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.


Do such matrices exist in 3D ?

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

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.

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    $\begingroup$ The same argument works in all odd dimensions, because any matrix in $SO_n(\Bbb{R})$, $n$ odd, has $\lambda=1$ as an eigenvalue. In even dimensions we can cook up examples by building rotation matrices using the ones you found as building blocks. $\endgroup$ – Jyrki Lahtonen Jun 27 '16 at 14:40
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    $\begingroup$ @Widawensis: If you are asking about the case $R_3=I_{2k}$, then there are infinitely many choices for $R_1$ and $R_2$ when $k>1$. You can first build $R_1$ from 2x2 diagonal blocks each rotating a 2-dimensional subspace by 60 degrees clockwise. Same for $R_2$ but rotate counterclockwise. Then, in place of $(R_1,R_2)$ you can use $(MR_1M^T, MR_2M^T)$ for any $M\in SO_{2k}$. Also you can mix cw/ccw in the above recipe. $\endgroup$ – Jyrki Lahtonen Jun 27 '16 at 14:53
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    $\begingroup$ Minor omission, the rotation is not necessarily proper, so you could have $R_1 u = -u$ (which would then contradict $R_2$ being unitary). $\endgroup$ – copper.hat Jun 27 '16 at 17:06
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    $\begingroup$ A good point @copper.hat! I was working inside $SO_3(\Bbb{R})$ all the time. $\endgroup$ – Jyrki Lahtonen Jun 27 '16 at 18:45
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    $\begingroup$ @JyrkiLahtonen: Unfortunately my world is disconnected :-). $\endgroup$ – copper.hat Jun 27 '16 at 18:48

So you want to find $\theta_1$, $\theta_2$, and $\theta_3$ such that:

$$\displaystyle\left[\begin{matrix}\cos\theta_1&-\sin\theta_1\\\sin\theta_1&\cos\theta_1\end{matrix}\right] + \left[\begin{matrix}\cos\theta_2&-\sin\theta_2\\\sin\theta_2&\cos\theta_2\end{matrix}\right] = \left[\begin{matrix}\cos\theta_3&-\sin\theta_3\\\sin\theta_3&\cos\theta_3\end{matrix}\right]$$

Meaning, equivalently:


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    $\begingroup$ How equations above are connected with 3D rotation matrix? I don't require that additive rotations are about the same axis. It is impossible. $\endgroup$ – Widawensen Jun 27 '16 at 14:24

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