Prove: if $\text{rank}(A)=\text{rank}(A^2)$ then $Ax=0$ and $A^2x=0$ have the same set of solutions.

Question

Let $A\in M_{n\times n}(\mathbb F)$. Prove that if $\text{rank}(A)=\text{rank}(A^2)$ then the systems $Ax=0$ and $A^2x=0$ have the same set of solutions.

I don't really have a clue where to start with this one. Multiplying the matrices didn't seem to help much. The only case that I could solve is if $A$ is an invertible matrix because $Ax=0\Leftrightarrow A^{-1}Ax=A^{-1}0\Leftrightarrow x=0$, similarly for $A^2x=0$.

Answer 1

$$n-\dim(\ker A^2)=\text{rank}~ A^2=\text{rank}~ A=n-\dim(\ker A)$$

we get

$$\dim(\ker A^2)=\dim(\ker A)\tag{1}$$

We also know:

$$x\in\ker A\Rightarrow Ax=0\Rightarrow A^2x=0\Rightarrow x\in \ker A^2$$ hence

$$\ker A\subseteq \ker A^2\tag{2}$$

By (1) and (2), what can you conclude?

Answer 2

Use the fact that ${\rm Ker} A \subseteq {\rm Ker} A^2$ which is true in general. (To see why, if we have $x \in {\rm Ker} A$, then $A x = 0 \implies A^2 x = A (A x) = A (0) = 0$, so that $x \in {\rm Ker} A^2$.) Combined with the fact that ${\rm rank} A = {\rm rank} A^2 \implies {\rm nul} A = {\rm nul} A^2$, this shows that ${\rm Ker} A = {\rm Ker} A^2$.

Answer 3

One can easily check that $$\rm Im(A)=Im(A^2)\Longleftrightarrow Ker(A)=Ker(A^2)$$ Make use of the formula $$\rm dim(Ker(A))+dim(Im(A))=n$$