General form of rotation matrix In a few places I have seen general results for rotation matrices (such as finding their eigenvalues etc.) proved by proving them for the case where the matrix is 
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & -\sin\theta \\
0 & \sin\theta & \cos\theta
\end{bmatrix}
$$
Why is it possible to reduce all rotations in $\mathbb R^3$ to this matrix WLOG?
(for example on pg.6 here: https://www2.bc.edu/~reederma/Linalg17.pdf)
 A: Let $R$ be some rotation matrix, and $R^*$ the matrix you propose. Let $f_1$ be the direction that the rotation matrix did not change. Choose $f_2$ and $f_3$ perpendicular to $f_1$ and each other. Now transform the rotation in the standard basis to the orthogonal basis orthogonal basis $\langle f_1,f_2,f_3 \rangle$.
Choose $B = (f_1 | f_2 | f_3)$, then $B^{-1} = B^T$ (due to orthogonality) and $R^* = B R B^T$.
A: Rotations are a rigid transformation.
If we have a rigid transformation, the eigenvalues, including possible complex values, must be |z| = 1,  If they are not, then we have some kind of compression / dilation, and the transformation is not rigid.
In R^3 the characteristic polynomial to determine eigenvalues is a cubic polynomial.  Which means that it must have one real root.
and since |z| = 1 it must be 1 or -1.
If the product of the eigenvalues = -1, then the sphere is flipped inside-out.
and since the characteristic polynomial has real coefficients, complex eigenvalues are in conjugate pairs.
that means the eigenvalues are  (1, e^it, e^-it) which is equivalent to (1, cos t + i sin t, cos t - i sin t), and could include (1,1,1) and (1,-1,-1).
What does all this mean?  There is always some axis of rotation which is the eigenvector associated with the eigenvalue 1, and then there is the degree of rotation, t.
