Past exam of the mathematical methods I course of the University of Oxford / 2018 / 5

Write the matrix trace in Einstein summation convention:The entries are scalars, so their multiplication commutes, and the two summed indices may be relabelled. The argument works for square matrices of any finite size, and also for rectangular $A,B$ of compatible opposite shapes.

Let $Av=\lambda v$ with $v\ne 0$. A vanishing commutator gives $A(Bv)=B(Av)=\lambda Bv$, so $Bv$ belongs to the eigenspace $E_{\lambda }$. This is eigenspace invariance for commuting linear maps. A nondegenerate eigenvalue has a one-dimensional eigenspace, so $Bv=\beta v$ for a scalar $\beta$; $\beta =0$ is allowed. Therefore $v$ is also an eigenvector of $B$. Without the simple-eigenvalue condition, $B$ can mix vectors inside a larger eigenspace, and an arbitrarily chosen eigenvector of $A$ need not be an eigenvector of $B$.

For Hermitian matrices, $A^{†}=A$ and $B^{†}=B$, soThe commutator is an anti-Hermitian matrix. Consequentlymakes $C^{†}=C$. More generally, any nonzero purely imaginary $\alpha$ works. For a nonzero commutator this is also necessary if $C$ is to be Hermitian: comparing the Hermitian adjoint equations gives $(\overline{\alpha }+\alpha )C=0$.