Relationship between nilpotency index and rank of a matrix.

Question

Assume two n-dimensional nilpotent matrices A and B with respective ranks $r_{A}$ and $r_{B}$ such that $r_{A}>r_{B}$. Can we conclude that the nilpotency index of A is always greater than B?

Answer 1

This is incorrect. Here is an example $$ A=\left( \begin{array}{ccc|ccc} 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \hline 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{array}\right),\ B= \left( \begin{array}{cc|cc|cc} 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \hline 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \hline 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{array}\right). $$ We have $\operatorname{rank}(A)=2$, $\operatorname{rank}(B)=3$ and $B^2=0$, $A^2\neq0$, $A^3=0$.