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)
\begin{align} \quad \mathrm{Sp}(X, xy) = \mathrm{Sp} (X, yx) \quad \blacksquare \end{align}
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 $\mathrm{Sp}(Y, \pi(x)) \subseteq \mathrm{Sp}(X, x)$. $\blacksquare$
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