Past exam of the mathematical methods I course of the University of Oxford / 2019 / 6

Let $Av=\lambda v$ with $v\ne 0$, using the standard complex Hermitian inner product. ThenTaking the complex conjugate and using $A^{†}=-A$ yields $\overline{\lambda }{v}^{†}v=v^{†}{A}^{†}v=-\lambda v^{†}v$. Since $v^{†}v>0$,Thus every eigenvalue of an anti-Hermitian matrix lies on the imaginary axis, including the possible zero eigenvalue; each nonzero eigenvalue is purely imaginary.

Now take $Au=\lambda u$ and $Av=\mu v$. On one hand $u^{†}Av=\mu u^{†}v$. On the other handConsequently $(\mu -\lambda )u^{†}v=0$. Distinct eigenvalues therefore implywhich is the required orthogonality of the eigenvectors.