将特征向量乘以一个倍数 k 并不会改变其原本对应的特征值

一、题目

已知 $\boldsymbol{A}$ 是三阶矩阵，其特征值是 $1,3,-2$, 相应的特征向量依次为 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$, 若 $\boldsymbol{P}=\left[\boldsymbol{\alpha}_{1}, 2 \boldsymbol{\alpha}_{3},-\boldsymbol{\alpha}_{2}\right]$, 则 $\boldsymbol{P}^{-1} \boldsymbol{A P}=?$

难度评级：

二、解析

快速解法

由于：

$$
A \alpha_{1}=1 \cdot \alpha_{1}
$$

$$
A \alpha_{3}=-2 \alpha_{3} \Rightarrow A\left(2 \alpha_{3}\right)=-2\left(2 \alpha_{3}\right)
$$

$$
A \alpha_{2}=3 \alpha_{2} \Rightarrow A\left(-\alpha_{2}\right)=3\left(-\alpha_{2}\right)
$$

因此可知，$\alpha_{1}$ 对应的特征值为 $1$, $2\alpha_{3}$ 对应的特征值为 $-2$, $-\alpha_{2}$ 对应的特征值为 $3$, 因此：

$$
P^{-1} A P = \left[\begin{array}{lll}1 & & \\ & -2 & \\ & & 3\end{array}\right]
$$

常规解法

$$
P=\left(\alpha_{1}, 2 \alpha_{3},-\alpha_{2}\right) \Rightarrow
$$

$$
\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 2 & 0\end{array}\right] = P \Rightarrow
$$

令：

$$
B=\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)
$$

$$
C=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 2 & 0\end{array}\right]
$$

则：

$$
P=B C \Rightarrow P^{-1} A P=(B C)^{-1} A(B C)=
$$

$$
C^{-1} B^{-1} A B C=C^{-1}\left(B^{-1} A B\right) C=
$$

$$
C^{-1}\left[\begin{array}{lll}1 & & \\ & 3 & \\ & & -2\end{array}\right] C
$$

接着，求解 $C^{-1}$:

$$
\left[\begin{array}{cccccc}1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 & 0 & 1\end{array}\right]=\left[\begin{array}{llllll}1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 0 & -1 & 0\end{array}\right] \Rightarrow
$$

$$
C^{-1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \\ 0 & -1 & 0\end{array}\right]
$$

于是：

$$
C^{-1}\left[\begin{array}{ccc}1 & & \\ & 3 & \\ & & -2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \\ 0 & -1 & 0\end{array}\right]\left[\begin{array}{lll}1 & & \\ & 3 & \\ & & -2\end{array}\right]=
$$

$$
\left[\begin{array}{lll}1 & & \\ & & -1 \\ & -3 & \end{array}\right]
$$

接着：

$$
\left[\begin{array}{lll}1 & & \\ & -1 \\ & -3\end{array}\right] C \Rightarrow
$$

$$
\left[\begin{array}{lll}1 & & \\ & & -1 \\ & -3 & \end{array}\right]\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 2 & 0\end{array}\right]=\left[\begin{array}{lll}1 & & \\ & -2 & \\ & & 3\end{array}\right]
$$

即：

$$
P^{-1} A P = \left[\begin{array}{lll}1 & & \\ & -2 & \\ & & 3\end{array}\right]
$$

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