TITLE: Additive rotation matrices
QUESTION [8 upvotes]: Let's assume that we want to find a rotation matrix which added to a given rotation matrix  gives also a rotation matrix. I would name such matrix a rotation additive matrix for a given rotation matrix.
First consider a 2D case  for identity matrix.  It is relatively easy to find such matrix.
 $ 
        R= \begin{bmatrix}
          -\dfrac{1}{2} & -\dfrac {\sqrt{3}}{2}   \\  
          \dfrac{\sqrt{3}}{2}  & -\dfrac{1}{2}  \\ 
       \end{bmatrix}
$ 
Really we have
 $ 
          \begin{bmatrix}
          -\dfrac{1}{2} & -\dfrac { \sqrt{3}}{2}   \\  
          \dfrac{ \sqrt{3}}{2}  & -\dfrac{1}{2}  \\ 
       \end{bmatrix} + \begin{bmatrix}
           1 & 0    \\  
          0  & 1  \\ 
       \end{bmatrix} =  \begin{bmatrix}
           \dfrac{1}{2} & -\dfrac { \sqrt{3}}{2}   \\  
           \dfrac{\sqrt{3}}{2}  &  \dfrac{1}{2}  \\ 
       \end{bmatrix}  
$   
Also symmetrical matrix to R is additive for identity matrix, so we have at least 2 such matrices. 
If it exists  for identity matrix should, I believe, exist for other 2D rotation matrices.      
I was searching also  for a such matrices in 3D. However without positive effects.    
Question 
Do  such matrices exist in 3D ? 

If so how to find them.  
If not how to prove it.

REPLY [19 votes]: There are no such 3D rotations.
Assume contrariwise that for certain rotations $R_1,R_2,R_3$ the equation
$$
R_1\vec{x}+R_2\vec{x}=R_3\vec{x}\qquad(*)
$$
holds for all $\vec{x}\in\Bbb{R}^3$. If this works for the triple $(R_1,R_2,R_3)$ then multiplying $(*)$ from the left by $R_3^{-1}$ we see that it also works for the triple $(R_3^{-1}R_1,R_3^{-1}R_2,I_3)$. So without loss of generality we can assume that $R_3$ is the identity mapping.
But $R_1$ has an axis (or $\lambda=1$ is one of its eigenvalues), so there exists a non-zero vector $\vec{u}$ such that $R_1\vec{u}=\vec{u}$. Plugging in $\vec{x}=\vec{u}$ shows that $R_2\vec{u}=\vec{0}$. This is impossible, because
as a rotation $R_2$ is non-singular.