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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$

Statement

Let $V$ be a vector space over a field $F$ (Vector space over a field) and let $W$ be a nonempty set of linear subspaces of $V$ (Linear subspace of a vector space). Then

$U = ⋂_{W \in W} W = {x \in V : x \in W for every W \in W}$

is a linear subspace of $V$ . In particular the intersection of two linear subspaces is a linear subspace.

Facts & Assumptions

Given: A field $F$ , a vector space $V$ over $F$ , a nonempty set $W$ of linear subspaces of $V$ , and $U$ the intersection of the members of $W$ .

[L1]

Each $W \in W$ contains $0_{V}$ , is closed under $+$ , and is closed under scalar multiplication (Linear subspace of a vector space).

[L2]

One-step test: a nonempty $S \subseteq V$ with $λ u + v \in S$ for all $λ \in F$ and $u, v \in S$ is a linear subspace of $V$ (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 $u, v \in W$ ).

Proof

technique · direct

1.1givenL1

$U \subseteq V$ , since $W$ is nonempty and every member of it is a subset of $V$ .

1.2L1

$0_{V} \in U$ , since $0_{V} \in W$ for every $W \in W$ ; in particular $U$ is nonempty.

1.3givenL1

Let $λ \in F$ and $u, v \in U$ , and let $W \in W$ be arbitrary. Then $u, v \in W$ , so $λ u \in W$ by closure under scalar multiplication and $λ u + v \in W$ by closure under addition.

2.1step 1.3

Since $W$ was an arbitrary member of $W$ , the vector $λ u + v$ lies in every member of $W$ , that is $λ u + v \in U$ .

3.1step 1.1step 1.2step 2.1L2∎

$U$ is a nonempty subset of $V$ satisfying the one-step test, hence a linear subspace of $V$ ; taking $W$ to have two members gives the last sentence of the statement.

Remarks

The hypothesis that $W$ is nonempty is load bearing. The intersection of the empty family of subsets of $V$ is not a subset of $V$ by any convention used here, and step 1.1 is where the hypothesis is spent. The same hypothesis appears for the same reason in The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ .
This is what makes the span definable. The set of linear subspaces of $V$ containing a given subset $S$ is nonempty, since $V$ itself belongs to it, so its intersection is a linear subspace, and it is by construction the smallest linear subspace containing $S$ . That is Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , and the pattern is copied from The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups.
Unions behave quite differently. The union of two linear subspaces is almost never a linear subspace, and the companion examples page records the failure as a false statement.

The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The intersection of a nonempty family of linear subspaces of VV is a linear subspace of VV

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$