Suppose $V$ is a vector space over field $F$, $\operatorname{char}(F)\neq 2$. Show that $V=V^+ \oplus V^-$, details below

Question

Let $T$ be a linear transformation $T: V\rightarrow V, T^2=I$. Define $$V^+ =\{v\in V \mid T(V)= +v \}, V^-=\{ v\in V \mid T(v)= -v \}.$$

My understanding of fields is still weak, does $\operatorname{char}(F)\neq 2$ just mean we are not dealing with the finite field with two elements? Can someone provide an example of when what I am trying to show is true?

Answer 1

use $v=\frac{v+T(v)}{2}+\frac{v-T(v)}{2}$ since $T(\frac{v+T(v)}{2})=\frac{v+T(v)}{2}$ and $T(\frac{v-T(v)}{2})=(\frac{T(v)-T(T(v))}{2})=\frac{T(v)-v}{2}=\color{red}- \frac{v-T(v)}{2}$

And if $v_0\in V^+ \cap V^-$ we have $T(v_0)=v_0=-v_0$ it means $2v_0=0$ contradict with $F, char(F)\neq 2$