Commutativity of Spectrum in Algebra over Complex Numbers

Theorem
Let $A$ be an algebra over $\C$.

Let $x, y \in A$.

Let $\map {\sigma_A} {x y}$ and $\map {\sigma_A} {y x}$ be the spectrum of $a b$ and $b a$ respectively in $A$.

Then:
 * $\map {\sigma_A} {x y} \cup \set 0 = \map {\sigma_A} {y x} \cup \set 0$

Proof
Let $\map G A$ be the group of units of $A$.

, by replacing $A$ by its unitization if necessary, that $A$ is unital.

Let $x, y \in A$.

We show that ${\mathbf 1}_A - x y \in \map G A$ ${\mathbf 1}_A - y x \in \map G A$.

By swapping $x$ and $y$, it suffices to show that if ${\mathbf 1}_A - x y \in \map G A$, then ${\mathbf 1}_A - y x \in \map G A$.

Specifically, we show that $\paren { {\mathbf 1}_A - y x}^{-1} = 1 + y \paren {{\mathbf 1}_A - x y}^{-1} x$.

We have:

and:

Hence we have shown that for $x, y \in A$, we have ${\mathbf 1}_A - x y \in \map G A$ ${\mathbf 1}_A - y x \in \map G A$.

Now fix $x, y \in A$ and take $\lambda \ne 0$.

We have ${\mathbf 1}_A - \paren {\lambda^{-1} x} y \in \map G A$ ${\mathbf 1}_A - y \paren {\lambda^{-1} x} \in \map G A$.

That is, $\lambda {\mathbf 1}_A - x y \in \map G A$ $\lambda {\mathbf 1}_A - y x \in \map G A$ for $\lambda \ne 0$.

Hence we have $\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x} \setminus \set 0$.

That is, $\map {\sigma_A} {x y} \cup \set 0 = \map {\sigma_A} {y x} \cup \set 0$.