Find all linear operators such that $F^2 = F$ and $F(x,y) = (ax,bx+cy)$ I need to find all linear operators that match $F^2 = F$ and $F(x,y) = (ax,bx+cy)$ 
*where $F^2$ means $F$ composed with itself. 
So what I did:
$F(x,y) = (ax,bx+cy)\implies F(F(x,y)) = (a(ax),b(ax)+c(bx+cy)) = (a^2x, bax+bcx+c^2y)$ 
So $F^2 = F \implies (a^2x, bax+bcx+c^2y) = (ax, bx+cy)\implies\\a^2x = ax\\bax + bcx + c^2y = bx+cy$
But I don't know how to solve this system. If I divide both sides by $x$ in the first equation, I get $a^2 = a\implies a = 0, a = 1$, but I can't do it, because a linear operator must map the $0$ vector to the $0$ vector itself, so $(0,0,0)$ is in the domain, so $x=0$ must be considered. But for $x=0$ we have infinite solutions for $a$. By the way, what I do with the second equation?
 A: In this context, the eqution $a^2x=ax$ means that this equality holds for every $x$. For example, for $x=1$, so you can legitimately deduce that $a^2=a$.
A: Since $F^2=F$, your system
$$\left\{\begin{align*}
&a^2x=ax\\
&bax+bcx+c^2y=bx+cy
\end{align*}\right.$$
has to hold for all $\langle x,y\rangle\in\Bbb R^2$. In particular, it has to hold for $\langle 1,0\rangle$, so you do know that $a^2=a$ and hence that $a$ is $0$ or $1$. The second equation will give you another nice relationship involving $a,b$, and $c$. By considering the point $\langle 0,1\rangle$ you can limit the possibilities for $c$. And when you put those three pieces together, it’s not too hard to find all of the possibilities.
A: In these equations we are not looking for $x$ and $y$, but for $a,b,c$, assuming that they hold for all $x,y$.
Use several particular values of $x$ and $y$ to get new equations for $a,b,c$ until you can solve them.
And, yes, $(0,0)$ has to be mapped to $(0,0,0)$, but this already holds by the definition, whatever $a,b,c$ are.

Another way to start is to write up the matrix $[F]$ of $F$ (w.r.t. the standard basis of $\Bbb R^2$) and then work with this matrix, solving $[F]^2=[F]$.
$$[F]=\pmatrix{a&0\\b&c}$$

 Substitution $x=1,\ y=0$ gives the first column, and $x=0,\ y=1$ gives the second.

