= Solution
Let $Av_j=a_jv_j$. Commutation gives $A(Bv_j)=B(Av_j)=a_jBv_j$, so $B$ preserves each <eigenspace> of $A$. Since the <eigenvalues> of $A$ are nondegenerate, each <eigenspace> is one-dimensional and $Bv_j=b_jv_j$. The <spectral theorem> for a <Hermitian matrix> lets us choose these <vectors> as an <orthonormal eigenbasis>. With $U=(v_1,\ldots,v_n)$,
$$
\boxed{A=U\Lambda_AU^\dagger,\qquad B=U\Lambda_BU^\dagger,\qquad U^\dagger U=I.}
$$
This is <simple spectrum gives simultaneous diagonalization>. The common <eigenvectors> can be chosen up to their phases and ordering. In fact degeneracy does not prevent a common eigenbasis for commuting <Hermitian matrices>, but the one-dimensional argument already suffices here.
The convergent <matrix exponential> series respects similarity, so $e^A=Ue^{\Lambda_A}U^\dagger$ and likewise for $B$. Scalar exponentials multiply on each diagonal entry, giving
$$
\boxed{e^Ae^B=Ue^{\Lambda_A}e^{\Lambda_B}U^\dagger=Ue^{\Lambda_A+\Lambda_B}U^\dagger=e^{A+B}.}
$$
The essential hypothesis is commutation; Hermiticity supplies this convenient diagonalization proof.
Back to article page
