I'm revising for my final and I can't seem to prove this.
If llx+yll=llx-yll then
then Real parts are 0
hence <x,y>=0
But I can't prove it starting with <x,y>
any help appreciated
Since you mention the real part is 0 I assume that $\displaystyle x,y \in \mathbb{C}$
Remember that
$\displaystyle |z|^2=z \bar{z}$
so if we square both sides we get
$\displaystyle |x+y|^2=|x-y|^2$ and using the above identity we get
$\displaystyle (x+y)\overline{(x+y)}= (x-y)\overline{(x-y)}$
$\displaystyle (x+y)(\bar {x}+\bar{ y})= (x-y)(\bar{ x}- \bar {y})$
$\displaystyle |x|^2+x\bar{y}+y\bar{x}+|y|^2=|x|^2-x\bar{y}-y\bar{x}+|y|^2$
$\displaystyle x\bar{y}+y\bar{x}=0 \iff x\bar{y}+\overline{x\bar{y}}=0 \iff 2Re[{x\bar{y}}]=0$
Suppose $\displaystyle Re<x,y>=0$
$\displaystyle ||x+y||^2=||x||^2+||y||^2+2Re<x,y>=||x||^2+||y||^2-2Re<x,y>=||x-y||^2$
Suppose $\displaystyle ||x+y||=||x-y||$
$\displaystyle <x+y,x+y>=<x-y,x-y>$
....
$\displaystyle 4Re<x,y>=0$
You're right, it is not true that the $\displaystyle <x,y>=0$, but it is always true that the Real part of $\displaystyle <x,y>$ is zero.