Rodrigues exponential mapping, norm and inequality I've been trying to understand application of Rodrigues rotation to prove error boundness from a paper I'm reading. However, I've stumbled upon a paragraph which I've spent a lot of time trying to figure out.
Given $R, R_d \in SO(3)$ then $R_d^T R \in SO(3)$. We also have $e_\Omega \in R^3$. Consider an error function:
$e_r = \frac{1}{2} \left( tr(R^TR_d)I_{3\times3} - R^TR_d \right) e_{\Omega} $
Rewrite this as: $e_r = C(R_d^TR)e_\Omega$ (which I'm not sure why they did). Now the original authors show that, for any $R_d^TR \in SO(3), ||C(R_d^TR)||_2 \leq 1$. The way they did (which was not shown in details) was to use Rodrigues' formula to show the eigenvalues of $C^T(exp (\hat x))C(exp (\hat x))$ (where the hat indicates skew-symmetric matrix) are $cos^2||x||, \frac{1}{2} (1+cos||x||), \frac{1}{2} (1+cos||x||)$, all are less or equal to 1 for any $x \in R^3$. Then they come to conclusion that $||C(R_d^TR)||_2 \leq 1$ and this implies that $||e_r|| \leq ||e_\Omega||$.
Originally I thought the notation $|| \bullet ||_2 $ was the Frebonius norm, but considering this norm also "acts" on $e_\Omega$ (as in the final inequality), this is an unlikely case. I also don't get to see why they put $C$ in the equations and the eigenvalues part.
I've come to this field from engineering, thus I had a lot of difficulty grasping the above content. Would you mind show me the details of the proof? Your help is greatly appreciated.
 A: I am going to explain, in a rather simple manner, why the eigenvalues have these expressions. 
If I have some time more, I will have a look at other aspects.
For a better understanding, I will simplify the setting.
Let $C$ be the matrix operation such that $R \mapsto tr(R)I-R$. Said otherwise: 
$$\tag{0}C(R)=tr(R)I-R.$$ 
(I have dropped, WLOG, the $\tfrac12$ and the product of two rotations has become a single rotation).
I am going to prove (without needing Rodrigues formula) that, being given the rotation matrix $R=R_{x,\theta}$ ($x$ being its axis, and $\theta$ its rotation angle), 
$$\tag{1} \text{The eigenvalues of} \ \ C(R)^TC(R) \ \ \text{are} \ \ 
\lambda_1 = 4 \cos(\theta)^2 \ \ \text{and } \ \ \lambda_2=\lambda_3=2(1+\cos(\theta))$$ 
Explanation of (1): it is identical to the result we want to prove because


*

*$\exp(\hat x)=R_{x,\|x\|}$ (See Appendix below),

*all eigenvalues have been multiplied by $4$ because of the dropping of factor $\tfrac12$ in the definition of operator $C$.
Proof of (1): A first remark is that the eigenvalues of $R$ are 
$$\tag{2}1, e^{i \theta}, e^{-i \theta}. \ \ \text{Thus} \ \ tr(R)= 1+2\cos(\theta)=:t$$ 
As a consequence, we have, for a certain matrix $P$:
$$\tag{3}R = P \Delta P^{-1} \ \ \text{with} \ \ \Delta:=diag(1,e^{i \theta}, e^{-i \theta})$$
Another remark is that, by orthogonality property :
$$\tag{4} R^T=R^{-1}$$
Let us first expand, using definition (0) and property (4) :
$C(R)^TC(R)=(tI_3-R)^T(tI_3-R)=(tI_3-R^T)(tI_3-R)=(1+t^2)I_3-t(R+R^T)$
Using (2) (and (4) again), we get:
$$C(R)^TC(R)=(1+t^2)I_3-tP (\Delta+\Delta^{-1})P^{-1}$$
i.e., 
$$C(R)^TC(R)=P\underbrace{\left[(1+t^2) I_3-t(\Delta+\Delta^{-1})\right]}_D P^{-1}$$
with
$$D=diag(1+t^2-2t, 1+t^2-2t \cos(\theta),  1+t^2-2t \cos(\theta))$$
Taking the expression of $t$ in (2) into account, the eigenvalues of $D$ are those presented in (1). 
We have finished because similar matrices $D$ and $PDP^{-1}$ have the same eigenvalues.
Appendix : if $x$ has coordinates $x_1,x_2,x_3$, 
$$\hat x := \begin{pmatrix}0&-x_3&x_2\\
x_3&0&-x_1\\-x_2&x_1&0\end{pmatrix}$$
Furthermore $exp(\hat x)=R_{x,\|x\|}$, i.e., $x$ is the rotation axis and $\|x\|$ the rotation angle (in radians of course). See for that the answer by Damien to this question.
