Spectral Mapping Theorem for Polynomials

Theorem

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:

\(\ds \map {\sigma_A} {\map p x}\)

\(=\)

\(\ds \map {\sigma_A} {\lambda {\mathbf 1}_A}\)

\(\ds \)

\(=\)

\(\ds \set {\mu \in \C : \mu {\mathbf 1}_A - \lambda {\mathbf 1}_A \not \in \map G A}\)

\(\ds \)

\(=\)

\(\ds \set {\mu \in \C : \paren {\mu - \lambda} {\mathbf 1}_A \not \in \map G A}\)

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:

$\ds \mu - \map p x \in \map G A$ if and only if $\zeta_j {\mathbf 1}_A - x \in \map G A$ for all $1 \le j \le n$

so that:

$\ds \mu - \map p x \not \in \map G A$ if and only if $\zeta_j {\mathbf 1}_A - x \not \in \map G A$ for some $1 \le j \le n$.

This is equivalent to:

$\mu - \map p x \not \in \map G A$ if and only if $\zeta_j \in \map {\sigma_A} x$ 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:

$\mu - \map p x \not \in \map G A$ if and only if there exists $z \in \map {\sigma_A} x$ such that $\map p z = \mu$.

That is:

$\mu - \map p x \in \map G A$ if and only if $\mu \in p \sqbrk {\map {\sigma_A} x}$

So we obtain:

$\map {\sigma_A} {\map p x} = p \sqbrk {\map {\sigma_A} x}$

$\blacksquare$