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According to Wikipedia, $(V, +, \cdot)$ is a vector space over field $\mathbb{F}$ if a list of axioms hold. But from my experience, what we usually do is to check whether closure under addition and scalar multiplication holds. I have two questions:

  1. I'm looking for a counterexample for which closure under addition and scalar multiplication holds but it is not a vector space. Any idea?
  2. Why is it usually enough to check closure under addition and scalar multiplication?
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  • $\begingroup$ What do you mean by "check addition and scalar multiplication"? Do you mean check whether the compatibility axioms hold, like $(\lambda + \mu)v = (\lambda \cdot v) + (\mu \cdot v)$ $\endgroup$ – Joppy Oct 4 '17 at 22:37
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    $\begingroup$ Please explain what you mean by "addition and scalar multiplication holds". $\endgroup$ – bof Oct 4 '17 at 22:42
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    $\begingroup$ It is enough to check addition and scalar multiplication to verify a subspace. $\endgroup$ – A.Γ. Oct 4 '17 at 22:48
  • $\begingroup$ More examples $\endgroup$ – A.Γ. Oct 4 '17 at 22:59
  • $\begingroup$ By closure under addition, I mean $v_1 + v_2 \in V$ for any $v_1, v_2 \in V$. By closure under scalar multiplication, I mean $\alpha v \in V$ for any $\alpha \in \mathbb{F}$ and $v \in V$. $\endgroup$ – Mehdi Jafarnia Jahromi Oct 5 '17 at 6:07
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Consider the empty set, which is vacuously closed under any operation, but has no identity under addition.

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Here's an example that has everything except for an identity element of scalar multiplication. Let $F$ be any field, and let $V = F^2$, with addition defined by member-wise addition. Define scalar multiplication by $ a(x, y) = (ax, 0) $.

In general, you do have to check all of the axioms, but when you have something that you think might be vector space, it usually has a particular structure. Almost always, $V$ is a subset of $F^I$, where $I$ is some index set, and vector addition and scalar multiplication are done element-wise.

Lemma: Suppose $F$ is a field, $I$ is a set, and $V \subseteq F^I$. Define:

  • $\mathbf{u} + \mathbf{v} = (u_i + v_i)_i$ for all $\mathbf{u}, \mathbf{v} \in V$
  • $a \mathbf{u} = (au_i)_i$ for all $\mathbf{u} \in V$ and $a \in F$

If $V$ is closed under both addition and scalar multiplication, then it is a vector space.

Proof: Define the identity element of addition by $\mathbf{0} = (0)_i$. $0\mathbf{u} = \mathbf{0}$ for every $\mathbf{u} \in V$, so $\mathbf{0} \in V$ by closure. Define the inverse elements of addition by $-\mathbf{u} = -1 \cdot \mathbf{u}$, which exists in $V$ by closure. All other properties follow from the definitions and some algebra.

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  • $\begingroup$ Thanks for the answer, Do you have an idea about the second question as well? $\endgroup$ – Mehdi Jafarnia Jahromi Oct 7 '17 at 20:09
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You can't get "Identity element of scalar multiplication" (i.e. $1 \cdot \vec v = \vec v$) from closure under addition and scalar multiplication.

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