蓝以中上册 双线性函数与二次型 370页 习题四10 解答

castelu

习题四10：
　　证明：
（1）如果
$$f=\sum\limits_{i=1}^n\sum\limits_{j=1}^na_{ij}x_ix_j(a_{ij}=a_{ji})$$
　　是正定二次型，那么
$$g(y_1,y_2,\cdots,y_n)=\left| {\begin{array}{*{20}{c}}
{a_{11}}&{a_{12}}&{\cdots}&{a_{1n}}&{y_1}\\
{a_{21}}&{a_{22}}&{\cdots}&{a_{2n}}&{y_2}\\
{\vdots}&{\vdots}&{}&{\vdots}&{\vdots}\\
{a_{n1}}&{a_{n2}}&{\cdots}&{a_{nn}}&{y_n}\\
{y_1}&{y_2}&{\cdots}&{y_n}&{0}
\end{array}} \right|$$
　　是负定二次型。
（2）如果$A$是正定矩阵，那么
$$|A| \le a_{nn} \cdot P_{n-1}$$
　　其中$P_{n-1}$是$A$的$n-1$阶顺序主子式
（3）如果$A$是正定矩阵，那么
$$|A| \le a_{11}a_{22} \cdots a_{nn}$$
（4）如果$T=(t_{ij})$是$n$阶实可逆矩阵，那么
$$|T|^2 \le \prod\limits_{i=1}^n(t_{1i}^2+t_{2i}^2+\cdots+t_{ni}^2)$$

---

解：
（1）$A$正定，故存在$n$阶实可逆矩阵$T$，使
$$T'AT=E$$
　　现作分块矩阵运算
$$\left( {\begin{array}{*{20}{c}}
{T'}&{0}\\
{0}&{1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{A}&{Y}\\
{Y'}&{0}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{T}&{0}\\
{0}&{1}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
{T'AT}&{T'Y}\\
{Y'T}&{0}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
{E}&{Z}\\
{Z'}&{0}
\end{array}} \right)$$
　　这里
$$Z=\left( {\begin{array}{*{20}{c}}
{z_1}\\
{z_2}\\
{\vdots}\\
{z_n}
\end{array}} \right)=T'\left( {\begin{array}{*{20}{c}}
{y_1}\\
{y_2}\\
{\vdots}\\
{y_n}
\end{array}} \right)$$
　　是$R$上一可逆线性变数替换。上面矩阵等式两边取行列式即得
$$\begin{eqnarray*}
|T|^2g(y_1,y_2,\cdots,y_n)&=&\left| {\begin{array}{*{20}{c}}
{1}&{0}&{0}&{\cdots}&{0}&{z_1}\\
{0}&{1}&{0}&{\cdots}&{0}&{z_2}\\
{0}&{0}&{1}&{\ddots}&{\vdots}&{z_3}\\
{\vdots}&{\vdots}&{\ddots}&{\ddots}&{0}&{\vdots}\\
{0}&{0}&{\cdots}&{0}&{1}&{z_n}\\
{z_1}&{z_2}&{z_3}&{\cdots}&{z_n}&{0}
\end{array}} \right|\\
&=&\left| {\begin{array}{*{20}{c}}
{1}&{0}&{\cdots}&{0}&{z_2}\\
{0}&{1}&{\ddots}&{\vdots}&{z_3}\\
{\vdots}&{\ddots}&{\ddots}&{0}&{\vdots}\\
{0}&{\cdots}&{0}&{1}&{z_n}\\
{z_2}&{z_3}&{\cdots}&{z_n}&{0}
\end{array}} \right|+(-1)^nz_1\left| {\begin{array}{*{20}{c}}
{0}&{1}&{0}&{0}&{\cdots}&{0}\\
{0}&{0}&{1}&{0}&{\cdots}&{0}\\
{0}&{0}&{0}&{1}&{\ddots}&{0}\\
{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\ddots}&{\vdots}\\
{0}&{0}&{0}&{\cdots}&{0}&{1}\\
{z_1}&{z_2}&{z_3}&{\cdots}&{z_{n-1}}&{z_n}
\end{array}} \right|\\
&=&h(z_2,\cdots,z_n)-z_1^2
\end{eqnarray*}$$
　　现在对$n$作数学归纳法
　　当$n=1$时
$$g(y_1)=\left| {\begin{array}{*{20}{c}}
{a_{11}}&{y_1}\\
{y_1}&{0}
\end{array}} \right|=-y_1^2$$
　　显然是负定的
　　设对$n-1$个变元，题中结论已成立，即
$$h(z_2,\cdots,z_n)=\left| {\begin{array}{*{20}{c}}
{1}&{0}&{\cdots}&{0}&{z_2}\\
{0}&{1}&{\ddots}&{\vdots}&{z_3}\\
{\vdots}&{\vdots}&{\ddots}&{0}&{\vdots}\\
{0}&{0}&{\cdots}&{1}&{z_n}\\
{z_2}&{z_3}&{\cdots}&{z_n}&{0}
\end{array}} \right|$$
　　是负定的，对任意非零$Y$，$Z=T'Y$也非零，于是
$$g(y_1,\cdots,y_n)=\frac{1}{|T|^2}(h(z_2,\cdots,z_n)-z_1^2)<0$$
　　这表明
$$g(y_1,\cdots,y_n)$$
　　是负定二次型
（2）设
$$A_{n-1}=\left( {\begin{array}{*{20}{c}}
{a_{11}}&{\cdots}&{a_{1,n-1}}\\
{\vdots}&{}&{\vdots}\\
{a_{n-1,1}}&{\cdots}&{a_{n-1,n-1}}
\end{array}} \right),B_{n-1}=(a_{n-1,1},\cdots,a_{n-1,n-1})$$
　　那么
$$A=\left( {\begin{array}{*{20}{c}}
{A_{n-1}}&{B'_{n-1}}\\
{B_{n-1}}&{a_{nn}}
\end{array}} \right)$$
$$|A|=\left| {\begin{array}{*{20}{c}}
{A_{n-1}}&{B'_{n-1}}\\
{B_{n-1}}&{a_{nn}}
\end{array}} \right|=\left| {\begin{array}{*{20}{c}}
{A_{n-1}}&{0}\\
{B_{n-1}}&{a_{nn}}
\end{array}} \right|+\left| {\begin{array}{*{20}{c}}
{A_{n-1}}&{B'_{n-1}}\\
{B_{n-1}}&{0}
\end{array}} \right|$$
　　根据$Hurwitz$定理，$A_{n-1}$为正定矩阵，再由上面小题知
$$\left| {\begin{array}{*{20}{c}}
{A_{n-1}}&{B'_{n-1}}\\
{B_{n-1}}&{0}
\end{array}} \right| \le 0$$
　　从而
$$|A| \le \left| {\begin{array}{*{20}{c}}
{A_{n-1}}&{0}\\
{B_{n-1}}&{a_{nn}}
\end{array}} \right|=a_{nn}|A_{n-1}|$$
（3）根据第（2）小题，结论显然成立。
（4）现在$T'T=T'ET$为正定矩阵，而$T'T$主对角线上元素为
$$t_{1i}^2+t_{2i}^2+\cdots+t_{ni}^2$$
　　根据第（3）小题，有
$$|T|^2=|T'T| \le \prod\limits_{i=1}^n(t_{1i}^2+t_{2i}^2+\cdots+t_{ni}^2)$$