# Cauchy-Schwartz inequality

1. Oct 10, 2006

### Logarythmic

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

$$a = \lambda b$$,

then the Cauchy-Schwartz inequality

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

is an equality.

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

2. Oct 10, 2006

### 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.

3. Oct 11, 2006

### Logarythmic

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

4. Oct 11, 2006

### StatusX

Do you know how to project one vector on another?

5. Oct 11, 2006

### Logarythmic

In this case,

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

right?

6. Oct 11, 2006

### 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.

7. Oct 11, 2006

### Logarythmic

Like

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

?

8. Oct 11, 2006

### StatusX

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

9. Oct 11, 2006

### Logarythmic

Then I'm stuck...

10. Oct 11, 2006

### 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).

11. Oct 12, 2006

### 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

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

??

12. Oct 12, 2006

### 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 $\lambda \in \mathbb{R}[/tex]. Now what you can you say about the inner product of [itex]a-\lambda b$ with itself?

13. Oct 12, 2006

### Logarythmic

I don't know, what can I say?

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

?

14. Oct 12, 2006

### 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.

Last edited: Oct 12, 2006
15. Oct 12, 2006

### Galileo

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.)

16. Oct 17, 2006

### Logarythmic

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

17. Oct 17, 2006

### 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?