$$|\lambda E-A|=\left|\begin{array}{ccc} \lambda-1 & -2 & -2 \\ -2 & \lambda-1 & 2 \\ 2 & 2 & \lambda-1 \end{array}\right|=\left|\begin{array}{ccc} \lambda-1 & \lambda-1 & \lambda-1 \\ -2 & \lambda-1 & 2 \\ 2 & 2 & \lambda-1 \end{array}\right|$$

\begin{aligned} & =(\lambda-1)\left|\begin{array}{ccc} 1 & 1 & 1 \\ -2 & \lambda-1 & 2 \\ 2 & 2 & \lambda-1 \\ 1 & 0 & 1 \\ -2 & \lambda-3 & 2 \\ 0 & 0 & \lambda+1 \end{array}\right|=(\lambda-1)\left|\begin{array}{ccc} 1 & 1 & 1 \\ -2 & \lambda-1 & 2 \\ 0 & \lambda+1 & \lambda+1 \end{array}\right| \\ & =(\lambda-1-1)(\lambda+1)\left|\begin{array}{cc} 1 & 0 \\ -2 & \lambda-3 \end{array}\right| \\ & =(\lambda+1)(\lambda-1)(\lambda-3) \end{aligned}

\begin{aligned} -E-A & =\left(\begin{array}{ccc} -2 & -2 & -2 \\ -2 & -2 & 2 \\ 2 & 2 & -2 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & -1 \end{array}\right) \\ & \rightarrow\left(\begin{array}{lll} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right) \rightarrow\left(\begin{array}{lll} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right) \end{aligned}

$$E-A=\left(\begin{array}{ccc} 0 & -2 & -2 \\ -2 & 0 & 2 \\ 2 & 2 & 0 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 0 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 0 & 1 & 1 \\ -1 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right)$$

$$3 E-A=\left(\begin{array}{ccc} 2 & -2 & -2 \\ -2 & 2 & 2 \\ 2 & 2 & 2 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & -1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right) \rightarrow\left(\begin{array}{lll} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right) .$$

\begin{aligned} & \Lambda=\left(\begin{array}{lll} 1 & & \\ & -1 & \\ & & 3 \end{array}\right) . \\ & \text { 令 } P=\left(\alpha_1, \alpha_2, \alpha_3\right)=\left(\begin{array}{ccc} 1 & 1 & 0 \\ -1 & -1 & 1 \\ 1 & 0 & -1 \end{array}\right) \text {, 由求逆矩阵的方法可知: } \\ & P^{-1}=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & -1 & -1 \\ 1 & 1 & 0 \end{array}\right) \\ & \end{aligned}

$$\left(P^{-1} A P\right)^n=P^{-1} A^n P=\Lambda^n=\left(\begin{array}{cc} 1 \text { ath }) & \\ (-1)^n & \\ & 3^n \end{array}\right) .$$

\begin{aligned} & A^n=P \Lambda^n P^{-1} \\ & =\left(\begin{array}{ccc} 1 & 1 & 0 \\ -1 & -1 & 1 \\ 1 & 0 & -1 \end{array}\right)\left(\begin{array}{ccc} 1 & & \\ & (-1)^n & \\ & & 3^n \end{array}\right)\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & -1 & -1 \\ 1 & 1 & 0 \end{array}\right) . \\ & =\left(\begin{array}{ccc} 1 & 1-(-1)^n & 1-(-1)^n \\ 3^n-1 & (-1)^n+3^n-1 & (-1)^n-1 \\ 1-3^n & 1-3^n & 1 \end{array}\right) \\ & \end{aligned}