# 2013年考研数二第23题解析：二次型、二次型的标准型

## 题目

$(Ⅰ)$ 证明：二次型 $f$ 对应的矩阵为 $2\alpha \alpha ^{\top}+\beta \beta ^{\top}$

$(Ⅱ)$ 若 $\alpha$, $\beta$ 正交且均为单位向量，证明：$f$ 在正交变换下的标准形为 $2y_{1}^{2}+y_{2}^{2}$.

## 解析

### 第 $(Ⅰ)$ 问

#### 方法一

$$f ( x_{1},x_{2},x_{3} ) \Rightarrow$$

$$2 ( a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3} ) ^{2}+ ( b_{1}x_{1}+b_{2}x_{2}+b_{3}x_{3} )^{2} \Rightarrow$$

$$2[(a_{1}x_{1} + a_{2}x_{2})^{2} + a_{3}^{2} x_{3}^{2} + 2a_{3}x_{3} (a_{1}x_{1} + a_{2}x_{2})] +$$

$$[(b_{1}x_{1} + b_{2}x_{2})^{2} + b_{3}^{2}x_{3}^{2} + 2b_{3}x_{3}(b_{1}x_{1} + b_{2}x_{2})] \Rightarrow$$

$$2[a_{1}^{2}x_{1}^{2} + a_{2}^{2}x_{2}^{2} + 2a_{1}a_{2}x_{1}x_{2} + a_{3}^{2}x_{3}^{2} + 2a_{1}a_{3}x_{1}x_{3} + 2a_{2}a_{3}x_{2}x_{3}] +$$

$$[b_{1}^{2}x_{1}^{2} + b_{2}^{2}x_{2}^{2} + 2b_{1}b_{2}x_{1}x_{2} + b_{3}^{2}x_{3}^{2} + 2b_{1}b_{3}x_{1}x_{3} + 2b_{2}b_{3}x_{2}x_{3}] \Rightarrow$$

$$2a_{1}^{2}x_{1}^{2} + 2a_{2}^{2}x_{2}^{2} + 4a_{1}a_{2}x_{1}x_{2} + 2a_{3}^{2}x_{3}^{2} + 4a_{1}a_{3}x_{1}x_{3} + 4a_{2}a_{3}x_{2}x_{3} +$$

$$b_{1}^{2}x_{1}^{2} + b_{2}^{2}x_{2}^{2} + 2b_{1}b_{2}x_{1}x_{2} + b_{3}^{2}x_{3}^{2} + 2b_{1}b_{3}x_{1}x_{3} + 2b_{2}b_{3}x_{2}x_{3} \Rightarrow$$

$$(2a_{1}^{2} + b_{1}^{2})x_{1}^{2} + (2a_{2}^{2} + b_{2}^{2})x_{2}^{2} + (2a_{3}^{2} + b_{3}^{2}) x_{3}^{2} +$$

$$(4a_{1}a_{2} + 2b_{1}b_{2})x_{1}x_{2} + (4a_{1}a_{3} + 2b_{1}b_{3})x_{1}x_{3} + (4a_{2}a_{3} + 2b_{2}b_{3})x_{2}x_{3}.$$

$$2\alpha \alpha ^{\top}+\beta \beta ^{\top} \Rightarrow$$

$$2 \begin{bmatrix} a_{1}\\ a_{2}\\ a_{3} \end{bmatrix} \begin{bmatrix} a_{1} & a_{2} & a_{3}\\ \end{bmatrix}+ \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix} \begin{bmatrix} b_{1} & b_{2} & b_{3}\\ \end{bmatrix} \Rightarrow$$

$$2 \begin{bmatrix} a_{1}^{2} & a_{1}a_{2} & a_{1}a_{3}\\ a_{1}a_{2} & a_{2}^{2} & a_{2}a_{3}\\ a_{1}a_{3} & a_{2}a_{3} & a_{3}^{2}\\ \end{bmatrix}+ \begin{bmatrix} b_{1}^{2} & b_{1}b_{2} & b_{1}b_{3}\\ b_{1}b_{2} & b_{2}^{2} & b_{2}b_{3}\\ b_{1}b_{3} & b_{2}b_{3} & b_{3}^{2}\\ \end{bmatrix} \Rightarrow$$

$$\begin{bmatrix} 2a_{1}^{2} + b_{1}^{2} & 2a_{1}a_{2} + b_{1}b_{2} & 2a_{1}a_{3} + b_{1}b_{3}\\ 2a_{1}a_{2} + b_{1}b_{2} & 2a_{2}^{2} + b_{2}^{2} & 2a_{2}a_{3}+b_{2}b_{3}\\ 2a_{1}a_{3} +b_{1}b_{3} & 2a_{2}a_{3} + b_{2}b_{3} & 2a_{3}^{2} + b_{3}^{2}\\ \end{bmatrix}.$$

#### 方法二

$$x =\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$$

$$f(x_{1}, x_{2}, x_{3}) \Rightarrow$$

$$2(a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3})^{2} + (b_{1}x_{1} + b_{2}x_{2} + b_{3}x_{3})^{2} \Rightarrow$$

$$2(a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3}) \cdot (x_{1}a_{1} + x_{2}a_{2}+x_{3}a_{3}) +$$

$$(b_{1}x_{1} + b_{2}x_{2} + b_{3}x_{3}) \cdot (x_{1}b_{1} + x_{2}b_{2} + x_{3}b_{3}) \Rightarrow$$

$$2 \left\{ \right.\begin{bmatrix} a_{1}\\ a_{2}\\ a_{3} \end{bmatrix} x^{\top}\left. \right \} \cdot \left\{ \right.x \begin{bmatrix} a_{1} & a_{2} & a_{3}\\ \end{bmatrix} \left. \right \}+$$

$$\left\{ \right. \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix} x^{\top}\left. \right \} \cdot \left\{ \right.x \begin{bmatrix} b_{1} & b_{2} & b_{3}\\ \end{bmatrix} \left. \right \} \Rightarrow$$

$$2 x^{\top} \left\{ \right. \begin{bmatrix} a_{1}\\ a_{2}\\ a_{3} \end{bmatrix} \begin{bmatrix} a_{1} & a_{2} & a_{3}\\ \end{bmatrix} \left. \right \} x + x^{\top} \left\{ \right. \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix} \begin{bmatrix} b_{1} & b_{2} & b_{3}\\ \end{bmatrix} \left. \right \} x \Rightarrow$$

$$x^{\top} \left\{ \right. 2\begin{bmatrix} a_{1}\\ a_{2}\\ a_{3} \end{bmatrix} \begin{bmatrix} a_{1} & a_{2} & a_{3}\\ \end{bmatrix}+ \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix} \begin{bmatrix} b_{1} & b_{2} & b_{3}\\ \end{bmatrix} \left. \right \} x \Rightarrow$$

$$x^{\top} (2 \alpha \alpha^{\top} + \beta \beta^{\top}) x.$$

### 第 $(Ⅱ)$ 问

$$(\alpha, \beta) = \alpha^{\top} \beta = 0;$$

$$(\beta, \alpha) = \beta^{\top} \alpha = 0.$$

$$\alpha \alpha^{\top} = E; \alpha^{\top} \alpha = 1;$$

$$\beta \beta^{\top} = E; \beta^{\top} \beta = 1.$$

$$A = 2 \alpha \alpha^{\top} + \beta \beta^{\top}.$$

$$A \alpha = (2 \alpha \alpha^{\top} + \beta \beta^{\top}) \alpha \Rightarrow$$

$$A \alpha = 2 \alpha (\alpha^{\top} \alpha) + \beta (\beta^{\top} \alpha) \Rightarrow$$

$$A \alpha = 2 \alpha \cdot 1 + \beta \cdot 0.$$

$$A \alpha = 2 \alpha.$$

$$A \beta = (2 \alpha \alpha^{\top} + \beta \beta^{\top}) \beta \Rightarrow$$

$$A \beta = 2 \alpha (\alpha^{\top} \beta) + \beta (\beta^{\top} \beta) \Rightarrow$$

$$A \beta = 2 \alpha \cdot 0 + \beta \cdot 1 \Rightarrow$$

$$A \beta = \beta.$$

$\alpha$ 为矩阵 $A$ 属于特征值 $\lambda_{1}=2$ 的特征向量；

$\beta$ 为矩阵 $A$ 属于特征值 $\lambda_{2}=1$ 的特征向量。

$$r(A) = r(2 \alpha \alpha^{\top} + \beta \beta^{\top}) \leqslant [r(2 \alpha \alpha^{\top}) + r(\beta \beta^{\top})].$$

$$[r(2 \alpha \alpha^{\top}) + r(\beta \beta^{\top})] = [r(E) + r(E)] = 1+1 = 2.$$

$$r(A) = 2 < 3.$$