Basic Theorems Regarding the Spectrum of an Element in an Algebra

# Basic Theorems Regarding the Spectrum of an Element in an Algebra

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

• If $X$ is an algebra with unit then:
(1)
\begin{align} \quad \mathrm{Sp}(X, x) = \{ \lambda \in \mathbb{C} : \lambda - x \in \mathrm{Sing}(X) \} \end{align}
• And if $X$ is an algebra without unit then:
(2)
\begin{align} \quad \mathrm{Sp}(X, x) = \{ 0 \} \cup \left \{ \lambda \in \mathbb{C} \setminus \{ 0 \} : \frac{1}{\lambda} x \in \mathrm{q-Sing}(X) \right \} \end{align}

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

(3)
\begin{align} \quad \mathrm{Sp}(X, x) = \mathrm{Sp}(X + \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 $X$ be an algebra over $\mathbb{C}$ and let $x, y \in X$. Then $\mathrm{Sp}(X, xy) \setminus \{ 0 \} = \mathrm{Sp}(X, yx) \setminus \{ 0 \}$.
• Proof: Suppose that $1 \in \mathrm{Sp}(X, xy)$. Then $1 - xy \in \mathrm{Sing}(X)$. From one of the propositions on the Basic Theorems Regarding Quasi-Invertible Elements in an Algebra we have that $xy$ is quasi-singular. But we know that $xy$ is quasi-singular if and only if $yx$ is quasi-singular (this is one of the propositions on the page linked above). Using the aforementioned theorem, this implies that $1 - yx \in \mathrm{Sing}(X)$ and thus $1 \in \mathrm{Sp}(X, yx)$.
• Let $\lambda \in \mathbb{C} \setminus \{ 0 \}$. Suppose that $\lambda \in \mathrm{Sp}(X, xy)$ Then $\lambda - xy \in \mathrm{Sing}(X)$ or equivalently:
(4)
\begin{align} \quad \lambda \left (1 - \frac{1}{\lambda} xy \right ) \in \mathrm{Sing}(X) \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 $1 - \frac{1}{\lambda}yx \in \mathrm{Sing}(X)$ and so $\lambda - yx \in \mathrm{Sing}(X)$. So $\lambda \in \mathrm{Sp}(X, yx)$. An analogous argument shows that $\lambda \in \mathbb{C} \setminus \{ 0 \}$ with $\lambda \mathrm{Sp}(X, yx)$ implies $\lambda \in \mathrm{Sp}(X, yx)$
• Lastly, suppose that $0 \in \mathrm{Sp}(X, xy)$. Then $-xy \in \mathrm{Sing}(X)$. But of course then $y(-x) = -yx \in \mathrm{Sing}(X)$ and so $0 \in \mathrm{Sp}(X, yx)$. From all of this we conclude that:
(5)
 Lemma 2: Let $X$ and $Y$ be algebras with unit over $\mathbb{C}$. Let $\pi : X \to Y$ be a homomorphism of $X$ into $Y$ such that $\pi (1_X) = 1_X$ where $1_X$ and $1_Y$ denote respectively the units in $X$ and $Y$. If $x \in \mathrm{Inv}(X)$ then $\pi(x) \in \mathrm{Inv}(Y)$ and moreover, $[\pi(x)]^{-1} = \pi(x^{-1})$.
• Proof: Suppose that $x \in \mathrm{Inv}(X)$. Then $x^{-1}$ exists and is such that $xx^{-1} = 1_X = x^{-1}x$. Thus:
(6)
\begin{align} \quad \pi(xx^{-1}) = \pi(1_X) = \pi(x^{-1}x) \end{align}
• Which is equivalent to:
(7)
\begin{align} \quad \pi(x) \pi(x^{-1}) = 1_Y = \pi(x^{-1})\pi(x) \end{align}
• So $\pi(x) \in \mathrm{Inv}(Y)$ and $[\pi(x)]^{-1} = \pi(x^{-1})$. $\blacksquare$
 Proposition 3: Let $X$ and $Y$ be algebras with unit over $\mathbb{C}$. Let $\pi : X \to Y$ be a homomorphism of $X$ into $Y$ such that $\pi (1_X) = 1_X$ where $1_X$ and $1_Y$ denote respectively the units in $X$ and $Y$. Let $x \in X$. Then $\mathrm{Sp}(Y, \pi(x)) \subseteq \mathrm{Sp} (X, x)$.
• Proof: Let $\lambda \in \mathrm{Sp}(Y, \pi(x))$. Then $\lambda - \pi(x) \in \mathrm{Sing}(Y)$. But:
(8)
\begin{align} \quad \lambda - \pi(x) = \lambda 1_Y - \pi(x) = \pi (\lambda 1_X - x) \in \mathrm{Sing}(Y) \end{align}
• The contrapositive of lemma 2 says that since $\pi(\lambda 1_X - x) \in \mathrm{Sing}(Y)$ we must have that $\lambda 1_X - x \in \mathrm{Sing}(X)$. Thus $\lambda \in \mathrm{Sp}(X, x)$. We conclude that $\mathrm{Sp}(Y, \pi(x)) \subseteq \mathrm{Sp}(X, x)$. $\blacksquare$