Idempotent linear transformation from $V$ to $V$ is the direct sum of $\operatorname{range}(T)$ and $\operatorname{null}(T)$ The problem statement is:

If $T\in\mathcal{L}(V)$ and $T^2=T$ (idempotent), prove that $V=\operatorname{range}(T)\oplus \operatorname{null}(T)$.

I'm not exactly sure where to start with this problem. 
 A: Hint: Notice that $v=Tv+(v-Tv)$...
A: We need to show that
$V = \text{Range}(T) + \text{Null}(T) \tag 1$
with
$\text{Range}(T) \cap \text{Null}(T) = \{0\}; \tag 2$
Now with our colleague Parcly Taxis we write, for $v \in V$,
$v = Tv + (I - T)v; \tag 3$
we may then use the hypothesis
$T^2 = T \iff T - T^2 = 0 \iff T(I - T) = (I - T)T = 0, \tag 4$
to see that for
$w = (I - T)v \tag 5$
we have
$Tw = T(I - T)v = 0, \tag 6$
that is,
$w \in \text{Null}(T); \tag 7$
thus we see that (1) binds; we have also established that 
$\text{Range}(I - T) \subset \text{Null}(T); \tag 8$
we observe that
$(I - T)^2 = I - 2T + T^2 = I - 2T + T = I - T; \tag 9$
therefore with $w$ as in (7) we find
$w = w - Tw = (I - T)w = (I - T)^2w = (I - T)(I - T)w, \tag{10}$
and thus we have
$w \in \text{Range}(I - T), \tag{11}$
so that
$\text{Range}(I - T) = \text{Null}(T); \tag{12}$
thus to validate (2) we need show
$\text{Range}(T) \cap \text{Range}(I - T) = \{0\}; \tag {12}$
if
$w \in \text{Range}(T) \cap \text{Range}(I - T), \tag{13}$
then there are $x, y \in V$ such that
$Tx = w =(I - T)y, \tag{14}$
whence
$w = Tx = T^2x = T(I - T)y = 0; \tag{15}$
thus (2) is affirmed and with it
$V = \text{Range}(T) \oplus \text{Null}(T). \tag {15}$
Remark:  It is worth noting that there is no restriction on $\dim V$ in the statement of this result; $\dim V = \infty$ is within the purview of this proposition.  End of Remark.
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