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Cauchy Schwarz proof with alternative dot product definition

  1. Aug 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Does the Cauchy Schwarz inequality hold if we define the dot product of two vectors [itex] A,B \in V_n [/itex] by [itex] \sum_{k=1}^n |a_ib_i| [/itex]? If so, prove it.

    2. Relevant equations

    The Cauchy-Schwarz inequality: [itex] (A\cdot B)^2 \leq (A\cdot A)(B\cdot B) [/itex]. Equality holds iff one of the vectors is a scalar multiple of the other.

    3. The attempt at a solution

    The result is trivial if either vector is the zero vector. Assume [itex] A,B\neq O [/itex].

    Let [itex] A',B'\in V_n [/itex] s.t. [itex] a'_i = |a_i| [/itex] and [itex] b'_i = |b_i| [/itex] for each [itex] i=1,2,...,n [/itex]. Let [itex] \theta [/itex] be the angle between [itex] A' [/itex] and [itex] B' [/itex]. Notice that [itex] A\cdot B = A'\cdot B' [/itex], [itex] ||A|| = ||A'|| [/itex], [itex] ||B|| = ||B'|| [/itex], and [itex] A\cdot B > 0 [/itex]. Then [tex] A\cdot B = ||A||\,||B||\,\cos\theta \implies A\cdot B = ||A||\,||B||\,|\cos\theta| \leq ||A||\,||B|| [/tex] since [itex] 0\leq |\cos\theta|\leq 1 [/itex]. Then [itex] (A\cdot B)^2 \leq ||A||^2||B||^2 = (A\cdot A)(B\cdot B) [/itex]. This proves the inequality.

    We now show that equality holds iff [itex] B = kA [/itex] for some [itex] k\in\mathbb{R} [/itex].

    ([itex] \Longrightarrow [/itex]) We prove this direction by the contrapositive. If [itex] B\neq k A [/itex] for any [itex] k\in\mathbb{R} [/itex], then [itex] B' \neq |k| A' [/itex]. Hence [itex] \theta \neq \alpha \pi [/itex] for any [itex] \alpha \in \mathbb{Z} [/itex]. Thus [itex] 0 < |\cos\theta| < 1 [/itex] and therefore [itex] A\cdot B < ||A||\,||B|| [/itex]. In other words, equality does not hold.

    ([itex] \Longleftarrow [/itex]) If [itex] A = kB [/itex] for some [itex] k\in\mathbb{R} [/itex], then [itex] B' = |k| A' [/itex]. Hence [itex] \theta = \alpha \pi [/itex] for some [itex] \alpha \in \mathbb{Z} [/itex]. Therefore [itex] \cos\theta = \pm 1 [/itex]. Thus equality holds.

    Does this look okay?
     
    Last edited: Aug 7, 2013
  2. jcsd
  3. Aug 7, 2013 #2

    verty

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    I may have missed something but I think you redefine ##A \cdot B## then try to use the fact that ##A \cdot B = ||A||\,||B||\,\cos\theta##, but this is negative if the angle between them is obtuse.
     
  4. Aug 7, 2013 #3
    But don't [itex] A',B' [/itex] still satisfy [itex] A'\cdot B' = ||A'||\,||B'||\,\cos\theta [/itex], where [itex] \theta [/itex] is the angle between [itex] A', B' [/itex]? As well, [itex] A\cdot B = A'\cdot B' [/itex], [itex] ||A|| = ||A'|| [/itex], and [itex] ||B|| = ||B'|| [/itex].
     
  5. Aug 7, 2013 #4

    verty

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    Oh, you should call it ##\theta'## then, it is the angle between ##A'## and ##B'##, not between ##A## and ##B##.
     
  6. Aug 7, 2013 #5
    Okay. I thought it was clear. Thanks for pointing that out!

    Does the proof look okay, then?
     
  7. Aug 7, 2013 #6

    verty

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    Well, you say the inequality is proved, but the original inequality has a square on the left. Something is amiss.
     
  8. Aug 7, 2013 #7
    Edited. What do you tink?
     
  9. Aug 7, 2013 #8

    verty

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    I think you're done. You were asked to prove the equality and I see no gaps. My work here is done :).
     
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