0_{V}$ .

L1L2L3

1.3

Condition (W2) says $u + v \in W$ for all $u, v \in W$ , which is condition (S2) for $(V, +, 0_{V})$ , whose operation is $+$ .

L2L3

1.4

Let $v \in W$ . By (W3) with $λ = - 1_{F}$ we get $(- 1_{F}) v \in W$ , and $(- 1_{F}) v = - v$ , so $- v \in W$ ; since the inverse of $v$ in the group $(V, +, 0_{V})$ is $- v$ , this is condition (S3).

L1L2L3L4

1.5

Conversely, let $W$ be a subgroup of $(V, +, 0_{V})$ with $λ v \in W$ for all $λ \in F$ and $v \in W$ . Condition (S1) gives $0_{V} \in W$ , which is (W1); condition (S2) gives closure under $+$ , which is (W2); and the hypothesis is (W3).

L2L3

2.1

By steps 1.2, 1.3 and 1.4 the subset $W$ satisfies (S1), (S2) and (S3), so it is a subgroup of $(V, +, 0_{V})$ ; by the properties of a subgroup it is then a group under the restricted addition, with identity $0_{V}$ and with the inverses of $V$ . This is claim 2.

step 1.2step 1.3step 1.4L2

2.2

By step 1.5 the subset $W$ of that step satisfies (W1), (W2) and (W3), so it is a linear subspace of $V$ . This is claim 3.

step 1.5L3

3.1

Claim 1 is step 1.1, claim 2 is step 2.1 and claim 3 is step 2.2; together they say that the linear subspaces of $V$ are exactly the subgroups of $(V, +, 0_{V})$ closed under scalar multiplication.

step 1.1step 2.1step 2.2∎

Remarks

What the two directions cost. Going from a linear subspace to a subgroup uses one fact about vector spaces and no group theory: closure under additive inverses is not assumed but derived, from closure under scalar multiplication at the scalar $- 1_{F}$ . Going back is pure bookkeeping, since (W1) and (W2) are literally (S1) and (S2).
Why this is worth an item. Every statement the library proves about subgroups applies to linear subspaces at once. In particular the intersection of a nonempty family of subgroups is a subgroup (The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ ), which is the group-theoretic shadow of The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$ below.
The hypothesis in claim 3 is not decoration. Conditions (S1)–(S3) do not mention the scalars at all, so a subgroup of $(V, +, 0_{V})$ is required only to contain $0_{V}$ and to be closed under addition and under negation; closure under multiplication by an arbitrary $λ \in F$ is a further condition, and claim 3 assumes it rather than deriving it. Claim 2 says that in the other direction nothing extra is needed, because (W3) is already one of the three defining conditions of a linear subspace.

Depends on

Used by

The first quadrant of ℝ² contains 0 and is closed under addition and is not a linear subspace, since it is not closed under multiplication by -1 Counterexample

Dependency tree · two levels

14 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)
Subgroup (Wikipedia) (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed. (free PDF, CC BY-NC) (standard reference, not scraped)