The Spectrum of an Element in a Maximal Com. Subset of an Algebra

The Spectrum of an Element in a Maximal Commutative Subset of an Algebra

Recall from the Maximal Commutative Subsets of an Algebra page that if $X$ is an algebra then a subset $E \subseteq X$ is said to be a maximal commutative subset of $X$ if it is a commutative subset of $X$ and it is not contained in any other different commutative subset of $X$.

We proved by Zorn's Lemma that every algebra $X$ has a maximal commutative subset $M$ and that:

  • 1. $M$ is a commutative subalgebra of $X$.
  • 2. If $X$ is an algebra with unit then $M$ is a commutative subalgebra of $X$ with unit.
  • 3. If $X$ is a normed algebra then $M$ is closed.

Now let $x \in M \subseteq X$. If $X$ is an algebra over $\mathbb{C}$ with unit then we can consider the spectrum of $x$ in both $M$ and $X$. The following proposition tells us that the spectrums are the same.

Proposition 1: Let $X$ be an algebra over $\mathbb{C}$ with unit. If $M$ is a maximal commutative subset of $X$ and $x \in M$ then:
a) $\mathrm{Sp}(M, x) = \mathrm{Sp}(X, x)$.
b) If $z \in \mathbb{C} \setminus \mathrm{Sp}(X, x)$ then $(z - x)^{-1} \in M$.
  • Proof: Let $\lambda \in \mathbb{C} \setminus \mathrm{Sp}(X, x)$. Then $\lambda - x \not \in \mathrm{Sing}(X)$ and so $\lambda - x \in \mathrm{Inv}(X)$. Thus $(\lambda - x)^{-1} \in \{ \lambda - x \}^{cc} \subseteq \{ x \}^{cc}$
  • To see this, note that since $(\lambda - x) \in \mathrm{Inv}(X)$ we have by one of the propositions on The Commutant (Centralizer) of a Set in an Algebra page that $(\lambda - x)^{-1} \in \{ \lambda - x \}^{cc}$. Note that if $y \in \{ x \}^c$ then $xy = yx$. So $(\lambda - x)y = \lambda y - xy = \lambda y - yx = y(\lambda - x)$ and so $y \in \{ \lambda - x \}^c$. By one of the propositions on the Commutative Subsets of an Algebra page we have that $\{ x \}^{cc} \supseteq \{ \lambda - x \}^{cc}$. Thus $(\lambda - x)^{-1} \in \{ x \}^{cc}$.
  • Now we claim that $(\lambda - x)^{-1} \cup M$ is a commutative subset of $X$. To see this, all we need to show is that $(\lambda - x)^{-1}$ commutes with every $m \in M$ (since pairs of elements in $M$ already commute with each other by definition).
  • Let $m \in M$. Since $x \in M$ we have that $mx = xm$. Therefore $m \in \{ x \}^c$. But since $(\lambda - x)^{-1} \in \{ x \}^{cc}$ we have that $(\lambda - x)^{-1}m = m(\lambda - x)^{-1}$. Since this is true for every $m \in M$ we see that $M \cup \{ (\lambda - x)^{-1} \}$ is a commutative subset of $X$. By $M$ is a maximal commutative subset of $X$, and so:
(1)
\begin{align} \quad M \cup \{ (\lambda - x)^{-1} \} \subseteq M \end{align}
  • Thus $(\lambda - x)^{-1} \in M$. $\blacksquare$
  • Proof of b) Note that since $M$ is a subalgebra of $X$ we have that if $\lambda \in \mathrm{Sp}(X, x)$ then $\lambda - x \in \mathrm{Sing}(X)$. Note that this implies that $\lambda - x \in \mathrm{Sing} (M)$. To see why, note that if $\lambda - x \in \mathrm{Inv}(M)$ then there exists a $x' \in M$ such that $(\lambda - x)x' = 1 = x'(\lambda - x)$. But $x' \in X$ too, contradicting $\lambda - x$ being singular in $X$. Thereefore we immediately have that:
(2)
\begin{align} \quad \mathrm{Sp}(X, x) \subseteq \mathrm{Sp}(M, x) \quad (*) \end{align}
  • Since $(\lambda - z)^{-1} \in M$ by (a) we have that $\lambda - z \in \mathrm{Inv}(M)$. Thus $\lambda \in \mathrm{C} \setminus \mathrm{Sp}(M, x)$. Thus $\mathbb{C} \setminus \mathrm{Sp}(X, x) \subseteq \mathbb{C} \setminus \mathrm{Sp}(M, x)$ implying that:
(3)
\begin{align} \quad \mathrm{Sp}(M, x) \subseteq \mathrm{Sp}(X, x) \quad (**) \end{align}
  • With $(*)$ and $(**)$ we see that $\mathrm{Sp}(M, x) = \mathrm{Sp}(X, x)$. $\blacksquare$
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