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Thread: 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

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

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

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

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

    $\displaystyle \begin{aligned}
    |x^*\alpha x| \leq& ||x|| ||\alpha x|| \\
    |\alpha| |x^*x| \leq& |\alpha| ||x||^2\\
    |x^*x| =& ||x||^2
    \end{aligned}$

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

    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:

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

    You have proven $\displaystyle \Leftarrow)$ .

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



    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:

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

    You have proven $\displaystyle \Leftarrow)$ .

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



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

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

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

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

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

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


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