Prove that if $p$ and $q$ are projections with the same kernel, then $p\circ q=p$ and $q\circ p=q$

Question

Let $K$ be a field, and let $E$ be $K$-vector space. Let $p$ and $q$ be two endomorphisms of $E$.

Prove the following proposition: ($p$ and $q$ are projections, and $\ker{p}$ = $\ker{q}$) $\Leftrightarrow$ ($p\circ q=p$, and $q\circ p=q$)

For the $\Rightarrow$ direction, it's easy to prove that: $$\forall x\in\ker{q},\quad p(q(x))=p(0_E)=0_E=p(x)$$ And similarly for $q \circ p$.

However, I don't see how I can go about proving equality for $x\notin \ker{p}$, nor how to prove the proposition in the $\Leftarrow$ direction.

Answer 1

• If $p\circ q=p$ and $q\circ p=q$ then $$p\circ p=(p\circ q)\circ p=p\circ(q\circ p)=p\circ q=p$$ and similarly, $q\circ q=q$.

• Let now $p,q:E\to E$ be arbitrary projections. Then, $\ker q={\rm im}({\rm id}_E-q)$ hence $$\ker q\subseteq\ker p\iff p\circ({\rm id}_E-q)=0\iff p\circ q=p$$ and similarly, $\ker p\subseteq\ker q\iff q\circ p=q$.