蓝以中上册 双线性函数与二次型 360页 习题二9 解答

castelu

习题二9：
　　给定数域$K$上的二次型$f=X'AX$。设$f$的秩为$r$
（1）证明$f$可用三角形变换化为
$$g=\lambda_1y_1^2+\lambda_2y_2^2+\cdots+\lambda_ry_r^2(\lambda_i \ne 0,i=1,2,\cdots,r)$$
　　的充分必要条件是
$$D_k=A\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\} \ne 0 (k=1,2,\cdots,r)$$
　　而
$$A\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\}=0 (k=r+1,\cdots,n)$$
（2）证明上题中的标准形$g$的系数满足
$$\lambda_k=\frac{D_k}{D_{k-1}}(k=1,2,\cdots,r;D_0=1)$$

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解：
　　若$f$经三角形变换$X=TY$化为
$$g=\lambda_1y_1^2+\lambda_2y_2^2+\cdots+\lambda_ry_r^2$$
　　$g$的矩阵为
$$\left( {\begin{array}{*{20}{c}}
{\lambda_1}&{}&{}&{}&{}&{}&{}\\
{}&{\lambda_2}&{}&{}&{}&{}&{}\\
{}&{}&{\ddots}&{}&{}&{}&{}\\
{}&{}&{}&{\lambda_r}&{}&{}&{}\\
{}&{}&{}&{}&{0}&{}&{}\\
{}&{}&{}&{}&{}&{\ddots}&{}\\
{}&{}&{}&{}&{}&{}&{0}
\end{array}} \right)$$
　　按照《蓝以中上册 双线性函数与二次型 359页 习题二8 解答》，我们有
$$D_k=A\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\}=G\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\}=\lambda_1\lambda_2\cdots\lambda_k \ne 0(k=1,2,\cdots,r)$$
　　而
$$A\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\}=G\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\}=0(k=r+1,\cdots,n)$$
　　根据上面的结果，我们显然有（令$D_0=1$）
$$\lambda_k=\frac{D_k}{D_{k-1}}(k=1,2,\cdots,r)$$
　　现在来证明充分性。对变量数$n$作数学归纳法
　　$n=1$时$f$已是标准形，只需作恒等变数替换
$$X=EY$$
　　设对$n-1$个变元的二次型充分性已成立。则当
$$f=\sum\limits_{i=1}^n\sum\limits_{j=1}^na_{ij}x_ix_j=X'AX$$
　　时，设
$$A\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\} \ne 0(k=1,2,\cdots,r)$$
$$A\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\}=0(k=r+1,\cdots,n)$$
　　现在
$$a_{11}=A\left\{\begin{array}{*{20}{c}}
{1}\\
{1}
\end{array}\right\} \ne 0$$
　　应用配方法，即作三角形变换
$$\begin{eqnarray*}
X&=&\left( {\begin{array}{*{20}{c}}
{x_1}\\
{x_2}\\
{\vdots}\\
{x_n}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
{1}&{-\frac{a_{12}}{a_{11}}}&{\cdots}&{-\frac{a_{1n}}{a_{11}}}\\
{}&{1}&{}&{}\\
{}&{}&{\ddots}&{}\\
{}&{}&{}&{1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{y_1}\\
{y_2}\\
{\vdots}\\
{y_n}
\end{array}} \right)\\
&=&TY
\end{eqnarray*}$$
　　则$f$化做
$$g=a_{11}y_1^2+\sum\limits_{i=2}^n\sum\limits_{j=2}^nb_{ij}y_iy_j=Y'BY$$
　　其矩阵为
$$B=\left( {\begin{array}{*{20}{c}}
{a_{11}}&{}\\
{}&{B_1}
\end{array}} \right)$$
　　若
$$r=1$$
　　则
$$1=r(B)=r(B_1)+1$$
　　故
$$r(B_1)=0$$
　　即
$$B_1=0$$
　　命题已证。下面设
$$r>1$$
　　现在
$$\begin{eqnarray*}
B_1\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\}&=&\frac{1}{a_{11}}B\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k+1}\\
{1}&{2}&{\cdots}&{k+1}
\end{array}\right\}\\
&=&\frac{1}{a_{11}}A\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k+1}\\
{1}&{2}&{\cdots}&{k+1}
\end{array}\right\} \ne 0
\end{eqnarray*}$$
　　这里
$$k=1,2,\cdots,r-1$$
　　而
$$\begin{eqnarray*}
B_1\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k}\\
{1}&{2}&{\cdots}&{k}
\end{array}\right\}&=&\frac{1}{a_{11}}B\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k+1}\\
{1}&{2}&{\cdots}&{k+1}
\end{array}\right\}\\
&=&\frac{1}{a_{11}}A\left\{\begin{array}{*{20}{c}}
{1}&{2}&{\cdots}&{k+1}\\
{1}&{2}&{\cdots}&{k+1}
\end{array}\right\}=0
\end{eqnarray*}$$
　　这里
$$k=r,r+1,\cdots,n-1$$
　　按归纳假设，有三角形变换
$$\left( {\begin{array}{*{20}{c}}
{y_2}\\
{y_3}\\
{\vdots}\\
{y_n}
\end{array}} \right)=T_1\left( {\begin{array}{*{20}{c}}
{z_2}\\
{z_3}\\
{\vdots}\\
{z_n}
\end{array}} \right)$$
　　使
$$\sum\limits_{i=2}^n\sum\limits_{j=2}^nb_{ij}y_iy_j=\lambda_2z_2^2+\cdots+\lambda_rz_r^2$$
　　作三角形变换
$$Y=\left( {\begin{array}{*{20}{c}}
{y_1}\\
{y_2}\\
{\vdots}\\
{y_n}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
{1}&{0}\\
{0}&{T_1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{z_1}\\
{z_2}\\
{\vdots}\\
{z_n}
\end{array}} \right)=T_2Z$$
　　则
$$\begin{eqnarray*}
g&=&Y'BY=Z'(T'_2BT_2)Z\\
&=&a_{11}z_1^2+\lambda_2z_2^2+\cdots+\lambda_rz_r^2
\end{eqnarray*}$$
　　现在$TT_2$仍是主对角线上为$1$的上三角矩阵，在三角变换
$$X=(TT_2)Z$$
　　下
$$\begin{eqnarray*}
f&=&X'AX=Z'(T'_2T'ATT_2)Z\\
&=&Z'(T'_2BT_2)Z\\
&=&a_{11}z_1^2+\lambda_2z_2^2+\cdots+\lambda_rz_r^2
\end{eqnarray*}$$