Requirements of a subspace I was recently taught that a subset W is a subspace of V if and only if:

*

*W is non-empty.

*W is closed under vector addition.

*W is closed under scalar multiplication.

So we only need to prove 3 out of the 10 vector space axioms; why is this? Is it because it's redundant to prove the other axioms once those 3 specific axioms are proven?
 A: The other axioms all assert the truth of some identities on calculations with vectors and scalars. The vectors in the subset are in the given larger space so the truth of the axioms there implies it when you happen to think of those vectors as in the subspace.
A: This is a very general property of algebraic structures.
One way of defining an "algebraic structure" is as a set $V$, together with some operations on $S$ and some identities involving those operations.
For vector spaces, the operations are $0 \in V$, $+ : V^2 \to V$, and $s \cdot : V \to V$, where $s$ is a scalar.
A "substructure" of $V$ is a subset $U \subseteq V$ such that all operations $o : V^n \to V$ restrict to operations $o|_{U^n} : U^n \to U$. In other words, we require that for every $n$-ary operation $o : V^n \to V$, for all $u_1, ..., u_n \in U$, $o(u_1, ..., u_n) \in U$.
It turns out that $U$, together with these restricted operations in inherited from $V$, must satisfy all of the identities that are satisfied on all of $V$.
To be specific, suppose that $(V, 0, +, \cdot)$ is a vector space. Suppose that $U \subseteq V$ is closed under all vector space operations. This means, in particular, that $0 \in U$, that $\forall a, b \in U$, $a + b \in U$, and that for all scalars $s$, for all $u \in U$, $s \cdot u \in U$.
Then $(U, 0, +|_{U^2}, \cdot|_U)$ is also a vector space, and the inclusion map $U \subseteq V$ is a vector space homomorphism (commonly called a linear function).
Let's look through some of the axioms to see why. We see that $(a +|_{U^2} b) +|_{U^2} c = (a + b) + c = a + (b + c) = a +|_{U^2} (b +|_{U^2} c)$. Thus, associativity holds.
We also see that $s_1 \cdot|_U (s_2 \cdot|_U u) = s_1 \cdot (s_2 \cdot u) = (s_1 \cdot s_2) \cdot u = (s_1 \cdot s_2) \cdot_U u$. Thus, multiplicative associativity holds.
You can run through all other identities and verify that they hold. The important thing is that $U$ inherits all the identities that hold in $V$ automatically.
The last note is that although your text only requires verifying that $U$ is nonempty, this actually implies that $0$ specifically is an element of $U$. This is because given $u \in U$, we have $0 = 0 \cdot u \in U$.
