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：

$$\begin{array}{rl}\mathit{A}& \left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\\ \\ \left(\begin{array}{ccc}1& 1& 1\\ 2& –1& 1\\ 0& 1& –1\end{array}\right)& \left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}{x}_1+x_2+x_3\\ 2x_1–x_2+x_3\\ x_2–x_3\end{array}\right)\end{array}$$

所以：

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

第 (2) 问

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

Note

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

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$$\begin{array}{r}|\lambda E–A|\\ \\ & =\left|\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right|–\left|\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 0& 1& –1\end{array}\right|\\ \\ & =\left|\begin{array}{ccc}\lambda –1& –1& –1\\ –2& \lambda +1& –1\\ 0& –1& \lambda +1\end{array}\right|\\ \\ & =(\lambda –1)\cdot \left|\begin{array}{cc}\lambda +1& –1\\ –1& \lambda +1\end{array}\right|+2\cdot \left|\begin{array}{cc}–1& –1\\ –1& \lambda +1\end{array}\right|\\ \\ & =(\lambda –1)({\lambda }^2+2\lambda )+2[(-\lambda –1)–1]\\ \\ & =(\lambda –1)({\lambda }^2+2\lambda )–2(\lambda +2)\\ \\ & =(\lambda –1)\lambda (\lambda +2)–2(\lambda +2)\\ \\ & =(\lambda +2)(\lambda –2)(\lambda +1)=0\\ \\ & \Rightarrow \{\begin{array}{l}{\lambda }_1=-2\\ {\lambda }_2=2\\ {\lambda }_3=-1\end{array}\end{array}$$

也就是说，矩阵 $\mathit{A}$ 的特征值为：

$$\begin{array}{rl}{\lambda }_1& =–2\\ {\lambda }_2& =2\\ {\lambda }_3& =–1\end{array}$$

Note

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

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[✓]. 当 $\lambda$ $=$ ${\lambda }_1$ $=$ $-2$ 时

$$\begin{array}{r}{\lambda }_{1}E–A\\ \\ & =\left(\begin{array}{ccc}–3& –1& –1\\ –2& –1& –1\\ 0& –1& –1\end{array}\right)\\ \\ & \to \left(\begin{array}{ccc}3& 1& 1\\ –2& –1& –1\\ 0& –1& –1\end{array}\right)\\ \\ & \to \left(\begin{array}{ccc}1& 0& 0\\ –2& –1& –1\\ 0& –1& –1\end{array}\right)\\ \\ & \to \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& –1& –1\end{array}\right)\\ \\ & \to \left(\begin{array}{ccc}{x}_1& x_2& x_3\\ 1& 0& 0\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)\end{array}$$

于是可知，若令自由未知数 $x_3$ $=$ $1$, 则非自由未知数 $x_1$ $=$ $0$, $x_2$ $=$ $-1$, 即 ${\lambda }_1$ 对应的特征向量为：

$${\mathit{\alpha }}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{,}–\mathbf{1}\mathbf{,}\mathbf{1}{\mathbf{)}}^{\mathbf{\top }}$$

[✓]. 当 $\lambda$ $=$ ${\lambda }_2$ $=$ $2$ 时

$$\begin{array}{r}{\lambda }_{2}E–A\\ \\ & =\left(\begin{array}{ccc}1& –1& –1\\ –2& 3& –1\\ 0& –1& 3\end{array}\right)\\ \\ & =\left(\begin{array}{ccc}1& –1& –1\\ 0& 1& –3\\ 0& –1& 3\end{array}\right)\\ \\ & \to \left(\begin{array}{ccc}{x}_1& x_2& x_3\\ 1& 0& –4\\ 0& 1& –3\\ 0& 0& 0\end{array}\right)\end{array}$$

于是可知，若令自由未知数 $x_3$ $=$ $1$, 则非自由未知数 $x_1$ $=$ $4$, $x_2$ $=$ $3$, 即 ${\lambda }_1$ 对应的特征向量为：

$${\mathit{\alpha }}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{1}{\mathbf{)}}^{\mathbf{\top }}$$

[✓]. 当 $\lambda$ $=$ ${\lambda }_3$ $=$ $-1$ 时

$$\begin{array}{r}{\lambda }_{3}E–A\\ \\ & =\left(\begin{array}{ccc}–2& –1& –1\\ –2& 0& –1\\ 0& –1& 0\end{array}\right)\\ \\ & =\left(\begin{array}{ccc}2& 1& 1\\ –2& 0& –1\\ 0& –1& 0\end{array}\right)\\ \\ & \to \left(\begin{array}{ccc}2& 0& 1\\ 0& 1& 0\\ 0& -1& 0\end{array}\right)\\ \\ & \to \left(\begin{array}{ccc}{x}_1& x_2& x_3\\ 2& 0& 1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)\end{array}$$

于是可知，若令自由未知数 $x_1$ $=$ $1$, 则非自由未知数 $x_2$ $=$ $0$, $x_3$ $=$ $-2$, 即 ${\lambda }_1$ 对应的特征向量为：

$${\mathit{\alpha }}_{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}–\mathbf{2}{\mathbf{)}}^{\mathbf{\top }}$$

因此，若令 $\mathit{P}$ $=$ $\mathbf{(}{\mathit{\alpha }}_{\mathbf{1}}\mathbf{,}{\mathit{\alpha }}_{\mathbf{2}}\mathbf{,}{\mathit{\alpha }}_{\mathbf{3}}\mathbf{)}$, 则 ${\mathit{P}}^{–\mathbf{1}}\mathit{A}\mathit{P}$ $=$ $\left(\begin{array}{ccc}–2& 0& 0\\ 0& 2& 0\\ 0& 0& –1\end{array}\right)$ $=$ $\mathbf{\Lambda }$, 即此时 ${\mathit{P}}^{–\mathbf{1}}\mathit{A}\mathit{P}$ $=$ $\mathbf{\Lambda }$ 成立

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