= Solution
Since $A^{\mathsf T}=A^{-1}$, each <eigenvector> $\mathbf v_k$ from part (d) remains an <eigenvector>, with reciprocal <eigenvalue>:
$$
\boxed{\mu_k=\lambda_k^{-1}=e^{-2\pi ik/3},\qquad
A^{\mathsf T}\mathbf v_k=\mu_k\mathbf v_k.}
$$
The set of <eigenvalues> is again $\{1,e^{2\pi i/3},e^{4\pi i/3}\}$, with the two nonreal values exchanged relative to the same <eigenvectors>. Equivalently, for $\mu=e^{2\pi ik/3}$ one can choose the <eigenvector> $(1,\mu,\mu^2)^{\mathsf T}$. Pairing each <eigenvalue> with its <eigenvector> matters in the next part.
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