Theorem: Spectral Mapping Theorem for Polynomials

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$. 
Let $p : \C \to \C$ be a polynomial with:
:$\ds \map p z = \sum_{j \mathop = 0}^n a_j z^j$
for all $z \in \C$, for some $a_0, \ldots, a_n \in \C$. 
Define:
:$\ds \map p y = a_0 {\mathbf 1}_A + \sum_{j \mathop = 1}^n a_j y^j$
for each $y \in A$. 
Let $x \in A$.
Let $\map {\sigma_A} x$ and $\map {\sigma_A} {\map p x}$ be the spectrum of $x$ and $\map p x$ respectively.
Then we have:
:$\map {\sigma_A} {\map p x} = p \sqbrk {\map {\sigma_A} x}$

Proof:
Let $\map G A$ be the group of units of $A$.
Suppose that $p$ is a constant polynomial, so that:
:$\map p z = \lambda$ 
for each $z \in \C$, and:
:$\map p x = \lambda {\mathbf 1}_A$
From Spectrum of Element of Banach Algebra is Non-Empty, we have:
:$\map {\sigma_A} x \ne \O$
so that:
:$p \sqbrk {\map {\sigma_A} x} = \set \lambda$
As to $\map {\sigma_A} {\map p x}$, we have:
{{begin-eqn}}
{{eqn	| l = \map {\sigma_A} {\map p x}
		| r = \map {\sigma_A} {\lambda {\mathbf 1}_A}
}}
{{eqn	| r = \set {\mu \in \C : \mu {\mathbf 1}_A - \lambda {\mathbf 1}_A \not \in \map G A}
}}
{{eqn	| r = \set {\mu \in \C : \paren {\mu - \lambda} {\mathbf 1}_A \not \in \map G A}
}}
{{end-eqn}}
Note that if $\mu \in \C$ is such that $\mu \ne \lambda$, we have:
:$\paren {\mu - \lambda} {\mathbf 1}_A \in \map G A$
with:
:$\ds \paren {\paren {\mu - \lambda} {\mathbf 1}_A}^{-1} = \frac 1 {\mu - \lambda} {\mathbf 1}_A$
If $\mu = \lambda$, we have:
:$\paren {\mu - \lambda} {\mathbf 1}_A = {\mathbf 0}_A$
Since ${\mathbf 0}_A x = {\mathbf 0}_A$ for each $x \in A$, we have:
:$\paren {\mu - \lambda} {\mathbf 1}_A \not \in \map G A$ for $\mu = \lambda$
Hence we conclude that:
:$\set {\mu \in \C : \paren {\mu - \lambda} {\mathbf 1}_A \not \in \map G A} = \set \lambda = p \sqbrk {\map {\sigma_A} x}$
in the case that $p$ is a constant polynomial.
Now suppose that $p$ is not a constant polynomial.
Let $\mu \in \C$. 
From Polynomial Factor Theorem: Corollary: Complex Numbers, there exists $\zeta_1, \ldots, \zeta_n \in \C$ and $\alpha \in \C \setminus \set 0$ such that:
:$\ds \mu - \map p z = \alpha \prod_{j \mathop = 1}^n \paren {\zeta_j - z}$
Then, we have:
:$\ds \mu - \map p x = \alpha \prod_{j \mathop = 1}^n \paren {\zeta_j {\mathbf 1}_A - x}$
From Product of Commuting Elements in Monoid is Unit iff Each Element is Unit, we have that:
:$\ds \mu - \map p x \in \map G A$ {{iff}} $\zeta_j {\mathbf 1}_A - x \in \map G A$ for all $1 \le j \le n$
so that:
:$\ds \mu - \map p x \not \in \map G A$ {{iff}} $\zeta_j {\mathbf 1}_A - x \not \in \map G A$ for some $1 \le j \le n$.
This is equivalent to:
:$\mu - \map p x \not \in \map G A$ {{iff}} $\zeta_j \in \map {\sigma_A} x$ for some $1 \le j \le n$.
Note that:
:$\set {z \in \C : \map p z = \mu} = \set {\zeta_1, \ldots, \zeta_n}$
So, we have:
:$\mu - \map p x \not \in \map G A$ {{iff}} there exists $z \in \map {\sigma_A} x$ such that $\map p z = \mu$.
That is:
:$\mu - \map p x \in \map G A$ {{iff}} $\mu \in p \sqbrk {\map {\sigma_A} x}$
So we obtain:
:$\map {\sigma_A} {\map p x} = p \sqbrk {\map {\sigma_A} x}$
{{qed}}
Category:Spectra (Spectral Theory)
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