Prove $V^+\oplus V^-=V \Longleftrightarrow f^2=1_V$ Let $V$ be a $\mathbb{R}$ vector space, let $f\in \operatorname{End}(V)$. We define subsets of $V$ as follows:
$V^+=\{v\in V:f(v)=v\}$ and $V^-=\{v\in V:f(v)=-v\}$
We know that $V^+$ and $V^-$ are vector subspaces of V, and that their intersection is the zero vector.
Prove:
$V^+\oplus V^-=V \Longleftrightarrow f^2=1_V$

Proving $\Longrightarrow$
Hypothesis: $V^+\oplus V^-=V$
Then: $\forall v\in V:v=v^++v^- $ where $ v^+\in V^+ ,v^-\in V^-$
$f^2(v)=v\Rightarrow f(f(v))=f(f(v^++v^-))=f(f(v^+))+f(f(v^-))$ and because of the definition of the subsets (subspaces) $V^+$ and $V^-$ we get that:
$f(f(v^+))+f(f(v^-))=f(v^+)+f(-v^-)=v^++v^-=v$
$f^2(v)=v$ We have our proof.

Proving $\Longleftarrow$
Hypothesis: $f^2=1_V$
This means that $f^2(v)=f(f(v))=v$ ($f^2$ is bijective) therefore $f$ is bijective.

Well I don't know how to keep going, we are given a hint but I don't know how to apply it.
Hint we are given:
To prove $\Longleftarrow$, we must see if $\forall v\in V$ can be written as $v^++v^-$ where $ v^+\in V^+$ and $v^-\in V^-$. To get $v^+$ and $v^-$ we suppose that we have $v=v^++v^-$, then apply $f$ to the equality and we would get the second quation for $v^+$ and $v^-$
Not sure what to make out of this.

Assume you have found the wanted $v^+$ and $v^-$; then
$$
f(v)=f(v^+)+f(v^-)=v^+-v^-
$$
Therefore
$$
v+f(v)=2v^+,\qquad v-f(v)=2v^-
$$
This gives necessary conditions: if $v^+$ and $v^-$ exist, then
$$
v^+=\frac{1}{2}(v+f(v)),\qquad v^-=\frac{1}{2}(v-f(v))
$$
Do these vectors satisfy the requirements under the assumption that $f^2$ is the identity?
