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Math Help - Cauchy-Schwarz inequality

  1. #1
    Member Mollier's Avatar
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    Cauchy-Schwarz inequality

    Hi,

    one of my books states the inequality as

    |x^*y| \leq ||x|| ||y|| for all x,y\in\mathbb{C}^{n\times 1},

    and says that equality holds if and only if y=\alpha x for

    \alpha =x^*y/x^*x.

    To me it looks like equality holds if y=\alpha x for any \alpha\in\mathbb{C}:

    \begin{aligned}<br />
|x^*\alpha x| \leq& ||x|| ||\alpha x|| \\<br />
|\alpha| |x^*x| \leq& |\alpha| ||x||^2\\<br />
|x^*x| =& ||x||^2<br />
\end{aligned}

    I do not understand why it has to be \alpha =x^*y/x^*x.

    Could someone please clear it up a bit? Thanks.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The result is the following:

    |x^*y|= \left\|{x}\right\| \left\|{y}\right\|\Leftrightarrow \{x,y\}\textrm{\;lin.\;dep.}

    You have proven \Leftarrow) .

    To prove \Rightarrow) you'll need to choose a particular \alpha .



    Fernando Revilla
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    The result is the following:

    |x^*y|= \left\|{x}\right\| \left\|{y}\right\|\Leftrightarrow \{x,y\}\textrm{\;lin.\;dep.}

    You have proven \Leftarrow) .

    To prove \Rightarrow) you'll need to choose a particular \alpha .



    Fernando Revilla
    So if |x^*y|=||x|| ||y|| then ||\alpha x-y||=0 and so \alpha x-y=0.

    Thanks.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    \Rightarrow)

    If x=0 the equality is trivial and \{x,y\} is linearly dependent . If x\neq 0 decompose:

    y=\alpha x+u,\quad (x\perp{u})

    Using |<x,y>|= \left\|{x}\right\| \left\|{y}\right\| , prove that <x,y>=\alpha <x,x> and 0=<u,u>.

    That is u=0 or equivalently y=\alpha x being \alpha=<x,y>/<x,x> .


    Fernando Revilla
    Last edited by FernandoRevilla; February 9th 2011 at 05:39 AM.
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