# 2024年考研数二第10题解析：相似对角化、矩阵的特征值与特征向量

## 二、解析

### 充分性的证明

#### 证明方法一

$$A \alpha=\lambda \alpha$$

$$B A \alpha = B \lambda \alpha \Rightarrow$$

$$B A \alpha = \lambda B \alpha$$

$$\textcolor{pink}{A} \textcolor{orangered}{\boldsymbol{B \alpha}} = \textcolor{springgreen}{\lambda} \textcolor{orangered}{\boldsymbol{B \alpha}}$$

(1) 若 $B \alpha \neq 0$, 则 $B \alpha$ 就是矩阵 $A$ 对应于 $\lambda$ 的特征向量。若 $B \alpha = k \alpha$ $(k \neq 0)$, 则 $\alpha$ 就是矩阵 $B$ 对应于 $\lambda=k$ 的特征向量；

(2)若 $B \alpha=0$, 则 $B \alpha$ 仍然是矩阵 $A$ 对应于 $\lambda$ 的特征向量。且此时只能有 $B \alpha = 0 \cdot \alpha$, 即 $\alpha$ 为矩阵 $B$ 对应于 $\lambda=0$ 的特征向量。

#### 证明方法二

$$\boldsymbol{P}=\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}\right)$$

$$\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}=\left(\begin{array}{ll}\lambda_{1} & \\ & \lambda_{2}\end{array}\right)$$

$$\boldsymbol{A B}=\boldsymbol{B A}$$

$$\boldsymbol{A} \boldsymbol{ \textcolor{springgreen}{B \alpha}}_{\textcolor{springgreen}{i}}=\boldsymbol{B} \boldsymbol{A} \boldsymbol{\alpha}_{i} = \boldsymbol{B} \lambda_{i} \boldsymbol{\alpha}_{i}=\lambda_{i} \boldsymbol{\textcolor{springgreen}{B \alpha}}_{\textcolor{springgreen}{i}}$$

$$\boldsymbol{B} \boldsymbol{\alpha}_{i}=k_{i} \boldsymbol{\alpha}_{i}$$

\begin{aligned} \boldsymbol{B P} \\ & = \boldsymbol{B}\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}\right) \\ & = \left(\boldsymbol{ B \alpha}_{1}, \boldsymbol{B \alpha}_{2}\right) \\ & = \left(k_{1} \boldsymbol{\alpha}_{1}, k_{2} \boldsymbol{\alpha}_{2}\right) \\ & = \left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}\right)\left(\begin{array}{ll} k_{1} & \\ & k_{2} \end{array}\right) \\ & = \boldsymbol{P}\left(\begin{array}{ll} k_{1} & \\ & k_{2} \end{array}\right) \end{aligned}

$$\boldsymbol{P}^{-1} \boldsymbol{B P}=\left(\begin{array}{ll}k_{1} & \\ & k_{2}\end{array}\right)$$

### 特别专题

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