这道题中的矩阵虽然很“宽”，但其实是一个单列矩阵

一、题目

已知，矩阵 $\boldsymbol{A}$ 满足对任意 $x_{1}, x_{2}, x_{3}$ 均有 $\boldsymbol{A}\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right)=\left(\begin{array}{l}x_{1}+x_{2}+x_{3} \\ 2 x_{1}-x_{2}+x_{3} \\ x_{2}-x_{3}\end{array}\right)$. 请求解以下两个问题：
[1]. 求 $A$;

[2]. 求可逆矩阵 $\boldsymbol{P}$ 与对角矩阵 $\boldsymbol{\Lambda}$, 使得 $\boldsymbol{P}^{-1} \boldsymbol{A P}=\boldsymbol{\Lambda}$.

难度评级：

二、解析

[1]

$$
A\left[\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right]=\left[\begin{array}{l}
x_{1}+x_{2}+x_{3} \\
2 x_{1}-x_{2}+x_{3} \\
x_{2}-x_{3}
\end{array}\right] \Rightarrow
$$

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

这里需要注意的是，$\left[\begin{array}{l}
x_{1}+x_{2}+x_{3} \\
2 x_{1}-x_{2}+x_{3} \\
x_{2}-x_{3}
\end{array}\right]$ 是一个只有一列的矩阵。

[2]

首先求解特征值：

$$
|\lambda E – A|=0 \Rightarrow\left|\begin{array}{ccc}
\lambda-1 & -1 & -1 \\
-2 & \lambda+1 & -1 \\
0 & -1 & \lambda+1
\end{array}\right|=0 \Rightarrow
$$

$$
(\lambda+1)^{2}(\lambda-1)-2-2(\lambda+1)-(r-1)=0 \Rightarrow
$$

$$
(\lambda+1)[(\lambda+1)(\lambda-1)-2]-(\lambda-1)-2=0
$$

$$
(\lambda+1)\left(\lambda^{2}-3\right)-(r+1)=0 \Rightarrow
$$

$$
(\lambda+1)\left(\lambda^{2}-4\right)=0 \Rightarrow
$$

$$
(\lambda+1)(\lambda+2)(\lambda-2)=0 \Rightarrow
$$

$$
\textcolor{orange}{
\lambda_{1}=-1, \quad \lambda_{2}=-2, \quad \lambda_{3}=2
}
$$

接着，分别求解特征值对应的特征向量。

求解 $\textcolor{orange}{ \lambda_{1} = -1 }$ 对应的特征向量 $\textcolor{orange}{ \alpha_{1} }$:

$$
\left(\lambda_{1} E-A\right) x=0 \Rightarrow\left(\lambda_{1} E-A\right)=
$$

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

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

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

$$
\textcolor{orange}{
\alpha_{1}=\left(\begin{array}{c}\frac{-1}{2} \\ 0 \\ 1\end{array}\right)
}
$$

求解 $\textcolor{orange}{ \lambda_{2} = -2 }$ 对应的特征向量 $\textcolor{orange}{ \alpha_{2} }$:

$$
\left(\lambda_{2} E-A\right)=\left[\begin{array}{ccc}-3 & -1 & -1 \\ -2 & -1 & -1 \\ 0 & -1 & -1\end{array}\right] \Rightarrow
$$

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

$$
\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{array}\right] \Rightarrow
$$

$$
\textcolor{orange}{
\alpha_{2}=\left(\begin{array}{cc}0 \\ -1 \\ 1\end{array}\right)
}
$$

求解 $\textcolor{orange}{ \lambda_{3} = 2 }$ 对应的特征向量 $\textcolor{orange}{ \alpha_{3} }$:

$$
\left(\lambda_{3} E – A \right)=\left[\begin{array}{ccc}1 & -1 & -1 \\ -2 & 3 & -1 \\ 0 & -1 & 3\end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{ccc}1 & -1 & -1 \\ 0 & 1 & -3 \\ 0 & 0 & 0\end{array}\right] \Rightarrow
$$

$$
\left[\begin{array}{ccc}1 & 0 & -4 \\ 0 & 1 & -3 \\ 0 & 0 & 0\end{array}\right] \Rightarrow
$$

$$
\textcolor{orange}{
\alpha_{3}=\left(\begin{array}{l}4 \\ 3 \\ 1\end{array}\right)
}
$$

综上可得：

$$
\textcolor{springgreen}{P} = \left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right) = \textcolor{springgreen}{ \left[\begin{array}{ccc}\frac{-1}{2} & 0 & 4 \\ 0 & -1 & 3 \\ 1 & 1 & 1\end{array}\right] } \Rightarrow
$$

$$
\textcolor{springgreen}{ P^{-1} A P} = \textcolor{springgreen}{ \left[\begin{array}{cc}-1 & & \\ & -2 & \\ & & 2 \end{array}\right] }
$$

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