Linear Algebra proof relating to linear transformations 
*

*Let $V$ be a vector space over a field $F$, and $T: V \to V$ a linear transformation. 


(a) Show that $T^2=T$ iff $N(T)=\{x−Tx|x\in V\}$.
(b) If $T^2=T$, show that $V=N(T)+R(T)$.
I am not sure where to start for this proof because I am not sure exactly what it implies if $T^2=T$.
 A: Note first that
$R(I - T) = \{x - Tx = (I - T)x, \; x \in V \}, \tag 1$
and
$T^2 = T \Longleftrightarrow T(I - T) = T - T^2 = 0. \tag 2$
For (a):
we have
$y \in N(T) \Longleftrightarrow Ty = 0 \Longleftrightarrow (I - T)y = y - Ty = y$ $\Longrightarrow y \in R(I - T)\Longrightarrow N(T) \subset R(I - T) ,\tag 3$
and
$y \in R(I - T) \Longleftrightarrow \exists z \in V, \; y = (I - T)z$
$\Longrightarrow Ty = T(I - T)z = (T - T^2)z = 0z = 0$
$\Longrightarrow y \in N(T) \Longrightarrow R(I - T) \subset N(T); \tag 4$
(4) and (5) together yield
$N(T) = R(I - T). \tag 5$
For (b):
$v = v - Tv + Tv = (I - T)v + Tv, \tag 6$
and
$T(I - T)v = (T - T^2)v = 0v = 0$
$\Longrightarrow (I - T)v \in N(T); \tag 7$
trivially,
$Tv \in R(T), \tag 8$
whence 
$v \in N(T) + R(T); \tag 8$
we thus may write
$N(T) + R(T) \subset V \subset N(T) + R(T)$
$\Longleftrightarrow V = N(T) + R(T).  \tag 9$
We can actually take things one step further: if
$y \in N(T) \cap R(T), \tag{10}$
then
$Ty = 0, \tag{11}$
and
$y = Tz, \; z \in V; \tag{12}$
thus,
$y = Tz = T^2z = Ty = 0, \tag{13}$
which shows that
$N(T) \cap R(T) = \{0\}, \tag{14}$
and in fact we have
$V = N(T) \oplus R(T), \tag{15}$
the direct sum or $N(T)$ and $R(T)$.
A: Since a) is suggested by these hints. So i shall prove b). 
Let's show first,  $ N(T)\cap R(T)=\{0\}$.
To prove this,  let's assume $ T(v) \in N(T), T(v)\in R(T)$,  such that this $T(v) $ is non zero.  (Hence $v$ is not in $N(T)$....$(*)$). But see that by hypothesis,
$ T(T(v))=T^2(v)=0$ (since $ T(v)\in N(T)$).
But see that $N (T)=N (T^2)$,    so $T^2(v)=0\implies v\in N (T^2)= N(T)$, 
 and so contradiction to $(*)$.
Now, 
$1)$.   $ N(T)\cap R(T)=\{0\}$.
$2.)$ $\dim N(T)+\dim R(T)= dim V$ (Rank- nulity theorem ).
$3.)$$N(T),  R(T)$ are the subspaces of $V$.
So now if $W=R(T)+ N(T)$ then $\dim W= \dim R(T)+\dim N(T)=\dim V $ and $W\subset V $.  So,  $W=V$. Proving the claim. 
