= Solution
If $\alpha v=\lambda v$ for nonzero $v$, then
$$
\alpha^3v=\lambda^3v,
$$
so $\lambda^3$ is an <eigenvalue>. Let $\omega=e^{2\pi i/3}$. Since $\lambda\ne0$, the three factors of
$$
x^3-\lambda^3=(x-\lambda)(x-\omega\lambda)(x-\omega^2\lambda)
$$
are pairwise coprime. The <kernel decomposition for coprime polynomials> therefore gives
$$
\boxed{
E_{\lambda^3}(\alpha^3)
=E_\lambda(\alpha)\oplus E_{\omega\lambda}(\alpha)
\oplus E_{\omega^2\lambda}(\alpha)},
$$
where an absent eigenspace is interpreted as zero.
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