# Math Help - Prove <x,y>=0 iff llx+yll=llx-yll

1. ## Prove <x,y>=0 iff llx+yll=llx-yll

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

2. Originally Posted by Roland25
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 $x,y \in \mathbb{C}$

Remember that
$|z|^2=z \bar{z}$

so if we square both sides we get

$|x+y|^2=|x-y|^2$ and using the above identity we get

$(x+y)\overline{(x+y)}= (x-y)\overline{(x-y)}$

$(x+y)(\bar {x}+\bar{ y})= (x-y)(\bar{ x}- \bar {y})$

$|x|^2+x\bar{y}+y\bar{x}+|y|^2=|x|^2-x\bar{y}-y\bar{x}+|y|^2$

$x\bar{y}+y\bar{x}=0 \iff x\bar{y}+\overline{x\bar{y}}=0 \iff 2Re[{x\bar{y}}]=0$

3. Originally Posted by Roland25
Prove <x,y>=0 iff llx+yll=llx-yll
This is true if the scalar field is the real numbers, but not if it is the complex numbers. For example, |1+i| = |1–i|, but $\langle1,i\rangle = \overline{i}\ne0$.

4. Originally Posted by Roland25
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
Suppose $Re=0$
$||x+y||^2=||x||^2+||y||^2+2Re=||x||^2+||y||^2-2Re=||x-y||^2$

Suppose $||x+y||=||x-y||$
$=$
....
$4Re=0$

Originally Posted by Opalg
This is true if the scalar field is the real numbers, but not if it is the complex numbers. For example, |1+i| = |1–i|, but $\langle1,i\rangle = \overline{i}\ne0$.
You're right, it is not true that the $=0$, but it is always true that the Real part of $$ is zero.