Cauchy Schwaz inequality

**charikaar** · April 4th 2010, 05:40 AM

$0 \leq \left\| u-\delta v \right\|^2<br /> = \langle u-\delta v,u-\delta v \rangle = \langle u,u \rangle - \bar{\delta} \langle u,v \rangle - \delta \langle v,u \rangle + |\delta|^2 \langle v,v\rangle. \,$

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)

thanks for your help.

**tonio** · April 4th 2010, 07:29 AM

Originally Posted by charikaar

$0 \leq \left\| u-\delta v \right\|^2<br /> = \langle u-\delta v,u-\delta v \rangle = \langle u,u \rangle - \bar{\delta} \langle u,v \rangle - \delta \langle v,u \rangle + |\delta|^2 \langle v,v\rangle. \,$

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)

thanks for your help.

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

Tonio

**charikaar** · April 4th 2010, 07:54 AM

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?

**tonio** · April 4th 2010, 12:10 PM

Originally Posted by charikaar

Does the bar mean complex conjugate?

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

Tonio

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