 Solution

For a skew-Hermitian matrix, $A^{†}=-A$, so both $A^{†}A$ and $A{A}^{†}$ equal $-A^2$. It is therefore a normal matrix. On an eigenvector with eigenvalue $\lambda$, part (a)(iii) givesThus $\overline{\lambda }=-\lambda$, and every eigenvalue lies in $i\mathbb{R}$, including the possible eigenvalue zero.