= Solution
Let $m_\alpha(x)$ be the monic <minimal polynomial>. The <Cayley-Hamilton theorem> says $\chi_\alpha(\alpha)=0$. Divide $\chi_\alpha$ by $m_\alpha$; its remainder also annihilates $\alpha$, and a nonzero remainder of smaller degree would contradict minimality. Thus $m_\alpha$ divides $\chi_\alpha$.
Every characteristic root $\lambda_i$ is an <eigenvalue>, with a nonzero <eigenvector> $v_i$. Since $m_\alpha(\alpha)v_i=m_\alpha(\lambda_i)v_i=0$, it must be a root of the <minimal polynomial> as well. This proves there are no possibilities beyond
$$
\boxed{m_\alpha(x)=\prod_i(x-\lambda_i)^{r_i},\qquad 1\le r_i\le n_i.}
$$
Every such choice really occurs. For each $i$, take one <Jordan block> of size $r_i$ at $\lambda_i$, together with $n_i-r_i$ scalar blocks at that <eigenvalue>, and form their direct sum. The <characteristic polynomial> has the prescribed multiplicities. The nilpotent part of a block of size $r_i$ has its $r_i$th power zero and its $(r_i-1)$st power nonzero, so this component's minimal exponent is exactly $r_i$. A polynomial annihilates the direct sum precisely when it annihilates every component, establishing the displayed <minimal polynomial> without any extra restrictions.
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