Can a linear isometry always be expressed in terms of an orthogonal matrix? Is the following true?
Let $S: \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation such that $||S(v)|| = ||v|| \ \text{for all} \ v \in \mathbb{R}^n$, where $||\cdot||$ denotes the  Euclidean norm. Then, for some $A \in O(n)$ and for all $v \in \mathbb{R}^n$, we have $S(v) = Av$. Here $O(n)$ denotes the set of all orthogonal $n \times n$ matrices.
If so, how can one prove it?

Let $A$ be the matrix of $S$. Since $\langle u,v\rangle=\frac14\lVert u+v\rVert^2-\frac14\lVert u-v\rVert^2$, we have that $\langle Au,Av\rangle=\langle u,v\rangle$ for all $u,v$. Id est, $u^t(A^tA)v=(Au)^t(Av)=u^tv=u^tIv$ for all $u,v$. Since $e_i^tXe_j=X_{ij}$, that identity implies $A^tA=I$.
