find the rank of a linear mapping such that $T^2=0$ Let $T:\Bbb R^6\to\Bbb R^6$ be a linear mapping such that $T^2=0$.Then which one is true?

a)Rank$(T)$ is less than or equal to 3

b)Rank$(T)$ is greater than  3

c)Rank$(T)$ is  equal to 5

d)Rank$(T)$ is less than or equal to 6


  I think, Rank of $T^2$ and Rank of $T$ are less than 6 since $T^2=0$ ...
  So d) may be false...Now How to proceed?Someone please explain...
$T^2=0$ implies $\text{Im}(T)\subseteq\text{ker}(T)$. Note that $\text{dim Im}(T)+\text{dim ker}(T)=6$ and $\text{dim Im}(T)\leq \text{dim ker}(T)$, we have that $\text{dim Im}(T)\leq 3$.
It could clearly seen that in Transformation $T^2$ that the Range of $T$ acts as the Domain for the other $T$. It's given that $T^2=0$, hence the $\operatorname{Rank}(T) \subseteq \operatorname{NullSpace}(T)$. Thus we know that 
$$\operatorname{Basis}\big(\operatorname{Rank}(T)\big) \subseteq \operatorname{Basis}\big(\operatorname{Nullspace}(T)\big)
$$ 
and hence $$\dim(\operatorname{Rank}(T)) \le \dim(\operatorname{Nullspace}(T))$$
And now by the Rank-Nulity Theorem we know that 
$$
\begin{split}
\operatorname{Rank}(T)+&amp;\operatorname{Nullspace}(T)=6\\
&amp;\qquad\Downarrow\\
&amp;\operatorname{Rank}(T) \le 3
\end{split}
$$