Basic Theorems: The Spectrum of an Element in an Algebra over C

Basic Theorems Regarding the Spectrum of an Element in an Algebra over C

Recall from The Spectrum of an Element in an Algebra over C page that if $\mathfrak{A}$ is an algebra over $\mathbb{C}$ and $x \in \mathfrak{A}$ then the spectrum of $x$ in $\mathfrak{A}$ denoted $\mathrm{Sp}(\mathfrak{A}, x)$ is defined as follows:

• If $\mathfrak{A}$ is an algebra with unit then:

(1)

\begin{align} \quad \mathrm{Sp}(\mathfrak{A}, x) = \{ \lambda \in \mathbb{C} : \lambda - x \in \mathrm{Sing}(\mathfrak{A}) \} \end{align}

• And if $\mathfrak{A}$ is an algebra without unit then:

(2)

\begin{align} \quad \mathrm{Sp}(\mathfrak{A}, x) = \{ 0 \} \cup \left \{ \lambda \in \mathbb{C} \setminus \{ 0 \} : \frac{1}{\lambda} x \in \mathrm{q-Sing}(\mathfrak{A}) \right \} \end{align}

We proved that if $\mathfrak{A}$ is an algebra without unit and $x \in X$ then:

(3)

\begin{align} \quad \mathrm{Sp}(\mathfrak{A}, x) = \mathrm{Sp}(\mathfrak{A} + \mathbb{C}, (x, 0)) \end{align}

We will now prove some basic results regarding the spectrum of a point in an algebra.

Proposition 1: Let $\mathfrak{A}$ be an algebra over $\mathbb{C}$ and let $x, y \in \mathfrak{A}$. Then $\mathrm{Sp}(\mathfrak{A}, xy) \setminus \{ 0 \} = \mathrm{Sp}(\mathfrak{A}, yx) \setminus \{ 0 \}$.

• Proof: Suppose that $1 \in \mathrm{Sp}(\mathfrak{A}, xy)$. Then $1 - xy \in \mathrm{Sing}(\mathfrak{A})$. By the proposition on the Algebras - x Has Quasi-Inverse y IFF (0, 1) - (x, 0) Has Inverse (0, 1) - (y, 0) page we have that $xy$ is quasi-singular. But we know by the proposition on the Algebras - xy is Quasi-Invertible IFF yx is Quasi-Invertible page that $xy$ is quasi-singular if and only if $yx$ is quasi-singular. Using the aforementioned theorem, this implies that $1 - yx \in \mathrm{Sing}(\mathfrak{A})$ and thus $1 \in \mathrm{Sp}(\mathfrak{A}, yx)$.

• Let $\lambda \in \mathbb{C} \setminus \{ 0 \}$. Suppose that $\lambda \in \mathrm{Sp}(\mathfrak{A}, xy)$ Then $\lambda - xy \in \mathrm{Sing}(\mathfrak{A})$ or equivalently:

(4)

\begin{align} \quad \lambda \left (1 - \frac{1}{\lambda} xy \right ) \in \mathrm{Sing}(\mathfrak{A}) \end{align}

• Therefore $\frac{1}{\lambda}xy$ is quasi-singular, which of course happens if and only if $\frac{1}{\lambda}yx$ is quasi-singular. Therefore $1 - \frac{1}{\lambda}yx \in \mathrm{Sing}(\mathfrak{A})$ and so $\lambda - yx \in \mathrm{Sing}(\mathfrak{A})$. So $\lambda \in \mathrm{Sp}(\mathfrak{A}, yx)$. An analogous argument shows that $\lambda \in \mathbb{C} \setminus \{ 0 \}$ with $\lambda \mathrm{Sp}(\mathfrak{A}, yx)$ implies $\lambda \in \mathrm{Sp}(\mathfrak{A}, yx)$

• From all of this we conclude that:

(5)

\begin{align} \quad \mathrm{Sp}(\mathfrak{A}, xy) \setminus \{ 0 \} = \mathrm{Sp} (\mathfrak{A}, yx) \setminus \{ 0 \} \quad \blacksquare \end{align}

Lemma 2: Let $\mathfrak{A}$ and $\mathfrak{B}$ be algebras with unit over $\mathbb{C}$. Let $\pi : \mathfrak{A} \to \mathfrak{B}$ be a homomorphism of $\mathfrak{A}$ into $\mathfrak{B}$ such that $\pi (1_{\mathfrak{A}}) = 1_{\mathfrak{B}}$ where $1_{\mathfrak{A}}$ and $1_{\mathfrak{B}}$ denote respectively the units in $\mathfrak{A}$ and $\mathfrak{B}$. If $x \in \mathrm{Inv}(\mathfrak{A})$ then $\pi(x) \in \mathrm{Inv}(\mathfrak{B})$ and moreover, $[\pi(x)]^{-1} = \pi(x^{-1})$.

• Proof: Suppose that $x \in \mathrm{Inv}(\mathfrak{A})$. Then $x^{-1}$ exists and is such that $xx^{-1} = 1_{\mathfrak{A}} = x^{-1}x$. Thus:

(6)

\begin{align} \quad \pi(xx^{-1}) = \pi(1_{\mathfrak{A}}) = \pi(x^{-1}x) \end{align}

• Which is equivalent to:

(7)

\begin{align} \quad \pi(x) \pi(x^{-1}) = 1_{\mathfrak{B}} = \pi(x^{-1})\pi(x) \end{align}

• So $\pi(x) \in \mathrm{Inv}(\mathfrak{B})$ and $[\pi(x)]^{-1} = \pi(x^{-1})$. $\blacksquare$

Proposition 3: Let $\mathfrak{A}$ and $\mathfrak{B}$ be algebras with unit over $\mathbb{C}$. Let $\pi : \mathfrak{A} \to \mathfrak{B}$ be a homomorphism of $\mathfrak{A}$ into $\mathfrak{B}$ such that $\pi (1_{\mathfrak{A}}) = 1_{\mathfrak{A}}$ where $1_{\mathfrak{A}}$ and $1_{\mathfrak{B}}$ denote respectively the units in $\mathfrak{A}$ and $\mathfrak{B}$. Let $x \in \mathfrak{A}$. Then $\mathrm{Sp}(\mathfrak{B}, \pi(x)) \subseteq \mathrm{Sp} (\mathfrak{A}, x)$.

• Proof: Let $\lambda \in \mathrm{Sp}(\mathfrak{B}, \pi(x))$. Then $\lambda - \pi(x) \in \mathrm{Sing}(\mathfrak{B})$. But:

(8)

\begin{align} \quad \lambda - \pi(x) = \lambda 1_{\mathfrak{B}} - \pi(x) = \pi (\lambda 1_{\mathfrak{A}} - x) \in \mathrm{Sing}(\mathfrak{B}) \end{align}

• The contrapositive of Lemma 2 says that since $\pi(\lambda 1_{\mathfrak{A}} - x) \in \mathrm{Sing}(\mathfrak{B})$ we must have that $\lambda 1_{\mathfrak{A}} - x \in \mathrm{Sing}(\mathfrak{A})$. Thus $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$. We conclude that $\mathrm{Sp}(\mathfrak{B}, \pi(x)) \subseteq \mathrm{Sp}(\mathfrak{A}, x)$. $\blacksquare$