已知二次型 $f\left(x_1, x_2, x_3\right)=x_1^2+2 x_2^2+2 x_3^2+2 x_1 x_2-2 x_1 x_3$,
$$
g\left(y_1, y_2, y_3\right)=y_1^2+y_2{ }^2+y_3^2+2 y_2 y_3
$$
(I) 求可逆变换 $x=P y$, 将 $f\left(x_1, x_2, x_3\right)$ 化为 $g\left(y_1, y_2, y_3\right)$ ;
(II) 是否存在正交变换 $x=Q y$, 将 $f\left(x_1, x_2, x_3\right)$ 化为 $g\left(y_1, y_2, y_3\right)$.
【答案】 (I) 利用配方法将 $f\left(x_1, x_2, x_3\right)$ 和 $g\left(y_1, y_2, y_3\right)$ 化为规范形, 从而建立两 者的关系.
先将 $f\left(x_1, x_2, x_3\right)$ 化为规范形.

$$
\begin{aligned}
& f\left(x_1, x_2, x_3\right)=x_1^2+2 x_2^2+x_3^2+2 x_1 x_2-2 x_1 x_3 \\
& =\left(x_1+x_2-x_3\right)^2+x_2^2+x_3^2+2 x_2 x_3=\left(x_1+x_2-x_3\right)^2+\left(x_2+x_3\right)^2
\end{aligned}
$$
令 $\left\{\begin{array}{l}z_1=x_1+x_2-x_3 \\ z_2=x_2+x_3 \\ z_3=x_3\end{array}\right.$,则 $f\left(x_1, x_2, x_3\right)=z_1^2+z_2^2$.
即 $\left(\begin{array}{l}z_1 \\ z_2 \\ z_3\end{array}\right)=\left(\begin{array}{ccc}1 & 1 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right)$, 使得 $f\left(x_1, x_2, x_3\right)=z_1^2+z_2^2$.
再将 $g\left(y_1, y_2, y_3\right)$ 化为规范形.
$$
g\left(y_1, y_2, y_3\right)=y_1^2+y_2^2+y_3^2+2 y_2 y_3=y_1^2+\left(y_2+y_3\right)^2
$$

令 $\left\{\begin{array}{l}z_1=y_1 \\ z_2=y_2+y_3 \\ z_3=y_3\end{array}\right.$, 则 $g\left(y_1, y_2, y_3\right)=z_1^2+z_2^2$.
即 $\left(\begin{array}{l}z_1 \\ z_2 \\ z_3\end{array}\right)=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{l}y_1 \\ y_2 \\ y_3\end{array}\right)$, 使得 $g\left(y_1, y_2, y_3\right)=z_1^2+z_2^2$.
从而有 $\left(\begin{array}{l}z_1 \\ z_2 \\ z_3\end{array}\right)=\left(\begin{array}{ccc}1 & 1 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right)=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{l}y_1 \\ y_2 \\ y_3\end{array}\right)$,
于是可得 $\left(\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right)=P\left(\begin{array}{l}y_1 \\ y_2 \\ y_3\end{array}\right)$, 其中 $P=\left(\begin{array}{ccc}1 & 1 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)^{-1}\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc}1 & -1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$ 为所求矩阵,
可将 $f\left(x_1, x_2, x_3\right)$ 化为 $g\left(y_1, y_2, y_3\right)$.

(II) 二次型 $f\left(x_1, x_2, x_3\right)$ 和 $g\left(y_1, y_2, y_3\right)$ 的矩阵分别为 $A=\left(\begin{array}{ccc}1 & 1 & -1 \\ 1 & 2 & 0 \\ -1 & 0 & 2\end{array}\right)$,
$$
B=\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 1 & 1
\end{array}\right) .
$$
由题意知,若存在正交变换 $x=Q y$, 则 $Q^T A Q=Q^{-1} A Q=B$, 可得 $A$ 和 $B$ 相似. 易知 $\operatorname{tr}(A)=5, \operatorname{tr}(B)=3$, 从而 $A$ 和 $B$ 不相似, 于是不存在正交变换 $x=Q y$, 使得 $f\left(x_1, x_2, x_3\right)$ 化为 $g\left(y_1, y_2, y_3\right)$.


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