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

Later:
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 $(\lambda-AB)^{-1}$.
