AB is conjugate to BA

Proposition 1.

Given square matrices $A$ and $B$ where one is invertible then $AB$ is conjugate to $BA$.

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 $AB=BA$. But in non-abelian groups $AB$ is only conjugate to $BA$. Thus the conjugacy class of a group are symmetric about the main diagonal.

Corollary 3.

If $A$ or $B$ is invertible then $AB$ and $BA$ 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 ABIsConjugateToBA 2013-03-22 16:00:40 2013-03-22 16:00:40 Algeboy (12884) Algeboy (12884) 4 Algeboy (12884) Theorem msc 15A04