Matrix having to be orthogonal, knowing it's norm-preserving If we know that a real $m \times m$ matrix $C$ is norm-preserving, $||C\textbf{v}|| = ||\textbf{v}||$, then $C$ has to be orthogonal. Why should this be the case?
 A: If $e_i$ is the $i$th vector of the standard basis, then $e_i^TXe_j=(X)_{ij}$ is the $(i,j)$-entry of a matrix $X$.
You can use the polarization identity 
$$
x^Ty=\frac{1}{4}(\|x+y\|_2^2+\|x-y\|_2^2).
$$
Since $C$ is norm preserving
$$
x^T(C^TC)y=(Cx)^T(Cy)=\frac{1}{4}\left[\|C(x+y)\|_2^2+\|C(x-y)\|_2^2\right]
=\frac{1}{4}\left[\|x+y\|_2^2+\|x-y\|_2^2\right]=x^Ty.
$$
Now select $x=e_i$ and $y=e_j$ to infer that
$$
(C^TC)_{ij}=e_i^T(C^TC)e_j=e_i^Te_j=\left\{\begin{array}{ll}1&\text{if $i=j$}\\0&\text{if $i\neq j$,}\end{array}\right\}
\iff C^TC=I.
$$

You can also forget about the polarization identity and, as in the other answers, start with 
$$\tag{1}x^TC^TCx=x^Tx\quad\text{for all $x$},$$ 
which is just the square of $\|Cx\|_2=\|x\|_2$.
Choosing $x=e_i$ in (1) implies that 
$$\tag{2}
(C^TC)_{ii}=e_i^TC^TCe_i=e_i^Te_i=1,
$$
so $C^TC$ has unit diagonal.
If $i\neq j$,
$$
\begin{split}
2&=(e_i+e_j)^T(e_i+e_j)\\&\stackrel{(1)}=(e_i+e_j)^TC^TC(e_i+e_j)
\\&=(C^TC)_{ii}+(C^TC)_{jj}+(C^TC)_{ij}+(C^TC)_{ji}
\\&\stackrel{(2)}=2+(C^TC)_{ij}+(C^TC)_{ji},
\end{split}
$$
so $(C^TC)_{ij}+(C^TC)_{ji}=0$. But $C^TC$ is symmetric ($(C^TC)_{ij}=(C^TC)_{ji}$) so this implies that $(C^TC)_{ij}=(C^TC)_{ji}=0$.
A: $$\|Cv\|^2=\|v\|^2\Rightarrow v^*(C^*C-I)v=0\qquad \forall v\in\mathbb{R}^n$$
Take as $v$ the eigenvectors of $C^*C-I$ to yield that all eigenvalues of $C^*C-I$ are zero and therefore 
$$C^*C=I$$
i.e. $C$ orthogonal.
