How to find rank of this linear transformation? 
$T : \mathbb R^4 \to \mathbb R^4$ is a linear transformation.
  And there exists $v \in \mathbb R^4$ s.t $v, T(v), T(T(v))$ is linearly independent and $T(T(T(v)))=0$ Find rank $T$


What only I can figure out from given sentence was only $T(T(v)) \in $ N($T$)$\cap$R($T$)
Please give me any hint or advice, not a solution.
 A: What does $T$ look like as a matrix in the basis $\{T^2v,Tv,v,u\}$? can you put a lower bound on the rank? Hint $$A=[c_1 c_2 \dots c_n] \Rightarrow Ae_i = c_i$$
A: A complete answer that establishes that there is not a single answer to your problem (as suspected by @Miguel Atencia): the rank may be 2 or 3.
First of all, vector $v$ cannot be $0$, because ... $0,0$ and $0$ are not independent... 
Since $T^3v=0$, $v \in Ker(T^3)$ with $v \neq 0$, thus $det(T^3)=0$, thus $det(T)=0$, implying $rank(T)<4$.
On the other hand, $rank(T)\geq2$, because $Tv$ and $T(Tv)$ belong to the range space of $T$ and are independent.
Now 2 examples: one can check that conditions $v,Tv,T(Tv)$ independent and $T^3v=0$ can be fulfilled with either a rank 3 matrix:
$$T=\begin{bmatrix}0 & 1 & 1 & 1\\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0\end{bmatrix} \ \ \text{and} \ \ v=\begin{bmatrix}0\\0\\1\\0 \end{bmatrix} \ \rightarrow \ Tv=\begin{bmatrix}1\\1\\0\\0 \end{bmatrix} \ T^2v=\begin{bmatrix}1\\0\\0\\0 \end{bmatrix} \  \ T^3v=\begin{bmatrix}0\\0\\0\\0 \end{bmatrix}$$
or a rank 2 matrix: 
$$T=\begin{bmatrix}0 & 0 & 1 & 1\\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0\end{bmatrix} \ \ \text{and} \ \ v=\begin{bmatrix}1\\1\\0\\1 \end{bmatrix} \ \rightarrow \ Tv=\begin{bmatrix}1\\1\\1\\0 \end{bmatrix} \ T^2v=\begin{bmatrix}1\\1\\0\\0 \end{bmatrix} \  \ T^3v=\begin{bmatrix}0\\0\\0\\0 \end{bmatrix}$$
