# 2016年考研数二第23题解析：相似对角化、特征值、特征向量、线性表示

## 题目

$(Ⅰ)$ 求 $A^{99}$;

$(Ⅱ)$ 设 $3$ 阶矩阵 $B=(\alpha_{1}, \alpha_{2}, \alpha_{3})$ 满足 $B^{2} = BA$. 记 $B^{100} = (\beta_{1}, \beta_{2}, \beta_{3})$, 将 $\beta_{1}$, $\beta_{2}$, $\beta_{3}$ 分别表示为 $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$ 的线性组合.

## 解析

### 第 $(Ⅰ)$ 问

$$|\lambda E – A| = 0 \Rightarrow$$

$${\color{Red} \begin{bmatrix} \lambda & 1 & -1\\ -2 & \lambda+3 & 0\\ 0 & 0 & \lambda \end{bmatrix} = 0}. \quad {\color{White}①} \Rightarrow$$

$$\lambda^{2} (\lambda + 3) + 2 \lambda = 0 \Rightarrow$$

$$\lambda[\lambda (\lambda + 3) + 2] = 0 \Rightarrow$$

$$\lambda(\lambda^{2} + 3 \lambda + 2) = 0 \Rightarrow$$

$$\left\{\begin{matrix} \lambda_{1} = 0;\\ \lambda_{2} = -1;\\ \lambda_{3} = -2. \end{matrix}\right.$$

$$\lambda_{1} E – A \Rightarrow$$

$$\begin{bmatrix} \lambda_{1} & 1 & -1\\ -2 & \lambda_{1}+3 & 0\\ 0 & 0 & \lambda_{1} \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} 0 & 1 & -1\\ -2 & 3 & 0\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 & 1 & -1\\ -2 & 0 & 3\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow$$

$$X_{1} = \begin{bmatrix} \frac{3}{2}\\ 1\\ 1 \end{bmatrix} \Rightarrow \begin{bmatrix} 3\\ 2\\ 2 \end{bmatrix}.$$

$$\lambda_{2} E – A \Rightarrow$$

$$\begin{bmatrix} \lambda_{2} & 1 & -1\\ -2 & \lambda_{2}+3 & 0\\ 0 & 0 & \lambda_{2} \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} -1 & 1 & -1\\ -2 & 2 & 0\\ 0 & 0 & -1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1 & -1 & 1\\ 0 & 0 & -2\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow$$

$$X_{2} = \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}.$$

$$\lambda_{3} E – A \Rightarrow$$

$$\begin{bmatrix} \lambda_{3} & 1 & -1\\ -2 & \lambda_{3}+3 & 0\\ 0 & 0 & \lambda_{3} \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} -2 & 1 & -1\\ -2 & 1 & 0\\ 0 & 0 & -2 \end{bmatrix} \Rightarrow \begin{bmatrix} -2 & 1 & -1\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow$$

$$X_{3} = \begin{bmatrix} \frac{1}{2}\\ 1\\ 0 \end{bmatrix} \Rightarrow \begin{bmatrix} 1\\ 2\\ 0 \end{bmatrix}.$$

$$P^{-1} A P = \Lambda \Rightarrow$$

$$A = P \Lambda P^{-1}.$$

$$(P \vdots E) \Rightarrow$$

$$\begin{bmatrix} 3 & 1 & 1 & \vdots & 1 & 0 & 0\\ 2 & 1 & 2 & \vdots & 0 & 1 & 0\\ 2 & 0 & 0 & \vdots & 0 & 0 & 1 \end{bmatrix} \Rightarrow$$

$$(E \vdots P^{-1}) = \begin{bmatrix} 1 & 0 & 0 & \vdots & 0 & 0 & \frac{1}{2}\\ 0 & 1 & 0 & \vdots & 2 & -1 & -2\\ 0 & 0 & 1 & \vdots & -1 & 1 & \frac{1}{2} \end{bmatrix}.$$

$$A = P \Lambda P^{-1} \Rightarrow$$

$$A^{99} = (P \Lambda P^{-1})^{99} \Rightarrow$$

$$A^{99} = (P \Lambda P^{-1})(P \Lambda P^{-1}) \cdots (P \Lambda P^{-1}) \cdots _{99个P \Lambda P^{-1}} \Rightarrow$$

$$A^{99} = P \Lambda (P^{-1} P) \Lambda (P^{-1} \cdots P) \Lambda P^{-1} \cdots _{99个P \Lambda P^{-1}} \Rightarrow$$

$$A^{99} = P \Lambda ^{99} P^{-1}.$$

$$A^{99} =$$

$$\begin{bmatrix} 3 & 1 & 1\\ 2 & 1 & 2\\ 2 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & & \\ & (-1)^{99} & \\ & & (-2)^{99} \end{bmatrix} \begin{bmatrix} 0 & 0 & \frac{1}{2}\\ 2 & -1 & -2\\ -1 & 1 & \frac{1}{2}. \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} 3 & 1 & 1\\ 2 & 1 & 2\\ 2 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & & \\ & -1 & \\ & & -2^{99} \end{bmatrix} \begin{bmatrix} 0 & 0 & \frac{1}{2}\\ 2 & -1 & -2\\ -1 & 1 & \frac{1}{2}. \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} 0 & -1 & -2^{99}\\ 0 & -1 & -2^{100}\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & \frac{1}{2}\\ 2 & -1 & -2\\ -1 & 1 & \frac{1}{2}. \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} -2+2^{99} & 1-2^{99} & 2-2^{98}\\ -2+2^{100} & 1-2^{100} & 2-2^{99}\\ 0 & 0 & 0 \end{bmatrix}.$$

### 第 $(Ⅱ)$ 问

$$B^{2} = BA \Rightarrow$$

$$B^{100} = B^{98}B^{2} \Rightarrow$$

$$B^{100} = B^{98}BA \Rightarrow$$

$$B^{100} = B^{99}A \Rightarrow$$

$$B^{100} = B^{97}B^{2}A \Rightarrow$$

$$B^{100} = B^{97}BAA \Rightarrow$$

$$B^{100} = B^{98}A^{2} \Rightarrow$$

$$\cdots \cdots$$

$$B^{100} = BA^{99} \Rightarrow$$

$$B^{100} = (\alpha_{1}, \alpha_{2}, \alpha_{3}) A^{99} \Rightarrow$$

$$(\beta_{1}, \beta_{2}, \beta_{3}) = (\alpha_{1}, \alpha_{2}, \alpha_{3}) A^{99} \Rightarrow$$

$$(\beta_{1}, \beta_{2}, \beta_{3}) = (\alpha_{1}, \alpha_{2}, \alpha_{3}) \begin{bmatrix} -2+2^{99} & 1-2^{99} & 2-2^{98}\\ -2+2^{100} & 1-2^{100} & 2-2^{99}\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow$$

$$\left\{\begin{matrix} \beta_{1} = (-2+2^{99}) \alpha_{1} + (-2+2^{100}) \alpha_{2} + 0 \alpha_{3};\\ \beta_{2} = (1-2^{99}) \alpha_{1} + ( 1-2^{100}) \alpha_{2} + 0 \alpha_{3};\\ \beta_{3} = (2-2^{98}) \alpha_{1} + (2-2^{99}) \alpha_{2} + 0 \alpha_{3}. \end{matrix}\right.$$