Determining the intersection of kernel and image when $\text{Ker}(T)=\text{Ker}(T^2)$ Let $T$ be a linear operator on a finite dimensional vector space $V$ such that $\text{rank}(T) = \text{rank}(T^2)$. I want to show that the kernel and image of $T$ are disjoint, i.e., that they have only the zero vector in their intersection. 
I am not able to use the information that the rank of $T$ and $T^2$ is same. I know that would imply that the nullity of $T$ and $T^2$ is also same. 
I started as follows: Let $v$ be a vector in the intersection of kernel and image or $T$. I wish to show that $v$ is the zero vector. I'm not able to do it. 
Any help is appreciated but hints are more appreciated than the answer itself. 
 A: A first thing to realise is that the image of $T^2$ is always contained in the image of $T$, since any vector of the form $T(T(v))$ is in particular of the form $T(w)$. Then the given fact that $T$ and $T^2$ have the same rank means that they actually have the same image; call this common image$~W$.
Now the restriction of $T$ to its own image $W$ defines a linear map $\tilde T:W\to W$ (one might call it $T|_W$, but note that both domain and codomain have changed). And the image of $\tilde T$ is by definition the image of $T^2$, which is all of$~W$, so $\tilde T$ is surjective. Then the kernel of $\tilde T$ is $\{0\}$, and by definition this kernel is the intersection of the kernel of$~T$ with its image$~W$.
A: Assume $v=Tw$ and $Tv=0$. Then $T^2w=0$. Hence $\ker(T)\subseteq \ker(T)+\langle w\rangle\subseteq\ker(T^2)$. From $\operatorname{rank}(T^2)=\operatorname{rank}(T)$, we conclude that $\ker(T)=\ker(T^2)$ and hence $w\in\ker T$, $v=0$.
A: Note that $T : \mathcal{R}(T)\rightarrow\mathcal{R}(T)$. You are told that the range of this restricted operator has full rank. So, the kernel of this restricted operator is $\{0\}$-- equivalently, $\mathcal{R}(T)\cap\mathcal{N}(T)=\{0\}$.
