= Solution
A column of <coordinates> transforms as $\mathbf x'=U^\dagger\mathbf x$, because the columns of $U$ are the new <orthonormal basis> vectors expressed in the old <coordinates>. A symmetric unitary choice is
$$
\boxed{U=\frac1{\sqrt2}\begin{pmatrix}1&0&i\\0&\sqrt2&0\\i&0&1\end{pmatrix},\quad
A'=\operatorname{diag}(-1,3,1),\quad
\mathbf x'=\begin{pmatrix}(1-i)/\sqrt2\\1\\(1-i)/\sqrt2\end{pmatrix}.}
$$
Its first and third columns are orthogonal <eigenvectors> of the $xz$ block, with <eigenvalues> $-1$ and $+1$ respectively; its second column is the $y$ axis with eigenvalue $3$. Direct multiplication checks $U^\dagger U=I$, while the displayed entries give $U^T=U$. The symmetry here means transpose symmetry, not <Hermitian> symmetry.
Back to article page
