2023年考研数二第22题解析：根据矩阵乘法凑出隐含的矩阵、矩阵的特征值和特征向量

一、题目

难度评级：

二、解析

第 (1) 问

由于下式对任意的 $x _{ 1 }$, $x _{ 2 }$, $x _{ 3 }$ 均成立：

$$
\begin{aligned}
\boldsymbol{\textcolor{orangered}{A}} & \textcolor{springgreen}{\left( \begin{array} { c } x _{ 1 } \\ x _{ 2 } \\ x _{ 3 } \end{array} \right) } = \\ \\
\textcolor{orangered}{\left( \begin{array} { c c c } 1 & 1 & 1 \\ 2 & – 1 & 1 \\ 0 & 1 & – 1 \end{array} \right) } & \textcolor{springgreen}{\left( \begin{array} { c } x _{ 1 } \\ x _{ 2 } \\ x _{ 3 } \end{array} \right) } = \left( \begin{array} { c c } x _{ 1 } + x _{ 2 } + x _{ 3 } \\ 2 x _{ 1 } – x _{ 2 } + x _{ 3 } \\ x _{ 2 } – x _{ 3 } \end{array} \right)
\end{aligned}
$$

所以：

$$
\boldsymbol{A} = \left( \begin{array} { c c c } 1 & 1 & 1 \\ 2 & – 1 & 1 \\ 0 & 1 & – 1 \end{array} \right)
$$

第 (2) 问

满足 $\boldsymbol{P ^{-1} A P}$ $=$ $\boldsymbol{\Lambda}$ 的对角矩阵 $\boldsymbol{\Lambda}$ 由矩阵的特征值组成，而矩阵 $\boldsymbol{P}$ 则由特征值对应的特征向量组成。所以，我们要求解矩阵 $\boldsymbol{P}$ 就要先求解出矩阵 $\boldsymbol{A}$ 的特征值 $\lambda$:

Note

事实上，表示对角矩阵的希腊字母 “$\Lambda$” 就是表示特征值的希腊字母 “$\lambda$” 的大写形式。

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$$
\begin{aligned}
|\lambda E – A | \\ \\
& = \left| \begin{array} { c c c } \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{array} \right| – \left| \begin{array} { c c c } 1 & 1 & 1 \\ 2 & -1 & 1 \\ 0 & 1 & – 1 \end{array} \right| \\ \\
& = \textcolor{blue}{\left| \begin{array} { c c c } \lambda – 1 & – 1 & – 1 \\ – 2 & \lambda + 1 & – 1 \\ 0 & – 1 & \lambda + 1 \end{array} \right|} \\ \\
& = ( \lambda – 1 ) \cdot \left| \begin{array} { c c } \lambda + 1 & – 1 \\ – 1 & \lambda + 1 \end{array} \right| + 2 \cdot \left| \begin{array} { c c } – 1 & – 1 \\ – 1 & \lambda + 1 \end{array} \right| \\ \\
& = ( \lambda – 1 ) \left( \lambda ^ { 2 } + 2 \lambda \right) + 2 [ (- \lambda – 1) – 1 ] \\ \\
& = ( \lambda – 1 ) \left( \lambda ^ { 2 } + 2 \lambda \right) – 2 ( \lambda + 2 ) \\ \\
& = ( \lambda – 1 ) \lambda \left( \lambda + 2 \right) – 2 ( \lambda + 2 ) \\ \\
& = ( \lambda + 2 ) ( \lambda – 2 ) ( \lambda + 1 ) = 0 \\ \\
& \Rightarrow \begin{cases}
\lambda_{1} = -2 \\
\lambda_{2} = 2 \\
\lambda_{3} = -1
\end{cases}
\end{aligned}
$$

也就是说，矩阵 $\boldsymbol{A}$ 的特征值为：

$$
\begin{aligned}
\lambda _{ 1 } & = – 2 \\
\lambda _{ 2 } & = 2 \\
\lambda _{ 3 } & = – 1
\end{aligned}
$$

Note

接下来的计算需要用到前面已经得到的矩阵：$\textcolor{blue}{\left| \begin{array} { c c c } \lambda – 1 & – 1 & – 1 \\ – 2 & \lambda + 1 & – 1 \\ 0 & – 1 & \lambda + 1 \end{array} \right|}$

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[✓]. 当 $\lambda$ $=$ $\lambda _{ 1 }$ $=$ $- 2$ 时

$$
\begin{aligned}
\lambda _{ 1 } E – A \\ \\
& = \left( \begin{array} { c c c } – 3 & – 1 & – 1 \\ – 2 & – 1 & – 1 \\ 0 & – 1 & – 1 \end{array} \right) \\ \\
& \rightarrow \left( \begin{array} { c c c } 3 & 1 & 1 \\ – 2 & – 1 & – 1 \\ 0 & – 1 & – 1 \end{array} \right) \\ \\
& \rightarrow \left( \begin{array} { c c c } 1 & 0 & 0 \\ – 2 & – 1 & – 1 \\ 0 & – 1 & – 1 \end{array} \right) \\ \\
& \rightarrow \left( \begin{array} { c c c } 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & – 1 & – 1 \end{array} \right) \\ \\
& \rightarrow \left( \begin{array} { c c c } \textcolor{pink}{x_{1}} & \textcolor{pink}{x_{2}} & \textcolor{springgreen}{x_{3}} \\ 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right)
\end{aligned}
$$

于是可知，若令自由未知数 $\textcolor{springgreen}{x_{3}}$ $\textcolor{springgreen}{=}$ $\textcolor{springgreen}{1}$, 则非自由未知数 $\textcolor{pink}{x_{1}}$ $\textcolor{pink}{=}$ $\textcolor{pink}{0}$, $\textcolor{pink}{x_{2}}$ $\textcolor{pink}{=}$ $\textcolor{pink}{-1}$, 即 $\lambda_{1}$ 对应的特征向量为：

$$
\textcolor{tan}{
\boldsymbol{
\alpha _{ 1 } = ( 0 , – 1 , 1 ) ^ { \top }
}
}
$$

[✓]. 当 $\lambda$ $=$ $\lambda _{ 2 }$ $=$ $2$ 时

$$
\begin{aligned}
\lambda _{ 2 } E – A \\ \\
& = \left( \begin{array} { c c c } 1 & – 1 & – 1 \\ – 2 & 3 & – 1 \\ 0 & – 1 & 3 \end{array} \right) \\ \\
& = \left( \begin{array} { c c c } 1 & – 1 & – 1 \\ 0 & 1 & – 3 \\ 0 & – 1 & 3 \end{array} \right) \\ \\
& \rightarrow \left( \begin{array} { c c c } \textcolor{pink}{x_{1}} & \textcolor{pink}{x_{2}} & \textcolor{springgreen}{x_{3}} \\ 1 & 0 & – 4 \\ 0 & 1 & – 3 \\ 0 & 0 & 0 \end{array} \right)
\end{aligned}
$$

于是可知，若令自由未知数 $\textcolor{springgreen}{x_{3}}$ $\textcolor{springgreen}{=}$ $\textcolor{springgreen}{1}$, 则非自由未知数 $\textcolor{pink}{x_{1}}$ $\textcolor{pink}{=}$ $\textcolor{pink}{4}$, $\textcolor{pink}{x_{2}}$ $\textcolor{pink}{=}$ $\textcolor{pink}{3}$, 即 $\lambda_{1}$ 对应的特征向量为：

$$
\textcolor{tan}{
\boldsymbol{
\alpha _{ 2 } = ( 4 , 3 , 1 ) ^ { \top }
}
}
$$

[✓]. 当 $\lambda$ $=$ $\lambda _{ 3 }$ $=$ $- 1$ 时

$$
\begin{aligned}
\lambda _{ 3 } E – A \\ \\
& = \left( \begin{array} { c c c } – 2 & – 1 & – 1 \\ – 2 & 0 & – 1 \\ 0 & – 1 & 0 \end{array} \right) \\ \\
& = \left( \begin{array} { c c c } 2 & 1 & 1 \\ – 2 & 0 & – 1 \\ 0 & – 1 & 0 \end{array} \right) \\ \\
& \rightarrow \left( \begin{array} { c c c } 2 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & -1 & 0 \end{array} \right) \\ \\
& \rightarrow \left( \begin{array} { c c c } \textcolor{springgreen}{x_{1}} & \textcolor{pink}{x_{2}} & \textcolor{pink}{x_{3}} \\ 2 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right)
\end{aligned}
$$

于是可知，若令自由未知数 $\textcolor{springgreen}{x_{1}}$ $\textcolor{springgreen}{=}$ $\textcolor{springgreen}{1}$, 则非自由未知数 $\textcolor{pink}{x_{2}}$ $\textcolor{pink}{=}$ $\textcolor{pink}{0}$, $\textcolor{pink}{x_{3}}$ $\textcolor{pink}{=}$ $\textcolor{pink}{-2}$, 即 $\lambda_{1}$ 对应的特征向量为：

$$
\textcolor{tan}{
\boldsymbol{
\alpha _{ 3 } = ( 1 , 0 , – 2 ) ^ { \top }
}
}
$$

因此，若令 $\boldsymbol{P}$ $=$ $\boldsymbol{(\alpha_{1}, \alpha_{2}, \alpha_{3}) }$, 则 $\boldsymbol{P ^ { – 1 } A P}$ $=$ $\left( \begin{array} { c c c } – 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & – 1 \end{array} \right)$ $=$ $\boldsymbol{\Lambda}$, 即此时 $\boldsymbol{P ^ { – 1 } A P}$ $=$ $\boldsymbol{\Lambda}$ 成立

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