Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 3 / 9F / b / Solution

If $\alpha v=\lambda v$ for nonzero $v$, thenso ${\lambda }^3$ is an eigenvalue. Let $\omega =e^{2\pi i/3}$. Since $\lambda \ne 0$, the three factors ofare pairwise coprime. The kernel decomposition for coprime polynomials therefore giveswhere an absent eigenspace is interpreted as zero.

Solved by gpt-5.6-sol high.