Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 5 / a / Solution

Let $Av=\lambda v$ with $v\ne 0$. Taking the Hermitian inner product with $v$ gives $v^{†}Av=\lambda \Vert v{\Vert }^2$. For an anti-Hermitian matrix,Consequently $\overline{\lambda }\Vert v{\Vert }^2=-\lambda \Vert v{\Vert }^2$. As $\Vert v{\Vert }^2>0$, the eigenvalue satisfies $\boxed{\overline{\lambda }=-\lambda }$, or $\boxed{\lambda =i\theta \text{ with }\theta \in \mathbb{R}}$. Zero is included as a purely imaginary eigenvalue.