# Form of rotations in three dimensional spaces

The Euler's theorem says that for every rotation $f\in SO(3)$ of three dimensional Euclidean space there exists a orthogonal basis $e_1, e_2,e_3$ and $\theta \in [0,\pi)$ such that $$M(f)= \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0&\cos \theta & -\sin \theta \\ 0& \sin \theta & \cos \theta \end{array} \right ] ,$$ where $M(f)$ is a matrice of $f$ in basis $e_1,e_2,e_3$.

How to obtain, in algebraic way, that there are $\phi, \psi \in [0,2\pi)$
such that $$M(f)=\left [ \begin{array}{rrr} \cos \phi & -\sin \phi &0 \\ \sin \phi & \cos \phi &0 \\ 0 & 0 & 1 \\ \end{array} \right ] \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0&\cos \theta & -\sin \theta \\ 0& \sin \theta & \cos \theta \end{array} \right ] \left [ \begin{array}{rrr} \cos \psi & -\sin \psi &0 \\ \sin \psi & \cos \psi &0 \\ 0 & 0 & 1 \\ \end{array} \right ]?$$

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I think, you have some misunderstanding. I would say, the $M(f)$ what you wrote, without $C$ and $C^{-1}$, is the matrix of $f$ for some orthonormal basis. And then, writing the basis vectors $e_1,e_2,e_3$ as columns with their original coordinates into a matrix, that will give $C$, so the matrix of $f$ in the original basis (that is: $\begin{pmatrix} 1\\0\\0 \end{pmatrix}$, $\begin{pmatrix} 0\\1\\0 \end{pmatrix}$, $\begin{pmatrix} 0\\0\\1 \end{pmatrix}$) is $C\cdot$(rotation)$\cdot C^{-1}$.
For the second part, why do you think, such $\phi,\psi$ angles exist?
Anyway, because of the matrix with $\psi$-s should correspond to $C^{-1}$ at the same time when the other matrix (with $\phi$-s) correspond to $C$. That is, $\psi=-\phi$ must hold..