# 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:

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. $$(a+b)v=av+bv\;\forall v \in V \;and \; a,b\in\mathbb{K}$$
• 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. Sep 7, 2021 at 17:31
• What is the definition of a vector space that you're using? Sep 7, 2021 at 17:35
• Added it to the question, did not know there were more definitions. Sep 7, 2021 at 17:56
• It would be much better if you gave a specific example of a proof that is causing your concern. Sep 7, 2021 at 20:50
• 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). Sep 7, 2021 at 21:36

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.