= Solution
Suppose first that $AB=BA$, and take an <eigenvector> $v_j$ of $A$, with $Av_j=a_jv_j$. Commutation gives
$$
A(Bv_j)=BAv_j=a_jBv_j.
$$
Since the <eigenvalue> $a_j$ is simple, its <eigenspace> is one-dimensional, so $Bv_j=b_jv_j$ 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 $v_1,\ldots,v_n$ are a common <eigenbasis> with $Av_j=a_jv_j$ and $Bv_j=b_jv_j$, then
$$
(AB-BA)v_j=(a_jb_j-b_ja_j)v_j=0
$$
for every basis vector. Hence $AB-BA=0$. Therefore
$$
\boxed{\text{common eigenvector directions}\quad\Longleftrightarrow\quad AB=BA.}
$$
This 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.
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