There is a problem in section 6 of Bill Kahan's document here, as follows:
Let $\mathbf{R}$ be a $3 \times 3$ real orthogonal matrix with $\operatorname{det}(\mathbf{R}) = 1$, and let $\mathbf{v}$ be a unit vector that is an "axis of rotation" for $\mathbf{R}$, in the sense that $\mathbf{R}\mathbf{v} = \mathbf{v}$. Then show that $$ \operatorname{adj}(\mathbf{R} - \mathbf{I}) = \big( 3 - \operatorname{trace}(\mathbf{R}) \big)\mathbf{v}\mathbf{v}^T $$ He says the problem is "hard".
I suppose I could prove it by brute force just by expanding everything, perhaps using a computer algebra system. But I'd like to find a tidier proof. I'm guessing that there's some trick using eigenvectors and/or the Cayley-Hamilton theorem.
As @hardmath commented, the author says that "$\operatorname{adj}(\mathbf{R} - \mathbf{I})$ must be some nonzero scalar multiple of $\mathbf{v}\mathbf{v}^T\,$". I can't prove that, either. If I could, then what's left is just to show that the multiplier factor is $3 - \operatorname{trace}(\mathbf{R})$.
Presumably, $R$ is real orthogonal. As it has size 3 and determinant 1, it is a rotation matrix. Therefore, by a change of orthonormal basis (you may use the identity $\operatorname{adj}(P^{-1}AP)=P^{-1}\operatorname{adj}(A)P$), we may assume that $$ R=\pmatrix{1&0&0\\ 0&\cos t&-\sin t\\ 0&\sin t&\cos t}. $$ Now you can easily verify the problem statement by calculating $\operatorname{adj}(R-I)$ directly.