Theorem: Spectrum of Product of Elements of Banach Algebra

Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $\sigma_A$ denote spectrum in $A$.
Let $x, y \in A$.
Then we have:
:$\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x} \setminus \set 0$

Proof:
First take $A$ to be unital.
We show that ${\mathbf 1}_A - x y$ is invertible {{iff}} ${\mathbf 1}_A - y x$ is invertible. 
Swapping $x$ and $y$ it suffices to show that if ${\mathbf 1}_A - x y$ is invertible then ${\mathbf 1}_A - y x$ is invertible.
Suppose that ${\mathbf 1}_A - x y$ is invertible.
We have:
{{begin-eqn}}
{{eqn	| l = \paren { {\mathbf 1}_A - y x} \paren { {\mathbf 1}_A + y \paren { {\mathbf 1}_A - x y}^{-1} x}
		| r = {\mathbf 1}_A - y x + y \paren { {\mathbf 1}_A - x y}^{-1} x - y x y \paren { {\mathbf 1}_A - x y}^{-1} x
}}
{{eqn	| r = {\mathbf 1}_A + y \paren {-\mathbf 1_A + \paren { {\mathbf 1}_A - x y}^{-1} - x y \paren { {\mathbf 1}_A - x y}^{-1} } x
}}
{{eqn	| r = {\mathbf 1}_A + y \paren {-\mathbf 1_A + \paren { {\mathbf 1}_A - x y}^{-1} \paren { {\mathbf 1}_A - x y} } x
}}
{{eqn	| r = {\mathbf 1}_A + y \paren { -\mathbf 1_A {\mathbf 1}_A} x
}}
{{eqn	| r = {\mathbf 1}_A + y {\mathbf 0}_A x
}}
{{eqn	| r = {\mathbf 1}_A 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \paren { {\mathbf 1}_A + y \paren { {\mathbf 1}_A - x y}^{-1} x} \paren { {\mathbf 1}_A - y x}
		| r = {\mathbf 1}_A + y \paren { {\mathbf 1}_A - x y}^{-1} x - y x - y \paren { {\mathbf 1}_A - x y}^{-1} x y x
}}
{{eqn	| r = {\mathbf 1}_A + y \paren { \paren { {\mathbf 1}_A - x y}^{-1} - {\mathbf 1}_A - \paren { {\mathbf 1}_A - x y}^{-1} x y}
}}
{{eqn	| r = {\mathbf 1}_A + y \paren { \paren { {\mathbf 1}_A - x y}^{-1} \paren { {\mathbf 1}_A - x y} } x
}}
{{eqn	| r = {\mathbf 1}_A + y \paren { -\mathbf 1_A {\mathbf 1}_A} x
}}
{{eqn	| r = {\mathbf 1}_A + y {\mathbf 0}_A x
}}
{{eqn	| r = {\mathbf 1}_A 
}}
{{end-eqn}}
So ${\mathbf 1}_A - y x$ is invertible with inverse ${\mathbf 1}_A y \paren { {\mathbf 1}_A - x y}^{-1} x$.
Hence ${\mathbf 1}_A - x y$ is invertible {{iff}} ${\mathbf 1}_A - y x$ is invertible.
Now take $\lambda \in \C \setminus \set 0$. 
Take $\mu \in \C \setminus \set 0$ such that $\lambda = \mu^2$. 
Then applying the above, we have that ${\mathbf 1}_A - \paren {x/\mu} \paren {y/\mu}$ is invertible {{iff}} ${\mathbf 1}_A - \paren {y/\mu} \paren {x/\mu}$ is invertible.
Hence ${\mathbf 1}_A - \lambda^{-1} x y$ is invertible {{iff}} ${\mathbf 1}_A - \lambda^{-1} y x$ is invertible if $\lambda \ne 0$. 
Hence $\lambda {\mathbf 1}_A - x y$ is invertible {{iff}} $\lambda {\mathbf 1}_A - y x$ invertible if $\lambda \ne 0$. 
This is precisely the statement:
:$\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x}$
Now suppose that $A$ is not unital.
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $A$.
By the unital case we have:
:$\map {\sigma_{A_+} } {\tuple {x, 0} \tuple {y, 0} } \setminus \set 0 = \map {\sigma_{A_+} } {\tuple {y, 0} \tuple {x, 0} } \setminus \set 0$
That is:
:$\map {\sigma_{A_+} } {\tuple {x y, 0} } \setminus \set 0 = \map {\sigma_{A_+} } {\tuple {y x, 0} } \setminus \set 0$
Hence by the definition of the spectrum in a non-unital algebra:
:$\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x} \setminus \set 0$ 
{{qed}}
Category:Spectrum (Spectral Theory)
Category:Banach Algebras
cad98r4kcz24uzvzikuwd5mug7d8cw9