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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:

  1. additive commutativity
  2. additive associativity
  3. existence of zero vector
  4. existence of additive inverse (i.e -v for all v in set)
  5. multiplicative identity
  6. scalar multiplication is associative
  7. $a(u+v)=au+av \;\forall u,v \in V \;and \; a\in\mathbb{K}$
  8. $(a+b)v=av+bv\;\forall v \in V \;and \; a,b\in\mathbb{K}$
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    $\begingroup$ I don't know if I've understood the question, but I try to help you. If the set is a subset of a vector space then the closure under the two operations it's sufficient. $\endgroup$
    – LuckyS
    Sep 7, 2021 at 17:31
  • $\begingroup$ What is the definition of a vector space that you're using? $\endgroup$ Sep 7, 2021 at 17:35
  • $\begingroup$ Added it to the question, did not know there were more definitions. $\endgroup$
    – Geeniee
    Sep 7, 2021 at 17:56
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    $\begingroup$ It would be much better if you gave a specific example of a proof that is causing your concern. $\endgroup$
    – Rob Arthan
    Sep 7, 2021 at 20:50
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    $\begingroup$ One can never show that some set, in and of itself, is a vector space. What one has to show is that a certain set together with a certain "addition" $+$ and a certain "scalar multiplication" $a \cdot$ for elements $a$ from a certain field, satisfies the axioms of a vector space. In contexts where that is clear, one uses sloppy language, but it confuses beginners. If no $+$ and $\cdot$ are given explicitly, they must be given implicitly (when it's obvious that something is a subset of some vector space, then you try to use the $+$ and $\cdot$ from that bigger space). $\endgroup$ Sep 7, 2021 at 21:36

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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.

  1. 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.

  2. 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 $U$".

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

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

  5. 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.

  6. 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:

  1. 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

  1. 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.

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