If $T$ and $T^2$ have equal rank then $V=\ker T\oplus {\rm im}\, T$ for $V$ finite dimensional. I am trying to prove the following:
Let $V$ be a finite-dimensional vector space. Consider an operator $T$ on $V$ such that $\text{dim range}(T)=\text{dim range}(T^2)$. Show that $V=\text{null}(T)\oplus \text{range}(T)$.
I have been given the following hint:
$\textit{Hint:}$ Show that there does not exist a $y\in V$ such that $y\notin \text{null}(T)$ and $y\notin \text{range}(T)$.
I decided to try and prove the hint by contradiction...
Suppose that there exists a $y\in V$ such that $y\notin \text{null}(T)$ and $y\notin \text{range}(T)$. Then 
$$Ty\neq 0$$
and for every $v\in V$ we have $$Tv\neq y.$$
I have NO idea what to do with this information! I would greatly appreciate some hints/help! Thank you.
 A: Because the range of $T$ contains the range of $T^2$, what you get is that the range of $T$ and of $T^2$ coincide. Because of the rank nullity formula, you also know the kernel of $T$ and $T^2$ coincide.
Pick an arbitrary vector $v\in V$. Because $Tv$ is in the image of $T$, there is some $w$ such that $Tv=T^2w$. This reads $T(v-Tw)=0$. It follows that you can write $$v=(v-Tw)+Tw\in \ker T+\operatorname{im}\, T$$
so it suffices to show the sum is direct. 
Suppose then that $Tv=w$ and that $Tw=0$. Then $T^2v=0$, but because of the remark in the first paragraph, $v\in \ker T$ too, so that $Tv=w=0$.
For a personal opinion, I find the hint quite misleading -- it is completely unnecessary to argue by contradiction here, and might even obscure what is really going on. Note that you can really "chase the hypotheses" here to get a solution: you have a vector $v$ in $V$, and you want to write it as a direct sum of two things. You know something about the image of $T$ and $T^2$. It is natural to look at $Tv$. Now you can use what you know, and you get what you want, immediately. Similarly, to show the sum is direct you start with the information that $w\in \ker T\cap {\rm im}\, T$, and naturally a condition relating $\ker T$ and $\ker T^2$ pops up. But you know something about $\ker T$ and $\ker T^2$, and you can conclude.
A: Let $V_0 = \text{Range}(T)$ and consider the restriction $T_1$ of $T$ to $V_0$, i.e.,
$T_1 : V_0 \to V,$ where $T_1(x) = T(x)$ for all $x \in V_0$. 
The range of $T_1$ is the range of $T^2$ and the null-space of $T_1$ is $\text{Range}(T) \cap \text{Kernel}(T)$, the rank + nullity theorem implies $\text{dim}(\text{Range}(T)) - \text{dim}(\text{Range}(T^2)) = \text{dim}(\text{Range}(T) \cap \text{Kernel}(T)),$ so $ \text{Range}(T) \cap \text{Kernel}(T) = \{0\},$ the rank + nullity theorem also implies $\text{dim}(\text{Range}(T)) + \text{dim}(\text{Kernel}(T)) = \text{dim}(V)$ so the result follows.
A: By rank-nullity $ \dim V = \dim \ker T + \dim \textrm{Im}\, T $, so it suffices to check that $ \ker T \cap \textrm{Im}\, T $ is trivial. Assume that $ x $ is in the intersection, then $ x = Tv $ for some $ v $, and $ Tx = T^2 v = 0 $. Since the given condition implies $ \ker T = \ker T^2 $ ($ \ker T $ is a subspace of $ \ker T^2 $ of equal dimension), $ v \in \ker T $ and $ x = 0 $. Thus, the desired conclusion $ V = \ker T \oplus \textrm{Im}\, T $ follows.
