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[parent] AB is conjugate to BA (Theorem)
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$. $ \qedsymbol$

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. 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)\vert _E$ is conjugate to $ (BA)\vert _E$ and so indeed the two transforms have identical non-zero eigenvalues.



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Cross-references: Transforms, eigenspaces, non-invertible, proof, eigenvalues, conjugacy class, non-abelian groups, diagonal, symmetric about, abelian group, Cayley table, group, ring, conjugate, invertible, square matrices

This is version 1 of AB is conjugate to BA, born on 2006-06-15.
Object id is 8045, canonical name is ABIsConjugateToBA.
Accessed 870 times total.

Classification:
AMS MSC15A04 (Linear and multilinear algebra; matrix theory :: Linear transformations, semilinear transformations)

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