【证明】不同特征值对应的线性无关的特征向量合并后仍然线性无关

推论　设 $\lambda_1$ 和 $\lambda_2$ 是方阵 $\boldsymbol{A}$ 的两个不同特征值，$\boldsymbol{\xi}_1,\boldsymbol{\xi}_2,\cdots,\boldsymbol{\xi}_s$ 和 $\boldsymbol{\eta}_1,\boldsymbol{\eta}_2,\cdots,\boldsymbol{\eta}_t$ 分别是对应于 $\lambda_1$ 和 $\lambda_2$ 的线性无关的特征向量，则 $\boldsymbol{\xi}_1,\boldsymbol{\xi}_2,\cdots,\boldsymbol{\xi}_s,\boldsymbol{\eta}_1,\boldsymbol{\eta}_2,\cdots,\boldsymbol{\eta}_t$

证明　设有
$x_1 \boldsymbol{\xi}_1 + x_2 \boldsymbol{\xi}_2 + \cdots + x_s \boldsymbol{\xi}_s + y_1 \boldsymbol{\eta}_1 + y_2 \boldsymbol{\eta}_2 + \cdots + y_t \boldsymbol{\eta}_t = \boldsymbol{0} \tag{1}$

用 $\boldsymbol{A}$ 左乘上式，得
$x_1 \boldsymbol{A} \boldsymbol{\xi}_1 + x_2 \boldsymbol{A} \boldsymbol{\xi}_2 + \cdots + x_s \boldsymbol{A} \boldsymbol{\xi}_s + y_1 \boldsymbol{A} \boldsymbol{\eta}_1 + y_2 \boldsymbol{A} \boldsymbol{\eta}_2 + \cdots + y_t \boldsymbol{A} \boldsymbol{\eta}_t = \boldsymbol{0} \tag{2}$
即
$x_1 \lambda_1 \boldsymbol{\xi}_1 + x_2 \lambda_1 \boldsymbol{\xi}_2 + \cdots + x_s \lambda_1 \boldsymbol{\xi}_s + y_1 \lambda_2 \boldsymbol{\eta}_1 + y_2 \lambda_2 \boldsymbol{\eta}_2 + \cdots + y_t \lambda_2 \boldsymbol{\eta}_t = \boldsymbol{0} \tag{3}$

$(3)$ 式减去 $(1)$ 式的 $\lambda_2$ 倍，得
$x_1 (\lambda_1 - \lambda_2) \boldsymbol{\xi}_1 + x_2 (\lambda_1 - \lambda_2) \boldsymbol{\xi}_2 + \cdots + x_s (\lambda_1 - \lambda_2) \boldsymbol{\xi}_s = \boldsymbol{0} \tag{4}$
因为 $\boldsymbol{\xi}_1,\boldsymbol{\xi}_2,\cdots,\boldsymbol{\xi}_s$ 线性无关，所以 $x_i (\lambda_1 - \lambda_2) = 0 \ (i=1,2,\cdots,s)$。又因为 $\lambda_1 \ne \lambda_2$，于是得
$x_i = 0 \hspace{1em} (i=1,2,\cdots,k-1) \tag{5}$
同理可得
$y_i = 0 \hspace{1em} (i=1,2,\cdots,t) \tag{6}$
将 $(5)$ 式和 $(6)$ 式代入 $(1)$ 可知，向量组 $\boldsymbol{\xi}_1,\boldsymbol{\xi}_2,\cdots,\boldsymbol{\xi}_s,\boldsymbol{\eta}_1,\boldsymbol{\eta}_2,\cdots,\boldsymbol{\eta}_t$