性质 1.4 (零向量 负向量)

1. $0\alpha =0$
2. $k0=0$
3. $\left(-1\right)\alpha =-\alpha$.

证明 (1) 0 为 $F$ 的零元素,若 $\alpha \in V$,则

$$\alpha +0\alpha =1\alpha +0\alpha =\left(1+0\right)\alpha =1\alpha =\alpha +0.$$

故由性质 1.3 知 $0\alpha =0$.

(2) 任取 $k\in F$. 若 $k=0$,则由 $\left(1\right),k0=0$. 今设 $k\ne 0$. 则 $k-1$ 存在,于具对任意 $\alpha \in V$,

$$\alpha +k0=k\left(k^{-1}\alpha \right)+k0=k\left(k^{-1}\alpha +0\right)=k\left(k^{-1}\alpha \right)=\left(k{k}^{-1}\right)\alpha =1\alpha =\alpha .$$

故由消去律知 $k0=0$.

(3) $\alpha +\left(-1\right)\alpha =1\alpha +\left(-1\right)\alpha =\left(1+\left(-1\right)\right)\alpha =0\alpha =0$. 由负向量的唯一性

知, $\left(-1\right)\alpha =-\alpha$.