Vector space decomposed by operator $T$ such that $T^2=I$. 
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.
 A: Since $T^2=I$ we have that $m_T\mid (X-1)(X+1)$. By what your wrote, you seem to have assumed $T\neq -{\rm id},\rm id$. Thus $m_T=(X-1)(X+1)$.  It follows $T$ is diagonalizable. Since the only eigenvalues of $T$ are $1,-1$, we have that $V=V(1)\oplus V(-1)$, for

THM A transformation $T:V\to V$ over a finite dimensional vector space is diagonalizable iff its minimal polynomial factors linearly, and all its roots are simple, iff $V=\bigoplus\limits_{i=1}^r V(\lambda_i)$ where $\lambda_i;i=1,2,\ldots,r$ are the distinct eigenvalues of $T$.

The proof is not too complicated, but we can use it for your special case. Since $\gcd(X-1,X+1)=1$, we know we can write $$\frac 1 2 (X+1)-\frac 1 2 (X-1)=1$$ 
It follows that, for any $v\in V$, $$\frac 1 2 (T+1)v-\frac 1 2 (T-1)v=v$$ 
ADD Careful! As Marc has noted, if $2=0$ in $K$, we're in trouble. See his answer.
Now, note that $(T+1)(T-1)v=0$ for any $v$. It follows that the first term is in $V(-1)$ and the second term is in $V(1)$. Hence $V=V(1)\oplus V(-1)$.

PROOF (of the theorem)
We use that if $B=\{v_1,\ldots,v_n\}$ is a basis for $V$ and $T\in{\rm end}\; V$ then $m_V={\rm lcm}(m_{v_1},\ldots,m_{v_n})$ where $m_{v_i}$ denotes the minimal polynomial of the vector $v_i$. If $T$ is diagonalizable, then it admits a basis  $B=\{v_1,\ldots,v_n\}$ of eigenvectors. It is immediate $m_{v_{i_j}}=X-\lambda_{i_j}$, where $\lambda_{i_j}$ is the corresponding eigenvalue. Thus $m_T={\rm lcm}(m_{v_1},\ldots,m_{v_n})=(X-\lambda_1)\cdots (X-\lambda_r)$ where $\lambda_i:i=1,2,\ldots,r$ are the distinct eigenvalues of $T$. Suppose conversely that $m_A=(X-\lambda_1)\cdots (X-\lambda_r)$. Since $\lambda_i\neq \lambda_j$ for $i\neq j$,  the polynomials $P_j=\prod_{i\neq j}(X-\lambda_i)$ are all coprime in pairs. Writing $$1=P_1Q_1+\cdots+P_rQ_r$$ we will have for any $v$ that $$v=P_1(T)Q_1(T)v+\cdots+P_r(T)Q_r(T)v$$
But $w_i=P_i(T)Q_i(T)v$ is in $V(\lambda_i)$, for $(X-\lambda_i)P_i=m_T$ by construction. This means  $V=\bigoplus\limits_{i=1}^r V(\lambda_i)$ where $\lambda_i;i=1,2,\ldots,r$ are the distinct eigenvalues of $T$, so $T$ is diagonalizable. $\blacktriangleleft$
A: Somehow the naive way seems to fail. Looking at the example $T=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, the following decomposition seems to be the right one:
$$
v=\frac{v+Tv}2+\frac{v-Tv}2.
$$
The first term is an eigenvector for $1$ and the second one for $-1$. 
A: 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.
