# On an inclusion of ranges involving the commutator of two operators

I will begin this question with an example. Let $$A,B \in B(H)$$ for some complex infinite dimensional Hilbert space $$H$$. Then, we can easily show that, if $$\operatorname{rank}([A,B])=1$$ and there is some nonzero $$x \in \ker B$$ such that $$A x \not \in \ker B$$, we have: $$Im [A,B] \subseteq Im B$$ where $$[A,B] := AB - BA$$ is the commutator of the two operators, and $$Im(\cdot)$$ denotes the range. Indeed, consider any $$x$$ as above. Then: $$[A,B] x = ABx - BAx = 0 - BA x = - BA x \neq 0$$ and since $$Im[A,B]$$ is $$1$$-dimensional, we can write: $$[A,B] x = -BA x = \alpha y$$ for some nonzero complex $$\alpha$$ and for any nonzero vector $$y$$ in $$Im [A,B]$$. By linearity, it is then clear that $$-BAx \in ImB$$ spans $$Im[A,B]$$, from which we get: $$Im[A,B] \subseteq ImB$$ I was wondering if there is any known condition under which the range of the commutator is contained in the range of the second operator involved, without assumptions on the rank of $$[A,B]$$ (that is, for any $$\operatorname{rank}([A,B]) \geq 1$$).