Rank is less than $\frac n2$

Question

Let $A \in \mathbb{R}^{n \times n}$ such that $A^2 =0$. Prove that $\mbox{rank}(A) \leq \frac n2$.

With Cayley-Hamilton, the characteristic polynomial is $\chi_A=X^2$. I also know $\dim A = \dim(Im(A)) + \dim(Kernel(A))$ so $n = \dim(Im(A)) + \dim(Kernel(A))$.

Answer 1

Hint:

$Im(A) \subset ker A$ and see what will happen!

Answer 2

Hint: Characteristic poly is $(x^2)^{n/2}$, as the order of the matrix $A$ is $n$.

$x^2$ is an anihilating polynomial. If $A$ is non-zero then $x^2$ is minimal polynomial of $A$.