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

1. ## 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

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

CB

3. Hi, thnx for your reaction.

I forgot to mention that it is about a complex space. Otherwise it would be very obvious.
The inproduct is not defined yet, so this statement has to be proved for every possible inproduct in the complex space.

Does anyone have a clue?

Eva