# Banach Algebra: $\sigma(xy)\cup\{0\} = \sigma(yx)\cup\{0\}$

It is Rudin excercise 10.4 where we aim to prove $\sigma(xy)\cup\{0\} = \sigma(yx)\cup \{0\}$ for elements $x,y\in A$ a Banach-algebra.( $\sigma$ being the spectrum)

In (a) we prove that $e-yx$ invertible $\Leftrightarrow e-xy$ invertible.

Following the hint: Put $z= (e-xy)^{-1}$, write $z$ as geometric series (assume $\left\|x\right\| < 1, \left\|y\right\|< 1$), and use the identity $(xy)^n = x(yx)^{n-1}y$ to obtain a finite formula $(e-yx)^{-1}$ in terms of $x,y,z$. Then show that this formula works without any restrictions on $\left\|x\right\|$ or $\left\|y\right\|$.

Ok, so lets put

\begin{align*} z = (e-xy)^{-1} &= \sum_{k=0}^{\infty} (xy)^k = e+\sum_{k=1}^{\infty} (xy)^k\\ & =e + \sum_{k=1}^{\infty}x(yx)^{k-1}y = e+ x(e-yx)^{-1}y \end{align*}

Is this right? But I dont see how i can single out out $(e-yx)^{-1}$. And how to make this work without restrictions on $\left\|x\right\|$ or $\left\|y\right\|$?

And second (b) I want to show that if $\lambda\neq 0$ and $\lambda \in \sigma(xy)$ then $\lambda \in \sigma(yx)$. Thus $\sigma(xy)\cup\{0\} = \sigma(yx)\cup \{0\}$. Then also, how do we see that $\sigma(xy)$ doesn't have to be equal to $\sigma(yx)$.

Thanks for tips and suggestions.

-
See this answer – Norbert Oct 28 '13 at 16:02

Suppose you know $e-yx$ is invertible with inverse $w$. You want to show that $(e-xy)$ has an inverse that equals $e+xwy$. You can check this directly. For example,
$(e-xy)(e+xwy)=e-xy+xwy-xyxwy=e-x(e-w+yxw)y=e-x(e-(e-yx)w)y=e.$
We do not need to place any restrictions on the norms of $x$ and $y$.
Suppose $\lambda\in \sigma (xy)$ with $\lambda\neq 0$. Then $xy-\lambda e$ is not invertible, so $(1/\lambda)xy-1$ is not invertible, which means $(1/\lambda)yx-1$ is not invertible by the above, so $\lambda\in \sigma (yx)$. Repeating this argument for the other direction, we find that $\sigma(xy)\cup \{0\}=\sigma(yx) \cup \{0\}$.
To show that $\sigma(xy)$ and $\sigma(yx)$ can differ, you should construct an example of where this is the case. The above work tells you that this can only happen for $\lambda =0$. Consider the space $l^2$, the unilateral shift $S$, and its adjoint $S^*$.
Then how does one use the identity $(xy)^{n} = x(yx)^{n-1)y$ to show that $\rho(xy)= \rho(yx)$, where $\rho$ denotes the spectral radius? This is Exercise 10.1 also in Rudin's text. – Libertron Nov 12 '13 at 17:23