Spectrum of Element in Unital Subalgebra
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Theorem
Let $A$ be a unital algebra over $\C$.
Let $B$ be a unital subalgebra of $A$.
Let $x \in B$.
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively.
Then:
- $\map {\sigma_A} x \subseteq \map {\sigma_B} x$
Corollary
Let $A$ be a non-unital algebra over $\C$.
Let $B$ be a subalgebra of $A$.
Let $x \in B$.
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively.
Then:
- $\map {\sigma_A} x \subseteq \map {\sigma_B} x$
Proof
Let $\map G A$ and $\map G B$ be the group of units of $A$ and $B$ respectively.
From Group of Units of Submonoid is Subgroup, we have:
- $\map G B \subseteq \map G A$
From Set Complement inverts Subsets, we have:
- $A \setminus \map G A \subseteq A \setminus \map G B$
Then, we have:
\(\ds \map {\sigma_A} x\) | \(=\) | \(\ds \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in A \setminus \map G A}\) | Definition of Spectrum (Spectral Theory) | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in A \setminus \map G B}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in B \setminus \map G B}\) | since $B$ is a unital subalgebra | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\sigma_B} x\) | Definition of Spectrum (Spectral Theory) |
$\blacksquare$