Table of Contents
Gram-Schmidt and Products of Linear Forms
对三项系列 $ \{x, y, z\} $使用Gram-Schmidt过程显示在实数内积空间有
$ <x, y><x, z> \le \frac{1}{2} (<y, z> + || y | | | | z | | ) | | x | |^{2} $
解答：不失一般性我们可假设x, y和z是线性无关的且 $ | | x | | = 1 $，这样Gram-Schmidt关系可写为 $ x = e_ {1}, y = \mu_ {1}e_ {2} + \mu_ {2} e_ {2} $, z = \upsilon_ {1}e_ {1} + \upsilon_ {2}e_ {2} + \upsilon_ {3}e_ {3} $，从中我们有 $ <x, x> = 1, <x, y> = \mu_ {1}, <x, z> = \upsilon_ {1}, <y, z> = \mu_ {1}\upsilon_ {1} + \mu_ {2}\upsilon_ {2} $。问题不等式的边界断言 $ \mu_ {1}\upsilon_ {1} \le \frac{1}{2} (\mu_ {1}\upsilon_ {1} + \mu_ {2}\upsilon_ {2} + (\mu^{2}_ {1} + \mu^{2}_ {2})^{\frac{1}{2}} (\upsilon^{2}_ {1} + \upsilon^{2}_ {2} + \upsilon^{2}_ {3})^{\frac{1}{2}}) $或 $ \mu_ {1} \upsilon_ {1} - \mu_ {2} \upsilon_ {2} \le (\mu^{2}_ {1} + \mu^{2}_ {2})^{\frac{1}{2}} (\upsilon^{2}_ {1} + \upsilon^{2}_ {2} + \upsilon^{2}_ {3})^{\frac{1}{2}} $，这从柯西不等式立即可得到
A Gram-Schmidt Finale
如果x, y, z是内积空间V的元素且如果 $ | | x | | = | | y | | = | | z | | = 1 $，则有不等式
$ | <x, x><y, z> - <x, y><x, z> | \le \{<x, x>^{2} - | <x,y> |^{2}\} \{<x, x>^{2} - | <x, z> |^{2} \} $
且不等式
$ <x, x>^{2} (| <y, z> |^{2} + |<y, x>|^{2} + | <x, z> |^{2}) \\ \le <x, x>^{4} + <x, x><z, y><y, x><x, z> \\ + <x, x><y, z><x, y><z, x> $
我们使用上一题解答的正定性和记号，左边L的边界可写为
$ | <x, x><y, z> - <x,y><x,z> | = | \{(\mu_ {1} \bar{\upsilon_ {1}} + \mu_ {2} \bar{\upsilon_ {2}} \} |^{2} = | \mu_ {2} \bar{\upsilon_ {2}} |^{2} $
且右边R可写为
$ \{<x,x>^{2} - | <x,y> |^{2} \} \{<x,x>^{2} - | <x, z> |^{2} \} \\ = (1 - | \mu_ {1} |^{2}) (1 - | \upsilon_ {1} |^{2}) = | \mu_ {2}|^{2} (| \upsilon_ {2} |^{2} + | \upsilon_ {3} |^{2}) $
因为我们有 $ 1 = | | y | | = | \mu_ {1} |^{2} + | \mu_ {2} |^{2}, 1 = | | z | | = | \upsilon_ {1} |^{2} + | \upsilon_ {2} |^{2} + | \upsilon_ {3} |^{2} $
这使得 $ L \le R $
现在为证明第二个不等式，它可以缩减为
$ | \mu_ {1} \bar{\upsilon_ {1}} + \mu_ {2}\bar{\upsilon_ {2}} |^{2} + | \mu_ {1} |^{2} + | \upsilon_ {1} |^{2} \\ le 1 + (\bar{\mu}_ {1} \upsilon_ {1} + \bar{\mu}_ {2}\upsilon_ {2})\mu_ {1} \bar{\upsilon_ {1}} + (\mu_ {1}\bar{\upsilon}_ {1} + \mu_ {2}\bar{\upsilon}_ {2}) \bar{\mu}_ {1}\upsilon_ {1} $
且通过扩展，有
$ | \mu_ {1} |^{2} + | \upsilon_ {1} |^{2} + | \mu_ {1} \upsilon_ {1} |^{2} + | \mu_ {2} \upsilon_ {2} |^{2} + 2 \Re\{\mu_ {1}\bar{\upsilon}_ {1} \bar{\mu}_ {1} \upsilon_ {2}\} \\ \le 1 + 2 | \mu_ {1}\upsilon_ {1} |^{2} + 2 Re \{\mu_ {1}\bar{\upsilon}_ {1}\bar{\mu}_ {2}\upsilon_ {2}\} $
在取消项目后，我们看到
$ L \equiv | \mu_ {1} |^{2} + | \upsilon_ {1} |^{2} + | \mu_ {2}\upsilon_ {2} |^{2} \le 1 + | \mu_ {1}\upsilon_ {1} |^{2} $
但替代 $ | \mu_ {2} \upsilon_ {2} |^{2} = (1 - | \mu_ {1} |^{2})(1 - | \upsilon_ {1} |^{2} - | \upsilon_ {3} |^{2}) $给了我们 $ L = 1 + | \mu_ {1}\upsilon_ {1} |^{2} + | \upsilon_ {3} |^{2} (| \mu_ {1} |^{2} - 1) \le 1 + | \mu_ {1}\upsilon_ {1} |^{2} $因为 $ | \mu_ {1} |^{2} \le 1 $