if rank$(T)$=rank$(T^2)$, then $V=R(T)+N(T)$? $\newcommand{\rank}{\operatorname{rank}}$
The statement above isn't exactly what the problem says; rather, I'm asked to show if $\rank(T)=\rank(T^2)$ then $V=R(T)+N(T)$ (where $R$ is the range and $N$ is the kernel), and $R(T)\cap N(T)=0$.
I've shown we must have $R(T)=R(T^2)$, and that $R(T)\cap N(T)=0$. I'm stuck on 
showing $V=R(T)+N(T)$. I'm trying to show it as such:
Let $v\in V$. Then write $v=T(v)+(v-T(v))$. The first is in $R(T)$, so I need to show $v-T(v)\in N(T)$, so that
$$T(v-T(v))=T(v)-T^2(v)=T(v)-T(v)=0$$
But clearly I need $T^2=T$ for this to work. I'm really struggling to show this, any hints/suggestions? Maybe I'm going about this the wrong way.
Edit: I changed the question because my assumption was wrong. So now I'm even more lost on how to show $V=R(T)+N(T)$.
 A: I assume from context that $T$ is a linear transformation (since we're talking about rank and kernel), and that $T:V\to V$ (since otherwise $T^2$ wouldn't make sense).
I further assume that $V$ is a finite-dimensional vector space. This turns out to be crucial, as the result need not hold without it.
For example, consider $\Bbb R^{\Bbb N},$ the set of sequences $\vec x=\langle x_n\rangle_{n\in\Bbb N},$ where $\Bbb N$ is the set of positive integers and $x_n\in\Bbb R$ for all $n\in\Bbb N.$ We can define addition and scalar multiplication in a natural way as $$\vec x+\vec y=\langle x_n+y_n\rangle_{n\in\Bbb N}$$ and $$c\vec x=\langle cx_n\rangle_{n\in\Bbb N}$$ to make $\Bbb R^{\Bbb N}$ a vector space over $\Bbb R.$
For any $k\in\Bbb N,$ we let $\vec e^{(k)}$ be the vector having $1$ as its $k$th entry and $0$ for all other entries. Readily, the vectors $\vec e^{(k)}$ comprise a basis for $\Bbb R^{\Bbb N}$ over $\Bbb R.$
Now, consider the linear transformation $T:V\to V$ given by $\vec e^{(1)}\mapsto\vec 0$ and $\vec e^{(k)}\mapsto\vec e^{(k-1)}$ for all other $k\in\Bbb N.$ What does this look like in general, though? Well, with a bit of work, we can show that $$\langle x_1,x_2,x_3,x_4,\dots\rangle\mapsto\langle x_2,x_3,x_4,x_5,\dots\rangle.$$
Thus, $R(T)=R(T^2)=\Bbb R^{\Bbb N},$ and $N(T)$ is the subspace generated by $\vec e^{(1)}.$
On the other hand, if we assume that $V$ is finite-dimensional, then by showing  $R(T)\cap N(T)=0,$ we have $$\dim(R(T)+N(T))=\dim V-\dim R(T)\cap N(T)=\dim V$$ by rank-nullity. Thus, since $V$ is finite-dimensional and $R(T)+N(T)\subseteq V,$ then....
