Do any two hermitian matrices A and B commute with the support of their commutator? 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 $PBP$.
 A: 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.
A: Short answer from trying random matrix inputs: it is not true for
$A = \begin{bmatrix}0.25634 &  0.417943&   0.696104 \\
 0.417943 &  0.0327021&  0.921007 \\
 0.696104 &  0.921007  & 0.0685762 \end{bmatrix}$
and
$
B = \begin{bmatrix}1.64721 &   0.29856 &  0.455064 \\
 0.29856 &  0.448668 & 1.56321 \\
 0.455064 & 1.56321  & 0.646562 \end{bmatrix}
$
But it still does often seem to hold for matrices that have some specific commutation rules between them. Also, for random matrices of even dimension (sampling elements between 0 and 1), Q is almost always the unit matrix and P is almost always zero, so it is very easy to trick oneself into thinking this is true.
