【证明】线性映射不影响向量组的线性组合

前置定义 1 设 V n V_n Vn U m U_m Um 分别是 n n n 维和 m m m 维线性空间, T T T 是一个从 V n V_n Vn U m U_m Um 的映射,如果映射 T T T 满足:

(i)任给 α 1 , α 2 ∈ V n \boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2 \in V_n α1,α2Vn(从而 α 1 + α 2 ∈ V \boldsymbol{\alpha}_1 + \boldsymbol{\alpha}_2 \in V α1+α2V),有
T ( α 1 + α 2 ) = T ( α 1 ) + T ( α 2 ) T(\boldsymbol{\alpha}_1 + \boldsymbol{\alpha}_2) = T(\boldsymbol{\alpha}_1) + T(\boldsymbol{\alpha}_2) T(α1+α2)=T(α1)+T(α2)
(ii)任给 α ∈ V n \boldsymbol{\alpha} \in V_n αVn λ ∈ R \lambda \in \R λR(从而 λ α ∈ V n \lambda \boldsymbol{\alpha} \in V_n λαVn),有
T ( λ α ) = λ T ( α ) T(\lambda \boldsymbol{\alpha}) = \lambda T(\boldsymbol{\alpha}) T(λα)=λT(α)
那么, T T T 就称为从 V n V_n Vn U m U_m Um线性映射,或称为 线性变换


性质 1 若 β = k 1 α 1 + k 2 α 2 + ⋯ + k m α m \boldsymbol{\beta} = k_1 \boldsymbol{\alpha}_1 + k_2 \boldsymbol{\alpha}_2 + \cdots + k_m \boldsymbol{\alpha}_m β=k1α1+k2α2++kmαm,则 T ( β ) = k 1 T ( α 1 ) + k 2 T ( α 2 ) + ⋯ + k m T ( α m ) T(\boldsymbol{\beta}) = k_1 T(\boldsymbol{\alpha}_1) + k_2 T(\boldsymbol{\alpha}_2) + \cdots + k_m T(\boldsymbol{\alpha}_m) T(β)=k1T(α1)+k2T(α2)++kmT(αm)

证明 根据前置定义 1,有
T ( β ) = T ( k 1 α 1 + k 2 α 2 + k 3 α 3 + ⋯ + k m α m ) = T ( k 1 α 1 ) + T ( k 2 α 2 + k 3 α 3 + ⋯ + k m α m ) = T ( k 1 α 1 ) + T ( k 2 α 2 ) + T ( k 3 α 3 ⋯ + k m α m ) = T ( k 1 α 1 ) + T ( k 2 α 2 ) + ⋯ + T ( k m α m ) = k 1 T ( α 1 ) + k 2 T ( α 2 ) + ⋯ + k m T ( α m ) \begin{aligned} T(\boldsymbol{\beta}) & = T(k_1 \boldsymbol{\alpha}_1 + k_2 \boldsymbol{\alpha}_2 + k_3 \boldsymbol{\alpha}_3 + \cdots + k_m \boldsymbol{\alpha}_m) \\ & = T(k_1 \boldsymbol{\alpha}_1) + T(k_2 \boldsymbol{\alpha}_2 + k_3 \boldsymbol{\alpha}_3 + \cdots + k_m \boldsymbol{\alpha}_m) \\ & = T(k_1 \boldsymbol{\alpha}_1) + T(k_2 \boldsymbol{\alpha}_2) + T(k_3 \boldsymbol{\alpha}_3 \cdots + k_m \boldsymbol{\alpha}_m) \\ & = T(k_1 \boldsymbol{\alpha}_1) + T(k_2 \boldsymbol{\alpha}_2) + \cdots + T(k_m \boldsymbol{\alpha}_m) \\ & = k_1 T(\boldsymbol{\alpha}_1) + k_2 T(\boldsymbol{\alpha}_2) + \cdots + k_m T(\boldsymbol{\alpha}_m) \\ \end{aligned} T(β)=T(k1α1+k2α2+k3α3++kmαm)=T(k1α1)+T(k2α2+k3α3++kmαm)=T(k1α1)+T(k2α2)+T(k3α3+kmαm)=T(k1α1)+T(k2α2)++T(kmαm)=k1T(α1)+k2T(α2)++kmT(αm)
得证。

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