Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 5 / c / ii / Solution

For a possibly complex eigenvector $v\ne 0$ with $Av=\lambda v$, invertibility of $I-A$ gives $(I-A{)}^{-1}v=(1-\lambda {)}^{-1}v$. ThereforeThus every eigenvector of $A$ remains an eigenvector of $N$. For a real skew-symmetric matrix, its eigenvalues have the form $\lambda =it$ with $t$ real, so $\nu =e^{2i\arctan t}$ lies on the unit circle, consistently with orthogonality.