Can a basis together with the zero vector form a vector space?

Question

Let $V$ be a vector space and let $B$ be a basis of $V$. Since every vector space has a zero vector and since every basis is linearly independent it can't contain the zero vector and therefore the basis can't be a vector space. So let's examine a new set $U=B\cup\{0\}$ and check if $U$ can be a vector space. First of all it's clear that $U$ need not be a vector space. Second, if it is a vector space then $V=U$ because $\text{Span}(U)=\text{Span}(B)=\text{Span}(V)$. Furthermore, we must have $|B|=1$ because if $|B|>1$ then there exist $v_1,v_2\in B$ such that $v_1+v_2\in V=U$ but $v_1+v_2\neq 0$ (because $v_1,v_2$ are linearly independent) therefore $v_1+v_2\in B$ and an element of the basis is a linear combination of vectors from the basis which is a contradiction. It's also clear that the field of the vector space $V$ is finite. With all the information we gathered we don't have many options. I think that the only sets that satisfy this problem are $V=\mathbb{Z_2}$ and $B=\{1\}$ over the field $\mathbb{Z_2}$. I'm not sure if this is true, or how to prove it.

Answer 1

It looks like you've mostly got it, there are just a few minor details missing.

First, I agree with your deduction that $B$ cannot contain more than 1 element. So we have two possibilities - either $B = \{v\}$ is a singleton, or $B = \{\}$ is empty.

If $B$ is empty, then $V = \{0\}$ which is the zero space aka the trivial space. In this instance, the underlying field can be absolutely anything and $\dim(V) = 0$.

If $B$ is non-empty, then as we've established it is a singleton set containing one vector, and as a result $V = \{0\} \cup \{v\} = \{0, v\}$. Since $V$ is a vector space, every vector has an additive inverse, so in other words there must exist $v' \in V$ such that $v + v' = 0$. The only possibilities are $v' = v$ and $v' = 0$. Having $v' = 0$ leads to the whole vector space collapsing under itself (since we get $v = v + 0 = v + v' = 0$), so clearly $v' = v$. You can then show that the underlying field must be (isomorphic to) $\mathbb{Z}_2$ just because that's the only way to have a vector field with only two distinct elements.