AB is conjugate to BA
ABIsConjugateToBA
2006-06-15 14:38:06
2006-06-15 14:38:06
Theorem
$AB$ and $BA$ are almost isospectral
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\begin{prop}
Given square matrices $A$ and $B$ where one is invertible then $AB$ is conjugate to $BA$.
\end{prop}
\begin{proof}
If $A$ is invertible then $A^{-1} ABA=BA$. Similarly if $B$ is invertible then
$B$ serves to conjugate $BA$ to $AB$.
\end{proof}
The result of course applies to any ring elements $a$ and $b$ where one is invertible. It also holds for all group elements.
\begin{remark}
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.
\end{remark}
\begin{coro}
If $A$ or $B$ is invertible then $AB$ and $BA$ have the same eigenvalues.
\end{coro}
This leads to an alternate proof of \PMlinkname{$AB$ and $BA$ being almost isospectral.}{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.