Exercice 58

1. On a $p = \mathrm{id} - q$, donc $0 = pq = q - q^{2}$ : $q^{2} = q$. Ensuite, $p^{2} = p(\mathrm{id} - q) = p - pq = p$, et $qp = q(\mathrm{id} - q) = q - q^{2} = 0$.

2. Comme $qp = 0$, on a $\operatorname{Im} p \subset \mathrm{Ker}\, q$ ; si $q(x) = 0$, alors $x = p(x) + q(x) = p(x) \in \operatorname{Im} p$. De même, $pq = 0$ et $x = p(x) + q(x)$ donnent $\operatorname{Im} q = \mathrm{Ker}\, p$. $$\boxed{p, q \text{ projecteurs}, \quad qp = 0, \quad \operatorname{Im} p = \mathrm{Ker}\, q, \quad \operatorname{Im} q = \mathrm{Ker}\, p}$$