Showing that this inner product equality holds

Question

Let $E$ be a finite dimensional vector space over the real with inner product $\langle \cdot, \cdot \rangle$, and for $v \in E$ define $\sigma_v : E \rightarrow E$ via $\sigma_v(x) = x - \frac{2\langle x, v \rangle}{\langle v,v \rangle}v$, for $x \in E$.

Show that $\langle \sigma_v(x), \sigma_v(y)\rangle = \langle x,y \rangle$.

I've done the obvious steps so far:

\begin{align*} \langle \sigma_v(x), \sigma_v(y) \rangle &= \left\langle x - \frac{2\langle x, v \rangle}{\langle v,v \rangle}v, y - \frac{2\langle y, v \rangle}{\langle v,v \rangle}v \right\rangle\\ &= \left\langle x , y - \frac{2\langle y, v \rangle}{\langle v,v \rangle}v \right\rangle - \left\langle \frac{2\langle x, v \rangle}{\langle v,v \rangle}v, y - \frac{2\langle y, v \rangle}{\langle v,v \rangle}v \right\rangle\\ &= \left\langle y - \frac{2\langle y, v \rangle}{\langle v,v \rangle}v, x \right\rangle - \left\langle y - \frac{2\langle y, v \rangle}{\langle v,v \rangle}v, \frac{2\langle x, v \rangle}{\langle v,v \rangle}v \right\rangle\\ &= \langle y,x \rangle - \left\langle \frac{2\langle y, v \rangle}{\langle v,v \rangle}v, x \right\rangle - \left\langle y, \frac{2\langle x, v \rangle}{\langle v,v \rangle}v \right\rangle + \left\langle \frac{2\langle y, v \rangle}{\langle v,v \rangle}v, \frac{2\langle x, v \rangle}{\langle v,v \rangle}v \right\rangle \end{align*}

Ideally I somehow show that the last three terms are $0$, but I can't seem to get them to cancel or use the property of inner products saying that $\langle 0,0 \rangle = 0$.

Answer 1

Note that for scalars $a$ and $b$, we have $\langle ax, by\rangle=ab\langle x,y\rangle$.

Therefore your last three terms

$$- \left\langle \color{blue}{\frac{2\langle y, v \rangle}{\langle v,v \rangle}}v, x \right\rangle - \left\langle y, \color{red}{\frac{2\langle x, v \rangle}{\langle v,v \rangle}}v \right\rangle + \left\langle \color{green}{\frac{2\langle y, v \rangle}{\langle v,v \rangle}}v, \color{green}{\frac{2\langle x, v \rangle}{\langle v,v \rangle}}v \right\rangle$$

simplify to

$$-\color{blue}{2\frac{\langle y,v\rangle}{\langle v,v\rangle}}\langle v,x\rangle-\color{red}{2\frac{\langle x,v\rangle}{\langle v,v\rangle}}\langle y,v\rangle+\color{green}{4\frac{\langle x,v\rangle\langle y,v\rangle}{\langle v,v\rangle^2}}\langle v,v\rangle$$

Can you take it from here?