Spectrum of Element in Unital Subalgebra/Corollary

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:

$\map {\sigma_A} x \subseteq \map {\sigma_B} x$

$\blacksquare$