eigenvalues of an involution Let $V \neq  \{0\}$ be a K-vector space and let $P : V \rightarrow V $be linear. Furthermore, let P be an involution, i.e. $P (P(x)) = x $ for every $x \in V.$
Show that if $P \neq ±id,$ then $V = Eig (P,-1) \oplus Eig (P,1)$, i.e. one can write ever element $v$ of $V$ as the sum $v_1 + v_2$ of elements $v_1 \in Eig (P, -1)$ and $v_2 \in Eig(P, 1)$ and that $Eig (P, -1) \cap Eig (P,1) = \{0\}$
 A: $$
v = \color{blue}{\frac{1}{2}(v + P(v))} + \color{red}{\frac{1}{2}(v-P(v))}
$$
is the decomposition you're looking for, where the blue part has eigenvalue $1$ and the red part has eigenvalue $-1$. This of course assumes that $K$ doesn't have characteristic $2$.
There is no reason in this argument to assume that $P \neq \pm\operatorname{id}$, other than the fact that the decomposition becomes trivial.
As to why $\operatorname{Eig} (P, -1) \cap \operatorname{Eig} (P,1) = \{0\}$, think about what that statement means. The left-hand side is the set of all vectors $v$ where $P(v) = v$ and $P(v) = -v$ are both true, which means that $v = -v$.
A: let $\lambda$ be an eigenvalue of $P.$ that is $Px = \lambda x, x \neq 0$ we can apply $P$ to the equation and get $x = P^2x = \lambda Px = \lambda^2 x$ since $x \neq 0,$ this gives $\lambda = \pm 1$
if $P$ is not an identity, then both eigenspaces $ker(A - I), ker(A+I)$ are nontrivial. only thing we are left to do is to show that these two spaces makeup 
the whole space.
first for any $x,$ we have 
$P(x-Px) = Px - P^2x = Px - x = -(x-Px)$ and 
$P(x+Px) = Px + P^2x = Px + x = (x+Px)$ that means $x\pm Px$ is an eigenvector of $P$ corresponding to the eigenvalue $\pm 1$ provided that it is not zero. 
we can write any vector $x = (1/2)(x+Px) + 1/2(x-Px)$ that will conclude the proof.
i am not able to argue why $Px \pm x \neq 0$ for $x \neq 0.$ i will think more about it and post later.  
we can argue in a different way. claim: $ker(A-I) = ker(A-I)^2, ker(A+I) = ker(A+i)^2.$
proof of the claim: take $x \in ker(A-I)^2 \setminus \ker(A-I).$ that means $$0=(P-I)^2x = (P^2 - 2P + I)x = -2(P-I)x \neq 0$$ contradicting the assumption.
the proof of the second part of the claim is similar.
now we can use the fact $V$ is the direct sum of $ker(P_I)$ and $ker(P+I).$
A: Since $P^2(x)=x$, we now that $P^2-I=0$ and hence the minimal polinomial of $P$ divided $(x-1)(x+1)$. Since $P\neq \pm I$ we conclude that the minimal polinomial is indeed $(x-1)(x+1)$.
From here $P$ is diaganizble with eigenvalues $\pm 1$. Take it from here.
