Spectrum of an Element Applied to a Polynomial in an Algebra over C

# The Spectrum of an Element Applied to a Polynomial in an Algebra over C

Let $D$ be a nonempty open subset of $\mathbb{C}$ and let $P(D)$ be the collection of all complex-valued polynomials on $D$. Let $R(D)$ be the collection of all complex-valued rational functions on $D$. It is easy to check that with the operations of pointwise function addition, scalar multiplication, and pointwise function multiplication; both $P(D)$ and $R(D)$ become algebras.

Proposition 1: Let $\mathfrak{A}$ be an algebra over $\mathbb{C}$ with unit and let $D$ be a nonempty open subset of $\mathbb{C}$. Let $x \in \mathfrak{A}$. Then the map $\phi_x : P(D) \to \mathfrak{A}$ defined for all $p \in P(D)$ by $\phi_x(p) = p(x)$ is a homomorphism. |

**Proof:**Let $p, q \in P(D)$. Then:

\begin{align} \quad \phi_x(p + q) = (p + q)(x) = p(x) + q(x) \end{align}

- And:

\begin{align} \quad \phi_x(pq) = (pq)(x) = p(x)q(x) = \phi_x(p) \phi_x(q) \end{align}

- So $\phi_x$ is a homomorphism. $\blacksquare$

Proposition 2: Let $\mathfrak{A}$ be an algebra over $\mathbb{C}$ with unit, $x \in \mathfrak{A}$ and let $D$ be a nonempty open subset of $\mathbb{C}$ for which $\mathrm{Sp}(\mathfrak{A}, x) \subseteq D$ and let $p \in P(D)$ be a non-constant polynomial. Then $\mathrm{Sp} (\mathfrak{A}, p(x)) = \{ p(z) : z \in \mathrm{Sp}(\mathfrak{A}, x) \}$. |

*Here we require $\mathrm{Sp}(\mathfrak{A}, x) \subseteq D$ so that the set $\{ p(z) : z \in \mathrm{Sp}(X, x) \}$ is well-defined.*

**Proof:**Let $p \in P(D)$ be a polynomial of degree $n \geq 1$ (since $p$ is non-constant). For any $\lambda \in \mathbb{C}$ we have that $\lambda - p$ is also a polynomial of degree $n$ and for some $\alpha_1, \alpha_2, ..., \alpha_n, \beta \in \mathbb{C}$:

\begin{align} \quad \lambda - p(z) = \beta (\alpha_1 - z)(\alpha_2 - z)...(\alpha_n - z) \end{align}

- For $x \in \mathfrak{A}$ let $\phi_x : P(D) \to \mathfrak{A}$ be the homomorphism defined for all $p \in P(D)$ by $\phi_x(p) = p(x)$ (as defined earlier). Applying $\phi_x$ to both sides of the equality above gives us:

\begin{align} \quad \phi_x(\lambda - p(z)) &= \phi_x(\beta (\alpha_1 - z)(\alpha_2 - z)...(\alpha_n - z)) \\ \quad \lambda - p(x) &= \beta (\alpha_1 - x)(\alpha_2 - x)...(\alpha_n - x) \end{align}

- Note that $\beta \neq 0$, for if it were, this would imply that $p(x) = \lambda$ for all $x \in \mathfrak{A}$, i.e., $p$ would be a constant polynomial. Therefore we see that $\lambda - p(x)$ is invertible if and only if $(\alpha_k - x)$ is invertible for all $1 \leq k \leq n$. Equivalently, $\lambda - p(x)$ is singular if and only if $(\alpha_k - x)$ is singular for some $1 \leq k \leq n$. Suppose that $(\alpha_k - x) \in \mathrm{Sing}(\mathfrak{A})$. Then $\alpha_k \in \mathrm{Sp}(\mathfrak{A}, x)$.

- Therefore $\lambda \in \mathrm{Sp}(\mathfrak{A}, p(x))$ if and only if for some $1 \leq k \leq n$ we have that $\alpha_k \in \mathrm{Sp}(\mathfrak{A}, x)$.

- But if for some $1 \leq k \leq n$ we have that $\alpha_k \in \mathrm{Sp}(\mathfrak{A}, x)$ then since $\phi_x$ is a homomorphism $\alpha_k - x = \phi_x(\alpha_k - z) \in \mathrm{Sing}(\mathfrak{A})$ implies that $\alpha_k - z \in \mathrm{Sing} (P(D))$, which happens if and only if $z = \alpha_k$ in which case $\lambda - p(z) = 0$ where $z \in \mathrm{Sp}(\mathfrak{A}, x)$. Thus:

\begin{align} \quad \mathrm{Sp}(\mathfrak{A}, p(x)) = \{ p(z) : z \in \mathrm{Sp}(\mathfrak{A}, x) \} \end{align}

Proposition 3: Let $\mathfrak{A}$ be an algebra over $\mathbb{C}$ with unit and let $D$ be a nonempty open subset of $\mathbb{C}$. Let $x \in \mathfrak{A}$. If $\Phi_x : R(D) \to \mathfrak{A}$ is a homomorphism that is an extension of $\phi_x$ then $\mathrm{Sp}(\mathfrak{A}, x) \subseteq D$. |

**Proof:**Let $\Phi_x : R(D) \to \mathfrak{A}$ is a homomorphism that is an extension of $\phi_x$, that is, $\Phi_X$ is a homomorphism from $R(D)$ to $\mathfrak{A}$ and is such that $\Phi_x |_{P(D)} = \phi_x$. (In particular, $\Phi_x(1) = 1_{\mathfrak{A}}$.)

- Let $\lambda \in \mathbb{C} \setminus D$. Let $f \in R(D)$ defined for all $z \in D$ by:

\begin{align} \quad f(z) = \frac{1}{\lambda - z} \end{align}

- Note that since $\lambda \not \in D$ we have that $f$ is defined on all of $D$ and is nonzero everywhere. Observe that $(\lambda - z) \in R(D)$ is such that $f(z)(\lambda - z) = 1 = (\lambda - z)f(z)$. Applying $\Phi_x$ to both sides of the equality gives us that:

\begin{align} \quad f(x)(\lambda - x) = 1_X = (\lambda - x)f(x) \end{align}

- Therefore $\lambda - x \not \in \mathrm{Sing}(\mathfrak{A})$, therefore, $\lambda \not \in \mathrm{Sp}(\mathfrak{A}, x)$. So $\lambda \in \mathbb{C} \setminus \mathrm{Sp}(\mathfrak{A}, x)$.

- Hence $\mathbb{C} \setminus D \subseteq \mathbb{C} \setminus \mathrm{Sp}(\mathfrak{A}, x)$ which implies that $\mathrm{Sp}(\mathfrak{A}, x) \subseteq D$. $\blacksquare$

Proposition 4: Let $\mathfrak{A}$ be an algebra over $\mathbb{C}$ with unit, $x \in \mathfrak{A}$, and let $D$ be a nonempty open subset of $\mathbb{C}$ for which $\mathrm{Sp}(\mathfrak{A}, x) \subseteq D$. For each $f \in R(D)$, write $f = p/q$ where $p, q \in P(D)$. Define a function $\Phi_x : R(D) \to \mathfrak{A}$ defined for all $f \in R(D)$ by $\Phi_x(f) = [p(x)][q(x)]^{-1}$. Then $\Phi_x$ is a homomorphism from $R(D)$ to $\mathfrak{A}$ that extends the homomorphism $\phi_x : P(D) \to \mathfrak{A}$. |

**Proof:**We first show that $\Phi_x$ is well-defined for every $f \in R(D)$. Given $f \in R(D)$ with $f = p/q$. Then by definition, since $f$ is defined on all of $D$ we must have that $q(z) \neq 0$ for all $z \in D$. Proposition 2 tells us that $\mathrm{Sp}(\mathfrak{A}, p(x)) = \{ p(z) : z \in D \}$ and so $0 \not \in \mathrm{Sp}(\mathfrak{A}, p(x))$. Therefore $0 - p(x) \in \mathrm{Inv}(\mathfrak{A})$, i.e., $p(x) \in \mathrm{Inv}(\mathfrak{A})$. Thus $\Phi_x(f) = [p(x)][q(x)]^{-1}$ is a defined element of $\mathfrak{A}$.

- Moreover, the representatives $p, q \in P(D)$ for $f$ are unique up to common factors which can be cancelled out.

- Now it is clear that $\Phi_x \in \mathcal L(R(D), \mathfrak{A})$ and that for all $f, g \in R(D)$ with $f = p/q$, $g = s/t$ we have that:

\begin{align} \quad \Phi_x(fg) =[p(x)s(x)][q(x)t(x)]^{-1} = [p(x)s(x)][t(x)]^{-1}[q(x)]^{-1} = p(x)\Phi_x(g)[q(x)]^{-1} \overset{(*)} = p(x)[q(x)]^{-1}\Phi_x(g) = \Phi_x(f)\Phi_x(g) \end{align}

- Where the equality at $(*)$ comes from the fact that polynomials in $P(D)$ commute with one another.

- Lastly, if $f \in P(D)$ then $f = f/1$ and $\Phi_x(f) = f(x)[1]^{-1} = f(x) = \phi_x(f)$. So $\Phi_x : R(D) \to X$ is a homomorphism of $R(D)$ to $\mathfrak{A}$ that extends $\phi_x$. $\blacksquare$