Let $A\in\mathcal{M}_4(\mathbb C)$ such that $\operatorname{rank}(A)=2$ and $A^{3}=A^2$ $\neq0$. Suppose that $A$ is not diagonalizable.
Then
1. One of the Jordan blocks of the Jordan cannonical form of $A$ is
$$A=
\begin{bmatrix}
0 & 1 \\
0 & 0 \\
\end{bmatrix}
$$
2. $A^2=A\neq0$
3. There exists a vector $v$ such that $Av\neq0$ but $A^2v=0$.
4. The characteristic polynomial of $A$ is $x^4-x^3$.
Please someone give me some useful link on complex matrix its rank and diagonalizability.I want to solve this question on my own ,please tell me what theory I should read to solve it
Since $A^3=A^2$ then the polynomial $P=x^3-x^2=x^2(x-1)$ annihilates $A$ and since $A$ isn't diagonalisable then the polynomial $x(x-1)$ with simple roots doesn't annihilate $A$ so $P$ is the minimal polynomial of $A$ because $A^2\ne0$ and then $A$ is similar to $$\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&1\end{pmatrix}$$ because $\operatorname{rank}A=2$ and $1$ is a simple roots of the minimal polynomial. Can you take it from here?