I'm studying differential geometry and need confirmation on the solution of the following problem.
A rotation is an orthogonal transformation $C$ such that det $C=+1$. Prove that $C$ does, in fact, rotate $\mathbb{R^3}$ around an axis. Explicitly, given a rotation $C$, show that there exists a $\theta$ and points $\mathbf{e_1}, \mathbf{e_2}, \mathbf{e_3}$ with $\mathbf{e_i}\cdot \mathbf{e_j}=\delta_{ij}$ such that
$$C(\mathbf{e_1})=\operatorname{cos}\theta \mathbf{e_1}+\operatorname{sin}\theta \mathbf{e_2},$$ $$C(\mathbf{e_2})=-\operatorname{sin}\theta \mathbf{e_1}+\operatorname{cos}\theta \mathbf{e_2},$$ $$C(\mathbf{e_3})=\mathbf{e_3}$$
Soltuion: Since the dimension is $3$, there must be an eigenvector of $C$ with eigenvalue $1$, and hence we can name that eigenvector, normalized, to be $\mathbf{e_3}$. Now consider the plane through the origin having $e_3$ as the normal vector. Then we can take any unit vector in the plane and name it $e_1$. Next, we can naturally define $e_2$ to be $e_3\times e_1$, ensuring that ${e_1,e_2,e_3}$ be positively oriented. Then since $C$ is an isometry, it carries the plane through the origin orthogonal to $e_3$, to the plane through the origin orthogonal to $C(e_3)=e_3$. Hence, since $C(e_1)\cdot C(e_3)=0$, $C(e_1)=xe_1+ye_2$ for some $x$ and $y$, but by the fact that $||C(e_1)||=1$, for some $\theta$, $C(\mathbf{e_1})=\operatorname{cos}\theta \mathbf{e_1}+\operatorname{sin}\theta \mathbf{e_2}$. Now taking the cross product of $e_3\times e_1$, we get the desired equation for $e_2$.