$\text{A.}$ $\boldsymbol{\alpha}_1+2 \boldsymbol{\alpha}_3$ $\text{B.}$ $\boldsymbol{\alpha}_2+2 \boldsymbol{\alpha}_3$ $\text{C.}$ $\boldsymbol{\alpha}_1+4 \boldsymbol{\alpha}_3$ $\text{D.}$ $\boldsymbol{\alpha}_2+4 \boldsymbol{\alpha}_3$
【答案】 C

【解析】 【解】由 $\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2\end{array}\right)$ 得 $\boldsymbol{P}^{-1} \boldsymbol{A}^2 \boldsymbol{P}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 4\end{array}\right), \boldsymbol{A}^2 \boldsymbol{P}=\boldsymbol{P}\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 4\end{array}\right)$,
$$\boldsymbol{A}^2\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3\right)=\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3\right)\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 4 \end{array}\right)=\left(\boldsymbol{\alpha}_1, \boldsymbol{0}, 4 \boldsymbol{\alpha}_3\right),$$

$$\boldsymbol{A}^2 \boldsymbol{\alpha}_1=\boldsymbol{\alpha}_1, \boldsymbol{A}^2 \boldsymbol{\alpha}_2=\mathbf{0}, \boldsymbol{A}^2 \boldsymbol{\alpha}_3=4 \boldsymbol{\alpha}_3 .$$