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)

$f命题：$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

$f命题：$设$W$为$n$维欧氏空间$V$的子空间，且$alpha in V$，则存在唯一的$eta in W$，使得$alpha - eta ot W$

参考答案

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

1

$(13华东师大八)$设$left( {alpha ,eta } ight)$为欧氏空间$V$的内积函数，且对任给$gamma in V$，定义$V$的函数${f_gamma }left( alpha ight) = left( {alpha ,gamma } ight)$
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${eta _1},{eta _2}, cdots ,{eta _m}$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${eta _i} = {a_i}{gamma _i},1 leqslant i leqslant m$

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$(06中南六)$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$sumlimits_{i = 1}^m {{{left( {alpha ,{alpha _i}} ight)}^2}}  leqslant {left| alpha   ight|^2}$

1

$f命题：$

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 {f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 = Qleft( {{E_p}, - {E_q},left( {egin{array}{*{20}{c}} {cos { heta _1}}&{sin { heta _1}} \ { - sin { heta _1}}&{cos { heta _1}} end{array}} ight), cdots ,left( {egin{array}{*{20}{c}} {cos { heta _s}}&{sin { heta _s}} \ { - sin { heta _s}}&{cos { heta _s}} end{array}} ight)} ight){Q^{ - 1}}]一方面注意到[{E_2} = {left( {egin{array}{*{20}{c}} { - 1}&0 \ 0&1 end{array}} ight)^2}, - {E_2} = left( {egin{array}{*{20}{c}} { - 1}&0 \ 0&1 end{array}} ight)left( {egin{array}{*{20}{c}} 1&0 \ 0&{ - 1} end{array}} ight)]是镜像矩阵，另一方面[left( {egin{array}{*{20}{c}} {cos { heta _i}}&{sin { heta _i}} \ { - sin { heta _i}}&{cos { heta _i}} end{array}} ight) = left( {egin{array}{*{20}{c}} {cos frac{3}{2}{ heta _i}}&{ - sin frac{3}{2}{ heta _i}} \ { - sin frac{3}{2}{ heta _i}}&{ - cos frac{3}{2}{ heta _i}} end{array}} ight)left( {egin{array}{*{20}{c}} {cos frac{1}{2}{ heta _i}}&{ - sin frac{1}{2}{ heta _i}} \ { - sin frac{1}{2}{ heta _i}}&{ - cos frac{1}{2}{ heta _i}} end{array}} ight)]其中[left( {egin{array}{*{20}{c}} {cos frac{3}{2}{ heta _i}}&{ - sin frac{3}{2}{ heta _i}} \ { - sin frac{3}{2}{ heta _i}}&{ - cos frac{3}{2}{ heta _i}} end{array}} ight) = {E_2} - 2left( {egin{array}{*{20}{c}} {sin frac{3}{4}{ heta _i}} \ {cos frac{3}{4}{ heta _i}} end{array}} ight)left( {sin frac{3}{4}{ heta _i}{ ext{ }}cos frac{1}{4}{ heta _i}} ight)]与[left( {egin{array}{*{20}{c}} {cos frac{1}{2}{ heta _i}}&{ - sin frac{1}{2}{ heta _i}} \ { - sin frac{1}{2}{ heta _i}}&{ - cos frac{1}{2}{ heta _i}} end{array}} ight) = {E_2} - 2left( {egin{array}{*{20}{c}} {sin frac{1}{4}{ heta _i}} \ {cos frac{1}{4}{ heta _i}} end{array}} ight)left( {sin frac{1}{4}{ heta _i}{ ext{ }}cos frac{1}{4}{ heta _i}} ight)]均为镜像矩阵，所以有结论成立