Proof that orthogonal reflection is an isometry Given $v\in V$, we consider the decomposition $v=\mathbf{u}+\tilde{\mathbf{u}}$, where $(\mathbf{u}, \tilde{\mathbf{u}}) \in U \times U^{\perp}$. 
So $\mathcal{R}_{U}(\mathbf{v})=\mathbf{u}-\tilde{\mathbf{u}}$, then we do that in order to prove that the orthogonal reflection is an isometry: $\left\|\mathcal{R}_{U}(\mathbf{v})\right\|^{2}=\|\mathbf{u}-\tilde{\mathbf{u}}\|^{2}=\|\mathbf{u}\|^{2}+\|\tilde{\mathbf{u}}\|^{2}=\|\mathbf{u}+\tilde{\mathbf{u}}\|^{2}=\|\mathbf{v}\|$
I struggle to understand each passage. Can somebody help me to understand it?
 A: $||\mathcal{R}(v)||^2=||u-u’||^2=\langle u-u’,u-u’
\rangle=$
$= ||u||^2-2\langle u, u’ \rangle +||u’||^2=||u||^2+||u’||^2=$
$=||u||^2+2\langle u,u’\rangle + ||u’||^2=\langle u+u’, u+u’\rangle =$
=$||u+u’||^2=||v||^2$ because $u\in U$ and $u’\in U^\perp$ so 
$\langle u, u’\rangle =0$ 
A: We have
$$\begin{align}
\|\mathbf{v}\|^2
&=\|\mathbf{u}+\tilde{\mathbf{u}}\|^{2}\\
&=\langle \mathbf{u}+\tilde{\mathbf{u}},\mathbf{u}+\tilde{\mathbf{u}}\rangle\\
&=\langle \mathbf{u},\mathbf{u}\rangle+\langle\tilde{\mathbf{u}},\mathbf{u}\rangle
+\langle\mathbf{u},\tilde{\mathbf{u}}\rangle+\langle\tilde{\mathbf{u}},\tilde{\mathbf{u}}\rangle\\
&=\|\mathbf{u}\|^{2}+\|\tilde{\mathbf{u}}\|^{2}\end{align}
$$
and also
$$\begin{align}\left\|\mathcal{R}_{U}(\mathbf{v})\right\|^{2}&=
\left\|\mathcal{R}_{U}(\mathbf{u}+\tilde{\mathbf{u}})\right\|^{2}\\
&=\|\mathbf{u}-\tilde{\mathbf{u}}\|^{2}\\
&=\langle \mathbf{u}-\tilde{\mathbf{u}},\mathbf{u}-\tilde{\mathbf{u}}\rangle\\
&=\langle \mathbf{u},\mathbf{u}\rangle-\langle\tilde{\mathbf{u}},\mathbf{u}\rangle
-\langle\mathbf{u},\tilde{\mathbf{u}}\rangle+\langle\tilde{\mathbf{u}},\tilde{\mathbf{u}}\rangle\\
&=\|\mathbf{u}\|^{2}+\|\tilde{\mathbf{u}}\|^{2}\end{align}$$
A: It's the (analogous of the) Pythagorean theorem for an inner product space:

If $\langle a, b\rangle =0$ (i.e. $a\perp b$), then
   $$\|a+b\|^2 =\|a\|^2+\|b\|^2$$

(Simply calculate $\langle(a+b), \, (a+b)\rangle$.) 
Apply it once with $a=u, b=\tilde u$, second with $a=u, b=-\tilde u$. 
