To prove this associative statement about spectral theory

Question

Show that $\sigma(AB) \cup \{0\} = \sigma(BA) \cup \{0\}$ in general, and that $\sigma(AB) = \sigma(BA)$ if $A$ is bijective.

I studied the associative statement of this somewhere but it did not include the zeroth element. If you assume the bijection, how can you show the first part?

My attempt

Let show that $A$ commutes with $B$ which is self-adjoint linear operatar such that $AB = BA$. \begin{equation*} AB = A^{\ast} B^{\ast} = (BA)^{\ast} = (ABA)^{\ast} = A^{\ast} B^{\ast} A^{\ast} = ABA = BA. \end{equation*} In general, the zeroth element follows directly in the left-hand-side if $A$ is bijective. $\square$

Comments

• not sure if $\sigma$ should be carried along; I was reading Kreuzig for the application

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How can you show the first part with the zeroth element?

Answer 1

Let $\lambda \in \sigma(AB)$, $\lambda \ne 0$ be given. Then, $AB - \lambda \, I$ is (boundedly) invertible. By a simple calculation, you can verify $$(BA - \lambda)^{-1} = \lambda^{-1} \, \Big(I - B \,(AB-\lambda)^{-1} \, A\Big),$$ see also the Sherman–Morrison-Woodbury formula. Hence, $\lambda \in \sigma(BA)$. This shows the first claim.

For the case that $A$ is invertible, you can use the argument by Martin Argerami.

Answer 2

I cannot understand what you are trying to do in your attempt.

If $A$ is a bijective operator between Banach spaces, then $A$ is invertible. Now $$ AB=A(BA)A^{-1}, $$ and $$ AB-\lambda I=A(BA-\lambda I)A^{-1}. $$ So $AB-\lambda I$ is invertible precisely when $BA-\lambda$ is, which shows that the resolvent of $AB$ equals to that of $BA$. Thus, $\sigma(AB)=\sigma(BA)$.