Let \(N_1\) and \(N_2\) be \(3 \times 3\) nilpotent matrices over the field \(F\). Prove that \(N_1\) and \(N_2\) are similar if and only if they have the same minimal polynomial.
Solution
Similarity always preserves the minimal polynomial, so only the converse needs proof.
For a nilpotent \(3\times3\) matrix, the Jordan blocks all have eigenvalue zero. If the largest Jordan block has size \(r\), then the minimal polynomial is \[ x^r \] [@HK71]. Since the total dimension is three, there are only three possibilities: \[ \begin{array}{c|c|c} m_N(x)&\text{Jordan block sizes}&\text{Jordan form}\\ \hline x&1+1+1&0_3\\ x^2&2+1&J_2(0)\oplus J_1(0)\\ x^3&3&J_3(0). \end{array} \] Thus the minimal polynomial uniquely determines the Jordan form. If \(N_1\) and \(N_2\) have the same minimal polynomial, they have the same Jordan form and are therefore similar. Hence \[ \boxed{N_1\sim N_2\iff m_{N_1}=m_{N_2}.} \]