Show that $\dim(\operatorname{range}(T)) = 1$. Let $T :\mathbb{R}^3 \to \mathbb{R}^3  $ be a linear transformation such that $T \neq 0$ but $T^2=0$. Show that $$\dim(\operatorname{range}(T))=1$$
 A: With the rank theorem, you have : $$\dim(\operatorname{range}(T))+\dim(\ker(T))=3$$
$T\neq 0$ so $\operatorname{range}(T) \neq \{0\}$, so $\dim(\operatorname{range}(T)) \geq1$. 
And $T^2 = 0$, so $\operatorname{range}(T) \subseteq \ker(T) $, so $\dim(\operatorname{range}(T)) \le \dim(\ker(T))$. 
So $\dim(\operatorname{range}(T)) \in \{0,1\} $.
Finally we have $\dim(\operatorname{range}(T))=1$.
A: Clearly  minimal polynomial of $T$ is $x^{2}$ so in Jordan Canonical form of $T$ there is one Jordan block of order $2$ and one block of order one and one Jordan block of order has only one $1$ hence rank is just one. 
A: We will use the rank-nullity theorem.
Clearly, $T$ may not be full-rank for otherwise we'd have $T^n \neq 0$ for all $n\in\mathbb{N}$. Of course, the rank of $T$ may not be $0$ for then we'd have $T=0$. So we need only show that $T$ does not have rank $2$.
Because $T²=0$, we know that $T$ maps all of $\text{range}(T)$ to $0$, so $\dim(\ker(T))\geq  \dim(\text{range}(T))$. By the rank nullity-theoerem, the only remaining valid option is $\dim(\text{range}(T))=1$.
