subspace

A subset $U\subseteq V$ is a subspace iff $U$ is also a vector space using the same addition and scalar multiplication as $V$

1. $0\in U$ (additive inverse exists)
2. $u,v\in V$ then $u+v\in U$ (closed under addition)
3. If $\alpha \in \mathbb{F}$ and $u\in U$ then $\alpha u\in U$ (closed under scalar multiplication)

Proof: [=>] If $U$ is a subspace then it is a vector space (def) thus 2 and 3 are automatic

But we should check that $0_U=0_V$. Note however

(0 is a scalar here). Because $U$ is closed under scalar multiplication, $0_V\in U$. Also $0_V$ is an additive identity of $U$. By uniqueness of additive identity $0_V=0_U$

[<=] Suppose $U$ satisfies 1, 2 and 3. If $u\in U$ then $(-1)\text{cdotu}=-u\in U$ using 3. Thus $U$ contains a additive inverse. The rest of the properties follow since addition and scalar multiplication are inherited from $V$