Spectrum of Product of Elements of Banach Algebra

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Theorem

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 if and only if ${\mathbf 1}_A - y x$ 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:

\(\ds \paren { {\mathbf 1}_A - y x} \paren { {\mathbf 1}_A + y \paren { {\mathbf 1}_A - x y}^{-1} x}\) \(=\) \(\ds {\mathbf 1}_A - y x + y \paren { {\mathbf 1}_A - x y}^{-1} x - y x y \paren { {\mathbf 1}_A - x y}^{-1} x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A + y \paren {-\mathbf 1_A + \paren { {\mathbf 1}_A - x y}^{-1} - x y \paren { {\mathbf 1}_A - x y}^{-1} } x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A + y \paren {-\mathbf 1_A + \paren { {\mathbf 1}_A - x y}^{-1} \paren { {\mathbf 1}_A - x y} } x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A + y \paren { -\mathbf 1_A {\mathbf 1}_A} x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A + y {\mathbf 0}_A x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A\)

and:

\(\ds \paren { {\mathbf 1}_A + y \paren { {\mathbf 1}_A - x y}^{-1} x} \paren { {\mathbf 1}_A - y x}\) \(=\) \(\ds {\mathbf 1}_A + y \paren { {\mathbf 1}_A - x y}^{-1} x - y x - y \paren { {\mathbf 1}_A - x y}^{-1} x y x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A + y \paren { \paren { {\mathbf 1}_A - x y}^{-1} - {\mathbf 1}_A - \paren { {\mathbf 1}_A - x y}^{-1} x y}\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A + y \paren { \paren { {\mathbf 1}_A - x y}^{-1} \paren { {\mathbf 1}_A - x y} } x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A + y \paren { -\mathbf 1_A {\mathbf 1}_A} x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A + y {\mathbf 0}_A x\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A\)

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 if and only if ${\mathbf 1}_A - y x$ 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 if and only if ${\mathbf 1}_A - \paren {y/\mu} \paren {x/\mu}$ is invertible.

Hence ${\mathbf 1}_A - \lambda^{-1} x y$ is invertible if and only if ${\mathbf 1}_A - \lambda^{-1} y x$ is invertible if $\lambda \ne 0$.

Hence $\lambda {\mathbf 1}_A - x y$ is invertible if and only if $\lambda {\mathbf 1}_A - y x$ 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:

$\map {\sigma_A} {x y} \setminus \set 0 = \map {\sigma_A} {y x} \setminus \set 0$

$\blacksquare$

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