Help with a proof.

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October 2nd 2011, 10:18 AM
Kuma

Help with a proof.

Hi, I am having trouble with this proof, I am wondering what to do because I keep getting the wrong answer.

I want to prove

||x-y|| = ||x|| ||y|| ||x bar - y bar||

where

x and y are vectors in Rn

and u bar is defined by u/(||u||^2)

The way i started it was expanding the right side completely by using the norm formula and then simplify it into making it equal to the left, but I end up with the wrong answer and a really messy calculation. I was wondering if there was a simpler way to do this.
October 2nd 2011, 11:30 AM
Plato

Re: Help with a proof.

Quote:

Originally Posted by Kuma

Hi, I am having trouble with this proof, I am wondering what to do because I keep getting the wrong answer.
I want to prove
||x-y|| = ||x|| ||y|| ||x bar - y bar||
where
x and y are vectors in Rn
and u bar is defined by u/(||u||^2)

It seems to me as if there is something missing here.
You write that $\overline{u}=\frac{u}{\|u\|^2}$

You are asked to prove that
$\|x-y\|=\|x\|\|y\|\|\overline{x}-\overline{y}\|$ .

Is that correct? What more is there to $\|x-y\|~?$
October 2nd 2011, 12:30 PM
Kuma

Re: Help with a proof.

Well this is what the question asks exactly

Given a nonzero vector u E R^n, put u bar as:

(1/||u||^2)*u

and the rest is the same, equality you put is correct, that is what it asks me to prove.

I figured that (1/||u||^2)*u would be the same as u/||u||^2, the way the question writes it is (1/||u||^2)*u.

Any ideas as to what is missing? When I try to do the proof by going the long hard way and expanding the norm of x bar - y bar, I get close to the left side but its not quite there after simplifying everything.
October 2nd 2011, 12:36 PM
Plato

Re: Help with a proof.

Quote:

Originally Posted by Kuma

When I try to do the proof by going the long hard way and expanding the norm of x bar - y bar,

What exactly does that mean?
October 2nd 2011, 12:51 PM
Kuma

Re: Help with a proof.

Quote:

Originally Posted by Plato

What exactly does that mean?

Ok so this is what i did for the norm of x bar - y bar

if x = (x1,...,xn) and y = (y1,...,yn)

then i have

||(x1,...,xn)/[sqrt(x1^2+...+xn^2)]^2 - (y1,...,yn)/[sqrt(y1^2+...+yn^2)]^2||

im just going to denote sqrt(x1^2+...+xn^2) = c and sqrt(y1^2+...+yn^2) = d

=sqrt[ (x1/(c^2) - y1/(d^2))^2 +...+(xn/(c^2) - yn/(d^2))^2]

and I went from there onward by simplifying. Is that the right way to do it?
October 3rd 2011, 02:45 PM
Deveno

Re: Help with a proof.

$||x||||y||||\overline{x}-\overline{y}|| = ||x||||y||\sqrt{(\overline{x}-\overline{y})\cdot(\overline{x}-\overline{y})}$

$=||x||||y||\sqrt{(\overline{x}\cdot\overline{x}) - 2(\overline{x} \cdot\overline{y}) + (\overline{y}\cdot\overline{y})}$

$=||x||||y||\sqrt{(\frac{x}{||x||^2}\cdot \frac{x}{||x||^2}) - 2(\frac{x}{||x||^2}\cdot \frac{y}{||y||^2}) + (\frac{y}{||y||^2} \cdot \frac{y}{||y||^2})}$

$=||x||||y||\sqrt{\frac{1}{||x||^2} - 2(\frac{x\cdot y}{||x||^2||y||^2}) + \frac{1}{||y||^2}}$ .

$=||x||||y||\sqrt{\frac{||y||^2 - 2(x \cdot y) + ||x||^2}{||x||^2||y||^2}}$

$=\sqrt{(||x||^2 - 2(x \cdot y) + ||y||^2}$

$=\sqrt{(x \cdot x) - 2(x \cdot y) + (y \cdot y)} = \sqrt{(x-y)\cdot(x-y)}$

$=||x-y||$