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标题: 蓝以中下册 带度量的线性空间 38页 习题二3 解答 [打印本页]

作者: castelu    时间: 2016-7-14 19:31
标题: 蓝以中下册 带度量的线性空间 38页 习题二3 解答
习题二3:
  设
$$\alpha_1,\alpha_2,\cdots,\alpha_s$$
  和
$$\beta_1,\beta_2,\cdots,\beta_s$$
  是$n$维欧式空间$V$中两个向量组,证明存在一个正交变换$A$,使
$$A\alpha_i=\beta_i(i=1,2,\cdots,s)$$
  的充分必要条件是
$$(\alpha_i,\alpha_j)=(\beta_i,\beta_j)(i,j=1,2,\cdots,s)$$



解:
  必要性
  设$A$是正交变换,且
$$A\alpha_i=\beta_i(i=1,2,\cdots,n)$$
  所以
$$(\beta_i,\beta_j)=(A\alpha_i,A\alpha_j)=(\alpha_i,\alpha_j),(i,j=1,2,\cdots,n)$$
  充分性
  设
$$(\alpha_i,\alpha_j)=(\beta_i,\beta_j)(i,j=1,2,\cdots,n)$$
  设
$$\sum\limits_{i=1}^nk_i\alpha_i=0$$
  则
$$\left(\sum\limits_{i=1}^nk_i\beta_i,\beta_j\right)=\sum\limits_{i=1}^nk_i(\beta_i,\beta_j)=\sum\limits_{i=1}^nk_i(\alpha_i,\alpha_j)=\left(\sum\limits_{i=1}^nk_i\alpha_i,\alpha_j\right)=0$$
  从而
$$\sum\limits_{i=1}^nk_i\beta_i=0$$
  同理由
$$\sum\limits_{i=1}^nk_i\beta_i=0$$
  可得
$$\sum\limits_{i=1}^nk_i\alpha_i=0$$
  所以
$$\alpha_1,\alpha_2,\cdots,\alpha_n$$
  与
$$\beta_1,\beta_2,\cdots,\beta_n$$
  有相同的线性关系
  不妨设
$$\alpha_1,\alpha_2,\cdots,\alpha_r$$
  是
$$\alpha_1,\alpha_2,\cdots,\alpha_n$$
  的极大无关组,则
$$\beta_1,\beta_2,\cdots,\beta_r$$
  是
$$\beta_1,\beta_2,\cdots,\beta_n$$
  的极大无关组,把
$$\alpha_1,\alpha_2,\cdots,\alpha_r$$
  施行正交化过程得标准正交组
$$(\epsilon_1,\epsilon_2,\cdots,\epsilon_r)=(\alpha_1,\alpha_2,\cdots,\alpha_r)\left( {\begin{array}{*{20}{c}}
{t_{11}}&{t_{12}}&{\cdots}&{t_{1r}}\\
{0}&{t_{22}}&{\cdots}&{t_{2r}}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{0}&{0}&{\cdots}&{t_{nr}}
\end{array}} \right)$$
  设
$$T=\left( {\begin{array}{*{20}{c}}
{t_{11}}&{t_{12}}&{\cdots}&{t_{1r}}\\
{0}&{t_{22}}&{\cdots}&{t_{2r}}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{0}&{0}&{\cdots}&{t_{nr}}
\end{array}} \right)$$
  则有
$$(\eta_1,\eta_2,\cdots,\eta_n)=(\beta_1,\beta_2,\cdots,\beta_r)T$$
  也是
$$\beta_1,\beta_2,\cdots,\beta_r$$
  施行正交化过程而得的标准正交基
  把
$$\epsilon_1,\epsilon_2,\cdots,\epsilon_r$$
  扩充为$V$的标准正交基
$$\epsilon_1,\epsilon_2,\cdots,\epsilon_n$$
  把
$$\eta_1,\eta_2,\cdots,\eta_r$$
  扩充为$V$的标准正交基
$$\eta_1,\eta_2,\cdots,\eta_n$$
  令
$$A\epsilon_i=\eta_i(i=1,2,\cdots,n)$$
  则$A$是一个正交变换,且
$$(\beta_1,\beta_2,\cdots,\beta_r)T=(\eta_1,\eta_2,\cdots,\eta_r)=(A\epsilon_1,A\epsilon_2,\cdots,A\epsilon_r)=(A\alpha_1,A\alpha_2,\cdots,A\alpha_r)T$$
  又由于$T$可逆可知
$$(\beta_1,\beta_2,\cdots,\beta_r)=(A\alpha_1,A\alpha_2,\cdots,A\alpha_r)$$
  即
$$A\alpha_i=\beta_i(i=1,2,\cdots,r)$$
  又有
$$A\alpha_s=\sum\limits_{i=1}^rk_{is}A\alpha_i=\sum\limits_{i=1}^nk_{is}\beta_i=\beta_s(s=1,2,\cdots,n)$$




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