Proposition 6.2.3 (Cyclic Permutation of Inverse).label If $A$ is a unital algebra over $\com$, then for any $x,y\in A$, we have
\begin{align*}\spec{A}{xy}\cup\curl{0}=\spec{A}{yx}\cup\curl{0}\end{align*}
In particular for every $x,y\in A$
\begin{align*}\norm{xy}{sp}=\norm{yx}{sp}\end{align*}
Proof. Suppose $\lam\nin \spec{A}{xy}\cup\curl{0}$ then $(xy-\lam)\inv=u$ exists.
\begin{align*}(xy-\lam)u=u(xy-\lam)=1\implies xyu=uxy=1+\lam u\end{align*}
from which we conclude
\begin{align*}(yx-\lam)(yux-1)=y(xyu)x-yx-\lam y(u)x+\lam=\lam&&(yux-1)(yx-\lam)=y(uxy)x-yx-\lam y(u)x+\lam=\lam\end{align*}
which shows $yx-\lam$ is invertible and $\lam\nin \spec{A}{yx}\cup\curl{0}$. By symmetry, $\spec{A}{xy}\cup\curl{0}=\spec{A}{yx}\cup\curl{0}$. Lastly we have
\begin{align*}\norm{xy}{sp}&=\sup\curl{\abs{\lam}:\lam\in\qspec{A}{xy}}=\sup\curl{\abs{\lam}:\lam\in\qspec{A}{xy}\cup\curl{0}}=\sup\curl{\abs{\lam}:\lam\in\qspec{A}{yx}\cup\curl{0}}\\&=\sup\curl{\abs{\lam}:\lam\in\qspec{A}{yx}}=\norm{yx}{sp}\end{align*}
noting that we are taking supremum over non-negative values.$\square$