Criteria for showing a set is a vector space seems ambiguous? I have attempted a lot of exercises where you are supposed to show that a given set is a vector space. Having done that I've also read other people's attempts as well. However, sometimes we go on to show several criteria for a vector space and sometimes it seems to suffice just to show closure under vector addition and scalar multiplication, and the existence of a zero vector. Do these criteria imply the others or does it depend on what field we're proving a given set is a vector space over to determine which requirements need to be showed? Because of this, I've realized that I don't really have a "clean" approach to these types of problems.
The criteria I've been using are:

*

*additive commutativity

*additive associativity

*existence of zero vector

*existence of additive inverse (i.e -v for all v in set)

*multiplicative identity

*scalar multiplication is associative

*$a(u+v)=au+av \;\forall u,v \in V \;and \; a\in\mathbb{K}$

*$(a+b)v=av+bv\;\forall v \in V \;and \; a,b\in\mathbb{K}$
 A: There are several theorems about vector spaces that allow us to prove a set, together with some operations, is a vector space without having to run through all the axioms. Here are a few of them.


*Let $(V, +_V, \cdot_V, 0_V)$ be a vector space, and consider $(U, +_U, \cdot_U, 0_U)$. If there is a bijection $f : U \to V$ satisfying the identities $f(x +_U y) = f(x) +_V f(y)$ and $f(r \cdot_U x) = r \cdot_V f(x)$ for all $x, y \in U$, then $U$ is a vector space.


*Let $(U, +, \cdot, 0)$ be a vector space. Suppose that $V \subseteq U$ satisfies the following three properties: (1) $0 \in V$, (2) if $x \in V$ and $r$ is a scalar then $r \cdot x \in V$, and (3) if $x, y \in V$ then $x + y \in V$. Then $(V, +, \cdot, 0)$ is also a vector space. We describe this situation as "$V$ is a subspace of $U$".


*Let $f : U \to V$ be a linear map between vector spaces. Then $\ker f \subseteq U$ is a subspace of $U$.


*Let $f : U \to V$ be a linear map between vector spaces. Then $im(f) \subseteq V$ is a subspace of $V$.


*Suppose that for each $i \in I$, $V_i$ is a vector space. Then $\prod\limits_{i \in I} V_i$ is a vector space with operators $(f + g)(i) = f(i) + g(i)$, $(r \cdot f)(i) = r \cdot f(i)$, and $0(i) = 0$. As a special case, given a set $A$ and a vector space $V$, the set $\{f : A \to V\} = \prod\limits_{a \in A} V$ forms a vector space.


*Suppose $U, V$ are vector spaces. The space $\{f : U \to V \mid f$ linear$\}$ is a subspace of $\{f : U \to V\}$.
This gives you many, many ways of forming vector spaces. In the particular case where the underlying field $\mathbb{F}$ is a topological ring, we have another great lemma:


*Suppose $X$ is a topological space. Then $\{f : X \to \mathbb{F} \mid f$ continuous$\}$ is a subspace of $\{f : X \to \mathbb{F}\}$.

If $\mathbb{F}$ is a topological ring and an $n$-differentiable manifold, we have


*Suppose $X$ is an $n$-differentiable manifold. Then $\{f : X \to \mathbb{F} \mid f$ is $n$-times differentiable$\}$ is a subspace of $\{f : X \to \mathbb{F}\}$.

Taken together, most common cases of proving that a given set is a vector space fall under one of these theorems. There are a few other constructions (quotient spaces and free spaces) that I have omitted.
