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

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

Recall from The Spectrum of an Element in a Normed or Banach Algebra 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 $X$ 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}(X, x) \} \end{align}

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

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

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

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

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

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