【ID】1847 【题型】解答题 【类型】考研真题 【来源】2011年全国硕士研究生招生考试试题

$$A\left(\begin{array}{rr} 1 & 1 \\ 0 & 0 \\ -1 & 1 \end{array}\right)=\left(\begin{array}{rr} -1 & 1 \\ 0 & 0 \\ 1 & 1 \end{array}\right) .$$
(I) 求 $\boldsymbol{A}$ 的所有特征值与特征向量;
(II) 求矩阵 $\boldsymbol{A}$.

(II) 由于不同特征值的特征向量正交, 则只需将 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 单位化, 得 $\boldsymbol{r}_{1}=\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}}\end{array}\right), \boldsymbol{r}_{2}=\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}\end{array}\right), \boldsymbol{r}_{3}=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)$

$$=\left(\begin{array}{ccc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{array}\right)\left(\begin{array}{ccc} \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \end{array}\right)=\left(\begin{array}{lll} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right) .$$