The Spectrum of an Element in an Algebra over C

Definition: Let $\mathfrak{A}$ be an algebra over $\mathbf{C}$ and let $x \in \mathfrak{A}$. The Spectrum of $x$ in $\mathfrak{A}$ is the set $\mathrm{Sp} (\mathfrak{A}, x)$ of complex numbers defined by: a) If $\mathfrak{A}$ is an algebra with unit then $\mathrm{Sp} (\mathfrak{A}, x) = \left \{ \lambda \in \mathbb{C} : \lambda - x \in \mathrm{Sing} (\mathfrak{A}) \right \}$. b) If $\mathfrak{A}$ is an algebra without unit then $\displaystyle{\mathrm{Sp} (\mathfrak{A}, x) = \{ 0 \} \cup \left \{ \lambda \in \mathbb{C} \setminus \{ 0 \} : \frac{1}{\lambda} x \in \mathrm{q-Sing}(\mathfrak{A}) \right \}}$.

For example, let $M_{2,2}$ be the set of all $2 \times 2$ matrices with entries in $\mathbb{C}$. Note that $M_{2, 2}$ is an algebra with unit $I_2$ (the $2 \times 2$ identity matrix). If $A \in M_{2, 2}$ then:

(1)

\begin{align} \quad \mathrm{Sp}(M_{2,2}, A) &= \left \{ \lambda \in \mathbb{C} : \lambda - A \in \mathrm{Sing}(M_{2,2}) \right \} \\ &= \left \{ \lambda \in \mathbb{C} : \det (\lambda I_2 - A) = 0 \right \} \end{align}

Where the above equality comes from the fact that a square matrix is singular if and only if its determinant is zero. From above we see that the spectrum of $A$ in $M_{2, 2}$ is exactly the set of eigenvalues of $A$.

Theorem 1: Let $\mathfrak{A}$ be an algebra without unit and let $x \in \mathfrak{A}$. Then $\mathrm{Sp}(\mathfrak{A}, x) = \mathrm{Sp}(\mathfrak{A} + \mathbb{C}, (x, 0))$.

• Proof: Let $x \in X$. Observe that if $(y, \beta) \in \mathfrak{A} + \mathbb{C}$ then:

(2)

\begin{align} \quad (x, 0) \circ (y, \beta) &= (x, 0) + (y, \beta) - (x, 0)(y, \beta) \\ &= (x, 0) + (y, \beta) - (xy + \beta x, 0) \\ &= (x + y - xy - \beta x, \beta) \\ &= (x \circ y - \beta x, \beta) \end{align}

(3)

\begin{align} \quad (y \beta) \circ (x, 0) &= (y, \beta) + (x, 0) - (y, \beta)(x, 0) \\ &= (y, \beta) + (x, 0) - (yx + \beta x, 0) \\ &= (y + x - yx - \beta x, \beta) \\ &= (y \circ x - \beta x, \beta) \end{align}

• Suppose that $(x, 0) \circ (y, \beta) = (0, 0) = (y, \beta) \circ (x, 0)$. Then from above $(x \circ y - \beta x, \beta) = (0, 0) = (y \circ x - \beta x, \beta)$ which implies that $\beta = 0$ and $x \circ y = 0 = y \circ x$. Thus $(x, )$ is quasi-invertible in $\mathfrak{A} + \mathbb{C}$ if and only if $x$ is quasi-invertible in $\mathfrak{A}$, or equivalently, $(x, 0) \in \mathrm{q-Sing}(\mathfrak{A} + \mathbb{C})$ if and only if $x \in \mathrm{q-Sing}(\mathfrak{A})$.

• But $x \in \mathrm{q-Sing}(\mathfrak{A})$ if and only if $(0, 1) - (x, 0) \in \mathrm{Sing}(\mathfrak{A} + \mathbb{C})$.

• So suppose that $\lambda \in \mathbb{C} \setminus \{ 0 \}$ is such that $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$. Then $\frac{1}{\lambda}x \in \mathrm{q-Sing}(\mathfrak{A})$. From above this happens if and only if $(0, 1) - \left ( \frac{1}{\lambda}x, 0 \right ) \in \mathrm{Sing}(\mathfrak{A} + \mathbb{C})$, which happens if and only if $\lambda (0, 1) - (x, 0) \in \mathrm{Sing}(\mathfrak{A} + \mathbb{C})$, i.e., $\lambda \in \mathrm{Sp}(\mathfrak{A} + \mathbb{C}, (x, 0))$.

• Lastly, observe that $(x, 0) \in \mathrm{Sing}(\mathfrak{A} + \mathbb{C})$. If it were invertible then for some $(y, \beta)$ we would have that $(x, 0)(y, \beta) = (0, 1) = (y, \beta)(x, 0)$, i.e., $(xy + \beta x, 0) = (0, 1) = (yx + \beta x, 0)$ which cannot happen. Thus $(x, 0) \in \mathrm{Sing}(\mathfrak{A} + \mathbb{C})$ if and only if $(-x, 0) \in \mathrm{Sing}(\mathfrak{A} + \mathbb{C})$ if and only if $0(0, 1) - (x, 0) \in \mathrm{Sing}(\mathfrak{A} + \mathbb{C})$. That is, $0 \in \mathrm{Sp}(\mathfrak{A} + \mathbb{C})$. Thus:

(4)

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