Exercise 596Linear algebra › Vector spaces and subspacesHigh schoolAdded 3 Nov 2026

The subspace criterion

#subspace criterion#examples

Statement

Show that a nonempty subset $F$ of a vector space $E$ is a subspace if and only if it is closed under addition and scalar multiplication.

Hint

The forward direction is immediate from the axioms. For the converse, verify that the axioms of a vector space hold in $F$ .

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Solution

Forward direction. If $F$ is a subspace, it is a vector space under the induced operations, so it is closed under addition and scalar multiplication by definition.

Converse. Suppose $F$ is nonempty and closed under addition and scalar multiplication. We check the axioms:
- *Zero vector.* Since $F$ is nonempty, take $v\in F$ . Then $0\cdot v\in F$ by scalar closure, and $0\cdot v=0_E$ . So $0_E\in F$ .
- *Opposite.* For $v\in F$ , $(-1)\cdot v\in F$ by scalar closure, and $(-1)\cdot v=-v$ . So $F$ contains opposites.
- *Addition, associativity, commutativity, distributivity, scalar compatibility*: these hold in $F$ because they hold in $E$ and $F$ is closed under the relevant operations.

Conclusion. $F$ is a subspace.

The subspace criterion

Statement

Solution

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