Given that $-R_{u,\theta}=\frac13 \begin{bmatrix} 1 & 2 & 2\\ 2 & 1 & -2\\ 2 & -2 & 1 \end{bmatrix}$, find $u\in \mathbb{S}^2$ and $\theta \in [0,\pi]$.
I understand that one of the ways to solve for these is to get four equations from $PRP^T$, where $P$ is a change of basis (orthogonal) matrix. We can get one equation with $p_1, p_2, p_3$ from $PRP^T$, one equation from the fact that $u$ is a unit vector, then probably some equations from orthogonality. But there should be a much shorter and elegant way, although I have no idea about it.
Would appreciate your help with this problem.
For a $3$-dimensional rotation matrix $A$, the angle of rotation $\theta$ can be found by considering the trace of the matrix which is $1+2\cos\theta$. The axis of the rotation can be found by solving $(A-I)v=0$.