Geometry of the Cayley Transform

Actually we do not need quaternions, because we are working only with one rotation so we can assume that $R$ is rotation around z-axis. We can restrict ourselfs only to xy-plane. Rotations in 2d can be expressed by unit complex numbers and skew-symmetric matrices correspond to pure imaginary numbers.

Cayley transformation for skew-symmetric matrices: $$ \phi:S \longmapsto (I-S)(I+S)^{-1} $$ can be understood through Cayley transformation on complex plane: $$ \psi:i b \longmapsto \frac{1-ib}{1+ib} $$

Thus if you want to know what $I-S$ does you only need to know what does $1-ib$ to complex plane.

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edit - reverse answer I answered you question in "reversed" way too. But It doesn't matter because Cayley transformation $\psi$ from $i\mathbb{R}\cup \{\infty\}$ to $S^1$ is bijection. So for any rotation $e^{i \theta}$ there exists $\psi^{-1}(e^{i \theta})$.

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Actually my answer nicely scales to arbitrarily dimensions. By spectral theorem for skew-symmetric matrices, you can transform to some basis where your matrix takes form:

$$ \begin{bmatrix} \begin{matrix}0 & \lambda_1\\ -\lambda_1 & 0\end{matrix} & 0 & \cdots & 0 \\ 0 & \begin{matrix}0 & \lambda_2\\ -\lambda_2 & 0\end{matrix} & & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & \begin{matrix}0 & \lambda_r\\ -\lambda_r & 0\end{matrix} \\ & & & & \begin{matrix}0 \\ & \ddots \\ & & 0 \end{matrix} \end{bmatrix} $$

And than study each two dimensional subspace associated with block $\left[ \begin{matrix}0 & \lambda_r\\ -\lambda_r & 0\end{matrix}\right]$ with complex numbers.