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 $B^{\mathrm{-}\mathrm{1}}$ commute, and if $A$ and $B$ are invertible, then $A^{\mathrm{-}\mathrm{1}}$ and $B^{\mathrm{-}\mathrm{1}}$ commute.

Proof.

By assumption

$$AB=BA,$$

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

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

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 $B^{-1}$
Canonical name	IfAAndBCommuteSoDoAAndB1
Date of creation	2013-03-22 15:27:14
Last modified on	2013-03-22 15:27:14
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	7
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 15-00
Classification	msc 15A27