When $A$ is $3\times3$, prove that $A^2=0$ iff $A$ has rank less than or equal one and trace zero

Question

Let $A \in M_3$. Prove that $A^2=0\Leftrightarrow \operatorname{tr}(A)=0,\operatorname{rank}(A)\le 1$

I can easily prove $\operatorname{tr}(A)=0,\,\operatorname{rank}(A)\le 1 \Rightarrow A^2=0$ since $\operatorname{rank}(A)\le 1 \Rightarrow A^2=\operatorname{tr}(A)A$.

For $A^2=0 \Rightarrow \operatorname{tr}(A)=0, \operatorname{rank}(A)\le 1$, I prove $\operatorname{tr}(A)=0$ by proving eigenvalues of a nilpotent matrix are zeros, but for $\operatorname{rank}\le 1$ I have no idea.

Can anyone give me a hint?

Answer 1

Suppose that $A^2 = 0$; we must have $\operatorname{Im}(A) \subset \ker(A)$. By the rank-nullity theorem, we can therefore conclude that the rank of $A$ (the dimension of $\operatorname{Im}(A)$) is at most $1$. Now, it suffices to use the fact that $A^2 = \operatorname{trace}(A)\cdot A$, as you noted.

Answer 2

The co-rank of $A^2$ is at most twice the co-rank of $A$.