Rank of a linear operator. 
Let $V$ be a vector space of dimension 3 and  $T $ be a linear operator on  $V $ such that  $T^3=0 $ and $T^2 \neq 0$. The question is to show the rank of $T=2$.

My attempt : As I had no other idea, I tried using proof by contradiction. I succeeded in showing that a rank of 0 or 3 will lead to a contradiction. 
Now suppose rank($T$) = 1. Let $\{v_1,v_2,v_3\} $ be a basis of  $V $ and WLOG assume  $\{v_1,v_2\}$ spans the kernel of $T$ and $Tv_3$ spans the range of  $T$. Then I  can show that Range($T^2$) is spanned by $T^2(v_3)$ and rank($T^2$)= 1. I'm completely stuck after this. I'm not even sure this result has any importance. Any hint is appreciated. Thanks. 
 A: Since we know that $T^3=0$, we know that the null space of $T^3$ contains the image of $T^2$.  Since $T^2\not=0$, we know that $\operatorname{rank}(T^2)\geq 1$.  Since $T^2(v)=T(T(v))$, i.e., $T$ applied to some vector $T(v)$, we know that $\operatorname{rank}(T^2)\leq\operatorname{rank}(T)$.  Therefore, the rank of $T$ is at least one (but you already knew that).
On the other hand, since $T^2\not=0$, there is some vector $v$ so that $T^2(v)\not=0$, but since $T^3(v)=T(T^2(v))=0$, it follows that there is some nonzero vector, i.e., $T^2(v)$ which is mapped to zero by $T$.  This implies that the kernel of $T$ is at least $1$-dimensional.  This implies, by the rank-nullity theorem that the rank of $T$ is at most $2$ (but you already knew that).  If you don't know the rank-nullity theorem, keep a count of free and pivot variables and you'll get the same conclusions.
Now, suppose that the rank of $T$ is $1$.  Then, since $T^2$ has rank $1$ as well, this implies that $T^2$ and $T$ have the same image.  But, since $T$ applied to the image of $T^2$ is zero, then $T$ applied to the image of $T$ is also zero.  But then, $T^2$ is zero, but this contradicts the assumptions.
This shows that the rank of $T$ must be $2$.  To prove that this is even possible, an example would be 
$$
\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}.
$$
