# Cauchy Schwaz inequality

• Apr 4th 2010, 05:40 AM
charikaar
Cauchy Schwaz inequality
$0 \leq \left\| u-\delta v \right\|^2
= \langle u-\delta v,u-\delta v \rangle = \langle u,u \rangle - \bar{\delta} \langle u,v \rangle - \delta \langle v,u \rangle + |\delta|^2 \langle v,v\rangle. \,$

if $\delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}. \,$ what is $\bar{\delta}$ equal to? is $\langle u,u \rangle=u.u$ (dot product)

• Apr 4th 2010, 07:29 AM
tonio
Quote:

Originally Posted by charikaar
$0 \leq \left\| u-\delta v \right\|^2
= \langle u-\delta v,u-\delta v \rangle = \langle u,u \rangle - \bar{\delta} \langle u,v \rangle - \delta \langle v,u \rangle + |\delta|^2 \langle v,v\rangle. \,$

if $\delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}. \,$ what is $\bar{\delta}$ equal to? is $\langle u,u \rangle=u.u$ (dot product)

$\delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}=\langle u,v\rangle \|v\|^{-1\slash 2} \,\Longrightarrow \bar {\delta}=\overline{\langle u,v \rangle }\|v\|^{-1\slash 2}$ (why?)

Tonio
• Apr 4th 2010, 07:54 AM
charikaar
Quote:

Originally Posted by tonio
$\delta = \langle u,v \rangle \cdot \langle v,v \rangle^{-1}=\langle u,v\rangle \|v\|^{-1\slash 2} \,\Longrightarrow \bar {\delta}=\overline{\langle u,v \rangle }\|v\|^{-1\slash 2}$ (why?)

Tonio

Does the bar mean complex conjugate?
• Apr 4th 2010, 12:10 PM
tonio
Quote:

Originally Posted by charikaar
Does the bar mean complex conjugate?

I suppose it does....you wrote it, what did you mean by that?!

Tonio