2024年考研数二第22题解析：线性方程组、正交变换

第 (2) 问

由 $f(x_1,x_2,x_3)$ $=$ $x^{T}BAx$ 可知，二次型 $f$ 对应的二次型矩阵为 $BA$, 且，由第 (1) 问的计算结果可知：

$$BA=\left(\begin{array}{cc}1& 1\\ 1& 1\\ 2& 2\end{array}\right)\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\end{array}\right)=\left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 2\\ 2& 2& 4\end{array}\right)$$

若令 $C=BA$, 则求得特征值为：

$$\begin{array}{r}|\lambda E-C|\\ \\ & =\left|\begin{array}{ccc}\lambda -1& -1& -2\\ -1& \lambda -1& -2\\ -2& -2& \lambda -4\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}\lambda & -\lambda & 0\\ -1& \lambda –1& -2\\ -2& -2& \lambda –4\end{array}\right|=0\\ \\ & \Rightarrow \left|\begin{array}{ccc}\lambda & 0& 0\\ -1& \lambda –2& -2\\ -2& -4& \lambda –4\end{array}\right|=0\\ \\ & \Rightarrow \lambda (\lambda –2)(\lambda –4)–8\lambda =0\\ \\ & \Rightarrow \{\begin{array}{l}{\lambda }_1={\lambda }_2=0\\ {\lambda }_3=6\end{array}\end{array}$$

*当 ${\lambda }_1$ $=$ ${\lambda }_2$ $=$ $0$ 时, $(0E-C)x$ $=$ $-Cx$ $=$ $0$ $\rightleftarrows$ $Cx$ $=$ $0$, 即：

$$\begin{array}{rl}& \left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 2\\ 2& 2& 4\end{array}\right)x=0\Rightarrow \\ \\ & \left(\begin{array}{ccc}1& 1& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)x=0\end{array}$$

得基础解系为：

$${\eta }_1=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right),{\eta }_2=\left(\begin{array}{c}-1\\ -1\\ 1\end{array}\right)$$

一般情况下，如果我们令自由未知数分别为 $0$, $1$ 和 $1$, $0$ 的话，得到的基础解系应该是 ${\eta }_1=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)$, ${\eta }_2=\left(\begin{array}{c}-2\\ 0\\ 1\end{array}\right)$, 但是这样得到的基础解系不是正交的，还需要进行正交化的操作。若令自由位置为分别为 $1$, $0$ 和 $-1$, $1$, 不仅得到的基础解系线性无关，而且是正交的，无需进行正交化。

**当 ${\lambda }_3$ $=$ $6$ 时, $(6E-C)x$ $=$ $0$, 即：

$$\begin{array}{rl}& \left(\begin{array}{c}\left(\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 6\end{array}\right)–\left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 2\\ 2& 2& 4\end{array}\right)\end{array}\right)x=0\Rightarrow \\ \\ & \left(\begin{array}{ccc}5& -1& -2\\ -1& 5& -2\\ -2& -2& 2\end{array}\right)x=0\Rightarrow \\ \\ & \left(\begin{array}{ccc}1& -5& 2\\ 0& 2& -1\\ 0& 0& 0\end{array}\right)x=0\Rightarrow \\ \end{array}$$

得基础解系为：

$${\eta }_3=\left(\begin{array}{c}1\\ 1\\ 2\end{array}\right)$$

又由于：

$$\{\begin{array}{l}{\eta }_1^{\mathrm{\top }}{\eta }_2=0\\ {\eta }_1^{\mathrm{\top }}{\eta }_3=0\\ {\eta }_2^{\mathrm{\top }}{\eta }_3=0\end{array}$$

于是可知，${\eta }_1$, ${\eta }_2$, ${\eta }_3$ 两两正交。

接着，将 ${\eta }_1$, ${\eta }_2$, ${\eta }_3$ 单位化：

$${\gamma }_1=\frac{{\eta }_1}{\left|{\eta }_1\right|}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)$$

$${\gamma }_2=\frac{{\eta }_2}{\left|{\eta }_2\right|}=\frac{1}{\sqrt{3}}\left(\begin{array}{c}-1\\ -1\\ 1\end{array}\right)$$

$${\gamma }_3=\frac{{\eta }_3}{\left|{\eta }_3\right|}=\frac{1}{\sqrt{6}}\left(\begin{array}{c}1\\ 1\\ 2\end{array}\right)$$

故正交矩阵为：

$$Q=\left(\begin{array}{c}{\gamma }_1,{\gamma }_2,{\gamma }_3\end{array}\right)=\left(\begin{array}{ccc}-\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{6}}\\ \\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{6}}\\ \\ 0& \frac{1}{\sqrt{3}}& \frac{2}{\sqrt{6}}\end{array}\right)$$

此时二次型经正交变换 $x=Qy$ 可化为标准形为：

$$f=0\cdot y_1^2+0\cdot y_2^2+6\cdot y_3^2=\mathbf{6}{\mathit{y}}_{\mathbf{3}}^{\mathbf{2}}$$

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