A complex matrix $A$ such that. $A^3 = A^2 \neq 0$ Suppose $A$ is a $4×4$ matrix over $C$ s.t. $Rank(A)=2$ and $A^3=A^2\neq0$. If $A$ is not diagonalizable then how to prove that:
There exists a vector $v$ s.t. $Av\neq 0$ and $A^2v=0$.
I know it is to be proved that $Imsp(A)$ is contained in $Nullsp(A)$, but really got no clue how to approach.
My work:
$x^2(x-1)$ is  the annihilating polynomial for $A$, but I am stuck in finding the characteristic polynomial. The only two possibilities are $x^3(x-1)$ and $x^2(x-1)^2$, but how to reject the later one?
Thanks for any hint.
 A: I will use jordon canonical form to prove this..
Since $A^2(A-I)=0$, only possible eigen values are $0,1$. Since $dim KerA=2$ , there wil be exactly two blocks corresponding to $0$ of size $(1,1)$ or $(2,1)$. 
First case:
if  the blocksize  corresponding to $0$ is  (1,1) ,then because $A$ is not diagonalizable $A$ must be of this form
\begin{bmatrix}
0  &0 & 0& 0\\
0 & 0 & 0 &0\\
0 & 0& 1& 1\\
0&0&0&1
\end{bmatrix}
second case:
if  the blocksize  corresponding to $0$ is  (2,1) ,then because $A$ is not diagonalizable $A$ must be of this form
\begin{bmatrix}
0  &1 & 0& 0\\
0 & 0 & 0 &0\\
0 & 0&0& 0\\
0&0&0&1
\end{bmatrix}
 Since $A^2=A^3$ only second case is possible. In that case $e_2$ is the vector such that $Ae_2\neq0$ but $A^2e_2=0$
A: Hint: If $Au\ne u$ then $A^2(Au-u)=0$.

 We don't need the info about the size (given as $4\times 4$) and rank (given as $2$) of the matrix at all; we do not even need that the matrix is over an algebraically closed field (given as  $\Bbb C$). If $A^2=A$ then $A$ is a projection and can be diagonalized, which is excluded. Hence There exists $u$ with $A^2u\ne Au$. For such $u$ let $v= Au-u$. Then $ Av = A^2u-Au\ne 0$ and $ A^2v=A^3u-A^2u=A^2u-A^2u=0$ as desired.

