Decomposition of the image of a projection. Let $\mathbb K$ be a field and $E$ be a $\mathbb K-$vector space. Let $p$ and $q$ two linear endomorphisms such that $p^2=p$ and $q^2=q$ and $p\circ q=0$ and let $r=p+q-q\circ p$.
I want to show that $\operatorname{Im} r= \operatorname{Im} p\oplus \operatorname{Im} q$. 
It is clear that $\operatorname{Im}p\cap \operatorname{Im} q =\{0\}$
hence it suffices to show that $
 \operatorname{Im}r= \operatorname{Im} p + \operatorname{Im} q$. The inclusion $\operatorname{Im} r \subset \operatorname{Im} p + \operatorname{Im} q$ is clear since $\forall x\in E$ we have that $r(x)=p(x)+q[x-p(x)]$ but the other inclusion is not clear to me. Thank you for your help!
 A: This turns out to be remarkably easy, since every element of $\def\Im{\operatorname{Im}}\Im p + \Im q$ is in fact fixed by $r$, so in particular in the image or $r$. Here's the computation. Let $x\in\Im p + \Im q$ be given, so there exist $v,w\in E$ with $x=p(v)+q(w)$; then
$$
\begin{align}
  r(x)&=r(p(v))+r(q(w)) &&\text{since $r$ is linear}\\
   &=p^2(v)+q(1-p)p(v)+pq(w)+q(1-p)q(w) &&\text{since $r=p+q\circ(1-p)$}\\
   &=p^2(v) + +q(1-p)q(w) &&\text{since $(1-p)p=0=pq$}\\
   &=p^2(v) +q^2(w) &&\text{since $pq=0$}\\
   &=p(v)+q(w)=x &&\text{since $p,q$ are projections.}
\end{align}
$$
It might seem mysterious why $x$ can be chosen as its own pre-image under $r$, but the following reasoning leads to this choice.
Since $p$ is a projection, one has $E=E_0\oplus E_1$ where $E_\lambda$ is the eigenspace of $p$ for $\lambda\in\{0,1\}$. Clearly $\Im(p)=E_1$, and from $p\circ q=0$ one gets $\Im(q)\subseteq E_0$. To realise $x$ as image of some $y$ under $r=p+q\circ(1-p)$, one can decompose $y=y_0+y_1$ along the eigenspaces, then $p(y)=y_1$ and $(1-p)(y)=y_0$ so one wants $x=y_1+q(y_0)$. By the above the witnesses of  $x\in\Im p + \Im q$ must in fact be the same as components $x_1,x_0$ of $x$ in the eigenspaces, so one must have $x_0\in\Im(q)$, but then $q^2=q$ implies $q(x_0)=x_0$. So taking $y=x$ works.
