When is the Cauchy-Schwartz inequality an equality?

[Logarythmic]

I can show that if the vectors a and b are parallel,

$$a = \lambda b$$,

then the Cauchy-Schwartz inequality

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

is an equality.

But how do I show that it is an equality if and only if they are parallel?

 

[StatusX]

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.

 

[Logarythmic]

Sorry, I don't get it. Can you give me more details?

 

[StatusX]

Do you know how to project one vector on another?

 

[Logarythmic]

In this case,

$$\hat{b} = \frac{b \cdot a}{a \cdot a} a$$

right?

 

[StatusX]

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.

 

[Logarythmic]

Like

$$\hat{b} = \frac{b \cdot a}{a \cdot a} a = \lambda \frac{a \cdot a}{a \cdot a} a = \lambda a$$

?

 

[StatusX]

That shows that b equals its projection if b=la, but not the converse.

 

[Logarythmic]

Then I'm stuck...

 

[StatusX]

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).

 

[Logarythmic]

I don't get it. If b = Pb, then

$$|<b|a>|^2 = <a|b> <b|a> = <a|Pb><Pb|a>$$

and I want to show that this equals

$$<a|a><b|b>$$

??

 

[Galileo]

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

 

[Logarythmic]

I don't know, what can I say?

$$<a - \lambda b|a - \lambda b> = <a|a> - \lambda <a|b> - \lambda <b|a> + \lambda^2 <b|b>$$

?

 

[StatusX]

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.

 

[Galileo]

I don't know, what can I say?

$$<a - \lambda b|a - \lambda b> = <a|a> - \lambda <a|b> - \lambda <b|a> + \lambda^2 <b|b>$$

?

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.

 

[Logarythmic]

So <v|v> is constant and I get an expression for $$\lambda$$? My head is pretty messed up after this thread.

 

[Galileo]

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:
$$\langle v,v \rangle \geq 0 \mbox{ with equality only if v=0}$$

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?