Let $A$ and $B$ be Hermitian matrices. Let $[A,B] = AB - BA$ be their commutator and let $[A,B]^+$ be the Moore-Penrose pseudoinverse of $[A,B]$.


Is it then true that $A$ and $B$ both commute with the projector $Q = [A,B] [A,B]^+ = [A,B]^+ [A,B]$ ?


$P = 1 - Q$ is effectively a projector onto the subspace where our two matrices commute and we should always have that $PAP$ commutes with $PBP$.



Do any two hermitian matrices A and B commute with the support of their commutator?
To attack this more theoretically: if A and B have a common eigenvector, then that must lie in $\ker Q$ and obviously $[Q,A]$ and $[Q,B]$ act trivially on this guy. Thus, we can take the perp to this eigenvector and ask the same question about the leftover matrices.


Thus, we should assume that $A$ and $B$ have no common eigenvectors; in this case, you get that $[Q,A]=[Q,B]=0$ if and only if $Q$ is the identity, i.e. if and only if $[A,B]$ is invertible, which is definitely not always the case (though as the other answer indicates, it is true generically).


**EDIT:** This is true because if $[Q,A]=[Q,B]=0$, then $A$ and $B$ preserve the kernel $\ker Q$, and as observed in the question, they would commute on this subspace, and thus have a common eigenvector. So if there are no such eigenvectors, $Q$ must be injective and thus the identity.


