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 $n/2$.

Let $N = \ker A$. Choose a basis $e_n, ..., e_{r+1}$ of $N$, where $n-r = \dim N$. Take a completion of that basis to obtain a basis $e_1, ..., e_r, e_{r+1}, ..., e_n$ of $K^n$. Denote $M$ as the subspace spanned by the vectors $e_1, ..., e_r$.

We can then write $K^n = M\oplus N$. Suppose that $\text{rank}(A) = r = \dim M>n/2$, then $\dim N = n-r\leq \lfloor n/2\rfloor$. Then $Ae_1$, ..., $Ae_r$ all belong to $N$, but since there's $\geq\lceil n/2 \rceil$ amount of them (in case $n/2$ is an integer, we want to use the inequality $r>n/2$), some must be linearly dependent, say $Ae_r = a_1Ae_1+...+a_{r-1}Ae_{r-1} = A(a_1e_1+...+a_{r-1}e_{r-1})$. But $A$ is 1-1 on $M$, so we must have $e_r = a_1e_1+...+a_{r-1}e_{r-1}$, which is a contradiction with the fact that $e_1, ..., e_r$ is a basis of $M$. Hence we must have $\dim M\leq n/2$.

$A$ is injective on $M$, because if it were $Ax = Ay$, then $A(x-y) = 0$ hence $x-y\in M\cap N = \{0\}$, so that $x = y$.