Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 6

A Hermitian matrix satisfies $A^{†}={\overline{A}}^T=A$. If $Av=\lambda v$ with $v\ne 0$, then $v^{†}Av=\lambda v^{†}v$ is real because its conjugate equals $v^{†}{A}^{†}v=v^{†}Av$. Since $v^{†}v>0$, its eigenvalue $\lambda$ is real. If $Aw=\mu w$ with $\mu \ne \lambda$, thenso $v^{†}w=0$: distinct eigenspaces are orthogonal for the standard complex inner product.

The converse is false. For example $\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$ has the two real eigenvalues one and two but is not a Hermitian matrix. It can be embedded as a block in a $3\times 3$ example if desired.

For Hermitian $A,B$, $(AB{)}^{†}=B^{†}{A}^{†}=BA$. Thus $AB$ is Hermitian exactly when $AB=BA$, equivalently when their commutator vanishes. Also $(AB+BA{)}^{†}=AB+BA$, so a value that works for every such pair isThis sum is their anticommutator.