性质 1 零向量是唯一的。
证明 设 0 1 , 0 2 \boldsymbol{0}_1, \boldsymbol{0}_2 01,02 是线性空间 V V V 中的两个零向量，即对任何 α ∈ V \boldsymbol{\alpha} \in V α∈V，有
α + 0 1 = α α + 0 2 = α \begin{align*} \boldsymbol{\alpha} + \boldsymbol{0}_1 = \boldsymbol{\alpha} \tag{1} \\ \boldsymbol{\alpha} + \boldsymbol{0}_2 = \boldsymbol{\alpha} \tag{2} \\ \end{align*} α+01=αα+02=α(1)(2)
将 ( 1 ) (1) (1) 代入 ( 2 ) (2) (2) 有 ( α + 0 1 ) + 0 2 = ( α + 0 1 ) (\boldsymbol{\alpha} + \boldsymbol{0}_1) + \boldsymbol{0}_2 = (\boldsymbol{\alpha} + \boldsymbol{0}_1) (α+01)+02=(α+01)，将 ( 2 ) (2) (2) 代入 ( 1 ) (1) (1) 有 ( α + 0 2 ) + 0 1 = ( α + 0 2 ) (\boldsymbol{\alpha} + \boldsymbol{0}_2) + \boldsymbol{0}_1 = (\boldsymbol{\alpha} + \boldsymbol{0}_2) (α+02)+01=(α+02)，进而有
0 1 + 0 2 = 0 1 , 0 2 + 0 1 = 0 2 \boldsymbol{0}_1 + \boldsymbol{0}_2 = \boldsymbol{0}_1, \hspace{1em} \boldsymbol{0}_2 + \boldsymbol{0}_1 = \boldsymbol{0}_2 01+02=01,02+01=02
所以
0 1 = 0 1 + 0 2 = 0 2 + 0 1 = 0 2 \boldsymbol{0}_1 = \boldsymbol{0}_1 + \boldsymbol{0}_2 = \boldsymbol{0}_2 + \boldsymbol{0}_1 = \boldsymbol{0}_2 01=01+02=02+01=02
得证。
性质 2 任一向量的负向量是唯一的， α \boldsymbol{\alpha} α 的负向量记作 − α - \boldsymbol{\alpha} −α。
证明 设 β , γ \boldsymbol{\beta}, \boldsymbol{\gamma} β,γ 是 α \boldsymbol{\alpha} α 的负向量，即 α + β = 0 \boldsymbol{\alpha} + \boldsymbol{\beta} = \boldsymbol{0} α+β=0， α + γ = 0 \boldsymbol{\alpha} + \boldsymbol{\gamma} = \boldsymbol{0} α+γ=0。于是
β = β + 0 = β + ( α + γ ) = γ + ( α + β ) = γ + 0 = γ \boldsymbol{\beta} = \boldsymbol{\beta} + \boldsymbol{0} = \boldsymbol{\beta} + (\boldsymbol{\alpha} + \boldsymbol{\gamma}) = \boldsymbol{\gamma} + (\boldsymbol{\alpha} + \boldsymbol{\beta}) = \boldsymbol{\gamma} + \boldsymbol{0} = \boldsymbol{\gamma} β=β+0=β+(α+γ)=γ+(α+β)=γ+0=γ
得证。
性质 3 0 α = 0 0 \boldsymbol{\alpha} = \boldsymbol{0} 0α=0。
证明 因为 $\boldsymbol{\alpha} + 0 \boldsymbol{\alpha} = 1\boldsymbol{\alpha} + 0 \boldsymbol{\alpha} = (1 + 0) \boldsymbol{\alpha} = 1\boldsymbol{\alpha} = \boldsymbol{\alpha} $，所以 0 α = 0 0 \boldsymbol{\alpha} = \boldsymbol{0} 0α=0。得证。
性质 4 ( − 1 ) α = − α (-1)\boldsymbol{\alpha} = - \boldsymbol{\alpha} (−1)α=−α。
证明 因为 α + ( − 1 ) α = 1 α + ( − 1 ) α = [ 1 + ( − 1 ) ] α = 0 α = 0 \boldsymbol{\alpha} + (-1)\boldsymbol{\alpha} = 1 \boldsymbol{\alpha} + (-1)\boldsymbol{\alpha} = [1 + (-1)] \boldsymbol{\alpha} = 0 \boldsymbol{\alpha} = \boldsymbol{0} α+(−1)α=1α+(−1)α=[1+(−1)]α=0α=0，所以 ( − 1 ) α = − α (-1)\boldsymbol{\alpha} = - \boldsymbol{\alpha} (−1)α=−α。得证。
性质 5 λ 0 = 0 \lambda \boldsymbol{0} = \boldsymbol{0} λ0=0。
证明 根据性质 4，有 λ 0 = λ [ α + ( − 1 ) α ] = 0 α = 0 \lambda \boldsymbol{0} = \lambda [\boldsymbol{\alpha} + (-1)\boldsymbol{\alpha}] = 0 \boldsymbol{\alpha} = \boldsymbol{0} λ0=λ[α+(−1)α]=0α=0。得证。
性质 6 如果 λ α = 0 \lambda \boldsymbol{\alpha} = \boldsymbol{0} λα=0，则 λ = 0 \lambda = 0 λ=0 或 α = 0 \boldsymbol{\alpha} = \boldsymbol{0} α=0。文章来源：https://www.toymoban.com/news/detail-724047.html
证明 若 λ ≠ 0 \lambda \ne 0 λ=0，在 λ α = 0 \lambda \boldsymbol{\alpha} = \boldsymbol{0} λα=0 两边乘 1 λ \frac{1}{\lambda} λ1，得
1 λ ( λ α ) = 1 λ 0 = 0 \frac{1}{\lambda} (\lambda \boldsymbol{\alpha}) = \frac{1}{\lambda} \boldsymbol{0} = \boldsymbol{0} λ1(λα)=λ10=0
而
1 λ ( λ α ) = ( 1 λ λ ) α = 1 α = α \frac{1}{\lambda} (\lambda \boldsymbol{\alpha}) = (\frac{1}{\lambda} \lambda) \boldsymbol{\alpha} = 1 \boldsymbol{\alpha} = \boldsymbol{\alpha} λ1(λα)=(λ1λ)α=1α=α
所以有 α = 0 \boldsymbol{\alpha} = \boldsymbol{0} α=0。得证。文章来源地址https://www.toymoban.com/news/detail-724047.html
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