Past exam of the mathematical methods I course of the University of Oxford / 2022 / 10 / b

Take the adjoint matrix of $M^{†}{\mathbf{v}}_i={\lambda }_i{\mathbf{v}}_i$ to obtain ${\mathbf{v}}_i^{†}M={\lambda }_i^{\ast }{\mathbf{v}}_i^{†}$. Evaluating ${\mathbf{v}}_i^{†}M{\mathbf{u}}_j$ using this equation and $M{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ givesHence $\boxed{{\mathbf{v}}_i^{†}{\mathbf{u}}_j=0\text{ if }{\lambda }_i^{\ast }\ne {\lambda }_j}$. The proof does not assume that $M$ is a normal matrix or diagonalizable.

In general the spectrum of an adjoint matrix is the complex conjugate of the original spectrum, not necessarily the same set. The equality of the two sets in the question is an additional assumption, possible for a conjugation-invariant spectrum; the displayed orthogonality proof only uses the two stated eigenvalue equations.