Existence of the Eigenvalues of an Idempotent Mapping

There is no need for space decomposition. Let $vneq0$ be an eigenvector of $E$ for eignevalue $lambda$, namely $$ Ev=lambda v. $$ So $$ E^2v=lambda Ev=lambda^2v. $$ Since $E^2=E$, one has $$ lambda v=lambda^2v $$ which gives $lambda=lambda^2$. So $lambda=0$ or $1$, namely $E$ only has eigenvalues $0$ and $1$.

Suppose $V$ is a vector space over any field $mathbb{F}$. Suppose that $E:Vto V$ is an idempotent linear map, that is $E^2=E$. Then $V=ker E oplus ker (I-E)$, the direct sum of what is clearly the $0$-eigenspace with the $1$-eigenspace.

[The decomposition is trivial to prove: for $vin V$ we have $v=(I-E)v +Ev$, and $E(I-E)v=(I-E)Ev=0$; and if $vinker E cap ker (I-E)$ then $Ev=0, (I-E)v=0$ so add and get $v=Iv=0$.]