Linear Transformation proof that $T^2 = T_0$ iff $R(T)\subseteq N(T)$ An example in my linear algebra textbook asks the following:
Let V be a vector space, and let T : V → V be linear. Prove
that $T^2 = T_0$ if and only if $R(T)\subseteq N(T)$.
It gives the following answer:

Suppose $T_2$ = $T_0$. Then for all $u ∈ R(T)$, there exists some $v ∈ V$ such
  that $Tv = u.0 = T_2v = T(T(v)) = T(u)$, so $u ∈ N(T)$.
  Thus $R(T) ⊆ N(T)$.
  Suppose $R(T) ⊆ N(T)$. Then $∀v ∈ V , T(v) ∈ R(T)$ so $T(v) ∈ N(T)$. Thus
  $0 = T(T(v)) = T
^2v$.  So $T^2 = T_0$.

Firstly, I don't understand what it means for the Rank to be contained in the Nullspace. How is that possible?  Also, I don't how that would make the square of a transformation equal to the zero vector.
Can someone explain this proof in other terms?
 A: Here's a less rigorous, less symbol heavy version of that proof:
$R(T)$ is the set of all things you can possibly get out of $T$.  $N(T)$ is the set of all things which $T$ turns into 0.  
So if you apply $T$ twice, you'll start with something in $V$, turn it into something in $R(T)$, and then assuming $R(T) \subset N(T)$ you'll put that back into $T$ and get 0 out.  
To prove the reverse, assume $T^2 = T_0$.  Take anything in $V$ and apply $T$ to it once.  I'm not sure what you get out, but whatever it is if you put it back into $T$ you'll get 0.  Because everything is 0 after it goes through $T$ twice.  So the range of $T$ must be a subset of $N(T)$ because everything that can come out of $T$ will get sent to 0.  
A: From a matrix point of view, Let $T$ be a $N \times N$ matrix with the $N \times 1$ columns $c_i, \forall i=1,...,N$, so that $T=[c_1,c_2,\dots,c_N]$. Now $T^2=TT=T[c_1,c_2,\dots,c_N]=[Tc_1,Tc_2,\dots,Tc_N]$. Now $T^2=0$ implies $Tc_i=0, \forall i$. Hence the columns $c_i$ of T should lie in the null space of $T$. This also implies any of their linear combination (which is nothing but $R(T)$) should also lie in the null space of $T$. So you get $R(T) \subseteq N(T)$
