Let $K$ be a field and $A \in K^{n \times n}$, $A^2=0$. Show that $\operatorname{rank}(A) \leq \frac{n}{2}$.

Question

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:

$A = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$

See https://math.stackexchange.com/questions/441116/if-a-is-an-n-times-n-matrix-and-a2-0-then-textranka-le-n-2 instead. — StubbornAtom, Jul 14 '19 at 13:36
I don't see how $\chi_A=X^n$ follows. And if it's true that $A$ has an eigenvalue $0$ with multiplicity $n$ it does not follow that the kernel has dimension $n$. The geometric multiplicity is less than or equal to the algebraic multiplicity. — David C. Ullrich, Jul 14 '19 at 13:38
Hint: Say the rank is $r$. The rank-nullity equation shows the dimension of the nullspace is $n-r$. Now if $r>n/2$ then $r>n-r$, so the image cannot be contained in the nullspace... — David C. Ullrich, Jul 14 '19 at 13:42

Let $K$ be a field and $A \in K^{n \times n}$, $A^2=0$. Show that $\operatorname{rank}(A) \leq \frac{n}{2}$.

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