Past exam of the mathematical methods I course of the University of Oxford / 2013 / 4

Let $A\mathbf{v}=\lambda \mathbf{v}$ with $\mathbf{v}\ne 0$ and $A^{†}=-A$. Taking the conjugate of the scalar ${\mathbf{v}}^{†}A\mathbf{v}$ givesSince ${\mathbf{v}}^{†}\mathbf{v}>0$, it follows thatThus every nonzero eigenvalue of an anti-Hermitian matrix is purely imaginary; zero is allowed as well.

For $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, compute the same mixed scalar in two ways:Using ${\lambda }^{\ast }=-\lambda$ gives $(\mu -\lambda ){\mathbf{u}}^{†}\mathbf{v}=0$. ThereforeThe relevant orthogonality uses the complex inner product, not the unconjugated bilinear dot product. Repeated eigenvalues do not force every possible pair of eigenvectors within that one eigenspace to be orthogonal.