 Solution

Suppose first that $AB=BA$, and take an eigenvector $v_j$ of $A$, with $A{v}_j=a_j{v}_j$. Commutation givesSince the eigenvalue $a_j$ is simple, its eigenspace is one-dimensional, so $B{v}_j=b_j{v}_j$ for some scalar $b_j$. Thus every vector in an eigenbasis of $A$ is also an eigenvector of $B$. The simple spectrum of $B$ ensures these account for its eigenvector directions as well. Eigenvectors are understood up to nonzero scaling and permutation, rather than as literally identical numerical vectors.

Conversely, if $v_1,\dots ,v_n$ are a common eigenbasis with $A{v}_j=a_j{v}_j$ and $B{v}_j=b_j{v}_j$, thenfor every basis vector. Hence $AB-BA=0$. ThereforeThis is simple spectrum gives simultaneous diagonalization. Without simple spectra, commuting Hermitian matrices still possess a common orthonormal eigenbasis, but not every arbitrary eigenvector of one matrix must be an eigenvector of the other.