# AB is conjugate to BA

###### Proposition 1.

Given square matrices^{} $A$ and $B$ where one is invertible^{} then $A\mathit{}B$ is conjugate^{} to $B\mathit{}A$.

###### Proof.

If $A$ is invertible then ${A}^{-1}ABA=BA$. Similarly if $B$ is invertible then $B$ serves to conjugate $BA$ to $AB$. ∎

The result of course applies to any ring elements $a$ and $b$ where one is invertible. It also holds for all group elements.

###### Remark 2.

This is a partial generalization^{} to the observation that the Cayley table of an abelian group^{} is symmetric about the main diagonal. In abelian groups this follows because $A\mathit{}B\mathrm{=}B\mathit{}A$. But in non-abelian groups^{} $A\mathit{}B$ is only conjugate to $B\mathit{}A$. Thus the conjugacy class^{} of a group are symmetric about the main diagonal.

###### Corollary 3.

If $A$ or $B$ is invertible then $A\mathit{}B$ and $B\mathit{}A$ have the same eigenvalues^{}.

This leads to an alternate proof of $AB$ and $BA$ being almost isospectral. (http://planetmath.org/ABAndBAAreAlmostIsospectral) If $A$ and $B$ are both non-invertible, then we restrict to the non-zero eigenspaces^{} $E$ of $A$ so that $A$ is invertible on $E$. Thus ${(AB)|}_{E}$ is conjugate to ${(BA)|}_{E}$ and so indeed the two transforms have identical non-zero eigenvalues.

Title | AB is conjugate to BA |
---|---|

Canonical name | ABIsConjugateToBA |

Date of creation | 2013-03-22 16:00:40 |

Last modified on | 2013-03-22 16:00:40 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 4 |

Author | Algeboy (12884) |

Entry type | Theorem |

Classification | msc 15A04 |