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When does equality hold in Cauchy-Schwarz inequality

  1. Sep 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that if V is a vector space over [tex] \mathbb{C}^n [/tex] with the standard inner product, then

    [tex]
    |<x,y>| = ||x|| \cdot ||y||
    [/tex]

    implies one of the vectors x or y is a multiple of the other.


    3. The attempt at a solution
    Assume the identity holds and that y is not zero. Let

    [tex]
    a = \frac {<x,y>} {||y||^2}
    [/tex]

    and let z = x - ay. I've shown that y and z are orthogonal and want to show

    [tex]
    |a| = \frac {||x||} {||y||}
    [/tex]

    Well,

    [tex]
    |a| \cdot ||y|| = \frac {|<x,y>|} {||y||} = \frac {|\sum_{i=1}^n a_i \overline{b_i} |} {\sqrt{\sum_{i=1}^n |b_i|^2}}
    = \sqrt{\frac {\left(\sum_{i=1}^n a_i \overline{b_i} \right) \left(\sum_{i=1}^n \overline{a_i} b_i \right)} {\sum_{i=1}^n |b_i|^2} }
    [/tex]

    but now I don't see how to simply this further to get this equal to the norm of x.
     
  2. jcsd
  3. Sep 3, 2008 #2
    Never mind ... I think I've got it. I've totally neglected that by assumption,

    [tex]
    \frac { | \langle x,y \rangle | } { \| y \| } = \| x \|
    [/tex]

    so then it follows that

    [tex]
    |a| \cdot \| y \| = \| x \|
    [/tex]

    Then since

    [tex]
    \| x \|^2 = \| ay + z \|^2 = \| ay \|^2 + \| z \|^2 \Rightarrow \| z \| = 0
    [/tex]

    the result follows.

    Embarassing.
     
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