Let $T:V\rightarrow V$ be an idempotent linear transformation. Prove that $V=V_0\oplus V_1$ where $T(v_0)=0$ for all $v_0\in V_0$ and $T(v_1)=v_1$ for all $v_1\in V_1$.
EDIT:
My attempt:
Since $T(v_0)=0$ for all $v_0\in V_0$, we have that $V_0=\ker T$. Since $T(v_1)=v_1$ for all $v_1\in V_1$, we have that $V_1=ImT$.
If $T=0$, then the problem is trivial, so suppose $T\neq 0$.
Let $v\in V$. As $T=T^2\implies Tv=T^2v\implies T(v-Tv)=0$. So, $v-Tv=\theta\in \ker T$. Thus $v=\theta+Tv$. Now, as $T(Tv)=Tv$ for all $v\in V$, we have that $Tv=v$ whenever $v\in V_1$.
Thus, $v=\theta+Tv$ is the sum of two elements, where $\theta\in V_0$ and $Tv\in V_1$. So, $V=V_0+V_1$.
Let $v\in V_0\cap V_1$. Then, $v\in V_0$ implies that $Tv=0$ and $v\in V_1$ implies $Tv=v$. Thus $0=Tv=v$ and so $v=0$, thus $V_0\cap V_1=\{0\}$; hence $V=V_0\oplus V_1$.