# 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 |