# 2018年考研数二第14题解析

## 解析

$$A(\alpha_{1}, \alpha_{2}, \alpha_{3}) = (A \alpha_{1}, A \alpha_{2}, A \alpha_{3}).$$

$$A(\alpha_{1}, \alpha_{2}, \alpha_{3}) =$$

$$(2 \alpha_{1}+\alpha_{2}+\alpha_{3}, \alpha_{2}+2\alpha_{3},- \alpha_{2}+\alpha_{3}) =$$

$$(\alpha_{1}, \alpha_{2}, \alpha_{3}) \begin{bmatrix} 2& 0& 0\\ 1& 1& -1\\ 1& 2& 1 \end{bmatrix} \Rightarrow$$

$$A(\alpha_{1}, \alpha_{2}, \alpha_{3}) =$$

$$(\alpha_{1}, \alpha_{2}, \alpha_{3}) \begin{bmatrix} 2& 0& 0\\ 1& 1& -1\\ 1& 2& 1 \end{bmatrix} \Rightarrow$$

$$P = (\alpha_{1}, \alpha_{2}, \alpha_{3});$$

$$Q = \begin{bmatrix} 2& 0& 0\\ 1& 1& -1\\ 1& 2& 1 \end{bmatrix}.$$

$$AP=PQ.$$

$$P^{-1} A P = P^{-1} P Q \Rightarrow$$

$$P^{-1} A P = Q.$$

$Q$ 的特征方程为：

$$|\lambda E – Q| = 0.$$

$$\begin{vmatrix} \lambda& 0& 0\\ 0& \lambda& 0\\ 0& 0& \lambda \end{vmatrix} \begin{vmatrix} 2& 0& 0\\ 1& 1& -1\\ 1& 2& 1 \end{vmatrix} =0 \Rightarrow$$

$$\begin{vmatrix} \lambda-2& 0& 0\\ -1& \lambda-1& 1\\ -1& -2& \lambda-1 \end{vmatrix} =0 \Rightarrow$$

$$(\lambda -2)(\lambda -1)^{2} + 2(\lambda – 2) = 0 \Rightarrow$$

$$(\lambda – 2)[(\lambda – 1)^{2} + 2] = 0. ①$$

$$(\lambda – 1)^{2} + 2 > 0.$$

EOF