Assume that $F\colon V\to V$ is an operator on a real vector field $V$ such that $F^3=-F$. How to prove that $F$ projects $V$ on $W \subset V$ where it satisfies $F^2=-\operatorname{Id}$?
The only thing that I can notice is that if $F^2\cdot F = (-\operatorname{Id})F= -F$ then by multiplying it with $F^{-1}$ on the right we obtain the identity $-\operatorname{Id} = -\operatorname{Id}$.
Assume that $V$ is finite dimensional. Let us show that $V = \operatorname{ker}(F) \oplus \operatorname{im}(F)$. By rank-nullity, it is enough to show that $\ker(F) \cap \operatorname{im}(F) = \{ 0 \}$ so let $Fv \in \ker(F) \cap \operatorname{im}(F)$. Then $F(Fv) = F^2 v = 0$ but then $0 = F^3 v = -F v$ so $Fv = 0$. Now set $W = \operatorname{im}(F)$. If $u = Fv \in W$ then $F^2 u = F^3 v = -Fv = -u$ so $\left( F|_{W} \right)^2 = -\operatorname{Id}|_{W}$.