on a complex vector space: $\|x\| = \|y\| \Leftrightarrow x+y$ orthogonal to $x-y$

Question

I want to know if $\|x\| = \|y\|$ $\Leftrightarrow$ $x+y$ orthogonal to $x-y$.

I was able to prove it for $V$ being a real vector space. However, I'm not able to work it out for complex vector spaces.

Here is the proof for the real case:

(1) suppose $x+y$ and $x-y$ are orthogonal: $$ 0 = \langle x+y,x-y\rangle = \langle x,x\rangle - \langle x,y\rangle + \langle y,x\rangle - \langle y,y\rangle = \langle x,x\rangle - \langle y,y\rangle = \|x\|^2 - \|y\|^2 $$ $\Rightarrow$ $\|x\| = \|y\|$

(2)

The other direction is pretty much the same: $\|x\| = \|y\|$ $\Rightarrow$ $\|x\|^2 = \|y\|^2$ so: $0 = \|x\|^2 - \|y\|^2$ and now the result follows from the equation above in part (1)

Any hints on how to prove it for a complex vectorspace or maybe a counterexample?

Thank you very much.

Answer 1

Let $x \neq 0$ and $y=ix$. Then, $\|x\|=\|y\|$. But $\langle x+y, x-y \rangle=2i\|x\|^{2} \neq 0$. [You have to remember that inner product is conjugate linear in the second variable].

However, $\langle x+y, x-y \rangle= 0$ does imply $\|x\|=\|y\|$ in the complex case also.

Answer 2

Assume $x,y \neq 0$.

$$\lVert x\lVert = \lVert y\lVert \implies (x+y,x-y)=\lVert x\lVert^2 - \lVert y\lVert^2-(x,y)+(y,x)=(y,x)-(x,y)$$

The case of orthogonality $(x,y)=0 \iff (y,x)-(x,y)=0=(x+y,x-y)$

Else If $x,y$ are linearly dependent then exist:

$$\alpha x+ \beta y=0 \implies y=-\frac{\alpha}{\beta}x$$

And from the equal norm we get $$ |\frac{\alpha}{\beta}|=1 \iff \frac{\alpha}{\beta}=e^{i\theta}$$

And $$(y,x)-(x,y)=-(\frac{\alpha}{\beta}- {\frac{\overline\alpha}{\overline\beta} ) } \lVert x\lVert^2=-2i \sin(\theta)\lVert x\lVert^2$$

The case $y= \pm x$ is $ e^{i\theta}= \pm1 \iff \sin(\theta)=0 \implies (x+y,x-y)=0$

$$ (x+y,x-y)=0 \implies (\lVert y\lVert+\lVert x\lVert)(\lVert y\lVert-\lVert x\lVert) = \lVert y\lVert^2 - \lVert x\lVert^2= (x,y)-(y,x)$$

This is a real number equals an imaginary number. Both must be $0$. Thus, either $(x,y)=0$ or $x = \pm y$ $$\lVert y\lVert-\lVert x\lVert=0 \iff \lVert y\lVert=\lVert x\lVert$$

SO: If $(x,y)=0$ or $x= \pm y$ then $$\lVert x \lVert = \lVert y \lVert \iff (x+y,x-y)=0$$