$p,q,r$ three projections; show that if $p+q+r = 0$, then $p =q =r =0$

Question

Let $\mathbb F$ be a field of zero characteristic.

Let $E$ be a $\mathbb F$-vector-space.

Let $p,q,r$ be three projections of $E$.

Prove that if $p+q+r = 0$, then $p =q =r =0$.

The case of finite dimension is really easy using the trace (since the rank or a projection is equal to its trace).

How to deal with the case of infinite dimension?

Some thoughts:

One can see that $p(E) \cap q(E)= \{0\}$ since $-2$ is not an eigenvalue for $r$.

Likewise, $p(E) \cap r(E)= q(E) \cap r(E) =\{0\}$.

Otherwise, $\ker p \cap \ker q \subset \ker r$

$\ker p \cap \ker r \subset \ker q$ and $\ker q \cap \ker r \subset \ker p$

...

Answer 1

Let $v \in q(E)$. Then $$-v - r(v) = p(v) = -p(-p(v)) = -p(v + r(v)) = v + q(r(v)) + 2r(v).$$ Hence $r(v) = (-2v-q(r(v)))/3 \in q(E)$. But this means $q(r(v)) = r(v)$ and actually $v +2r(v) = 0$. But now $v \in r(E)$ and $v = 0$. Therefore $q = 0$

The same argument can now be repeated to prove that all three projections vanish. Alternatively $p = -r$ and the argument outlined by @lisyarus applies.