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Math Help - Cauchy-Schwartz Inequality

  1. #1
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    Cauchy-Schwartz Inequality

    This is a question that has been bugging me for quite some time. When I was a sophomore in Linear Algebra the Professor gave us a proof of the Cauchy Schwarz inequality like so,
    For any vectors u,v in R^n
    | u*v | = | ||u||*||v|| * cos (theta)|
    = ||u|| * ||v|| * |cos(theta)|
    <= ||u|| * ||v|| * 1 since |cos(theta)|<=1
    = ||u|| * ||v||

    Now here is my question, the first step of this proof requires knowing that u*v= ||u||*||v||*cos(theta), which is easily proven using the Law of Cosines. But, the Law of Cosines doesn't necessarily work in R^n for n>3. Therefore, this proof only works for the case when n<=3. Am I correct? Or does the first step of the proof work in any dimension? I'm trying to make the transition from R^n to general normed vector spaces and I gotta get this stuff clear in my head!! Thanks for any replies.
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  2. #2
    hpe
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    Here's a proof that doesn't require the law of cosines.

    Consider the expression f(t) = |u + tv|^2 = (u+tv)*(u+tv). Obviously, f(t) >=0 for all t.

    Expand:
    f(t) = u*u + tu*v + tv*u + t^2v*v
    = |u|^2 + 2tu*v + t^2|v|^2
    = c + bt + at^2

    with c = |u|^2, b = 2u*v, a = |u|^2.

    This is a quadratic function that is never negative (has at most one real zero).

    Thus the discriminant, b^2-4ac is non-positive, b^2-4ac <=0 and thus
    4(u*v)^2 < 4|u|^2|v|^2.
    Divide by 4 and take the square root, and you get the result:

    |u*v| <= |u||v|.

    The law of cosines (in any dimension) then is a consequence, or rather you can define the angle theta between two vectors u,v by u*v = |u||v|cos(theta).

    Hope this helps
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  3. #3
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    Thank you, that's the first time I ever saw that version of the proof.
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