Existence of a decomposition of an arbitrary rotation into three rotations about the $x,y,z$ axis respectively. In reading about Euler Angles from various sources on the internet, it seems the treatment of this subject usually assumes that for an arbitrary rotation $3 \times 3$ rotation matrix $R$ with real entries, that there exists various decompositions of $R=ABC$ where $A,B,C$ are rotations of three angles respective to three co-ordinate axes, and then proceeds to show how to find the angles.
Examples include, for three angles in radians, say $\psi, \theta, \phi$, a decomposition $R=R_x(\psi)R_y(\theta)R_z(\phi)$, i.e. rotations around the $x,y,z$ axis respectively. Wikipedia also includes in their description here, Proper Euler Angles a decomposition using these rotation axis':
Proper Euler angles (z-x-z, x-y-x, y-z-y, z-y-z, x-z-x, y-x-y), where the first decomposition reuses the z axis.
In the case where we wish to express $R=R_z(\psi)R_y(\theta)R_x(\phi)$, I am trying to write an existence proof for such a decomposition. If we assume this is true, then we can solve
$\small R = \begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{33}\\R_{31} & R_{32} & R_{33}\end{bmatrix} = \begin{bmatrix} \cos \psi & -\sin \psi & 0\\ \sin \psi & \cos \psi & 0 \\ 0 & 0 & 1  \end{bmatrix} \begin{bmatrix} \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta\\ \end{bmatrix} \cdot \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos \phi & -\sin \phi \\ 0 &  \sin \phi & \cos \phi \end{bmatrix}$
Giving
$R = \begin{bmatrix} \cos \theta \cos \phi & \sin \psi \sin \theta \cos \phi - \cos \psi \sin\phi & \cos \phi \sin \theta \cos \phi + \sin \psi \sin \phi\\ \cos\theta \sin\phi & \sin \psi \sin\theta \sin \phi + \cos \psi \cos \phi & \cos \psi \sin \theta \sin \phi - \sin \psi\cos \phi & \\-\sin \theta & \sin \psi \cos \theta & \cos \psi \cos \theta \end{bmatrix}$.
Then assuming existence, we can solve for each angle, where for example $\theta = - \sin ^{-1}(R_{31})$.
I am not sure why such a decomposition exists, a priori. Any insights appreciated.
 A: Consider the point on the unit sphere that you want to rotate to $(0,0,1)$. With the rotation about the $x$-axis you can rotate it to the $x$-$z$-plane. Then with the rotation about the $y$-axis you can rotate it to $(0,0,1)$. Then you can use the rotation about the $z$-axis to get all the other points on the unit sphere right, and thus all points.
A: I am assuming you understand how rotation about a single axis work. Now all the rotation matrices form what is known as a group. More specifically this is called special linear group of order 3 i.e $\mathrm{SL}(3,\mathrm R)$.
The property of groups is that they are closed or you can multiply any two members of the group and the product is a member of the group(product of two rotation matrices is a rotation matrix). Since all the matrices you mention in the question are members of this group hence we can be sure that when $R=R_z(\psi)R_y(\theta)R_x(\phi)$, $R$ is a member of $\mathrm{SL}(3,\mathrm R)$.
Rotation takes every point on the surface of sphere to another point. Imagine that you try to bring a certain point $A'$ to it's original position using only 2D rotations. Say $R_z(\psi), R_y(\theta), R_x(\phi)$ then $$RR_x(\phi)R_y(\theta)R_z(\psi)=I$$
Hence $$R=R_x^{-1}(\psi)R_y^{-1}(\theta)R_z^{-1}(\phi)=R'_z(\psi)R'_y(\theta)R'_x(\phi)$$
It should also be clear that this type of decomposition isn't unique and can be done in many other ways.
