# Proving that $T^{2} = T$ for a linear operator on $W$ implies that $V = \operatorname{Null} T \oplus \operatorname{Range} T$

How can I prove that $T^{2} = T$ for a linear operator on $W$ implies that $V = Null T \oplus Range T?$ I know that their dimensions add up to the dimension of $W$, how do I show I can represent any element in $W$ like that? Or could I start with a basis for null and then add vectors of a basis of the range and say that the list doesn't become dependent because vectors (besides o) that form a basis for image T can't be in the basis? Or is that hand-wavy?

-
You probably want $W$ and not $V$ in the first sentence? –  spin Sep 30 '11 at 6:19
Am I going insane? I can't understand this question at all! –  Alexei Averchenko Sep 30 '11 at 10:16

HINT: $\rm T(v-Tv) = Tv-T^2v = 0$ so $\rm v-Tv \in \ker T$, and $\rm v= (v-Tv) + Tv$
Let's examine what would happen if this were true.. for $v \in W$ you want $v = s + t$, where $s \in \text{null}(T)$ and $t \in \text{range}(T)$. Then $t = Pu$ and $Ps = 0$. Thus $Pv = Pu$. Figure out how to choose $u$, and you can solve $s$. Then just show that your choice satisfies $v = s + t$, $s \in \text{null}(T)$ and $t \in \text{range}(T)$. Then show that the representation is unique.