# 当特征值等于零的时候，求解特征值和特征向量的式子其实就是一个齐次线性方程组

## 一、题目

### 解题思路简图

graph TD
A[原式] --> |变形| B[特征值] --> |公式| C[特征向量];
D[秩为 1] --> E[只有一个非零特征值] --> F[0 为二重特征值] --> |实对称矩阵| G[特征值对应的特征向量正交];
C --> G;
G --> H[求解特征值] --> |变形| I[验证选项]


## 二、解析

$$\textcolor{springgreen}{ \boldsymbol { A } \boldsymbol { \alpha } = 2 \boldsymbol { \alpha } }$$

$$\boldsymbol { \alpha } _ { 1 } = \boldsymbol { \alpha } = ( – 1 , 1 , 1 ) ^ { \mathrm {\top} }$$

$$\begin{cases} \lambda _ { 2 } = 0 \\ \lambda _ { 3 } = 0 \end{cases}$$

$$\boldsymbol { \beta } ^ { \mathrm {\top} } \boldsymbol { \alpha } = 0 \Rightarrow$$

$$(x_{1}, x_{2}, x_{3}) \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} = 0 \Rightarrow$$

$$– x _ { 1 } + x _ { 2 } + x _ { 3 } = 0 \Rightarrow$$

$$\begin{cases} x_{1} = 1 \\ x_{2} = 1 \\ x_{3} = 0 \end{cases} \text{ 或者 } \begin{cases} x_{1} = 1 \\ x_{2} = 0 \\ x_{3} = 1 \end{cases}$$

$$\begin{cases} & \boldsymbol { \beta } _ { 2 } = ( 1 , 1 , 0 ) ^ { \mathrm {\top} } \\ \\ & \boldsymbol { \beta } _ { 3 } = ( 1 , 0 , 1 ) ^ { \mathrm {\top} } \end{cases}$$

\begin{aligned} ( 0 E – \boldsymbol { A } ) \boldsymbol { \beta } _ { 2 } = 0 \\ & \Rightarrow – \boldsymbol { A } \boldsymbol { \beta } _ { 2 } = 0 \\ & \Rightarrow \boldsymbol { A } \boldsymbol { \beta } _ { 2 } = 0 \end{aligned}

\begin{aligned} ( 0 E – \boldsymbol { A } ) \boldsymbol { \beta } _ { 3 } = 0 \\ & \Rightarrow – \boldsymbol { A } \boldsymbol { \beta } _ { 3 } = 0 \\ & \Rightarrow \boldsymbol { A } \boldsymbol { \beta } _ { 3 } = 0 \end{aligned}

$$\begin{cases} ( 1 , 1 , 0 ) ^ { \top } \\ ( 1 , 0 , 1 ) ^ { \mathrm {\top} } \end{cases}$$