The Cayley transform of a skew-symmetric matrix

Idea. The matrices $I+K$ and $I-K$ commute and are transposes of each other, which gives orthogonality; the formula is an involution.

Step 1: $I+K$ is invertible. If $(I+K)v=0$ then $0=v^T(I+K)v=|v|^2+v^TKv=|v|^2$, since $v^TKv=0$ for skew $K$ ($v^TKv=(v^TKv)^T=-v^TKv$). So $v=0$.

Step 2: $C(K)$ is orthogonal. Using $K^T=-K$: $(I+K)^T=I-K$, so
$$C(K)^T=\big((I+K)^{-1}\big)^T(I-K)^T=(I-K)^{-1}(I+K).$$
Since $I+K$ and $I-K$ commute,
$$C^TC=(I-K)^{-1}(I+K)(I-K)(I+K)^{-1}=(I-K)^{-1}(I-K)(I+K)(I+K)^{-1}=I.$$
Moreover $\det C=\frac{\det(I-K)}{\det(I+K)}=1$, because $\det(I-K)=\det((I+K)^T)=\det(I+K)$. So $C(K)\in SO(n)$.

Step 3: $-1$ is not an eigenvalue. If $C(K)v=-v$, let $w=(I+K)^{-1}v$. Then $(I-K)w=-(I+K)w$, so $w-Kw=-w-Kw$, hence $2w=0$, $v=0$.

Step 4: the inverse map. For $U\in SO(n)$ with $I+U$ invertible, define $K=(I-U)(I+U)^{-1}$.
- *Skew:* with $U^T=U^{-1}$, $K^T=(I+U^T)^{-1}(I-U^T)=\big(U^{-1}(U+I)\big)^{-1}U^{-1}(U-I)=(I+U)^{-1}(U-I)=-K$ (as $I\pm U$ commute).
- *$C(K)=U$:* $I+K=2(I+U)^{-1}$ and $I-K=2U(I+U)^{-1}$, so $C(K)=2U(I+U)^{-1}\cdot\frac12(I+U)=U$.
- Similarly the inverse formula applied to $C(K)$ returns $K$ (the roles of the two maps are symmetric: the Cayley formula is an involution).

Step 5: continuity. Both maps are composites of inversion (continuous, Problem 68), sums and products, hence continuous.

Example. $n=2$, $K=t\begin{pmatrix}0&-1\\1&0\end{pmatrix}$: $C(K)$ is the rotation by angle $-2\arctan t$. Numerically $t=0.7$ gives $\arctan2(C_{10},C_{00})=-1.22145=-2\arctan0.7$. The only rotation not reached is the one with $-1$ in its spectrum, namely $R_\pi=-I$ ($t\to\infty$).

Remark.
- For even $n$, orthogonal matrices with $\det=-1$ always have $-1$ as an eigenvalue, and for odd $n$ too; so the condition "$\in SO(n)$" is automatic.
- This shows $SO(n)$ minus a closed nowhere dense set is homeomorphic to $\mathbb{R}^{n(n-1)/2}$: a rational (algebraic) chart, an alternative to the exponential map.