 Solution

By Jordan normal form, each complex nilpotent matrix of size six can be put by matrix similarity into a block diagonal matrix whose Nilpotent Jordan blocks have sizesA Nilpotent Jordan block of size $s$ has matrix rank $s-1$ and minimal polynomial $t^s$. Therefore, for the whole matrix,Thus the given matrix rank determines the number $k$ of Jordan blocks, and the given minimal polynomial determines the largest size $s_1$. For size six these two data determine the entire integer partition. Here is the complete list, arranged by the largest size:To see that the list is exhaustive, choose the largest part $s_1$ between one and six and partition the remainder $6-s_1$ into parts at most $s_1$. The resulting rows all have distinct pairs $(s_1,k)$. Hence $A$ and $B$ have the same Jordan normal form $J$.

Explicitly, choose invertible matrices $X,Y$ such that $A=XJ{X}^{-1}$ and $B=YJ{Y}^{-1}$. ThenThis proves matrix similarity and illustrates the classification of six-dimensional nilpotent matrices by rank and nilpotency index.