Let \(N_1\) and \(N_2\) be \(6 \times 6\) nilpotent matrices over the field \(F\). Suppose that \(N_1\) and \(N_2\) have the same minimal polynomial and the same nullity. Prove that \(N_1\) and \(N_2\) are similar. Show that this is not true for \(7 \times 7\) nilpotent matrices.
Solution
For a nilpotent matrix, the Jordan block sizes form a partition of the dimension. The exponent of \(x\) in the minimal polynomial is the largest block size, while the nullity is the number of Jordan blocks [@HK71].
1. In dimension six these two data determine the partition uniquely.
The partitions of \(6\), recorded as \((\text{largest part},\text{number of parts})\), are \[ \begin{array}{c|c} 6&(6,1)\\ 5+1&(5,2)\\ 4+2&(4,2)\\ 4+1+1&(4,3)\\ 3+3&(3,2)\\ 3+2+1&(3,3)\\ 3+1+1+1&(3,4)\\ 2+2+2&(2,3)\\ 2+2+1+1&(2,4)\\ 2+1+1+1+1&(2,5)\\ 1+1+1+1+1+1&(1,6). \end{array} \] No two rows have the same ordered pair. Hence if two nilpotent \(6\times6\) matrices have the same minimal polynomial and nullity, they have the same Jordan block partition, hence the same Jordan form and are similar.
2. Dimension seven admits two distinct partitions with the same data.
Consider the nilpotent Jordan matrices \[ N_1=J_3(0)\oplus J_3(0)\oplus J_1(0), \] and \[ N_2=J_3(0)\oplus J_2(0)\oplus J_2(0). \] Both are \(7\times7\), both have minimal polynomial \(x^3\), and both have nullity \(3\) because each has three Jordan blocks. But their Jordan block partitions, \(3+3+1\) and \(3+2+2\), are different. Therefore they are not similar. This proves the asserted failure in dimension seven.