= Solution
Expanding gives $(I-E)^2-(I-E)=E^2-E$, so either operator is <idempotent> exactly when the other is. For an <idempotent> $E$,
$$
v=(v-Ev)+Ev,\qquad E(v-Ev)=0.
$$
If $w\in\ker E\cap\operatorname{Im}E$, write $w=Eu$; then $0=Ew=E^2u=Eu=w$. Hence $V=\ker E\oplus\operatorname{Im}E$.
Next,
$$
(E_1+E_2)^2-(E_1+E_2)=E_1E_2+E_2E_1.
$$
If the right side vanishes, left multiplication by $E_1$ gives $E_1E_2+E_1E_2E_1=0$, while right multiplication gives $E_1E_2E_1+E_2E_1=0$. Subtraction shows $E_1E_2=E_2E_1$, and their original sum now gives $2E_1E_2=0$. Over the real or complex numbers this implies \b[$E_1E_2=E_2E_1=0$].
For three <idempotents> summing to $I$, each pair sum equals the complementary <idempotent> $I-E_k$. Thus every product of distinct operators vanishes. Every <vector> equals $E_1v+E_2v+E_3v$. If $v_1+v_2+v_3=0$ with $v_i\in\operatorname{Im}E_i$, applying $E_j$ gives $v_j=0$. Therefore
$$
\boxed{V=\operatorname{Im}E_1\oplus\operatorname{Im}E_2\oplus\operatorname{Im}E_3.}
$$
Take a <basis> of each summand and unite the three <bases>. A <vector> in the $j$th <basis> has <eigenvalue> one for $E_j$ and <zero> for the other two, so this is the requested common <eigenbasis>. <Zero> summands simply contribute no <basis> <vectors>.
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