Prove that $\sigma(AB) \backslash \{0\} = \sigma(BA)\backslash \{0\} $. Where $A,\ B$ are bounded operators in Banach space and $\sigma$ denotes spectrum.


This is a classic (I learned this from Appendix A, A.1 of G.K. Pedersen, $C^{\ast}$-algebras and their automorphism groups, see also exercise 4.1.3 in his Analysis Now — thanks, w(ild)3life):

Let $\mathcal A$ be a $\mathbb{C}$-algebra with a unit (here $\mathcal{A}$ is the algebra of bounded linear operators on your Banach space). Then $\sigma(AB) \smallsetminus \{0\} = \sigma(BA) \smallsetminus \{0\}$.

If $\lambda \notin \sigma(AB) \cup \{0\}$ then there is $C$ such that $$ C(\lambda - AB) = 1 = (\lambda-AB)C. $$ Then verify that $\lambda^{-1}(1 + BCA)$ is the inverse of $(\lambda-BA)$ so that $\lambda \notin \sigma(BA) \cup \{0\}$: $$ (1 + BCA)(\lambda-BA) = \lambda = (\lambda-BA)(1+BCA), $$ as a straightforward computation shows.


There's a nice mnemonic on how to guess the inverse, also addressed in-depth in this MO-thread by Bill Dubuque:

Recall the geometric series $(1-q)^{-1} = 1 + q + q^2 + \cdots$, so formally $$\begin{align*} (1-BA)^{-1} &= 1+ BA + BABA + BABABA + \cdots \\ &= 1 + B(1+ AB+ ABAB + \cdots)A \\ &= 1 + B(1-AB)^{-1}A \\ & = 1+BCA \end{align*}$$ with $C = (1-AB)^{-1}$. Similarly with $(\lambda-AB)^{-1}$.

  • $\begingroup$ I'm pretty sure I've seen this problem here before, but I couldn't find it. $\endgroup$ – t.b. Nov 5 '11 at 16:07
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    $\begingroup$ You may also be interested in having a look at this MO thread started by Bill Dubuque. $\endgroup$ – t.b. Nov 5 '11 at 16:08
  • $\begingroup$ Is a $\mathbb C = C^*$? Never seen that notation. $\endgroup$ – Jonas Teuwen Nov 5 '11 at 16:56
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    $\begingroup$ @Jonas: A $\mathbb{C}$-algebra is an algebra over the field of complex numbers (as in "$k$-algebra over a commutative ring $k$"): a complex vector space equipped with a bilinear and associative multiplication. The spectrum of an element of a unital $\mathbb{C}$-algebra is $$\sigma(A) = \{\lambda \in \mathbb{C} : \lambda - A \text{ is not invertible}\}.$$ $\endgroup$ – t.b. Nov 5 '11 at 16:59
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    $\begingroup$ This is also exercise E.4.1.3 (page 133) of Pederson's Anaylsis NOW (his hint is the answer). $\endgroup$ – wildildildlife Nov 6 '11 at 0:58

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