Proving the intersection of the range and null space of a linear transformation equals {0} 
Let $V$ be a finite-dimensional vector space, and let $T: V \rightarrow V$ be linear. If $\text{rank}(T) = \text{rank}(T^2)$, prove that $R(T) \cap N(T) = {0}$. Deduce that $V = R(T) \oplus N(T)$

I understand that since $\text{rank}(T) = \text{rank}(T^2)$, then we also have that nullity(T) = nullity($T^2$). Now I understand that  $\forall y \in R(R^2), \exists x \in V$ such that $T^2(x) = y$, which also implies that $y = T(T(x))\in R(T)$. This to me suggests that $R(T^2)\subseteq R(T)$. 
Now it appears to me that since we are trying to prove that $R(T) \bigcap N(T) = {0}$, then we can take an arbitrary element say $z\in V$ and suppose that $z\in R(T) \cap N(T)$. After this however is where I'm getting confused.
Since $z$ is in the intersection of $R(T)$ and $N(T)$ must we simply consider $T(z)$, which if we assume T(z) = $0$ automatically falls in the intersection since $T(z)=0\in N(T)$ and $T(z)\in R(T)$? Or does this occur by definition i.e. does $R(T) \cap N(T)$ always equal $0$? I'm also unsure if I am supposed to utilize the transformation $T^2$ in finishing the proof.
 A: Since ${\cal R} T^2 \subset {\cal R} T$ and $\operatorname{rk} T^2 = \operatorname{rk} T$, we have ${\cal R} T^2 = {\cal R} T$.
Hence the rank nullity theorem gives $\dim \ker T^2 = \dim \ker T$, and since
$\ker T \subset \ker T^2$, we have $\ker T =\ker T^2$.
Now suppose $x \in {\cal R}T \cap \ker T$, then there is some $y$ such that
$x=Ty$ and $Tx = 0$. Hence $T^2 y = 0$ and so $Ty = 0$ and so $x = 0$. Hence
${\cal R}T \cap \ker T = \{0\}$.
Note that this works for finite dimensional spaces. Here is an example for 
an infinite dimensional space.
If you look at the linear operator $T(x_1,x_2,...) = (x_2,x_3,...)$ on the
space of sequences, we see that ${\cal R} T^2 = {\cal R} T$,
but with $x = (1,0,0,...)$ we have $x = T(0,1,0,0,...)$ and $T x = 0$,
hence ${\cal R}T \cap \ker T \neq \{0\}$.
A: $$x\in R(T)\cap N(T)\implies \exists\,v\in V\;\text{such that}\;Tv=x\;\text{ and also}\;\;Tx=0\implies$$
$$0=Tx=T(Tv)=T^2v\stackrel{\text{given}}=Tv=x\implies R(T)\cap N(T)=\{0\}$$
A: 
Or does this occur by definition i.e. does R(T)∩N(T) always equal 0?
I'm also unsure if I am supposed to utilize the transformation T2 in
finishing the proof.

This is not necessarily true for general matrices.
Consider the matrix
\begin{bmatrix}
1  & 1  \\
-1 & -1 
\end{bmatrix}
Then its range and nullspace are both the same i.e. the line $\lambda [1,-1]^T$.
Notice also that the matrix $A$ above is such that $A^2=0_{2x2}$, i.e. rank($T$) $\neq$ rank($T^2)$)
