A $4 \times 4$ matrix over $\mathbb{C}$ such that rank$(A)=2$ and $A^3 =A^2 \neq 0$ and $A$ is not diagonalizable This question was asked in my linear algebra quiz and I am having Difficulty in it . SO, I am asking for help here.
Edit : I am looking for an answer which solves it without using Jordan Forms. I have seen some Duplicate which use Jordan forms and which don't answer my 1 part. So, I asked it.

Let $A$  be  a $4 \times 4$ matrix over $\mathbb{C}$ such that rank (A)=2 and $A^3 =A^2 \neq 0$ and $A$ is not diagonalizable. (1) Then what is characteristic polynomial of $A$? (2) does there exists a vector v such that $Av\neq 0$ but $A^2 v=0$?

I have proved that minimal polynomial = $x^2 (x-1)$ but unable to deduce characteristic polynomial and (2) also.

Edit :For minimal polynomial I used that $A$ is not diagonalizable and $A^2 \neq 0$. So , (x) (x-1)( Diagonalizability) , x , $x^2$ can't be minimal polynomial. Also , rank =2 implies $x-1$ can't be .

Hope it's fine now.
Can you please help?
 A: Note that the minimal polynomial $m(x) = x^2(x-1)$ must divide the characteristic polynomial, and any zeros of the characteristic polynomial must also be zeros of the minimal polynomial. Since $A$ has size $4$, the characteristic polynomial must have degree $4$.
Thus, there are two possible characteristic polynomials: either we have $p(x) = x^2(x-1)^2$ or $p(x) = x^3(x-1)$. In order to eliminate the first possibility, we need to use the fact that $\operatorname{rank}(A) = 2$.
Suppose for contradiction that $x^2(x-1)^2$ is the characteristic polynomial. Because the exponent of $x$ is $2$, we note that $\dim \ker A^2 \leq 2 = \dim \ker A$. Since it always holds that $\ker A \subseteq \ker A^2$, we have $\ker A = \ker A^2$. However, the fact that $A^2(A-I) = 0$ implies that the image of $A - I$ is a subset of the kernel of $A^2$. Thus, the image of $A - I$ is a subset of the kernel of $A$, which means that $A(A-I) = 0$. This implies that the minimal polynomial is $m(x) = x(x-1)$, which means that $A$ is diagonalizable, contradicting the given information.
As we see in the above proof, it must hold that $\dim \ker A \subsetneq \dim \ker A^2$. This means that there exists a vector $v \in \ker A^2 \setminus \ker A$, which is to say that $Av \neq 0$ but $A^2 v = 0$.
