How can one prove that any rotation of a rigid object in 3-dimensional (3D) space can be represented by a sequence of three rotations around pre-fixed axes by 3 Euler angles? I see this statement in many textbooks, but so far I did not find a proof of the statement.
I do understand that 3 parameters are generally needed to represent rotation of a 3D object (e.g. from Goldstein, Poole, and Safko, Classical Mechanics, 3rd Ed. Ch. 4). However, I can not be sure that 3 Euler angles can be such 3 parameters.
For now, I accept that, for any 3D rotation, there exists a unique matrix in the group SO(3) that transforms the coordinates of a point in the rotated object. I also accept that such a matrix for the rotation around the $z$-axis by angle $\alpha$ is expressed as \begin{equation} R_z (\alpha) = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{equation} and that rotation around $y$-axis has a similar matrix representation. Therefore, I understand that for a given $A \in \mathrm{SO}(3)$, I can calculate the Euler angles by comparing 3 elements of $A$ (e.g. some of the elements in the 3rd row and 3rd columns in the convention adopted below) with the following product of 3 matrices corresponding to the rotations by 3 Euler angles, \begin{equation} R_z (\alpha) R_y (\beta) R_z (\gamma) = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 0 \\ \sin \beta & 0 & \cos\beta \end{pmatrix} \begin{pmatrix} \cos \gamma & \sin \gamma & 0 \\ -\sin \gamma & \cos \gamma & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{equation} However, I can not be sure if the calculated Euler angles always equate the other elements not used in the calculation.
Is there a simple proof without going through much algebraic manipulation?