$\Rightarrow$ 当 $\lambda \boldsymbol{E}-\boldsymbol{A}$ 中无零行, 可令其任意一行(成比例行优先)为零行且换到第 $n$ 行 (称为零行置换)

$$|\lambda \boldsymbol{E}-\boldsymbol{A}|=\left|\begin{array}{ccc} \lambda-2 & -1 & -1 \\ -1 & \lambda-2 & -1 \\ -1 & -1 & \lambda-2 \end{array}\right|$$
=
$$\left|\begin{array}{ccc} \lambda-4 & -1 & -1 \\ \lambda-4 & \lambda-2 & -1 \\ \lambda-4 & -1 & \lambda-2 \end{array}\right|$$
=
$$(\lambda-4)\left|\begin{array}{ccc} 1 & -1 & -1 \\ 1 & \lambda-2 & -1 \\ 1 & -1 & \lambda-2 \end{array}\right|$$
=
$$=(\lambda-4)\left|\begin{array}{ccc} 1 & -1 & -1 \\ 0 & \lambda-1 & 0 \\ 0 & 0 & \lambda-1 \end{array}\right|$$
=
$$(\lambda-4)(\lambda-1)^2,$$

$4E-A$= $\left(\begin{array}{ccc} x_1 & x_2 & x_3 \\ 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right),$

$$\left\{\begin{array}{l} x_1=x_3, \\ x_2=x_3 \end{array} \text { ( } x_3 \text { 为自由末知量), 取 } x_3=1 \text { 得矩阵 } \boldsymbol{A} \text { 对应于特征值 } \lambda_1=4 \text { 的特征向量: } \xi_1=\left(\begin{array}{l} 1 \\ 1 \\ 10 \end{array}\right)\right. \text {, }$$

$E-A$=$\left(\begin{array}{ccc} x_1 & x_2 & x_3 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$

$$\text { 取 }\left(\begin{array}{l} x_2 \\ x_3 \end{array}\right)=\left(\begin{array}{l} 1 \\ 0 \end{array}\right),\left(\begin{array}{l} 0 \\ 1 \end{array}\right) \text { 得矩阵 } \boldsymbol{A} \text { 对应于特征值 } \lambda_2=\lambda_3=1 \text { 的特征向量: } \xi_2=\left(\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right), \xi_3=\left(\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right) \text {; }$$

$$\Lambda^{20}=\left(\begin{array}{lll} 4 & & \\ & 1 & \\ & & 1 \end{array}\right)^{20}=\left(\begin{array}{ccc} 4^{20} & & \\ & 1 & \\ & & 1 \end{array}\right)$$

$$(\boldsymbol{P}, \boldsymbol{E})=\left(\begin{array}{rrrlll} 1 & -1 & -1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 \end{array}\right)$$

$$\boldsymbol{P}^{-1}=\left(\begin{array}{ccc} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{array}\right)=\frac{1}{3}\left(\begin{array}{rrr} 1 & 1 & 1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{array}\right)$$

$$\boldsymbol{A}^{20}=\boldsymbol{P} \boldsymbol{\Lambda}^{20} \boldsymbol{P}^{-1}= =\left(\begin{array}{rrr} 1 & -1 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right)\left(\begin{array}{lll} 4^{20} & & \\ & 1 & \\ & & 1 \end{array}\right) \frac{1}{3}\left(\begin{array}{rrr} 1 & 1 & 1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{array}\right)$$

$$=\frac{1}{3}\left(\begin{array}{ccc} 4^{20}+2 & 4^{20}-1 & 4^{20}-1 \\ 4^{20}-1 & 4^{20}+2 & 4^{20}-1 \\ 4^{20}-1 & 4^{20}-1 & 4^{20}+2 \end{array}\right) .$$