Proof that orthogonal reflection is an isometry

Question

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?

Answer 1

$||\mathcal{R}(v)||^2=||u-u’||^2=\langle u-u’,u-u’ \rangle=$

$= ||u||^2-2\langle u, u’ \rangle +||u’||^2=||u||^2+||u’||^2=$

$=||u||^2+2\langle u,u’\rangle + ||u’||^2=\langle u+u’, u+u’\rangle =$

=$||u+u’||^2=||v||^2$ because $u\in U$ and $u’\in U^\perp$ so

$\langle u, u’\rangle =0$

Answer 2

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 $$\begin{align}\left\|\mathcal{R}_{U}(\mathbf{v})\right\|^{2}&= \left\|\mathcal{R}_{U}(\mathbf{u}+\tilde{\mathbf{u}})\right\|^{2}\\ &=\|\mathbf{u}-\tilde{\mathbf{u}}\|^{2}\\ &=\langle \mathbf{u}-\tilde{\mathbf{u}},\mathbf{u}-\tilde{\mathbf{u}}\rangle\\ &=\langle \mathbf{u},\mathbf{u}\rangle-\langle\tilde{\mathbf{u}},\mathbf{u}\rangle -\langle\mathbf{u},\tilde{\mathbf{u}}\rangle+\langle\tilde{\mathbf{u}},\tilde{\mathbf{u}}\rangle\\ &=\|\mathbf{u}\|^{2}+\|\tilde{\mathbf{u}}\|^{2}\end{align}$$

Answer 3

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 $a=u, b=-\tilde u$.