If $A$ is an $n \times n$ matrix and $ A^2 = 0$, then $\text{rank}(A)\le n/2$.

$$\dim \ker A+rank A=n.$$ In addition, $A^2=0$, hence $im A\subset \ker A$, so $$\dim im A=\operatorname{rank} A\le \dim ker A.$$ These two expressions allow to conclude.

Hint: Let $k = \dim\ker A$. What can you say about $\dim\ker A^2$? Now use $\dim\ker A^2 + \mathop{\rm rank} A^2 = n$.

it's characteristic equation is $$x^2=0$$ which means the eigenvalues of $A$ are all zero.

Besides, $A$ is a nilpotent matrix and all the Jordan blocks have an order no more than 2. This observation leads to its rand no more than $n/2$.