The question of linear space

Question

Let $V$ be a vector space and $v \in V$ . Prove that

\begin{align*} -v = (-1) \cdot v \end{align*}

by only using the vector spaces axioms.

Answer 1

Let $V$ be a vector space over a field $\Bbb{K}$ and $v \in V$. We first prove that $0_{\Bbb{K}} \cdot v = 0_V$. We have that

\begin{align*} 0_{\mathbb{K}} \cdot v & = (0_{\Bbb{K}} + 0_{\Bbb{K}}) \cdot v & (0_{\Bbb{K}} \in \Bbb{K} \text{ identity for $+_{\Bbb{K}}$}) \\ & = 0_{\Bbb{K}} \cdot v + 0_{\Bbb{K}} \cdot v & (\text{distributive law}) \end{align*}

Next, note that you can add $- 0_{\Bbb{K}} \cdot v$ to each side of the above equality (Why?). From this you will get that $0_{\Bbb{K}} \cdot v = 0_V$. Then we conclude that

\begin{align*} v + (-1)\cdot v & = 1_{\Bbb{K}} \cdot v + (-1) \cdot v & (1_{\Bbb{K}} \in \Bbb{K} \text{ identity for } \cdot) \\ & = (1_{\Bbb{K}} + (-1)) \cdot v & (\text{distributive law}) \\ & = 0_{\Bbb{K}} \cdot v & (\text{existence of multiplicative inverse}) \\ & = 0_V \end{align*}

Hence, we have that $(-1) \cdot v \in V$ is a symmetric of $v$. I will assume that you already have proved that the symmetric element of $v$ is unique (if not, you can ask), so we conclude that $-v=(-1)\cdot v$. $\square$

Any doubt on this, please let me know.