# 伴随矩阵的特征值与原矩阵的特征值之间有什么关系？

## 二、解析

### 解法一：用特征值的定义求解

$$B = \alpha \boldsymbol { \beta } ^ { \mathrm { \top } }$$

$$B = \left[ \begin{array} { c c } 1 \\ 3 \\ 2 \end{array} \right] [ 1 , – 1 , 2 ] = \left[ \begin{array} { c c c } 1 & – 1 & 2 \\ 3 & – 3 & 6 \\ 2 & – 2 & 4 \end{array} \right]$$

\begin{aligned} | \lambda E – B | \\ \\ & = \lambda ^ { 3 } – 2 \lambda ^ { 2 } \\ \\ & = \lambda ^ { 2 } ( \lambda – 2 ) \\ \\ & = 0 \\ \\ & \Rightarrow \begin{cases} \lambda_{1} = 2 \\ \lambda_{2} = 0 \\ \lambda_{3} = 0 \end{cases} \end{aligned}

$$\begin{cases} \lambda_{a} = 2 \\ \lambda_{b} = 0 \\ \lambda_{c} = 0 \end{cases}$$

$$\begin{cases} \lambda_{ * } = 5 \\ \lambda_{ ** } = 1 \\ \lambda_{ *** } = 1 \end{cases}$$

$$| 2 A + E | = 5 \times 1 \times 1 = 5$$

$$\begin{cases} \bar{\lambda_{1}} = 1 \\ \bar{\lambda_{2}} = 5 \\ \bar{\lambda_{3}} = 5 \end{cases}$$

### 解法二：用秩为 1 的矩阵的性质

$$B = \alpha \beta ^ { \mathrm { \top } }$$

\begin{aligned} B \\ \\ & = \left[ \begin{array} { c c } 1 \\ 3 \\ 2 \end{array} \right] [ 1 , – 1 , 2 ] \\ \\ & = \left[ \begin{array} { c c c } 1 & – 1 & 2 \\ 3 & – 3 & 6 \\ 2 & – 2 & 4 \end{array} \right] \\ \\ & = \left[ \begin{array} { c c c } 1 & – 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \end{aligned}

$$r(B) = 1$$

$$\begin{cases} \lambda_{1} = 0 \\ \lambda_{2} = 0 \\ \lambda_{3} = \operatorname { t r } ( B ) = \boldsymbol { \beta } ^ { \mathrm { \top } } \boldsymbol { \alpha } = 1 – 3 + 4 = 2 \end{cases}$$

$$\begin{cases} \lambda_{1} = 0 \\ \lambda_{2} = 0 \\ \lambda_{3} = 2 \end{cases}$$