Given the following rotation matrix:
$$R_{BA} = \begin{bmatrix} 0.2362 & 0.9412 & 0.2414\\ 0.7558 & -0.3341 & 0.5631\\ 0.6107 & 0.0494 & -0.7903 \end{bmatrix} $$
Is it possible to reconstruct the above transformation with only two rotations?
Thoughts:
My intuition tells me no simply because there is no entry that contains a $0$ value so the rotations had to have been a symmetric/asymmetric 3 rotation sequence (ie. 3-1-3, 3-2-1, etc) but how can I show this mathematically? Is it enough to simply construct a 2 rotation DCM and show there's a $0$ value? For example:
3-1 rotation rotation would lead to the DCM $$R_{BA} = \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos(i) & \sin(i)\\ 0 & -\sin(i) & cos(i) \end{bmatrix} \begin{bmatrix} \cos(\omega) & \sin(\omega) & 0\\ -\sin(\omega) & \cos(\omega) & 0\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos(\omega) & \sin(\omega) & 0\\ -\cos(i)\sin(\omega) & \cos(i)\cos(\omega) & \sin(i)\\ \sin(i)\sin(\omega) & -\sin(i)\cos(\omega) & \cos(i) \end{bmatrix} $$
Thanks for your help!
The linear transformation corresponding to this matrix is a (one) rotation about the line through $(0,0,0)$ parallel to the eigenvector of the matrix.
I guess the OP might have meant Euler rotations about the coordinate axes.