Let $K$ be a field and $A \in K^{n \times n}$, $A^2=0$. Show that $\operatorname{rank}(A) \leq \frac{n}{2}$.
I have found similar posts here and here. Unfortunately it is still not clear to me how to solve that task. Can someone help me to find the correct proof?
What I have tried:
It seems like A is nilpotent.
If $A^2=0$ then $\det(A^2)=0$. Because $\det(A \cdot B) = \det(A) \cdot \det(B)$ it follows that:
$\det(A^2)=\det(A) \cdot \det(A) \Rightarrow \det(A)=0$.
$\Rightarrow \chi_A = \det(A - X \cdot E_n) = X^n$
$\Rightarrow A$ has Eigenvalue $0$ (n times).
$\Rightarrow A$ has a $n$-dimensional kernel.
$\Rightarrow$ by nullity theorem: $rank(A)=0 \leq \frac{n}{2}$.
Which is wrong. Counterexample:
$A = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$