Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3 / 13E / i / 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 beyondEvery 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.