I have stumbled upon this question and have a quetion to that. Short summary here. Say, I have an SO(2) rotation represeted by
$$R(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$
If I take a derivative of this matrix elementwise it is not skew symmetric. However, in the linked thread the answer suggests that I should evaluate the result in $\theta = 0$ which makes the resulting matrix skew symmetric. My question is why is that so? Is there a nice illustrative explanation on that? Or could you at least point me where to read about it. Thanks!
Taking the derivative works entry-wise, yielding $$R'(\theta) = \begin{pmatrix}-\sin \theta&-\cos\theta\\\cos\theta&-\sin\theta\end{pmatrix}.$$ In general this is indeed not skew-symmetric. However, if we want to investigate the tangential space at the identity of $SO(2)$, then we have to evaluate $R'(\theta)$ at the one $\theta$ that belongs to $R(\theta)=I$, that is $\theta=0$. In that case $R(\theta)$ is skew-symmetric. We conclude that the tangential space at the identity consists precisely of the skew-symmetric matrices.