Theorem: Spectrum of Element in Unitization of Unital Algebra

Let $A$ be a unital algebra over $\C$.
Let $A_+$ be the unitization of $A$.
Let $a \in A$. 
Then:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } = \map {\sigma_A} a \cup \set 0$
where $\sigma_{A_+}$ and $\sigma_A$ denote the spectrum in $A_+$ and $A$ respectively.

Proof:
First, we note that $0 \in \map {\sigma_{A_+} } {\tuple {a, 0} }$.
Note that for each $\tuple {b, \lambda} \in A_+$, we have $\tuple {a, 0} \tuple {b, \lambda} = \tuple {a b + \lambda a, 0}$.
This cannot be equal to $\tuple { {\mathbf 0}_A, 1}$.
Hence $\tuple {a, 0}$ is not invertible and $0 \in \map {\sigma_{A_+} } {\tuple {a, 0} }$.
Now take $\lambda \in \C \setminus \set 0$. 
We show that $\lambda \tuple {0, 1} - \tuple {a, 0}$ is invertible in $A_+$ {{iff}} $\lambda {\mathbf 1}_A - a$ is invertible in $A$ for $\lambda \ne 0$. 
We have:
:$\lambda \tuple {0, 1} - \tuple {a, 0} = \tuple {-a, \lambda}$
Suppose that $\tuple {-a, \lambda}$ is invertible in $A_+$ with inverse $\tuple {b, \mu}$. 
Then:
:$\tuple {-a, \lambda} \tuple {b, \mu} = \tuple {-a b - \mu a + \lambda b, \lambda \mu} = \tuple { {\mathbf 0}_A, 1}$
and:
:$\tuple {b, \mu} \tuple {-a, \lambda} = \tuple {-b a - \mu a + \lambda b, \lambda \mu} = \tuple { {\mathbf 0}_A, 1}$
Hence $\lambda \mu = 1$ and $-a b - \mu a + \lambda b = -b a - \mu a + \lambda b = {\mathbf 0}_A$.
We then have:
:$\paren {\lambda {\mathbf 1}_A - a} \paren {b + \mu {\mathbf 1}_A} = \lambda b - a b + \lambda \mu {\mathbf 1}_A - \mu a = \paren {-a c - \mu a + \lambda b} + \paren {\lambda \mu} {\mathbf 1}_A$
as above we have $-a b - \mu a + \lambda b = {\mathbf 0}_A$ and $\lambda \mu = 1$ so:
:$\paren {\lambda {\mathbf 1}_A - a} \paren {b + \mu {\mathbf 1}_A} = {\mathbf 1}_A$
Similarly we have:
:$\paren {b + \mu {\mathbf 1}_A} \paren {\lambda {\mathbf 1}_A - a} = \lambda b - b a + \lambda \mu {\mathbf 1}_A - \mu a = \paren {-b a - \mu a + \lambda b} + \paren {\lambda \mu} {\mathbf 1}_A = {\mathbf 1}_A$
Hence if $\tuple {-a, \lambda}$ is invertible in $A_+$ then $\lambda {\mathbf 1}_A - a$ is invertible in $A$.
Now suppose that $\lambda {\mathbf 1}_A - a$ is invertible in $A$ with inverse $c$. 
We show that $\tuple {-a, \lambda}$ is invertible in $A_+$.
So:
:$\lambda c - a c = \lambda c - c a = {\mathbf 1}_A$
Hence:
{{begin-eqn}}
{{eqn	| l = \tuple {c - \lambda^{-1} {\mathbf 1}_A, \lambda^{-1} } \tuple {-a, \lambda}
		| r = \tuple {-c a + \lambda^{-1} a + \lambda c - {\mathbf 1}_A - \lambda^{-1} a, 1}
}}
{{eqn	| r = \tuple {-c a + \lambda c - {\mathbf 1}_A, 1}
}}
{{eqn	| r = \tuple { {\mathbf 0}_A, 1}
		| c = since $-c a + \lambda c = {\mathbf 0}_A$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \tuple {-a, \lambda} \tuple {c - \lambda^{-1} {\mathbf 1}_A, \lambda^{-1} }
		| r = \tuple {-a c + \lambda^{-1} a + \lambda c - {\mathbf 1}_A - \lambda^{-1} a, 1}
}}
{{eqn	| r = \tuple {-a c + \lambda c - {\mathbf 1}_A, 1}
}}
{{eqn	| r = \tuple { {\mathbf 0}_A, 1}
		| c = since $-a c + \lambda c = {\mathbf 0}_A$
}}
{{end-eqn}}
So $\tuple {-a, \lambda}$ is invertible with inverse $\tuple {c - \lambda^{-1} {\mathbf 1}_A, \lambda^{-1} }$. 
Hence for $\lambda \ne 0$, $\lambda \tuple {0, 1} - \tuple {a, 0}$ is invertible in $A_+$ {{iff}} $\lambda {\mathbf 1}_A - a$ is invertible in $A$.
Hence we have:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } \setminus \set 0 = \map {\sigma_A} a$
Hence:
:$\paren {\map {\sigma_{A_+} } {\tuple {a, 0} } \setminus \set 0} \cup \set 0 = \map {\sigma_A} a \cup \set 0$ 
Since $0 \in \map {\sigma_{A_+} } {\tuple {a, 0} }$ as already established, we have:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } = \map {\sigma_A} a \cup \set 0$
{{qed}}
Category:Unitizations of Algebras over Fields
Category:Spectra (Spectral Theory)
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