Spectra of products variously permutated

Question

Let $x,y$ be elements of a Banach algebra $A$ and $\lambda\in\mathbb C\setminus\{0\}$. If $\lambda-xy $ is invertible, then $\frac1{\lambda}\big[1+y(\lambda-xy)^{-1}x \big]$ is clearly an inverse of $ \lambda-yx$. This implies $\sigma(xy)\setminus \{0\}= \sigma(yx)\setminus \{0\}$ (an example showing that $0$ may go in and out, is any pair $(L,R)$ of non-invertible left/right inverses on a Banach space: then $LR=I$ and $RL$ is a linear projector with $\ker RL=\ker L\neq(0)). $ In particular, $\rho(xy)=\rho(yx)$, which also follows quite directly from the spectral radius formula, since $\|(xy)^{n+1}\|\le\|x\|\|y\| \|(xy)^n\| $.

The above identity immediately generalizes to products of $n$ factors circularly permutated: $\sigma(x_n\dots x_2x_1)\setminus \{0\}= \sigma(x_1x_n\dots x_3x_2)\setminus \{0\}=\dots$ But what can be said for a generic permutation, $x_{\sigma_n} \cdots x_{\sigma_2}x_{\sigma_1}$? For instance, is it the case that $\sigma(xyz)\setminus \{0\}= \sigma(yxz)\setminus \{0\}$?

edit By Christophe Leuridan‘s answer, in general $\sigma(x_nx_{n-1}\dots x_1)\setminus\{0\}= \sigma(x_{\sigma_n}x_{\sigma_{n-1}}\dots x_ {\sigma_1})\setminus\{0\}$ is false for non-cyclic permutations of $\{1,2,\dots n\}$. But what can be said e.g. for permutations of factors with repetitions, i.e. of $x_1^{n_1}\dots x_r^{n_r}$?

Answer 1

This is false for other permutations. For example, call $(e_1,e_2,e_3)$ the canonical basis of $\mathbb{R}^3$ and let A,B,C be the $3 \times 3$ real matrices such that $Ae_1=e_2$, $Be_2=e_3$, $Ce_3=e_1$, and the other images of the $e_j$ by $A,B,C$ are null. Then $BC$ is null, so $ABC$ is null. Yet $CBAe_1=e_1$ and $CBAe_j=0$ for $j \in \{2,3\}$, so the trace of $CBA$ is $1$.