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Cauchy-Schwartz inequality

  1. Oct 10, 2006 #1
    I can show that if the vectors a and b are parallel,

    [tex]a = \lambda b[/tex],

    then the Cauchy-Schwartz inequality

    [tex]
    \newcommand{\braket}[2]{{<\!\!{#1|#2}\!\!>}}
    |\braket{a}{b}|^2 \leq \braket{a}{a} \braket{b}{b}
    [/tex]

    is an equality.

    But how do I show that it is an equality if and only if they are parallel?
     
  2. jcsd
  3. Oct 10, 2006 #2

    StatusX

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    You can write b=Pb+Qb, where Pb is the projection of b onto the subspace generated by a, ie, Pb is parellel to a, and Qb is perpendicular to a. Then a=lambda*b iff Qb=0. Plug the expansion of b into the inequality.
     
  4. Oct 11, 2006 #3
    Sorry, I don't get it. Can you give me more details?
     
  5. Oct 11, 2006 #4

    StatusX

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    Do you know how to project one vector on another?
     
  6. Oct 11, 2006 #5
    In this case,

    [tex]\hat{b} = \frac{b \cdot a}{a \cdot a} a[/tex]

    right?
     
  7. Oct 11, 2006 #6

    StatusX

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    Right, and b is equal to its projection onto a iff a and b are parellel. See if you can express (a,b) in terms of the projection of b onto a.
     
  8. Oct 11, 2006 #7
    Like

    [tex]\hat{b} = \frac{b \cdot a}{a \cdot a} a = \lambda \frac{a \cdot a}{a \cdot a} a = \lambda a[/tex]

    ?
     
  9. Oct 11, 2006 #8

    StatusX

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    That shows that b equals its projection if b=la, but not the converse.
     
  10. Oct 11, 2006 #9
    Then I'm stuck...
     
  11. Oct 11, 2006 #10

    StatusX

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    Never mind, the converse is trivial. Ok, so you've shown that b is a multiple of a iff Qb=b-Pb=0, where Pb is the projection of b onto a. Now plug b=Pb+Qb into (a,b).
     
  12. Oct 12, 2006 #11
    I don't get it. If b = Pb, then

    [tex]|<b|a>|^2 = <a|b> <b|a> = <a|Pb><Pb|a>[/tex]

    and I want to show that this equals

    [tex]<a|a><b|b>[/tex]

    ??
     
  13. Oct 12, 2006 #12

    Galileo

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    Just in case you want to try a different path (because it may be easier):
    If a and b are linearly independent, then [tex]a-\lambda b\not= 0[/tex] for all [itex]\lambda \in \mathbb{R}[/tex]. Now what you can you say about the inner product of [itex]a-\lambda b[/itex] with itself?
     
  14. Oct 12, 2006 #13
    I don't know, what can I say?

    [tex]<a - \lambda b|a - \lambda b> = <a|a> - \lambda <a|b> - \lambda <b|a> + \lambda^2 <b|b>[/tex]

    ?
     
  15. Oct 12, 2006 #14

    StatusX

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    That's a polynomial in lambda that has no real solutions. Look at the determinant.

    Galileo's way is probably easier (and nicer, as it doesn't require a projection operator), but all I wanted you to do was plug in |<a,b>|^2=|<a,Pb+Qb>|^2=|<a,Pb>+<a,Qb>|^2=..., and you'll eventually end up with something like |<a,b>|^2+c||Qb||^2=||a||^2||b||^2, so that equality holds iff Qb is zero.
     
    Last edited: Oct 12, 2006
  16. Oct 12, 2006 #15

    Galileo

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    True. Now, if v is a nonzero vector, what does that say about <v,v>?
    (You have a vector v=a-Lb that is nonzero for EVERY L.)

    Then look at your equation as a quadratic equation in L.
     
  17. Oct 17, 2006 #16
    So <v|v> is constant and I get an expression for [tex]\lambda[/tex]? My head is pretty messed up after this thread.
     
  18. Oct 17, 2006 #17

    Galileo

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    No, <v,v> is not constant wrt lambda. It's a quadratic equation in terms of lambda, as you have shown.

    The inner product has the property that:
    [tex]\langle v,v \rangle \geq 0 \mbox{ with equality only if v=0}[/tex]

    Since your v is not zero it is positive for any value of lambda. That means you have a quadratic equation which has no real roots. What does that tell you about its determinant?
     
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