Let $P,Q$ be linear transformations from $n$ dimensional vector space $V$ to $V$ such that $P^2=P$, $Q^2=Q$, $im P\cap im Q=\{0\}$. Show $V=\ker P\oplus \ker Q$.
It sounds easy geometrically. But how to give the proof? In particular, how to show $\ker P\cap \ker Q=\{0\}$?
I think a reasonable assumption is that $\text{im}P\oplus \text{im}Q=V$ and $PQ=QP.$ If so, then since $\ker P\oplus\text{im}P=V=\ker Q\oplus \text{im}Q$ since they are projections, $\dim\ker P + \dim\ker Q =n.$ Furthermore, if $v\in\ker P\cap\ker Q,$ then we can write $v=Pa+Qb,$ and hence $$0=Pv=Pa+PQb$$ and $$0 = Qv = QPa+Qb$$ so $$Qb=-QPa=QPQb=PQb,$$ i.e., $Qb\in\text{im} P.$ Thus $Qb\in\text{im}P\cap \text{im}Q=\{0\}.$ Likewise we can show $Pa=0.$ Thus $v=0$
I am not sure if the condition $PQ=QP$ is necessary.
Original answer:
It seems that you need other conditions to assure your conclusion.
Suppose $V=\mathbb R^3,$ $P$ the orthogonal projection onto the $x$-axis, and $Q$ the orthogonal projection onto the $y$-axis (where we view the coordinate $(x,y,z)\in\mathbb R^3$). Then their images have trivial intersection, but their kernel doesn't (since $(0,0,1)$ is in both of their kernels).