If we know a vector space V has a linear operator T from $V$ to itself. And this $T^2$ is an identity map. Then if we create two subspace $V+$ contains all $v$ such that $Tv=v$ and $V-$ contains all $Tv=-v$, now how to get those two subgroups are actually a direct sum of $V$?
To show their intersection only contain 0, we can quickly get that by contradiction. But how to get that all $v$ are in them. I tried to use contradiction again but don't get any progress.
For $v\in V$
$$v = \frac{1}{2}\left(v-T(v)\right)+\frac{1}{2}\left(v+T(v)\right)$$
Then $\frac{1}{2}\left(v-T(v)\right) \in V_-$ and $\frac{1}{2}\left(v+T(v)\right)\in V_+$. You also should assume that characteristic of the field is $\neq 2$.
Edit
In case when the field is of characteristic $2$ there are counterexamples. Consider a linear endomorphism $\phi:\mathbb{F}_2^{\oplus 2} \rightarrow \mathbb{F}_2^{\oplus 2}$ with matrix $$ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $$
Then $\phi^2$ is the identity, but $\mathbb{F}_2^{\oplus 2}$ could not be decomposed on plus and minus part of $\phi$.