Let $A, B \in M_{3}(\mathbb{R})$ two matrices so that $\mathrm{rank}(A) \gt \mathrm{rank}(B)$. Prove $\mathrm{rank}(A^2) \ge \mathrm{rank}(B^2)$
Suppose $B \ne 0_3$, otherwise it's obvious.
Of course, it $\mathrm{rank}(A)=3$ then $\mathrm{rank}(A^2)=3\ge \mathrm{rank}(B^2)$ so the only interesting case is when $rank(A)=2$. This case $rank(B)=1$.
I know $\mathrm{rank}(A^2) \le \mathrm{rank}(A) = 2$ and $\mathrm{rank}(B^2) \le \mathrm{rank}(B) = 1$. If $\mathrm{rank}(B^2) = 1$ I have to prove $\mathrm{rank}(A^2) \ge 1$.
In other words I have to prove: $\mathrm{rank}(A) = 2$ implies $A^2 \ne 0_3$ and here I've got stuck.
Any hint will be appreciated.
I only need an elementary solution.
Assume that $\operatorname{rank}(A) = 2$ and $\operatorname{rank}(B) = 1$. Then $\operatorname{rank}(B^2) \leq \operatorname{rank}(B) = 1$. If $\operatorname{rank}(B^2) = 0$, the result is clear so assume $\operatorname{rank}(B^2) = 1$. Then we need to show that $\operatorname{rank}(A^2) \geq 1$. If $\operatorname{rank}(A^2) = 0$ then $A^2 = 0$ and so $A$ is nilpotent of nilpotency index two. But then we have
$$ \operatorname{Im}(A) \subseteq \ker(A) \implies \dim \ker A \geq 2, \dim \operatorname{Im}(A) = 2$$
which contradicts $\dim \ker A + \dim \operatorname{Im}(A) = 3$.