In Sheldon Axler's "Linear Algebra Done Right," he writes that a subset $U \subset V$ is a subspace if and only if $U$ contains the identity, is closed under addition, and is closed under scalar multiplication.
Is the first of these conditions redundant? Suppose the latter two conditons hold. If $v \in V$, then $-v \in V$ by closure under scalar multiplication. Then, $v + (-v) = 0 \in V$ by closure under vector addition.
Is it enough, on this basis, to just prove closure?
Best Answer
If the first condition were omitted, the empty set would qualify as a subspace. Note that the other conditions can be phrased as "If $u_1 \in U \text { and } u_2 \in U, \text { then }..."$ and "If$u \in U \text { and } \alpha \in F, \text { then }..."$. Both of those "If" statements are true by vacuous implication in the case of U being the empty set.