Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 5 / ii / Solution

Suppose $AB=BA$ and $A{v}_j=a_j{v}_j$. ThenBecause the eigenvalues of $A$ are distinct, the eigenspace for $a_j$ is one-dimensional. It follows that $B{v}_j=b_j{v}_j$, including the possibility $b_j=0$. Thus an orthonormal eigenbasis of $A$ is also an eigenbasis of $B$. The simplicity of both spectra means their eigenvectors coincide as directions, up to ordering, phases and nonzero scalar multiples.

Conversely, if they share a complete eigenbasis, choose the unitary matrix $U$ whose columns are these common normalized eigenvectors. Then $U^{†}AU=D_A$ and $U^{†}BU=D_B$ are diagonal, so $D_A{D}_B=D_B{D}_A$ and $AB=BA$. HenceThis is simultaneous diagonalization of the two Hermitian matrices.