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[原创]问题解答57

1. 高等代数问题解答57

Example 1. (武汉大学2013)设$W$为$n$维欧氏空间$V$的子空间,$\alpha\in V.$定义$\alpha$到$W$的距离
$$d(\alpha,W)=|\alpha-\alpha'|,$$
其中$\alpha'$为$\alpha$在子空间$W$上的正交投影.证明:若$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$是$W$的一个基,则
$$d(\alpha,W)=\sqrt{\dfrac{|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m},\alpha)|}{|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})|}},$$
其中$G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$为向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$的Gram矩阵.

注:设$W$是欧式空间$V$的非平凡子空间,则存在惟一的子空间$W^{\bot},$使得$V=W\oplus W^{\bot}.$这样$\forall\alpha\in V,$可以惟一分解为
$$\alpha=\beta+\gamma,\beta\in W,\gamma\in W^{\bot}.$$
称$\beta$为$\alpha$在$W$上的内摄影或正交投影.

\textbf{证明:}(法1)由正交投影的定义,知$\forall\alpha\in V,$
$$\alpha=\alpha'+\beta,\alpha'\in W,\beta\in W^{\bot}.$$
由$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$是$W$的一个基,设
$$\alpha'=x_{1}\alpha_{1}+x_{2}\alpha_{2}+\cdots+x_{m}\alpha_{m}=(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})X,$$
其中$X=(x_{1},x_{2},\cdots,x_{m})^{T}.$
由$\alpha-\alpha'\bot W,$有
$$\left\{
\begin{aligned}%
0=&(\alpha_{1},\alpha-\alpha')=(\alpha_{1},\alpha)-x_{1}(\alpha_{1},\alpha_{1})-x_{2}(\alpha_{1},\alpha_{2})-\cdots-x_{m}(\alpha_{1},\alpha_{m})\\
0=&(\alpha_{2},\alpha-\alpha')=(\alpha_{2},\alpha)-x_{1}(\alpha_{2},\alpha_{1})-x_{2}(\alpha_{2},\alpha_{2})-\cdots-x_{m}(\alpha_{2},\alpha_{m})\\
\cdots&\cdots\\
0=&(\alpha_{m},\alpha-\alpha')=(\alpha_{m},\alpha)-x_{1}(\alpha_{m},\alpha_{1})-x_{2}(\alpha_{m},\alpha_{2})-\cdots-x_{m}(\alpha_{m},\alpha_{m})\\
\end{aligned}
\right.$$
此即
$$G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})X
=\begin{pmatrix}%
(\alpha_{1},\alpha)\\
(\alpha_{2},\alpha)\\
\vdots\\
(\alpha_{m},\alpha)
\end{pmatrix}.$$

若记$Y=\begin{pmatrix}%
(\alpha_{1},\alpha)\\
(\alpha_{2},\alpha)\\
\vdots\\
(\alpha_{m},\alpha)
\end{pmatrix},$
则$G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})X=Y.$
同时注意到$G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$是正定的,从而可逆,于是
$$\begin{aligned}%
&|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m},\alpha)|\\
=&\begin{vmatrix}%
(\alpha_{1},\alpha_{1})&(\alpha_{1},\alpha_{2})&\cdots&(\alpha_{1},\alpha_{m})&(\alpha_{1},\alpha)\\
(\alpha_{2},\alpha_{1})&(\alpha_{2},\alpha_{2})&\cdots&(\alpha_{2},\alpha_{m})&(\alpha_{2},\alpha)\\
\vdots&\vdots&\vdots&\vdots\vdots\\
(\alpha_{m},\alpha_{1})&(\alpha_{m},\alpha_{2})&\cdots&(\alpha_{m},\alpha_{m})&(\alpha_{m},\alpha)\\
(\alpha,\alpha_{1})&(\alpha,\alpha_{2})&\cdots&(\alpha,\alpha_{m})&(\alpha,\alpha)\\
\end{vmatrix}\\
=&\begin{vmatrix}%
G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})&Y\\
Y^{T}&(\alpha,\alpha)
\end{vmatrix}\\
=&\begin{vmatrix}%
G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})&Y\\
0&(\alpha,\alpha)-Y^{T}G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})^{-1}Y
\end{vmatrix}\\
=&\begin{vmatrix}%
G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})&Y\\
0&(\alpha,\alpha)-X^{T}G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})X
\end{vmatrix}\\
=&|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})|((\alpha,\alpha)-(\alpha',\alpha'))
\end{aligned}.$$
这样
$$\sqrt{\dfrac{|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m},\alpha)|}{|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})|}}=\sqrt{(\alpha\alpha)-(\alpha',\alpha')},$$
而$0=(\alpha-\alpha',\alpha')=(\alpha,\alpha')-(\alpha',\alpha'),$即$(\alpha,\alpha')=(\alpha',\alpha'),$于是
$$d(\alpha,W)=|\alpha-\alpha'|=\sqrt{(\alpha-\alpha',\alpha-\alpha')}=\sqrt{(\alpha,\alpha)-2(\alpha,\alpha')+(\alpha',\alpha,)}=\sqrt{(\alpha\alpha)-(\alpha',\alpha')}.$$
这样就完成了证明.

(法2)由正交投影的定义,知$\forall\alpha\in V,$
$$\alpha=\alpha'+\beta,\alpha'\in W,\beta\in W^{\bot}.$$
由$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$是$W$的一个基,设
$$\alpha'=x_{1}\alpha_{1}+x_{2}\alpha_{2}+\cdots+x_{m}\alpha_{m}=(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})X,$$
其中$X=(x_{1},x_{2},\cdots,x_{m})^{T}.$则
$$(\alpha_{i},\alpha')=x_{1}(\alpha_{i},\alpha_{1})+x_{2}(\alpha_{i},\alpha_{2})+\cdots+x_{m}(\alpha_{i},\alpha_{m}),i=1,2,\cdots,m.$$
$$(\alpha',\alpha')=x_{1}(\alpha',\alpha_{1})+x_{2}(\alpha',\alpha_{2})+\cdots+x_{m}(\alpha',\alpha_{m})$$
于是
$$
|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m},\alpha')|
=\begin{vmatrix}%
(\alpha_{1},\alpha_{1})&\cdots&(\alpha_{1},\alpha_{m})&(\alpha_{1},\alpha')\\
(\alpha_{2},\alpha_{1})&\cdots&(\alpha_{2},\alpha_{m})&(\alpha_{2},\alpha')\\
\vdots&\vdots&\vdots&\vdots\\
(\alpha_{m},\alpha_{1})&\cdots&(\alpha_{m},\alpha_{m})&(\alpha_{m},\alpha')\\
(\alpha',\alpha_{1})&\cdots&(\alpha',\alpha_{m})&(\alpha',\alpha')\\
\end{vmatrix}
=0.$$
这是因为最后一列是其余列的线性组合.注意到$(\alpha-\alpha',\alpha_{i})=0,i=1,2,\cdots,m,(\alpha-\alpha',\alpha')=0,$则
$$\begin{aligned}%
&|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m},\alpha)|\\
=&|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m},\alpha+(\alpha-\alpha'))|\\
=&\begin{vmatrix}%
(\alpha_{1},\alpha_{1})&\cdots&(\alpha_{1},\alpha_{m})&(\alpha_{1},\alpha'+(\alpha-\alpha'))\\
(\alpha_{2},\alpha_{1})&\cdots&(\alpha_{2},\alpha_{m})&(\alpha_{2},\alpha'+(\alpha-\alpha'))\\
\vdots&\vdots&\vdots&\vdots\\
(\alpha_{m},\alpha_{1})&\cdots&(\alpha_{m},\alpha_{m})&(\alpha_{m},\alpha'+(\alpha-\alpha'))\\
(\alpha'+(\alpha-\alpha'),\alpha_{1})&\cdots&(\alpha'+(\alpha-\alpha'),\alpha_{m})&(\alpha'+(\alpha-\alpha'),\alpha'+(\alpha-\alpha'))\\
\end{vmatrix}\\
=&\begin{vmatrix}%
(\alpha_{1},\alpha_{1})&\cdots&(\alpha_{1},\alpha_{m})&(\alpha_{1},\alpha'+(\alpha-\alpha'))\\
(\alpha_{2},\alpha_{1})&\cdots&(\alpha_{2},\alpha_{m})&(\alpha_{2},\alpha'+(\alpha-\alpha'))\\
\vdots&\vdots&\vdots&\vdots\\
(\alpha_{m},\alpha_{1})&\cdots&(\alpha_{m},\alpha_{m})&(\alpha_{m},\alpha'+(\alpha-\alpha'))\\
(\alpha',\alpha_{1})&\cdots&(\alpha',\alpha_{m})&(\alpha'+(\alpha-\alpha'),\alpha'+(\alpha-\alpha'))\\
\end{vmatrix}\\
=&\begin{vmatrix}%
(\alpha_{1},\alpha_{1})&\cdots&(\alpha_{1},\alpha_{m})&(\alpha_{1},\alpha')\\
(\alpha_{2},\alpha_{1})&\cdots&(\alpha_{2},\alpha_{m})&(\alpha_{2},\alpha')\\
\vdots&\vdots&\vdots&\vdots\\
(\alpha_{m},\alpha_{1})&\cdots&(\alpha_{m},\alpha_{m})&(\alpha_{m},\alpha')\\
(\alpha',\alpha_{1})&\cdots&(\alpha',\alpha_{m})&(\alpha',\alpha'+(\alpha-\alpha'))\\
\end{vmatrix}\\
&+
\begin{vmatrix}%
(\alpha_{1},\alpha_{1})&\cdots&(\alpha_{1},\alpha_{m})&(\alpha_{1},\alpha-\alpha')\\
(\alpha_{2},\alpha_{1})&\cdots&(\alpha_{2},\alpha_{m})&(\alpha_{2},\alpha-\alpha')\\
\vdots&\vdots&\vdots&\vdots\\
(\alpha_{m},\alpha_{1})&\cdots&(\alpha_{m},\alpha_{m})&(\alpha_{m},\alpha-\alpha')\\
(\alpha',\alpha_{1})&\cdots&(\alpha',\alpha_{m})&(\alpha-\alpha',\alpha'+(\alpha-\alpha'))\\
\end{vmatrix}\\
=&|G(\alpha_{1},\cdots,\alpha_{m},\alpha')|+\begin{vmatrix}%
(\alpha_{1},\alpha_{1})&\cdots&(\alpha_{1},\alpha_{m})&0\\
(\alpha_{2},\alpha_{1})&\cdots&(\alpha_{2},\alpha_{m})&0\\
\vdots&\vdots&\vdots&\vdots\\
(\alpha_{m},\alpha_{1})&\cdots&(\alpha_{m},\alpha_{m})&0\\
(\alpha',\alpha_{1})&\cdots&(\alpha',\alpha_{m})&(\alpha-\alpha',\alpha-\alpha')\\
\end{vmatrix}\\
=&|G(\alpha_{1},\cdots,\alpha_{m})|d(\alpha,W)^{2}.
\end{aligned}$$
所以结论成立.

 

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