关于欧氏空间的专题讨论(欧氏空间的定义，标准正交基，正交变换，对称变换)

$\bf命题：$设${V_1},{V_2}$为欧氏空间$V$的子空间，且$dim{V_1} < dim{V_2}$，则${V_2}$中一定含有非零向量与${V_1}$正交

参考答案

$\bf命题：$设$W$为$n$维欧氏空间$V$的子空间，且$\alpha \in V$，则存在唯一的$\beta \in W$，使得$\alpha - \beta \bot W$

参考答案

$\bf命题：$设$V$为一个欧氏空间，$W_1$为$V$的有限维子空间，令${W_2} = \left\{ {\alpha  \in V|\alpha  \bot {W_1}} \right\}$，证明：${W_2} = {W_1}^ \bot $

1

$(13华东师大八)$设$\left( {\alpha ,\beta } \right)$为欧氏空间$V$的内积函数，且对任给$\gamma \in V$，定义$V$的函数${f_\gamma }\left( \alpha \right) = \left( {\alpha ,\gamma } \right)$
证明：$(1)$${f_\gamma }$是$V$的线性函数     $(2)$$V$的线性函数都具有${f_\gamma }$的形式

1

$(02川大七)$设${\alpha _1},{\alpha _2}, \cdots ,{\alpha _m}$为欧氏空间$V$的一组线性无关的向量，而${\beta _1},{\beta _2}, \cdots ,{\beta _m}$和${\gamma _1},{\gamma _2}, \cdots ,{\gamma _m}$均可由${\alpha _1},{\alpha _2}, \cdots ,{\alpha _i}$线性表出，证明：存在$m$个实数${a_1},{a_2}, \cdots ,{a_m}$，使得${\beta _i} = {a_i}{\gamma _i},1 \leqslant i \leqslant m$

1

$(06中南六)$设${\alpha _1},{\alpha _2}, \cdots ,{\alpha _m}$为欧氏空间$V$的一个标准正交组，证明：对任意的向量$\alpha \in V$，一定有$\sum\limits_{i = 1}^m {{{\left( {\alpha ,{\alpha _i}} \right)}^2}}  \leqslant {\left| \alpha  \right|^2}$

1

$\bf命题：$

 

 

证明：$n$维欧氏空间$V$的任意正交变换$\sigma$都可以写成若干个镜面反射的乘积 {\bf分析一}\quad将题目转化为正交矩阵的正交相似分解问题. 由$A$为$n$阶正交阵知，存在正交阵$Q$，使得\[A = Q\left( {{E_p}, - {E_q},\left( {\begin{array}{*{20}{c}} {\cos {\theta _1}}&{\sin {\theta _1}} \\ { - \sin {\theta _1}}&{\cos {\theta _1}} \end{array}} \right), \cdots ,\left( {\begin{array}{*{20}{c}} {\cos {\theta _s}}&{\sin {\theta _s}} \\ { - \sin {\theta _s}}&{\cos {\theta _s}} \end{array}} \right)} \right){Q^{ - 1}}\]一方面注意到\[{E_2} = {\left( {\begin{array}{*{20}{c}} { - 1}&0 \\ 0&1 \end{array}} \right)^2}, - {E_2} = \left( {\begin{array}{*{20}{c}} { - 1}&0 \\ 0&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&{ - 1} \end{array}} \right)\]是镜像矩阵，另一方面\[\left( {\begin{array}{*{20}{c}} {\cos {\theta _i}}&{\sin {\theta _i}} \\ { - \sin {\theta _i}}&{\cos {\theta _i}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\cos \frac{3}{2}{\theta _i}}&{ - \sin \frac{3}{2}{\theta _i}} \\ { - \sin \frac{3}{2}{\theta _i}}&{ - \cos \frac{3}{2}{\theta _i}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\cos \frac{1}{2}{\theta _i}}&{ - \sin \frac{1}{2}{\theta _i}} \\ { - \sin \frac{1}{2}{\theta _i}}&{ - \cos \frac{1}{2}{\theta _i}} \end{array}} \right)\]其中\[\left( {\begin{array}{*{20}{c}} {\cos \frac{3}{2}{\theta _i}}&{ - \sin \frac{3}{2}{\theta _i}} \\ { - \sin \frac{3}{2}{\theta _i}}&{ - \cos \frac{3}{2}{\theta _i}} \end{array}} \right) = {E_2} - 2\left( {\begin{array}{*{20}{c}} {\sin \frac{3}{4}{\theta _i}} \\ {\cos \frac{3}{4}{\theta _i}} \end{array}} \right)\left( {\sin \frac{3}{4}{\theta _i}{\text{ }}\cos \frac{1}{4}{\theta _i}} \right)\]与\[\left( {\begin{array}{*{20}{c}} {\cos \frac{1}{2}{\theta _i}}&{ - \sin \frac{1}{2}{\theta _i}} \\ { - \sin \frac{1}{2}{\theta _i}}&{ - \cos \frac{1}{2}{\theta _i}} \end{array}} \right) = {E_2} - 2\left( {\begin{array}{*{20}{c}} {\sin \frac{1}{4}{\theta _i}} \\ {\cos \frac{1}{4}{\theta _i}} \end{array}} \right)\left( {\sin \frac{1}{4}{\theta _i}{\text{ }}\cos \frac{1}{4}{\theta _i}} \right)\]均为镜像矩阵，所以有结论成立