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Thread: Inner Product Space

  1. #1
    Super Member Deadstar's Avatar
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    Inner Product Space

    Could someone explain to me how to go from (1) to (2)?

    $\displaystyle ||x - \langle x,y\rangle y||^2 = \langle x - \langle x,y\rangle y,x - \langle x,y\rangle y\rangle $

    $\displaystyle = \langle x,x\rangle - \langle x,\langle x,y\rangle y\rangle - \langle \langle x,y\rangle y,x\rangle + \langle \langle x,y\rangle y,\langle x,y\rangle y\rangle$ (1)

    $\displaystyle ||x||^2 - \overline{\langle x,y\rangle}\langle x,y\rangle - \langle x,y\rangle\langle y,x\rangle + \langle x,y\rangle\overline{\langle x,y\rangle}$ (2)

    $\displaystyle ||x||^2 - |\langle x,y\rangle|^2$.

    (Just before this it states assume $\displaystyle ||y|| = 1$)
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  2. #2
    Senior Member
    Joined
    Nov 2008
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    Hello there!

    Quote Originally Posted by Deadstar View Post
    Could someone explain to me how to go from (1) to (2)?

    $\displaystyle ||x - \langle x,y\rangle y||^2 = \langle x - \langle x,y\rangle y,x - \langle x,y\rangle y\rangle $

    $\displaystyle = \langle x,x\rangle - \langle x,\langle x,y\rangle y\rangle - \langle \langle x,y\rangle y,x\rangle + \langle \langle x,y\rangle y,\langle x,y\rangle y\rangle$ (1)

    $\displaystyle ||x||^2 - \overline{\langle x,y\rangle}\langle x,y\rangle - \langle x,y\rangle\langle y,x\rangle + \langle x,y\rangle\overline{\langle x,y\rangle}$ (2)

    $\displaystyle ||x||^2 - |\langle x,y\rangle|^2$.

    (Just before this it states assume $\displaystyle ||y|| = 1$)
    There are some rules by definition of the dot product,

    like <x,x> = ||x||^2

    By definition it is

    $\displaystyle <x, \lambda y> = \overline{\lambda}<x,y>$, but

    $\displaystyle <\lambda x, y> = \lambda<x,y>,$

    So let $\displaystyle \lambda := <x,y>$

    $\displaystyle <x, <x,y>y > = \overline{<x,y>}<x,y>$

    And that is why $\displaystyle <<x,y>y,x > = <x,y><y,x>$

    And the last one

    $\displaystyle <<x,y>y,<x,y>y> = \overline{<x,y>}<x,y><y,y>$

    $\displaystyle = \overline{<x,y>}<x,y>||y||^2$

    and $\displaystyle ||y||^2 = 1 = ||y||$ so

    $\displaystyle = \overline{<x,y>}<x,y>$

    Yours
    Rapha
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