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Math Help - Cauchy Schwaz inequality

  1. #1
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    Cauchy Schwaz inequality

     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.
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  2. #2
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    Quote Originally Posted by charikaar View Post
     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
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  3. #3
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    Quote Originally Posted by tonio View Post
     \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?
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  4. #4
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    Quote Originally Posted by charikaar View Post
    Does the bar mean complex conjugate?

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

    Tonio
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