Vector space as a summation of 2 eigenvalues So let $V$ be a $\Bbb R-$ vector space of maybe unlimited dimension, and $f\in \textsf{End}(V)$.
I have to show that:
$f\circ f=f$ $\Longrightarrow$ $V=V_1\oplus V_0$
and: 
$f \circ f=id$ $\Longrightarrow$ $V=V_1\oplus V_{-1}$
I just don't even really know how to start. I would appreciate any help, maybe the first one would already be enough and then I can try the second one alone.
 A: Hints: For the first, note that any vector may be decomposed as
$$
v = f(v) + (v - f(v))
$$
For the second, we have
$$
v = \frac 12 (v + f(v)) + \frac 12 (v - f(v))
$$
or alternatively, note that $g = \frac {f + id}{2}$ satisfies $g \circ g = g$.
A: Let me give you some hints. Think of it this way:
$ f^{2} = f$
means that $f$ acts trivially on its image $W$. If $K = ker (f)$, then $W$ and $K$ are disjoint. Now if $w \notin ker(f)$, then $w$ is surely in the image? Can you find its preimage?
For the second problem, clearly, $f(v) = v, -v$ are two options that come to mind. You have to prove that these are the only two. Clearly they are disjoint and now prove that if $f(v) \neq v$, then it must be the other.
EDIT: The preimage of $w$ is, $f(w)$ as $f(f(w)) = w$. Moreover, in the second problem, you need to show that if $f(v) \neq v$, then it must be that $f(v) = -v$.
EDIT 2: For the first problem, we need to prove that $V = ker(f) \oplus im(f)$, they clearly intersect in $0$ because if $w \in im(f)$, then $w = f(v)$, but since $f^{2} = f$, we have that $f(w) = f^{2}(v) = f(v) = w$ and if in addition, $w \in ker(f)$, then $f(w) = 0$ and hence $w = 0$. Also, if $w \notin ker(f)$, then $w \in im(f)$ which proves they cover the whole of $V$.
