I want to show that on a Banach algebra can be $\sigma(xy) \neq \sigma(yx)$. I have thought to use the Right and Left shift operator, but I seem to be stuck on how to show this explicitly. I think I have to show that $\sigma(LR)=\{ 1 \}$ and $\sigma(RL)=\{ 0,1 \}$, but how?
Thanks for any help and/or advice.
Your choice of candidate operators is good. Define $L$ and $R$ on $\ell^2$ as
$$L: (a_1, a_2, a_3, \dots) \mapsto (a_2, a_3, a_4, \dots)\\ R: (a_1, a_2, a_3, \dots)\mapsto(0, a_1, a_2, \dots).$$
It is clear that $LR$ is the identity, so $\sigma(LR) = \{1\}$. On the other hand, $RL:(a_1, a_2, a_3, \dots)\mapsto(0, a_2, a_3, \dots)$. Let $A\subset\ell^2$ consist of the elements of the form $(x, 0, 0, \dots)$ and $B\subset\ell^2$ of the elements of the form $(0, x_1, x_2, \dots)$. We see that $A$ is the kernel of $RL$ and that $RL$ acts as identity on $B$. Therefore, $\sigma(RL)=\{0, 1\}$.