TITLE: Vector space decomposed by operator $T$ such that $T^2=I$.
QUESTION [6 upvotes]: Let $T$ be a linear operator on a finite dimensional vector space $V$ such that $T^2$ is the identity operator. Prove that for any $v\in V$, $v-T(v)$ is either $0_V$ or is an eigenvector with eigenvalue $-1$. Prove that $V=V(1)\oplus V(-1)$.

*$V(\lambda)$ denotes eigenspace
I have no problem showing that $v-T(v)$ is either $0_V$ or is an eigenvector with eigenvalue $-1$. From this we see that $v\in V(1)$ or $v-T(v)\in V(-1)$ for every $v\in V$.
Also, no problem showing that $V(1)$ and $V(-1)$ are independent. However, I am stuck on showing that $V=V(1)+V(-1)$. So far:
We know that either $v\in V(1)$ or $v\notin V(1)$. If the former, then $v=v+0\in V(1)+V(-1)$. If $v\notin V(1)$, then we know that $v-T(v)\in V(-1)$. Then we could write
$$
v = T(v)+v-T(v),
$$
and if $T(v)\in V(1)$, then we are done. But I haven't been able to prove nor disprove $T(v)\in V(1)$.
Am I on the right track? I feel like I am missing something completely obvious. Somehow I think I should use that $T$ is invertible, and hence has trivial kernel, but I do not see how.
And I would love to see different solutions for this problem, if they exist.

REPLY [6 votes]: Instead of using $v=T(v)+(v-T(v))$ you should try to use $2v=(v+T(v))+(v-T(v))$, and then show $v+T(v)\in V(1)$, which is easy. Then you know that $2v\in V(1)+V(-1)$, and if your field is not of characteristic$~2$ you can divide by$~2$ to conclude that $v\in V(v)+V(-1)$ as well. If you do have a field of characteristic$~2$ then you are out of luck, and the property you are trying to prove is actually false. But since you already showed that $V(1)$ and $V(-1)$ are "independent" (you must have meant that they form a direct sum), you must have been (implicitly) using that the field is not of characteristic$~2$, since if it were then $V(1)=V(-1)$ and their sum will not be direct.
Summary: in characteristic other than$~2$ the sum is direct and equal to the whole space as desired; in characteristic$~2$ the sum is neither direct nor fills the whole space.