# A counterexample that shows addition and scalar multiplication is not enough for a vector space?

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?
• What do you mean by "check addition and scalar multiplication"? Do you mean check whether the compatibility axioms hold, like $(\lambda + \mu)v = (\lambda \cdot v) + (\mu \cdot v)$ – Joppy Oct 4 '17 at 22:37
• Please explain what you mean by "addition and scalar multiplication holds". – bof Oct 4 '17 at 22:42
• It is enough to check addition and scalar multiplication to verify a subspace. – A.Γ. Oct 4 '17 at 22:48
• More examples – A.Γ. Oct 4 '17 at 22:59
• 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 $v \in V$. – Mehdi Jafarnia Jahromi Oct 5 '17 at 6:07

Consider the empty set, which is vacuously closed under any operation, but has no identity under addition.

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.

• Thanks for the answer, Do you have an idea about the second question as well? – Mehdi Jafarnia Jahromi Oct 7 '17 at 20:09

You can't get "Identity element of scalar multiplication" (i.e. $1 \cdot \vec v = \vec v$) from closure under addition and scalar multiplication.