# Proof of ||x||=||y|| implies that <x+y,x-y> = 0

• Sep 22nd 2009, 12:03 AM
Eva BSc
Proof of ||x||=||y|| implies that <x+y,x-y> = 0
Dear Mathematician,

I need to prove that:

norm ||x||=||y|| implies that inproduct < x+y,x-y > = 0

I thought about it for a long time, but I don't come any further.

I've got that:

< x+y,x-y > = <x,x> + <x,y> - <y,x> - <y,y>
= ||x||^2 + <x,y> - <y,x> - ||y||^2

So somehow ||x||=||y|| implies that <x,y> = <y,x>

I thought of using Cauchy-Schwartz, the triangle inequality or the parallelogram law..

Can someone finish the prove?

Cheers, Eva
• Sep 22nd 2009, 02:59 AM
CaptainBlack
Quote:

Originally Posted by Eva BSc
Dear Mathematician,

I need to prove that:

norm ||x||=||y|| implies that inproduct < x+y,x-y > = 0

I thought about it for a long time, but I don't come any further.

I've got that:

< x+y,x-y > = <x,x> + <x,y> - <y,x> - <y,y>
= ||x||^2 + <x,y> - <y,x> - ||y||^2

So somehow ||x||=||y|| implies that <x,y> = <y,x>

I thought of using Cauchy-Schwartz, the triangle inequality or the parallelogram law..

Can someone finish the prove?

Cheers, Eva

In what domain are we working?

Are these real vectors, or complex, or real or complex functions, or..

For $x, y \in \mathbb{R}^n$ and the usual inner product we have $\langle x,y \rangle=\langle y,x \rangle$ and you are there. This is generaly the case in a real space, but not in a complex space.

CB
• Sep 22nd 2009, 05:56 AM
Eva BSc