Why Must a Linear Subspace Contain the Additive Identity?

A linear subspace must contain the zero vector of its ambient space. This condition is often listed separately in the subspace test, but it also follows from nonemptiness and closure under scalar multiplication.

Let $V$ be a vector space over a field $F$, and let $W\subseteq V$ inherit the operations of $V$. Suppose that:

1. $W$ is nonempty;
2. $u+v\in W$ whenever $u,v\in W$;
3. $av\in W$ whenever $a\in F$ and $v\in W$.

Choose $w\in W$. Since $0\in F$, scalar closure gives $0w=0_V\in W$. Taking the scalar $-1$ gives $-w\in W$, so additive inverses are present too. The remaining vector-space identities are inherited from $V$.

Nonemptiness matters: the empty set satisfies both closure statements vacuously but is not a vector space. Addition alone is insufficient: the positive integers form a nonempty subset of $\mathbb R$ closed under addition, yet exclude zero and are not closed under all real scalar multiples.

Can the subset have a different identity?

Under inherited vector addition, no. If $e\in W$ were its additive identity, then for any $w\in W$,

Cancellation in the ambient additive group gives $e=0_V$.

A different phenomenon can occur with multiplication in a general monoid. The matrices

are closed under matrix multiplication and have an internal multiplicative identity $\operatorname{diag}(1,0)$, different from the ambient identity $I_2$. There is no contradiction: matrix multiplication does not have the cancellation property used above. Under the convention that a submonoid must contain the ambient identity, this $W$ is not a submonoid of $M_2(\mathbb R)$.

This example helps distinguish algebraic conventions. It does not change the subspace test: a nonempty subset closed under addition and all field scalar multiples is a vector subspace, and its zero is necessarily the ambient zero.