 Solution

Let $Av=\alpha v$. The commutation relation gives $A(Bv)=B(Av)=\alpha Bv$, so $B$ preserves the eigenspace of $A$ with eigenvalue $\alpha$. Because it is a nondegenerate eigenvalue, the eigenspace is one-dimensional, and $Bv=\beta v$ for some real $\beta$. This includes $Bv=0$. Thus the eigenvectors of $A$ form a common eigenbasis, andWriting the corresponding unit-modulus eigenvalue of $U$ as $\zeta =e^{i\theta }$ givesThe pairing refers to the same common eigenvector; independent reorderings of the two real spectra would lose this information. The simultaneous diagonalization can also be constructed without requiring simple eigenvalues by diagonalizing $B$ within each eigenspace of $A$.