r(xy) = r(yx) When A is a Banach Algebra Over C

# r(xy) = r(yx) When A is a Banach Algebra Over C

Recall from The Spectrum of an Element in a Normed or Banach Algebra over C page that we proved a very substantial theorem, that being that if $X$ is a Banach algebra over $\mathbb{C}$ then the spectrum of $x$ in $\mathfrak{A}$ is ALWAYS a nonempty compact subset of $\mathbb{C}$ and moreover, the spectral radius of $x$ is given to be:

(1)\begin{align} \quad r(x) = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(\mathfrak{A}, x) \} \end{align}

We will use this result to show that if $\mathfrak{A}$ is a Banach algebra over $\mathbb{C}$ and $x, y \in \mathfrak{A}$ then the spectral radii of $xy$ and $yx$ are the same.

Theorem 1: Let $\mathfrak{A}$ be a Banach algebra over $\mathbb{C}$ and let $x, y \in \mathfrak{A}$. Then $r(xy) = r(yx)$. |

**Proof:**Since $\mathfrak{A}$ is a Banach algebra over $\mathbb{C}$ we have by the theorem on The Spectrum of an Element in a Normed or Banach Algebra over C page that $\mathrm{Sp}(\mathfrak{A}, xy)$ is a nonempty subset of $\mathbb{C}$ for which:

\begin{align} \quad r(xy) = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(\mathfrak{A}, xy) \} \end{align}

- But by a proposition on the Basic Theorems Regarding the Spectrum of an Element in an Algebra over C page we know that $\mathrm{Sp}(\mathfrak{A}, xy) \setminus \{ 0 \} = \mathrm{Sp}(\mathfrak{A}, xy) \setminus \{ 0 \}$. Thus $\max \{ |\lambda| : \lambda \in \mathrm{Sp}(\mathfrak{A}, xy) \} = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(\mathfrak{A}, yx) \}$. From above we see that:

\begin{align} \quad r(xy) = r(yx) \quad \blacksquare \end{align}