$A\in M_4(\mathbb{C}) \ni A^3=A^2\ne 0$ and rank$(A)=2$, $A$ is not diagonalizable also. $A\in M_4(\mathbb{C}) \ni A^3=A^2\ne 0$ and rank$(A)=2$, $A$ is not diagonalizable also.
Then which of the following are (is) true?


*

*$A^2=A\ne 0$

*$\exists v\ni Av\ne 0$ but $A^2v=0$

*Char Poly of $A$ is $x^4-x^3$ 
Could anyone give me hints?
 A: Note that $A^3 = A^2$ means that $A^3 - A^2 = 0$.  So, taking $p(x) = x^3 - x^2 = x^2(x-1)$, we note that the minimal polynomial of $A$ divides $p$.
So, the only eigenvalues of $A$ are $0$ and $1$.  The maximum possible size of any $1$-block in the Jordan form is $1$, and the maximum size of any $0$-block is $2$.
We note that $\dim(\ker(A - 0I)) = \dim(\ker(A)) = n - \text{rank}(A) = 4 - 2 = 2$.  So, $A$ has two linearly independent eigenvectors associated with $0$.  So, $A$ has two Jordan blocks associated with $0$.
So, the possible Jordan forms (up to permutations of the blocks) are as follows:
$$
\pmatrix{0\\&0\\&&1\\&&&1},
\pmatrix{0&1\\&0\\&&0\\&&&1},
\pmatrix{0&1\\&0\\&&0&1\\&&&0}
$$
Noting that $A^2 \neq 0$, the last possibility doesn't work, so that we are left with
$$
\pmatrix{0\\&0\\&&1\\&&&1},
\pmatrix{0&1\\&0\\&&0\\&&&1}
$$
$A$ is not diagonalizable, so the only possibility we're left with is
$$
\pmatrix{0&1\\&0\\&&0\\&&&1}
$$
We see, then, that 2 and 3 are the correct answer.
