Proposition 1

Let A be an algebra and $x, y\in A$. One has Sp$'(xy) =$ Sp$'(yx)$.

By passing to $\widetilde{A}$, we reduce to the case where A possesses a unit element $e$. It is then enough to prove that, if $\lambda \not= 0$ is such that $xy-\lambda e$ admits an inverse $u$, then $yx-\lambda e$ is invertible. Put $z=yux-e$. Since $xyu=\lambda u+e$, one has

and analogously $z(yx-\lambda e) =\lambda e$. Since $\lambda \not= 0$, one sees that $yx-\lambda e$ is invertible.

If A is a unital algebra and if $x, y\in A$, the preceding proposition implies that Sp($xy$)$\cup \{0\}=$ Sp($yx$)$\cup \{0\}$, but one can have Sp($xy$)$\not=$ Sp($yx$) (cf. I, p. 153, exer. 3).