# If $A$ and $B$ commute so do $A$ and $B^{-1}$

###### Theorem 1.

Let $A$ and $B$ be commuting matrices. If $B$ is invertible, then $A$ and $B^{-1}$ commute, and if $A$ and $B$ are invertible, then $A^{-1}$ and $B^{-1}$ commute.

###### Proof.
 $AB=BA,$

multiplying from the left and from the right by $B^{-1}$ yields

 $B^{-1}A=AB^{-1}.$

The second claim follows similarly. ∎

The statement and proof of this result can obviously be extended to elements of any monoid. In particular, in the case of a group, we see that two elements commute if and only if their inverses do.

Title If $A$ and $B$ commute so do $A$ and $B^{-1}$ IfAAndBCommuteSoDoAAndB1 2013-03-22 15:27:14 2013-03-22 15:27:14 mathcam (2727) mathcam (2727) 7 mathcam (2727) Theorem msc 15-00 msc 15A27