Theorem: Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra with identity element ${\mathbf 1}_A$.
Let $B \subseteq A$ be a unital $\text C^\ast$-subalgebra of $A$. 
Let $\sigma_A$ and $\sigma_B$ be the spectrum in $A$ and $B$ respectively. 
Let $x \in B$. 
Then we have:
:$\map {\sigma_A} x = \map {\sigma_B} x$

Proof:
First take $x$ to be Hermitian. 
Let $B' \subseteq B$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x}$. 
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B'$ is commutative.
From Spectrum of Hermitian Element in Unital C*-Algebra is Real, we have $\map {\sigma_{B'} } x \subseteq \R$. 
From Spectrum of Element in Unital Subalgebra, we therefore have $\map {\sigma_B} x \subseteq \map {\sigma_{B'} } x \subseteq \R$. 
From Interior of Set of Real Numbers in Complex Numbers is Empty: Corollary, we have $\map {\sigma_B} x^\circ = \O$ where $\map {\sigma_B} x^\circ$ denotes the interior of $\map {\sigma_B} x$. 
From the definition of boundary, we have that $\partial \map {\sigma_B} x = \map {\sigma_B} x^- \setminus \map {\sigma_B} x^\circ$.
Since $\map {\sigma_B} x$ is closed from Spectrum of Element of Banach Algebra is Closed and $\map {\sigma_B} x^\circ = \O$, we then have that $\map {\sigma_B} x = \partial \map {\sigma_B} x$.
Then, from Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra we have:
:$\map {\sigma_B} x = \partial \map {\sigma_B} x \subseteq \partial \map {\sigma_A} x$
From Set is Closed iff it Contains its Boundary, we have $\partial \map {\sigma_A} x \subseteq \map {\sigma_A} x$. 
So $\map {\sigma_B} x \subseteq \map {\sigma_A} x$. 
From Spectrum of Element in Unital Subalgebra we have $\map {\sigma_A} x \subseteq \map {\sigma_B} x$. 
Hence we conclude $\map {\sigma_A} x = \map {\sigma_B} x$ in this case. 
In particular, $0 \in \map {\sigma_A} x$ {{iff}} $0 \in \map {\sigma_B} x$. 
Hence if $x$ is Hermitian, it is invertible in $B$ {{iff}} it is invertible in $A$.
Let $b \in B$. 
Suppose that $b$ is invertible in $A$. 
We show that it is invertible in $B$. 
Take $a \in A$ such that $a b = b a = {\mathbf 1}_A$.
From Inverse of Star of Element in Unital *-Algebra, we have $b^\ast a^\ast = a^\ast b^\ast = {\mathbf 1}_A$. 
We therefore have:
{{begin-eqn}}
{{eqn	| l = b b^\ast a^\ast a
		| r = b \paren {b^\ast a^\ast} a
}}
{{eqn	| r = b {\mathbf 1}_A a 
}}
{{eqn	| r = b a 
}}
{{eqn	| r = {\mathbf 1}_A 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = a^\ast a b b^\ast
		| r = a^\ast \paren {a b} b^\ast 
}}
{{eqn	| r = a^\ast {\mathbf 1}_A b^\ast 
}}
{{eqn	| r = a^\ast b^\ast
}}
{{eqn	| r = {\mathbf 1}_A 
}}
{{end-eqn}}
Hence $b b^\ast$ is invertible in $A$.
From Product of Element in *-Star Algebra with its Star is Hermitian, $b b^\ast$ is Hermitian.
Hence from what we have already shown, $b b^\ast$ is invertible in $B$. 
So there exists $c \in B$ such that $b b^\ast c = {\mathbf 1}_A$. 
Then:
{{begin-eqn}}
{{eqn	| l = a
		| r = a b b^\ast c
}}
{{eqn	| r = \paren {a b} b^\ast c
}}
{{eqn	| r = {\mathbf 1}_A b^\ast c
}}
{{eqn	| r = b^\ast c
}}
{{end-eqn}}
Since $b^\ast \in B$ and $c \in B$, we have $a \in B$. 
So $b$ is invertible in $B$. 
Clearly conversely if $b$ is invertible in $B$, then it is invertible in $A$.
Hence we have shown that if $b \in B$, then $b$ is invertible in $B$ {{iff}} it is invertible in $A$.
In particular, if $x \in B$ then ${\mathbf 1}_A - x$ is invertible in $B$ {{iff}} it is invertible in $A$.
This is precisely the statement $\map {\sigma_A} x = \map {\sigma_B} x$. 
{{qed}}