Linear subspace of a vector space

Let $V$ be a vector space over a field $F$ (Vector space over a field). A subset $W \subseteq V$ is a linear subspace of $V$ when

• (W1) $0_V \in W$;
• (W2) $W$ is closed under the vector addition: $u, v \in W$ implies $u + v \in W$;
• (W3) $W$ is closed under scalar multiplication: $\lambda \in F$ and $v \in W$ imply $\lambda v \in W$.

Every vector space $V$ has the two trivial linear subspaces $\{0_V\}$ and $V$ itself; a linear subspace $W$ with $W \ne V$ is called proper.

The restricted operations are the required data, and $W$ is a vector space. By (W2) the vector addition of $V$ restricts to a binary operation $W \times W \to W$, and by (W3) the scalar multiplication restricts to a map $F \times W \to W$. With these and the element $0_V$, the set $W$ is a vector space over $F$:

• axioms (V2)–(V5) are equations required of elements of $W$, which are in particular elements of $V$, so they are inherited from $V$; likewise associativity and commutativity of the restricted addition;
• $0_V$ lies in $W$ by (W1) and is a two-sided identity for the restricted addition, since it is one in $V$;
• for $v \in W$ the vector $(-1_F)v$ lies in $W$ by (W3), and $(-1_F)v = -v$ (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 $v = 0_V$), so $-v \in W$ and $v + (-v) = 0_V$ holds in $W$.

So $(W,+,0_V)$ is an abelian group, which is axiom (V1), and $W$ is a vector space over $F$ whose zero vector and whose additive inverses are those of $V$. In the language of Subgroup, the three displayed conditions (S1) $0_V \in W$, (S2) closure under addition and (S3) closure under additive inverses all hold, so $W$ is a subgroup of the abelian group $(V,+,0_V)$ (Group and abelian group); that reading, and its converse, are recorded as 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 and are cited from there rather than re-argued below.

• "Linear subspace", never bare "subspace", in this library. The word subspace is already in use here for the topological notion, a subset of a topological space carrying the induced topology, which is an unrelated idea. The names on this page therefore all say linear: def-linear-subspace, lem-linear-subspace-criterion, lem-intersection-of-linear-subspaces, lem-linear-subspace-is-a-subgroup, def-sum-of-linear-subspaces. Where the ambient vector space is fixed and no confusion is possible, the surrounding prose still writes the full phrase.

• Closure under negatives is not a fourth condition. It follows from (W3), because the additive inverse of a vector is the scalar multiple $(-1_F)v$. This is why the definition asks for three conditions where the definition of a subgroup asks for three of its own, and why the one-step test (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 $u, v \in W$) can compress them into one.

• (W1) cannot be replaced by "$W \ne \varnothing$" while dropping the others. It can be replaced by nonemptiness given (W3), since a nonempty $W$ closed under scalar multiplication contains $0_F v = 0_V$ (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 $v = 0_V$) for any of its elements $v$. Stated with (W1) the definition is checkable one condition at a time, and the economical single test is 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 $u, v \in W$.

• The field matters. A subset of $V$ may be closed under the scalars of a subfield $K \subseteq F$ without being closed under all of $F$, so "linear subspace" always means "linear subspace over the field named". Restriction of scalars (A field is a vector space over itself, and over any subfield $K \subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars) is what makes that distinction possible.