The following exercise is from Brown's book, A Second Course in Linear Algebra:
Let $V$ be a vector space over $F$, and $T\in\mathscr E(V)$ such that $T^3=T$. Show that $V=V_0 \oplus V_1\oplus V_2$ , where the $V_i$ are subspaces of $V$ with the following properties:
$(a)\ \alpha\in V_0\implies T(\alpha)=0$,
$(b)\ \alpha\in V_1\implies T(\alpha)=\alpha$,
$(c)\ \alpha\in V_2\implies T(\alpha)=-\alpha$.
Assume $2\ne0$ in $F$.
My Attempt:
I first showed $V_0 \cap V_1\cap V_2 =\varnothing$. Let $\alpha\in V_0 \cap V_1\cap V_2$. Then $\alpha\in V_0 \implies T(\alpha)=0$. Again, $\alpha\in V_1\implies T(\alpha)=\alpha$. Since $T$ is a linear transformation, this gives us $\alpha=0$.
Next, we need to show $V=V_0+V_1+V_2$ but I have no clue as to how! I've tried the following:
For $v\in V$, we have
\begin{align}
T^3(v)&=T(v)\\
\implies T(T^2(v)-v)&=0
\end{align}
But I don't see how this can be written as three summands to show that $V\subseteq V_0+V_1+V_2$.
Any hints/alternate approaches would be appreciated. TIA.