Eigenvector of two rotation matrices 
I am having difficulty in understanding a geometry problem which contains geometric-transformation, rotation and reflection.
Background
In this image, a camera with camera center $O_c$ is presented. There is a mirror with unit normal vector $n$. $d$ is the perpendicular distance from mirror surface to $O_c$. Therefore $dn$ is a vector pointing from $O_c$ to mirror surface and it is perpendicular to mirror surface. Point $p=[p_x, p_y, p_z]$ is a point to be observed by the camera, and it has its own coordinate system $O_pX_pY_pZ_p$. Assume $p_c$ is the coordinates of $p$ in $O_cX_cY_cZ_c$, R is a 3x3 rotational matrix and T is a translation vector to convert $p$ in $O_pX_pY_pZ_p$ to $p_c$ in $O_cX_cY_cZ_c$. The transformation can be written as
$p_c = R \cdot p + T ----------(1)$
If I view $p_c$ from the mirror, we are observing the virtual image of $p_c$. Assume that the virtual image of point $p$ is $p'$ and we create a new coordinate system  $O_p'X_p'Y_p'Z_p'$ such that $p' = [p_x, p_y, p_z]$. We also assume $p_c'$ to be the coordinates of $p'$ in $O_cX_cY_cZ_c$. By reflection and translation, we have
$p_c' = (I - 2nn^T)(R \cdot p + T) + 2dn --------(2)$
Consider a 3x3 rotation matrix $R_v$ and translation vector $T_v$ which converts $p'$ in $O_p'X_p'Y_p'Z_p'$ to $p_c'$ in $O_cX_cY_cZ_c$, we have
$p_c' = R_v \cdot p' + T_v --------------(3)$ 
From (2) and (3), we will have $R_v = (I - 2nn^T)R$ and $T_v = (I - 2nn^T)T + 2dn$
If I now rotate the mirror to a different position (i.e. new unit normal vector $n_2$), $p_c'$ will appear on a different position in $O_cX_cY_cZ_c$. With same derivation as the above, we will have
$R_{v2} = (I - 2n_2n_2^T)R$ and $T_v = (I - 2n_2n_2^T)T + 2d_2n_2$
Question
I saw an additional relation which says that if we assume $m$ as the unified cross product of $n$ and $n_2$. Then $R_vR_{v2}^Tm = m$. This essentially says that the eigenvector of the unit eigenvalue of $R_vR_{v2}^T$ equals to the cross product of $n$ and $n_2$. I don't quite understand this part. What is the geometric interpretation of this relation? Is there any topics/materials I could refer to?
Thank you very much!
PS: For your information, I found this relation from a paper Mirror-based Extrinsic Camera Calibration, page 7. 
 A: It looks like you’ve either misunderstood or misquoted this part of the paper. The product of two rotations is another rotation, but the corresponding axis has a rather complicated relationship to the two component rotation axes. A simple counterexample: rotate 90° about the $x$-axis and then 90° about the $z$-axis. The result is equivalent to a 120-degree rotation about $(1,1,1)$, which is certainly not parallel to $(0,0,1)\times(1,0,0)$. 
The identity in the paper is actually $A_jA_{j'}^T\mathbf m=\mathbf m$, where the $A$’s are reflections rather than rotations: $A=(I-2\mathbf n\mathbf n^T)R$. The rotations $R$ are the same for both $A$’s and they cancel, leaving a product of two pure reflections.  
What this identity says is that the composition of two reflections is a rotation about an axis that’s perpendicular to both of the normals of the reflecting planes. This is a standard result. In fact, the rotation axis is the intersection of these planes: each reflection fixes points that lie on its reflecting plane, so the fixed points of the composition are those points that lie on both planes. (If the planes are parallel, the composition is instead a translation, which might be viewed as a rotation about a line at infinity.) It turns out that the rotation angle is twice the dihedral angle of the planes, but that doesn’t appear to be used in the paper. The parenthetical remark in the paper about this product being a “special orthogonal matrix with 2 complex conjugate eigenvalues and 1 eigenvalue equal to 1” is just a long-winded way of saying that it is a rotation.
