Spectrum of Element in Unital Subalgebra

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)

$\blacksquare$