Use the result of Exercise 1 and the Jordan form to prove the following. Let \(A\) and \(B\) be \(n \times n\) matrices over the field \(F\) which have the same characteristic polynomial \[ f = (x - c_1)^{d_1} \cdots (x - c_k)^{d_k} \] and the same minimal polynomial. If no \(d_i\) is greater than 3, then \(A\) and \(B\) are similar.
Solution
For each eigenvalue \(c_i\), let the exponent of \((x-c_i)\) in the common minimal polynomial be \(e_i\). In Jordan form, the \(c_i\)-primary component has total dimension \(d_i\), and its largest Jordan block has size \(e_i\) [@HK71].
For \(d_i\le3\), the pair \((d_i,e_i)\) determines the partition of \(d_i\) into Jordan block sizes uniquely: \[ \begin{array}{c|c} d_i,e_i&\text{block sizes}\\ \hline 1,1&1\\ 2,1&1+1\\ 2,2&2\\ 3,1&1+1+1\\ 3,2&2+1\\ 3,3&3. \end{array} \] Thus the common characteristic polynomial determines each total multiplicity \(d_i\), while the common minimal polynomial determines each maximal block size \(e_i\); together they determine every Jordan block for every eigenvalue. Therefore \(A\) and \(B\) have identical Jordan forms up to block order, and hence \[ \boxed{A\sim B.} \] This is precisely the dimension-at-most-three nilpotent classification from Exercise 1 applied separately to the operators \(A-c_iI\) and \(B-c_iI\) on their primary components.