2022考研数二第22题解析：二次型、正交变换、标准型

一、题目

二、解析

第（Ⅰ）问

由二次型 $f$ 的表达式 $f\left(x_{1}, x_{2}, x_{3}\right) = 3 x_{1}^{2} + 4 x_{2}^{2} + 3 x_{3}^{2} + 2 x_{1} x_{3}$ 可知，二次型 $f$ 对应的矩阵 $\boldsymbol{A}$ 为：

$$
\boldsymbol{A} = \begin{pmatrix}
3 & 0 & 1 \\
0 & 4 & 0 \\
1 & 0 & 3
\end{pmatrix}
$$

接着，通过矩阵 $\boldsymbol{A}$ 的特征多项式求解矩阵 $\boldsymbol{A}$ 的特征值：

$$
\begin{aligned}
\begin{vmatrix}
\lambda \boldsymbol{E} – \boldsymbol{A}
\end{vmatrix} & = \begin{vmatrix}
\lambda – 3 & 0 & -1 \\
0 & \lambda – 4 & 0 \\
-1 & 0 & \lambda – 3
\end{vmatrix} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\text{在第二行展开这个行列式}} \\ \\
& = \left(\lambda – 4\right) \begin{vmatrix}
\lambda – 3 & -1 \\
-1 & \lambda – 3
\end{vmatrix} \\ \\
& = \left(\lambda – 4\right) \cdot \left[ \left(\lambda – 3 \right)^{2} – 1 \right] \\ \\
& = \left(\lambda – 4\right) \left(\lambda^{2} – 6 \lambda + 9 – 1\right) \\ \\
& = \left(\lambda – 4\right) \left(\lambda^{2} – 6 \lambda + 8\right) \\ \\
& \textcolor{lightgreen}{ \leadsto } \begin{cases}
\lambda – 4 = 0 \\
\lambda^{2} – 6 \lambda + 8 = 0
\end{cases} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{lightgreen}{ \lambda_{1} = 2, \, \lambda_{2} = 4, \, \lambda_{3} = 4 }
\end{aligned}
$$

于是，根据特征值的计算公式 $\left( \lambda_{i} \boldsymbol{E} – \boldsymbol{A} \right) \boldsymbol{\alpha}_{i} = \left( \boldsymbol{A} – \lambda_{i} \boldsymbol{E} \right) \boldsymbol{\alpha}_{i} = 0$, 可知：

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda = 2$ 时：

$$
2 \boldsymbol{E} – \boldsymbol{A} = \begin{pmatrix}
-1 & 0 & -1 \\
0 & -2 & 0 \\
-1 & 0 & -1
\end{pmatrix} \to \begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{pmatrix} \to \textcolor{lightgreen}{ \boldsymbol{\alpha}_{3} = \begin{pmatrix}
-1 \\
0 \\
1
\end{pmatrix}
}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $\lambda = 4$ 时：

$$
4 \boldsymbol{E} – \boldsymbol{A} = \begin{pmatrix}
1 & 0 & -1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} \to \textcolor{lightgreen}{ \boldsymbol{\alpha}_{1} = \begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix} }, \, \textcolor{lightgreen}{ \boldsymbol{\alpha}_{2} = \begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix} }
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 经过计算验证可知，向量 $\boldsymbol{\alpha}_{1}$, $\boldsymbol{\alpha}_{2}$, $\boldsymbol{\alpha}_{3}$ 已经是正交向量：

$$
\begin{aligned}
& \ \boldsymbol{\alpha}_{1}^{\top} \boldsymbol{\alpha}_{2} = \begin{pmatrix}
0, 1, 0
\end{pmatrix} \begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix} = 0 \\ \\
& \ \boldsymbol{\alpha}_{1}^{\top} \boldsymbol{\alpha}_{3} = \begin{pmatrix}
0, 1, 0
\end{pmatrix} \begin{pmatrix}
-1 \\
0 \\
1
\end{pmatrix} = 0 \\ \\
& \ \boldsymbol{\alpha}_{2}^{\top} \boldsymbol{\alpha}_{3} = \begin{pmatrix}
0, 1, 0
\end{pmatrix} \begin{pmatrix}
-1 \\
0 \\
1
\end{pmatrix} = 0 \\ \\
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 因此，直接对向量 $\boldsymbol{\alpha}_{1}$, $\boldsymbol{\alpha}_{2}$, $\boldsymbol{\alpha}_{3}$ 进行单位化即可：

$$
\boldsymbol{\beta}_{1} = \boldsymbol{\alpha}_{1} = \begin{pmatrix}
0 \\ 1 \\ 0
\end{pmatrix}, \, \boldsymbol{\beta}_{2} = \frac{\boldsymbol{\alpha}_{2}}{\left|\boldsymbol{\alpha}_{2}\right|} = \begin{pmatrix}
\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}
\end{pmatrix}, \, \boldsymbol{\beta}_{3} = \frac{\boldsymbol{\alpha}_{3}}{\left|\boldsymbol{\alpha}_{3}\right|} = \begin{pmatrix}
\frac{-1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}
\end{pmatrix}
$$

综上，可得：

$$
\boldsymbol{Q} = \begin{pmatrix}
0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\
1 & 0 & 0 \\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{pmatrix}
$$

同时可知，正交变换 $\boldsymbol{x} = \boldsymbol{Q} \boldsymbol{y}$ 对应的标准型为：

$$
f\left(x_{1}, x_{2}, x_{3}\right) = f = 4 y_{1}^{2} + 4 y_{2}^{2} + 2 y_{3}^{2}
$$

$\textcolor{orange}{\blacktriangleright}$ 注意 1：在上面的计算过程中，之所以由 $\boldsymbol{Q} = \begin{pmatrix} 0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$ 可以得出 $f = 4 y_{1}^{2} + 4 y_{2}^{2} + 2 y_{3}^{2}$, 是因为，根据二次型的定义可知：$\boldsymbol{Q}^{-1} \boldsymbol{A} \boldsymbol{Q} = \boldsymbol{Q}^{\top} \boldsymbol{A} \boldsymbol{Q} = \begin{pmatrix} 0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}^{\top} \begin{pmatrix} 3 & 0 & 1 \\ 0 & 4 & 0 \\ 1 & 0 & 3 \end{pmatrix} \begin{pmatrix} 0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 2 \end{pmatrix}$.

$\textcolor{orange}{\blacktriangleright}$ 注意 2：根据 $\lambda$ 的先后顺序不同，矩阵 $\boldsymbol{Q}$ 和对应的标准型 $f$ 也可以是如下这些形式：

$$
\begin{aligned}
& \ \textcolor{white}{\blacktriangleright} \boldsymbol{Q} = \begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & \frac{-1}{\sqrt{2}} \\
0 & 1 & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}
\end{pmatrix}, \, f = 4 y_{1}^{2} + 4 y_{2}^{2} + 2 y_{3}^{2} \\ \\
& \ \textcolor{white}{\blacktriangleright} \boldsymbol{Q} = \begin{pmatrix}
\frac{-1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & 1 & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}
\end{pmatrix}, \, f = 2 y_{1}^{2} + 4 y_{2}^{2} + 4 y_{3}^{2} \\ \\
& \ \textcolor{white}{\blacktriangleright} \boldsymbol{Q} = \begin{pmatrix}
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & 0 & 1 \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0
\end{pmatrix}, \, f = 2 y_{1}^{2} + 4 y_{2}^{2} + 4 y_{3}^{2} \\ \\
& \ \textcolor{white}{\blacktriangleright} \boldsymbol{Q} = \begin{pmatrix}
0 & \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
1 & 0 & 0 \\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{pmatrix}, \, f = 4 y_{1}^{2} + 2 y_{2}^{2} + 4 y_{3}^{2} \\ \\
& \ \textcolor{white}{\blacktriangleright} \boldsymbol{Q} = \begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0 \\
0 & 0 & 1 \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0
\end{pmatrix}, \, f = 4 y_{1}^{2} + 2 y_{2}^{2} + 4 y_{3}^{2} \\ \\
\end{aligned}
$$

第（Ⅱ）问

$\textcolor{lightgreen}{\blacktriangleright}$ 分析可知，第（Ⅱ）问要求解的式子 “$\min\limits_{\boldsymbol{x} \neq 0} \dfrac{f\left(\boldsymbol{x}\right)}{\boldsymbol{x}^{\top} \boldsymbol{x}} = 2$” 中向量 $\boldsymbol{x}$ 的内积 “$\boldsymbol{x}^{\top} \boldsymbol{x}$” 实际上就是向量长度的平方，因为向量的长度就是用根号下向量的内积表示的.

例如，向量 $\boldsymbol{x} = \begin{pmatrix} x_{1}, x_{2}, \cdots x_{n} \end{pmatrix}$ 的长度 $\lVert \boldsymbol{x} \rVert$ 为：

$$
\lVert \boldsymbol{x} \rVert = \sqrt{\boldsymbol{x}^{\top} \boldsymbol{x}} = \sqrt{x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2}}
$$

类似的，向量 $\boldsymbol{y} = \begin{pmatrix} y_{1}, y_{2}, \cdots y_{n} \end{pmatrix}$ 的长度 $\lVert \boldsymbol{y} \rVert$ 为：

$$
\lVert \boldsymbol{y} \rVert = \sqrt{\boldsymbol{y}^{\top} \boldsymbol{y}} = \sqrt{y_{1}^{2} + y_{2}^{2} + \cdots + y_{n}^{2}}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 接着，从「荒原之梦考研数学」的《全面理解正交向量与正交矩阵》这篇讲义可知，正交变换 $\boldsymbol{x} = \boldsymbol{Q} \boldsymbol{y}$ 只会改变向量的方向，不会改变向量的长度，所以：

$$
\begin{aligned}
& \ \lVert \boldsymbol{x} \rVert = \lVert \boldsymbol{y} \rVert \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lVert \boldsymbol{x} \rVert^{2} = \lVert \boldsymbol{y} \rVert^{2} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ \boldsymbol{x}^{\top} \boldsymbol{x} = \boldsymbol{y}^{\top} \boldsymbol{y} }
\end{aligned}
$$

当然，对于 $\textcolor{lightgreen}{ \boldsymbol{x}^{\top} \boldsymbol{x} = \boldsymbol{y}^{\top} \boldsymbol{y} }$ 这个结论，我们也可以通过下面的步骤完成证明：

$$
\begin{aligned}
& \ \boldsymbol{x} = \boldsymbol{Q} \boldsymbol{y} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{x}^{\top} \boldsymbol{x} = \left( \boldsymbol{Q} \boldsymbol{y} \right)^{\top} \left( \boldsymbol{Q} \boldsymbol{y} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{x}^{\top} \boldsymbol{x} = \boldsymbol{y}^{\top} \boldsymbol{Q}^{\top} \boldsymbol{Q} \boldsymbol{y} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{x}^{\top} \boldsymbol{x} = \boldsymbol{y}^{\top} \left( \boldsymbol{Q}^{\top} \boldsymbol{Q} \right) \boldsymbol{y} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{x}^{\top} \boldsymbol{x} = \boldsymbol{y}^{\top} \boldsymbol{E} \boldsymbol{y} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ \boldsymbol{x}^{\top} \boldsymbol{x} = \boldsymbol{y}^{\top} \boldsymbol{y} }
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 综上：

$$
\begin{aligned}
\frac{f}{\boldsymbol{x}^{\top} \boldsymbol{x}} & = \frac{f}{\boldsymbol{y}^{\top} \boldsymbol{y}} \\ \\
& = \frac{4 y_{1}^{2} + 4 y_{2}^{2} + 2 y_{3}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} \\ \\
& = \frac{2 y_{1}^{2} + 2 y_{2}^{2} + 2 y_{3}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} + \frac{2 y_{1}^{2} + 2 y_{2}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} \\ \\
& = 2 + \frac{2 y_{1}^{2} + 2 y_{2}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}}
\end{aligned}
$$

又因为，$\frac{2 y_{1}^{2} + 2 y_{2}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}}$ 中都是平方项，一定大于等于零，所以：

$$
\frac{f}{\boldsymbol{x}^{\top} \boldsymbol{x}} = 2 + \frac{2 y_{1}^{2} + 2 y_{2}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} \geqslant 2
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 因此可知，$\min_{\boldsymbol{x} \neq 0} \frac{f}{\boldsymbol{x}^{\top} \boldsymbol{x}} = 2$ 得证.

$\textcolor{orange}{\blacktriangleright}$ 注意 3：虽然下面的式子也是正确的，但是，仅凭下面的式子，我们只能直接得出 $\frac{4 y_{1}^{2} + 4 y_{2}^{2} + 2 y_{3}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}}$ 的取值范围在 $\left[ 0, 4 \right]$ 上，而不能直接说 $\frac{4 y_{1}^{2} + 4 y_{2}^{2} + 2 y_{3}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}}$ 的最小值为 $2$, 最大值为 $4$:

$$
\frac{4 y_{1}^{2} + 4 y_{2}^{2} + \textcolor{pink}{2} y_{3}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} \geqslant \frac{\textcolor{pink}{2} y_{1}^{2} + \textcolor{pink}{2} y_{2}^{2} + \textcolor{pink}{2} y_{3}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} = 2
$$

$$
\frac{\textcolor{lightblue}{4} y_{1}^{2} + \textcolor{lightblue}{4} y_{2}^{2} + 2 y_{3}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} \leqslant \frac{\textcolor{lightblue}{4} y_{1}^{2} + \textcolor{lightblue}{4} y_{2}^{2} + \textcolor{lightblue}{4} y_{3}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} = 4
$$

---