Cauchy-Schwartz inequality

  • #1
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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?
 

Answers and Replies

  • #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.
 
  • #3
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Sorry, I don't get it. Can you give me more details?
 
  • #4
StatusX
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Do you know how to project one vector on another?
 
  • #5
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In this case,

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

right?
 
  • #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.
 
  • #7
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Like

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

?
 
  • #8
StatusX
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That shows that b equals its projection if b=la, but not the converse.
 
  • #9
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Then I'm stuck...
 
  • #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).
 
  • #11
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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]

??
 
  • #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?
 
  • #13
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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]

?
 
  • #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:
  • #15
Galileo
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Logarythmic said:
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]

?
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.
 
  • #16
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So <v|v> is constant and I get an expression for [tex]\lambda[/tex]? My head is pretty messed up after this thread.
 
  • #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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