Proof of Elementary Identities in a Vector Space

Let $V$ be a vector space over $K$; conditions 1-8 below are those of that definition, and we use the field axioms in $K$.

Claim 1. Condition 3 provides an element $0_{V}$ with $v+0_{V}=v$ for every $v\in V$. If $0'\in V$ also satisfies $v+0'=v$ for every $v\in V$, then

using the property of $0_{V}$ at $v=0'$, condition 2, and the property of $0'$ at $v=0_{V}$.

Claim 2. Let $v\in V$. Condition 4 provides $w\in V$ with $v+w=0_{V}$. If $w'\in V$ also satisfies $v+w'=0_{V}$, then, using conditions 3, 1 and 2,

Claim 3. Since $0+0=0$ in $K$, condition 8 gives $0v+0v=(0+0)v=0v$. Adding the additive inverse $-(0v)$ of $0v$, which exists by condition 4 and is unique by claim 2, to both sides and using conditions 1 and 3,

Claim 4. Since $0_{V}+0_{V}=0_{V}$ by condition 3, condition 7 gives $\lambda 0_{V}+\lambda 0_{V}=\lambda(0_{V}+0_{V})=\lambda 0_{V}$, and the cancellation of claim 3, applied with $\lambda 0_{V}$ in place of $0v$, gives $\lambda 0_{V}=0_{V}$.

Claim 5. By conditions 6 and 8 and claim 3,

so $(-1)v$ is an additive inverse of $v$, and by the uniqueness in claim 2 it equals $-v$.

Claim 6. Suppose $\lambda v=0_{V}$ and $\lambda\neq0$. Then $\lambda$ has a multiplicative inverse $\lambda^{-1}$ in $K$, and by conditions 6 and 5 and claim 4,

Hence $\lambda=0$ or $v=0_{V}$.