 Solution

Since $A^{\mathsf{T}}=A^{-1}$, each eigenvector ${\mathbf{v}}_k$ from part (d) remains an eigenvector, with reciprocal eigenvalue: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.