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