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Proof of Cauchy-Shwartz Inequality

  1. Dec 12, 2007 #1
    How would you prove the following:

    let V be an inner product space. For v,w [itex]\epsilon[/itex] V we have:

    |<v,w>| [itex]\leq[/itex] ||v|| ||w||

    with equality if and only if v and w are linearly dependent.

    So far I know that the Cauchy-Schwartz inequality says |< v,w>| is less than or equal to ||v||||w|| for any two vectors v and w in an inner product space.

    I also realise i have to prove

    |< v, w>|< ||v||||w||

    that is, that there is a strict inequality, except in the case where one vector is a multiple of the other.

    any ideas on the proof for this please
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 12, 2007 #2
    Separate them into two cases, one where vector x = 0, and one where it does not. The first is simple to prove, the starting point for the other case is that the scalar product of a vector with itself is greater or equal to 0 (or is it? Why? Try to understand why this is the case!).

    x,y are vectors in [itex]R^n[/tex]; t is an arbitrary scalar.

    ? [tex]\leq (tx + y) \cdot (tx + y)[/tex] = ?...
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