AB is conjugate to BA

Proposition 1.

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

Proof.

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

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 $A B$ is only conjugate to $B A$ . Thus the conjugacy class of a group are symmetric about the main diagonal.

Corollary 3.

If $A$ or $B$ is invertible then $A B$ and $B A$ have the same eigenvalues.

This leads to an alternate proof of $A B$ and $B A$ 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.

单词	ABIsConjugateToBA
释义	AB is conjugate to BA Proposition 1. Given square matrices $A$ and $B$ where one is invertible then $A B$ is conjugate to $B A$ . Proof. If $A$ is invertible then $A^{-1}ABA=BA$ . Similarly if $B$ is invertible then $B$ serves to conjugate $B A$ to $A B$ .∎ 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 $A B$ is only conjugate to $B A$ . Thus the conjugacy class of a group are symmetric about the main diagonal. Corollary 3. If $A$ or $B$ is invertible then $A B$ and $B A$ have the same eigenvalues. This leads to an alternate proof of $A B$ and $B A$ 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.
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