Past exam of the mathematical methods I course of the University of Oxford / 2015 / 5

Taking the conjugate transpose reverses the product order: $(AB{)}^{†}=B^{†}{A}^{†}=BA$. ThusThe necessary and sufficient additional condition is that the two Hermitian matrices commute.

Complex conjugation changes $i$ to $-i$, and transposition reverses the factors. HenceThus the expression is a Hermitian matrix without any additional condition. Equivalently, the commutator of two Hermitian matrices is skew-Hermitian.

The diagonal entries of a Hermitian matrix satisfy $A_{jj}=\overline{A_{jj}}$, so each is real. Consequently their sum, the trace, is real. No additional condition is required. The same conclusion follows by summing the real eigenvalues.

If $Au=\lambda u$ and $Av=\mu v$, the Hermitian matrix identity givesbecause both eigenvalues are real. Thus eigenvectors belonging to distinct eigenvalues are automatically orthogonal. A simple spectrum guarantees this for an arbitrary choice of one eigenvector per eigenvalue.

Within a degenerate eigenspace, arbitrary independent eigenvectors need not be orthogonal. The Gram-Schmidt process always supplies an orthonormal basis of that eigenspace. Accordingly the existence of an orthonormal eigenbasis requires no additional condition, whereas automatic orthogonality of arbitrary independent eigenvectors is ensured by distinct eigenvalues. Scalar multiples of the same nonzero eigenvector are of course not mutually orthogonal.