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 $\mathfrak{A}$ is an algebra then a subset $E \subseteq \mathfrak{A}$ is said to be a maximal commutative subset of $\mathfrak{A}$ if it is a commutative subset of $\mathfrak{A}$ and it is not contained in any other different commutative subset of $\mathfrak{A}$.

We proved by Zorn's Lemma that every algebra $\mathfrak{A}$ has a maximal commutative subset $M$ and that:

**1.**$M$ is a commutative subalgebra of $\mathfrak{A}$.

**2.**If $\mathfrak{A}$ is an algebra with unit then $M$ is a commutative subalgebra of $\mathfrak{A}$ with unit.

**3.**If $\mathfrak{A}$ is a normed algebra then $M$ is closed.

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

Proposition 1: Let $\mathfrak{A}$ be an algebra over $\mathbb{C}$ with unit. If $M$ is a maximal commutative subset of $\mathfrak{A}$ and $x \in M$ then:a) $\mathrm{Sp}(M, x) = \mathrm{Sp}(\mathfrak{A}, x)$.b) If $z \in \mathbb{C} \setminus \mathrm{Sp}(\mathfrak{A}, x)$ then $(z - x)^{-1} \in M$. |

**Proof:**Let $\lambda \in \mathbb{C} \setminus \mathrm{Sp}(\mathfrak{A}, x)$. Then $\lambda - x \not \in \mathrm{Sing}(\mathfrak{A})$ and so $\lambda - x \in \mathrm{Inv}(\mathfrak{A})$. Thus $(\lambda - x)^{-1} \in \{ \lambda - x \}^{cc} \subseteq \{ x \}^{cc}$

- To see this, note that since $(\lambda - x) \in \mathrm{Inv}(\mathfrak{A})$ 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 $\mathfrak{A}$. 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 $\mathfrak{A}$. By $M$ is a maximal commutative subset of $\mathfrak{A}$, and so:

\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 $\mathfrak{A}$ we have that if $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$ then $\lambda - x \in \mathrm{Sing}(\mathfrak{A})$. 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 $\mathfrak{A}$. Therefore we immediately have that:

\begin{align} \quad \mathrm{Sp}(\mathfrak{A}, 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 $\mathbb{C} \setminus \mathrm{Sp}(\mathfrak{A}, x) \subseteq \mathbb{C} \setminus \mathrm{Sp}(M, x)$ implying that:

\begin{align} \quad \mathrm{Sp}(M, x) \subseteq \mathrm{Sp}(\mathfrak{A}, x) \quad (**) \end{align}

- With $(*)$ and $(**)$ we see that $\mathrm{Sp}(M, x) = \mathrm{Sp}(\mathfrak{A}, x)$. $\blacksquare$