Rank of a matrix from its order of nilpotency

Question

Let $A$ be an $n \times n$ nilpotent matrix of order $r$, i.e. $A^r = 0$ ,but $A^{r-1} \neq 0$.

Can we conclude from the above information about the rank of $A$?

Can we say rank $A$ is at most $r$?

Answer 1

Consider $Vect(e_1,e_2,e_3,e_4,e_5,e_6)$, $A(e_1)=A(e_2)=A(e_3)=0,A(e_4)=e_1, A(e_5)=e_2, A(e_6)=e_3$, $A^2=0$ and the rank of $A$ is $3$.

Answer 2

The order of nilpotency $r$ is precisely the size of the largest Jordan block of $A$.

On the other hand, the nullity of $A$ is the number of Jordan blocks.

Therefore you can trivially conclude that the Jordan form can have at most $n-r$ other blocks (since their size is $\ge 1$).

So $$\operatorname{null} (A) \le n - r + 1 \implies \operatorname{rank}(A) = n - \operatorname{null} (A) \ge r-1$$

with equality if there is only one block of size $r$, and all other blocks are of size $1$.