A linear operator $P: V \to V$ is a projection operator if and only if $P^2 =P$. I have a linear operator $P: V \to V$. I need to show that this is a projection operator if and only if  $P^2 = P$.
Suppose $P$ is a linear operator onto $W$ subspace of $V$. So by definition I have that $P(V) = W$ and $P(w) = w$, $\forall w \in W$.
So then I take $P(P(V)) = P(W)$ which implies that  $P^2 (V) =  P(W)$. And $P^2(w) = P(w)$, $\forall w \in W$. I don't know how to conclude from this that $P^2 = P$.
Now to prove the converse, suppose I say that $P^2 = P$. If I take some $v \in V$, my book says to write every $v$ as $v = P(v) +(v-P(v))$. So then if I take $P(v) = P^2(v) + P(v-P(v))$. My book says I should conclude that $V = V_0(P) + V_1(P)$ and that $P$ is a projection operator onto $V_1 (P)$. I'm having trouble reaching that conclusion.
Any help is appreciated.
 A: For the forward direction, suppose $P: V \to V$ is a projection operator and let $v \in V$. Then $P^2(v)=P(v)$ since $P(v) \in W$ and $P(w)=w$ for all $w \in W$. 
Now suppose we have a linear operator $P: V \to V$ such that $P^2=P$. Since $P$ is a linear operator, we have a decomposition of $V$ into a direct sum
$$V= \operatorname{ker}P \oplus \operatorname{im}P.$$
We show that $P$ is a projection onto the subspace $\operatorname{im}P$. By definition, $P(V)=\operatorname{im}P$. Secondly, for any $w$ in the image of $P$ we have $P(w)=P(P(v))$ for some $v \in V$. But by assumption $P^2=P$, so $P(w)=P(v)=w$. Hence $P$ restricts to the identity on $\operatorname{im}P$ and so $P$ is a projection operator. 
$\textbf{EDIT}:$ Note also that the hint in your book is essentially a proof of the fact that we have a direct sum decomposition $V=\operatorname{ker}P \oplus \operatorname{im}P$. The kernel of $P$ is given by vectors of the form $v-P(v)$, so every vector of $v$ can be expressed as the sum of a vector in the image of $P$ and a vector in the kernel of $P$. 
