Given a rotation A close to identity, how do you explicitly find a skew symmetric matrix approximating A-I?

Question

Suppose I have a rotation matrix $A \in SO(n)$ with $\|A - I_n\|$ small. Can I explicitly write down - in terms of the entries of $A$ - a skew symmetric matrix $S$ which approximates $A-I$? (i.e. the error should be $\|A-I\|^2$).

I sort of feel like this must be possible: If you consider the hyperplane $P'$ spanned by $Ae_1, \dots, Ae_{n-1}$. This is a graph over the hyperplane $P$ spanned by $e_1,\dots,e_{n-1}$. I feel like the linear function $L : P \to P^{\perp}$ whose graph is equal to $P'$ somehow should correspond to $S$. Can anyone help?

Answer 1

You may simply approximate $A-I$ by the skew-symmetric part of $A$.

Let $K=\log(A)$. Then $$ K=\log\left(I+(A-I)\right)=(A-I)-\frac{(A-I)^2}{2}+\frac{(A-I)^3}{3}-\frac{(A-I)^4}{4}+\cdots=O(\|A-I\|) $$ when $\|A-I\|\to0$. It follows that \begin{aligned} \frac{A-A^T}{2}-(A-I) &=I-\frac{A+A^T}{2}\\ &=I-\frac{e^K+e^{-K}}{2}\\ &=-\frac{K^2}{2!}-\frac{K^4}{4!}-\cdots\\ &=O(\|K\|^2)\\ &=O(\|A-I\|^2) \end{aligned} when $\|A-I\|\to0$.

Answer 2

By assumption, $A=(I+S)$ and $S$ is skew and small.

Therefore subtracting the transpose of $A$ isolates $S$, i.e. $$(A-A^T) = (I+S)-(I-S) \;=\; 2S$$ The matrix $A$ is orthogonal since $$A^TA = (I-S)(I+S) \;=\; (I-S^2) \;\to I$$