Spectrum of Element in Unital Subalgebra/Corollary
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Theorem
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 $A_+$ and $B_+$ be the unitizations of $A$ and $B$ respectively.
From Unitization of Algebra over Field preserves Subalgebra Relation, we have:
- $B_+$ is a unital subalgebra of $A_+$.
From Spectrum of Element in Unital Subalgebra, we have that:
- $\map {\sigma_{A_+} } {\tuple {x, 0} } \subseteq \map {\sigma_{B_+} } {\tuple {x, 0} }$
From the definition of the spectrum in a non-unital algebra, we have:
- $\map {\sigma_{A_+} } {\tuple {x, 0} } = \map {\sigma_A} x$
and:
- $\map {\sigma_{B_+} } {\tuple {x, 0} } = \map {\sigma_B} x$
So, we have:
- $\map {\sigma_A} x \subseteq \map {\sigma_B} x$
$\blacksquare$