$\|x\|=\|y\|$ implies that $\langle x+y,x-y\rangle=0$.

Question

Kreyszig - Introduction to Functional Analysis - Page 135 Exercise 4.

If an inner product space $X$ is real, show that the condition $\|x\|=\|y\|$ implies that $\langle x+y,x-y\rangle=0$.

Okay so $\|x\|=\|y\|\implies \|x^2\| = \|y\|^2\implies \def\l{\langle}\def\r{\rangle} \l x,x\r=\l y,y\r$

I tried playing around from here, but couldn't get it. Below is just a bunch of steps that might be in the right direction, not sure:

$$\begin{align*} \l x,x \r - \l y,y\r &=0\\ \l x,x\r+\l y,y\r &= 2\l y,y\r\\ \quad\quad\quad\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad\,2\l y,y \r &= \l x+y,x+y\r\,\,\,\text{ (by parallelogram equality) }\\ \l x,x\r+ \l y,y\r + 2\l x,y\r &= 2\l y,y\r\\ \l x,x\r + 2\l x,y\r &= \l y,y\r\\ \l x,x\r + \l x,y\r + \l x-y, y\r &= 0\\ \l x,x \r + \l y,x\r + \l x-y,y\r &= 0\\ \l x+y,x\r + \l x-y,y\r &=0 \end{align*}$$

I felt like I was getting close, but not sure from this point. Ideas?

Answer 1

Real inner products are bilinear (that is, linear in each argument) and symmetric. Therefore $$\begin{align*} \langle x+y,x-y\rangle&=\langle x+y,x\rangle-\langle x+y,y\rangle\\ &=\langle x,x\rangle+\langle y,x\rangle-\langle x,y\rangle-\langle y,y\rangle\\ &=\langle x,x\rangle +\langle x,y\rangle-\langle x,y\rangle -\langle y,y\rangle\\ &=\langle x,x\rangle-\langle y,y\rangle\\ &=\|x\|^2-\|y\|^2 \end{align*}$$ equals $0$ if and only if $\|x\|^2=\|y\|^2$, which is the case if and only if $\|x\|=\|y\|$.

Answer 2

$$ \langle x+y,x-y\rangle = \langle x,x\rangle + \langle y,x\rangle - \langle x,y\rangle - \langle y,y\rangle = \langle y,x\rangle - \langle x,y\rangle $$ Now since you are over the reals, $\langle y,x\rangle = \langle x,y\rangle$