avewright/proofwiki-math huggingface dataset:
Theorem: Element of Unital Commutative Algebra Invertible iff not Contained in Maximal Ideal

Let $R$ be a ring.
Let $A$ be an unital commutative $R$-algebra.
Let $a \in A$.
Then $a$ is invertible {{iff}} there does not exist a maximal ideal $M$ with $a \in M$.

Proof:
=== Necessary Condition ===
We prove the contrapositive.
Then, by Proof by Contraposition we will done. 
We will prove that if $a$ is in a maximal ideal then $a$ is not invertible.
Let $M$ be a maximal ideal with $a \in M$. 
{{AimForCont}} there exists $b \in A$ with $a b = {\mathbf 1}_A$.
Since $M$ is an ideal, we have $a b \in M$, hence ${\mathbf 1}_A \in M$.
This contradicts that $M$ is proper.
{{qed|lemma}}
Theorem: Characterization of Maximal Ideal of Unital Commutative Banach Algebra

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $M \subseteq A$.
Then $M$ is maximal {{iff}} there exists a character $\phi$ with $M = \ker \phi$.

Proof:
=== Necessary Condition ===
This is Maximal Ideal of Unital Commutative Banach Algebra is Kernel of Character.
{{qed|lemma}}
Theorem: Element of Unital Commutative Banach Algebra is Invertible iff not in Kernel of Character

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $a \in A$.
Then $x$ is invertible {{iff}}:
:for every character $\phi$ of $A$, we have $a \not \in \ker \phi$.

Proof:
From Element of Unital Commutative Algebra Invertible iff not Contained in Maximal Ideal, we have that $a$ is invertible {{iff}}:
:for every maximal ideal $M$ of $A$, we have $a \not \in M$. 
From Characterization of Maximal Ideal of Unital Commutative Banach Algebra, we have that:
:$M$ is a maximal ideal of $A$ {{iff}} there exists a character $\phi$ of $A$ with $M = \ker \phi$. 
Hence $a$ is invertible {{iff}}:
:for every character $\phi$ of $A$, we have $a \not \in \ker \phi$.
{{qed}}
Category:Commutative Banach Algebras
i61jr2e416v7aev1g04ehva0edtlw35
Theorem: Spectrum of Element of Unital Commutative Banach Algebra

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $\Phi_A$ be the spectrum of $A$.
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$. 
Then:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A}$

Proof:
We have $\lambda \in \map {\sigma_A} x$ {{iff}}:
:$\lambda {\mathbf 1}_A - x$ is not invertible.
From Element of Unital Commutative Banach Algebra is Invertible iff not in Kernel of Character, this is the case {{iff}} there exists a character $\phi$ such that:
:$\map \phi {\lambda {\mathbf 1}_A - x} = 0$
From linearity, we have:
:$\map \phi x = \lambda \map \phi { {\mathbf 1}_A}$ 
in this case. 
From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have $\map \phi { {\mathbf 1}_A} = 1$.
Hence we have:
:$\lambda \in \map {\sigma_A} x$ {{iff}} there exists a character $\phi$ with $\lambda = \map \phi x$.
This is precisely the statement:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A}$ 
{{qed}}
Category:Commutative Banach Algebras
Category:Unital Banach Algebras
Category:Spectrum of Element of Unital Commutative Banach Algebra
gjsqfbe3zushteq4hlimp0z5i5ya4go
Theorem: Spectral Radius of Element of Commutative Banach Algebra

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative Banach algebra over $\C$. 
Let $\Phi_A$ be the spectrum of $A$.
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$. 
Let $\map {r_A} x$ be the spectral radius of $x$.
Then:
:$\ds \map {r_A} x = \sup_{\phi \in \Phi_A} \cmod {\map \phi x}$

Proof:
From the definition of the spectral radius, we have:
:$\ds \map {r_A} x = \sup_{\lambda \in \map {\sigma_A} x} \cmod \lambda$ 
From Spectrum of Element of Unital Commutative Banach Algebra, we have:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A}$ if $A$ is unital
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$ if $A$ is not unital.
Hence we have:
:$\ds \map {r_A} x = \max \set {\sup_{\phi \in \Phi_A} \cmod {\map \phi x}, 0} = \sup_{\phi \in \Phi_A} \cmod {\map \phi x}$
{{qed}}
Category:Banach Algebras
Category:Spectral Radius
99gwawx8e1e5dmt8b3x23pbvkbtkpq0
Theorem: Regression Coefficients of Normally Distributed Random Variable

Let $X$ be a random variable.
Let $x$ be a given value of $X$.
Let $Y$ be a random variable with a normal distribution.
Let the variance $\sigma^2$ of $Y$ (usually unknown) be independent of $x$.
Let $S$ be a sample of $n$ independent pairs of observations $\tuple {x_i, y_i}$ for $i = 1, 2, \ldots, n$.
The maximum likelihood estimators $\beta_0$ and $\beta_1$ are given by:
{{begin-eqn}}
{{eqn | l = \beta_1
      | r = \dfrac {\ds \sum_i \paren {x_i - \overline x} \paren {y_i - \overline y} } {\ds \sum_i \paren {x_1 - \overline x}^2}
}}
{{eqn | l = \beta_0
      | r = \overline y - \beta_1 \overline x
}}
{{end-eqn}}
where:
{{begin-eqn}}
{{eqn | l = \overline x
      | r = \sum_i \dfrac {x_i} n
}}
{{eqn | l = \overline y
      | r = \sum_i \dfrac {y_i} n
}}
{{end-eqn}}

Proof:
The maximum likelihood estimators are obtained by the method of least squares.
That is, the aim is to minimize $\ds \sum_i \paren {y_i - \beta_0 - \beta_1 x_i}^2$.
{{ProofWanted}}
Theorem: Double Root is Root of First Derivative of Equation but not Second Derivative

Let $\EE$ be the equation:
:$\map f x = 0$
where $f: \R \to \R$ is a real function.
Let $\xi$ be a double root of $\EE$.
Then:
:$\xi$ is a root of the equation $\map {f'} x = 0$
:$\xi$ is ''not'' a root of the equation $\map {f' '} x = 0$
where:
:$f': \R \to \R$ is the first derivative of $f$ {{WRT|Differentiation}} $x$
:$f' ': \R \to \R$ is the second derivative of $f$ {{WRT|Differentiation}} $x$.

Proof:
{{DefinitionWanted|Definition about multiplicity of roots on arbitrary functions}}{{ProofWanted}}
Theorem: Zero Vector in *-Algebra is Hermitian

Let $\struct {A, \ast}$ be a $\ast$-algebra.
Let ${\mathbf 0}_A$ be the zero vector of $A$. 
Then we have:
:${\mathbf 0}_A^\ast = {\mathbf 0}_A$

Proof:
By $(\text C^\ast 2)$ in the definition of an involution, we have:
:${\mathbf 0}_A^\ast = \paren { {\mathbf 0}_A + {\mathbf 0}_A}^\ast = {\mathbf 0}_A^\ast + {\mathbf 0}_A^\ast$
Hence:
:${\mathbf 0}_A^\ast = {\mathbf 0}_A^\ast - {\mathbf 0}_A^\ast = {\mathbf 0}_A$
{{qed}}
Category:Hermitian Elements of *-Algebras
6hr0rg63u9lujabvw64pwnchb5hj2oc
Theorem: C* Identity implies Involution is Isometry

Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$ that is a Banach algebra when given the algebra norm $\norm {\, \cdot \,}$.
Suppose that the $\text C^\ast$ identity:
:$\norm {x x^\ast} = \norm x^2$
holds for all $x \in A$. 
Then:
:$\norm x = \norm {x^\ast}$
for each $x \in A$.

Proof:
Let $x \in A$. 
From Zero Vector in *-Algebra is Hermitian, we have: 
:${\mathbf 0}_A^\ast = {\mathbf 0}_A$
where ${\mathbf 0}_A$ is the zero vector of $A$.
Hence:
:$\norm x = \norm {x^\ast}$
holds in the case $x = {\mathbf 0}_A$.
Hence take $x \ne {\mathbf 0}_A$ so that $\norm x \ne 0$ by {{NormAxiomVector|1}}. 
We have: 
:$\norm {x x^\ast} \le \norm x \norm {x^\ast}$
from the definition of an algebra norm.
Hence we have:
:$\norm x^2 \le \norm x \norm {x^\ast}$
Hence, dividing by $\norm x$ we have:
:$\norm x \le \norm {x^\ast}$
for all $x \in A \setminus \set { {\mathbf 0}_A}$. 
Swapping $x$ for $x^\ast$ we have:
:$\norm {x^\ast} \le \norm {x^{\ast \ast} }$
From $(\text C^\ast 1)$ in the definition of an involution, we have:
:$x^{\ast \ast} = x$
and hence:
:$\norm {x^\ast} \le \norm x$
Since we have already shown that $\norm x \le \norm {x^\ast}$, we conclude:
:$\norm {x^\ast} = \norm x$ for each $x \in A$.
{{qed}}
Category:C*-Algebras
Category:*-Algebras
herh407ky6h2zjh2e5gm6894l0zxd9o
Theorem: Identity Mapping forms Galois Connection

Let $\struct {S, \preceq}$ be an ordered sets.
Let $\operatorname{id}_S : S \to S$ be the identity mapping on $S$.
Then:
:$\tuple {\operatorname{id}_S, \operatorname{id}_S}$ is a Galois connection.

Proof:
We have
{{begin-eqn}}
{{eqn | q = \forall x, y \in S 
      | l = x
      | o = \preceq
      | r = \map {\operatorname{id}_S} y
      | c = 
}}
{{eqn | ll= \iff
      | l = x
      | o = \preceq
      | r = y
      | c = {{Defof|Identity Mapping}}
}}
{{eqn | ll= \iff
      | l = \map {\operatorname{id}_S} x
      | o = \preceq
      | r = y
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
The result follows.
{{qed}}
Category:Galois Connections
pyd40jhirkzc1lvjbbwgkh9e8t0f8kq
Theorem: Explicit Form for Generated Subalgebra

Let $K$ be a field. 
Let $A$ be an algebra over $K$.
Let $S \subseteq A$ be a non-empty set. 
Let $K \sqbrk S$ be the subalgebra generated by $A$.
Then:
:$\ds K \sqbrk S = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$

Proof:
Let:
:$B = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
First, for each $x \in S$ we have:
:$x = {\mathbf 1}_K x^1 \in \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
Hence $S \subseteq B$.
We show that $B$ is a subalgebra.
From Linear Span is Linear Subspace, $B$ is a linear subspace of $A$.
Now let $x, y \in B$.
Since $B$ is a linear subspace, it suffices to consider $x, y$ of the form:
:$x = x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n}$
and:
:$y = y_1^{s_1} y_2^{s_2} \ldots y_m^{s_m}$
We have:
:$x y = x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} y_1^{s_1} y_2^{s_2} \ldots y_m^{s_m}$
By relabelling $x_{n + t} = y_t$ and $k_{n + t} = s_t$ for $1 \le t \le m$ we obtain that $x y \in B$.
Hence $B$ is subalgebra of $A$.
Since $S \subseteq B$, we obtain $K \sqbrk S \subseteq B$ from the definition of the subalgebra generated by $A$.
Conversely, note that for each $x_1, \ldots, x_n \in S$ we have $x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} \in S$ from the definition of a subalgebra.
Since $K \sqbrk S$ is a linear subspace of $A$, we have $B = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1} \subseteq K \sqbrk S$.
Hence $B = K \sqbrk S$. 
{{qed}}
Category:Algebras
b04onhmiapczdtefdqinwunqqquxxji
Theorem: Subalgebra Generated by Commuting Elements is Commutative

Let $K$ be a field. 
Let $A$ be an algebra over $K$.
Let $S \subseteq A$ be a non-empty set such that:
:for all $x, y \in S$ we have $x y = y x$. 
Let $K \sqbrk S$ be the subalgebra generated by $S$.
Then $K \sqbrk S$ is a commutative algebra.

Proof:
From Explicit Form for Generated Subalgebra, we have:
:$K \sqbrk S = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
and $K \sqbrk S$ is a subalgebra of $A$. 
Let $x, y \in K \sqbrk S$. 
Then:
:$\ds x = \sum_{j \mathop = 1}^N \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} }$
and:
:$\ds y = \sum_{i \mathop = 1}^M \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} }$
where each $x_{j, k}$ and $x_{i, k}$ is in $S$.
Then we have:
:$\ds x y = \sum_{j \mathop = 1}^N \sum_{i \mathop = 1}^M \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} }$
Since $K$ is a field and every element in $S$ commutes, we have:
:$\ds \sum_{j \mathop = 1}^N \sum_{i \mathop = 1}^M \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } = \sum_{i \mathop = 1}^M \sum_{j \mathop = 1}^N \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } }$
Then:
:$\ds \sum_{i \mathop = 1}^M \sum_{j \mathop = 1}^N \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } = \paren {\sum_{i \mathop = 1}^M \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\sum_{j \mathop = 1}^N \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } = y x$
Hence $K \sqbrk S$ is a commutative algebra.
{{qed}}
Category:Algebras
dldefsnem3duw8m0fmrjflm312ka7yp
Theorem: Existence of Maximal Commutative Subalgebra

Let $K$ be a field. 
Let $A$ be an algebra over $K$.
Let $S \subseteq A$ be a non-empty set such that:
:for all $x, y \in S$ we have $x y = y x$. 
Then there exists a commutative subalgebra of $A$ containing $S$ that is maximal (among all subalgebras) {{WRT}} set inclusion.

Proof:
Let $\SS$ be the set commutative subalgebras of $A$ containing $S$. 
We look to apply Zorn's Lemma to $\struct {\SS, \subseteq}$. 
From Subalgebra Generated by Commuting Elements is Commutative, we have $\SS \ne \O$. 
Let $\CC$ be a chain in $\struct {\SS, \subseteq}$.
From Union of Chain of Subalgebras is Subalgebra, $\bigcup \CC$ is a commutative subalgebras of $A$ containing $S$.
We can see that $\bigcup \CC$ is an upper bound for $\CC$ since $A \subseteq \bigcup \CC$ for each $A \in \CC$. 
Hence we have that every chain in $\struct {\SS, \subseteq}$ has an upper bound.
So by Zorn's Lemma, $\struct {\SS, \subseteq}$ has a maximal element $B$.
Now suppose that $C$ is a commutative subalgebras of $A$ with $B \subseteq C$.
Then $C$ in particular contains $S$, so $C \in \CC$. 
Then $C = B$. 
So $B$ is indeed maximal among all subalgebras.
{{qed}}
{{AoC|Zorn's Lemma}}
Category:Algebras
Category:Existence of Maximal Commutative Subalgebra
c3ntsv2u39hr04rrjot7stmaewosqsk
Theorem: Union of Chain of Subalgebras is Subalgebra

Let $K$ be a field. 
Let $A$ be an algebra over $K$.
Let $\BB$ be a $\subseteq$-chain of subalgebras of $A$. 
Then $\bigcup \BB$ is a subalgebra of $A$. 
Further, if each $A \in \BB$ is commutative, then so is $\bigcup \BB$.

Proof:
From Union of Chain of Submodules is Submodule, $\bigcup \BB$ is a linear subspace of $A$.
Now let $x, y \in \bigcup \BB$.
Then there exists $B_1, B_2 \in \BB$ such that $x \in B_1$ and $y \in B_2$.
Since $\BB$ is a chain, we either have $B_1 \subseteq B_2$ or $B_2 \subseteq B_1$. 
Hence we can pick $i \in \set {1, 2}$ such that $x, y \in B_i$. 
Since $B_i$ is a subalgebra, we have $x y \in B_i$ and hence $x y \in \bigcup \BB$.
Hence $\bigcup \BB$ is a subalgebra of $A$. 
Further, if each $A \in \BB$ is commutative, then in particular so is $B_i$.
So $x y = y x$.
{{qed}}
Category:Algebras
3klq6o87dlhsi8i5mzkqpwoh7s5sitm
Theorem: Spectral Radius is Subadditive on Commuting Elements in Banach Algebra

Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x, y \in A$ be commuting elements of $A$.
Let $r_A$ denote spectral radius in $A$.
Then we have:
:$\map {r_A} {x + y} \le \map {r_A} x + \map {r_A} y$

Proof:
First take $\struct {A, \norm {\, \cdot \,} }$ to be unital. 
From Existence of Maximal Commutative Subalgebra: Unital, there exists a maximal commutative unital subalgebra $B$ containing $x$ and $y$. 
From Spectrum of Element in Maximal Commutative Subalgebra of Unital Banach Algebra we have $\map {\sigma_B} x = \map {\sigma_A} x$.
Hence by the definition of spectral radius, we have $\map {r_B} x = \map {r_A} x$. 
Let $\Phi_B$ be the spectrum of $B$.
Then we have, from Spectral Radius of Element of Commutative Banach Algebra:
:$\ds \map {r_B} {x + y} = \sup_{\phi \in \Phi_B} \cmod {\map \phi {x + y} }$
By linearity we have:
:$\map \phi {x + y} = \map \phi x + \map \phi y$ 
Hence by the Triangle Inequality for Complex Numbers, we have:
:$\ds \cmod {\map \phi {x + y} } \le \cmod {\map \phi x} + \cmod {\map \phi y} \le \sup_{\phi \in \Phi_B} \cmod {\map \phi x} + \sup_{\phi \in \Phi_B} \cmod {\map \phi y}$
Taking supremums, we have:
:$\ds \sup_{\phi \in \Phi_B} \cmod {\map \phi {x + y} } \le \sup_{\phi \in \Phi_B} \cmod {\map \phi x} + \sup_{\phi \in \Phi_B} \cmod {\map \phi y}$
Hence:
:$\map {r_B} {x + y} \le \map {r_B} x + \map {r_B} y$ 
from Spectral Radius of Element of Commutative Banach Algebra.
Since again $\map {\sigma_B} x = \map {\sigma_A} x$ we deduce:
:$\map {r_A} {x + y} \le \map {r_A} x + \map {r_A} y$ 
Now suppose that $\struct {A, \norm {\, \cdot \,} }$ is not necessarily unital. 
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $A$.
By Unitization of Normed Algebra is Banach Algebra iff Original Algebra is Banach Algebra, $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital Banach algebra.
We then have:
:$\map {r_{A_+} } {\tuple {x + y, 0} } \le \map {r_A} {\tuple {x, 0} } + \map {r_A} {\tuple {y, 0} }$
From the definition of the spectrum, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
:$\map {\sigma_A} y = \map {\sigma_{A_+} } {\tuple {y, 0} }$
:$\map {\sigma_A} {x + y} = \map {\sigma_{A_+} } {\tuple {x + y, 0} }$
Hence:
:$\map {r_{A_+} } {\tuple {x + y, 0} } = \map {r_A} {x + y}$
:$\map {r_{A_+} } {\tuple {x, 0} } = \map {r_A} x$
:$\map {r_{A_+} } {\tuple {y, 0} } = \map {r_A} y$
Hence we obtain:
:$\map {r_{A_+} } {\tuple {x + y, 0} } \le \map {r_A} {\tuple {x, 0} } + \map {r_A} {\tuple {y, 0} }$
for $\struct {A, \norm {\, \cdot \,} }$ not necessarily unital.
{{qed}}
Category:Spectral Radius
Category:Banach Algebras
fm59mckscftij7scdr0r6qgclz4q4al
Theorem: Spectral Radius is Multiplicative on Commuting Elements in Banach Algebra

Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x, y \in A$ be commuting elements of $A$.
Let $r_A$ denote spectral radius in $A$.
Then we have:
:$\map {r_A} {x y} \le \map {r_A} x \map {r_A} y$

Proof:
First take $\struct {A, \norm {\, \cdot \,} }$ to be unital. 
From Existence of Maximal Commutative Subalgebra, there exists a maximal commutative subalgebra $B$ containing $x$ and $y$. 
From Maximal Subalgebra in Normed Algebra is Closed, $B$ is closed.
From Spectrum of Element in Maximal Commutative Subalgebra of Unital Banach Algebra we have $\map {\sigma_B} x = \map {\sigma_A} x$.
Hence by the definition of spectral radius, we have $\map {r_B} x = \map {r_A} x$. 
Let $\Phi_B$ be the spectrum of $B$.
Then we have, from Spectral Radius of Element of Commutative Banach Algebra:
:$\ds \map {r_B} {x y} = \sup_{\phi \in \Phi_B} \cmod {\map \phi {x y} }$
Since $\phi$ is a character we have:
:$\map \phi {x + y} = \map \phi x \map \phi y$ 
Hence by the Triangle Inequality for Complex Numbers, we have:
:$\ds \cmod {\map \phi {x y} } \le \cmod {\map \phi x} \cmod {\map \phi y} \le \sup_{\phi \in \Phi_B} \cmod {\map \phi x} \sup_{\phi \in \Phi_B} \cmod {\map \phi y}$
Taking supremums, we have:
:$\ds \sup_{\phi \in \Phi_B} \cmod {\map \phi {x y} } \le \sup_{\phi \in \Phi_B} \cmod {\map \phi x} \sup_{\phi \in \Phi_B} \cmod {\map \phi y}$
Hence:
:$\map {r_B} {x y} \le \map {r_B} x \map {r_B} y$ 
from Spectral Radius of Element of Commutative Banach Algebra.
Since again $\map {\sigma_B} x = \map {\sigma_A} x$ we deduce:
:$\map {r_A} {x y} \le \map {r_A} x \map {r_A} y$ 
Now suppose that $\struct {A, \norm {\, \cdot \,} }$ is not necessarily unital. 
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $A$.
By Unitization of Normed Algebra is Banach Algebra iff Original Algebra is Banach Algebra, $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital Banach algebra.
We then have:
:$\map {r_{A_+} } {\tuple {x y, 0} } \le \map {r_A} {\tuple {x, 0} } + \map {r_A} {\tuple {y, 0} }$
From the definition of the spectrum, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
:$\map {\sigma_A} y = \map {\sigma_{A_+} } {\tuple {y, 0} }$
:$\map {\sigma_A} {x y} = \map {\sigma_{A_+} } {\tuple {x y, 0} }$
Hence:
:$\map {r_{A_+} } {\tuple {x y, 0} } = \map {r_A} {x y}$
:$\map {r_{A_+} } {\tuple {x, 0} } = \map {r_A} x$
:$\map {r_{A_+} } {\tuple {y, 0} } = \map {r_A} y$
Hence we obtain:
:$\map {r_{A_+} } {\tuple {x y, 0} } \le \map {r_A} {\tuple {x, 0} } + \map {r_A} {\tuple {y, 0} }$
for $\struct {A, \norm {\, \cdot \,} }$ not necessarily unital.
{{qed}}
Category:Spectral Radius
Category:Banach Algebras
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Theorem: Character on Unital C*-Algebra is Real at Hermitian Elements

Let $\tuple {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\phi$ be a character on $A$.
Let $x \in A$ be Hermitian.
Then:
:$\map \phi x \in \R$

Proof:
Write:
:$\map \phi x = \alpha + i \beta$
with $\alpha, \beta \in \R$. 
Let:
:$x_t = x + i t {\mathbf 1}_A$ 
for each $t \in \R$. 
From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have $\map \phi { {\mathbf 1}_A} = 1$, and so:
:$\map \phi {x_t} = \map \phi x + i t = \alpha + i \paren {\beta + t}$ 
Hence, we have:
:$\cmod {\map \phi {x_t} }^2 = \alpha^2 + \paren {\beta + t}^2$
We now have: 
{{begin-eqn}}
{{eqn	| l = \cmod {\map \phi {x_t} }^2
		| o = \le
		| r = \norm {x_t}^2
		| c = Character on Banach Algebra is Continuous
}}
{{eqn	| r = \norm {x_t x_t^\ast}
		| c = $(\text C^\ast 5)$ in the definition of a involution
}}
{{eqn	| r = \norm {\paren {x + i t {\mathbf 1}_A} \paren {x - i t {\mathbf 1}_A} }
		| c = $(\text C^\ast 2)$, $(\text C^\ast 4)$ in the definition of a involution
}}
{{eqn	| r = \norm {x^2 + t^2 {\mathbf 1}_A}
}}
{{eqn	| o = \le
		| r = \norm {x^2} + t^2 
		| c = {{NormAxiomVector|3}}
}}
{{eqn	| o = \le
		| r = \norm x^2 + t^2
		| c = Bound on Norm of Power of Element in Normed Algebra
}}
{{end-eqn}}
So we have:
:$\alpha^2 + \beta^2 + 2 \beta t + t^2 \le \norm x^2 + t^2$
so that:
:$\alpha^2 + \beta^2 + 2 \beta t \le \norm x^2$
for each $t \in \R$. 
{{AimForCont}} that $\beta \ne 0$.
Suppose that $\beta > 0$.
Then taking:
:$\ds t > \frac 1 {2 \beta} \norm x^2$
we have:
:$\alpha^2 + \beta^2 + 2 \beta t > \alpha^2 + \beta^2 + \norm x^2 \ge \norm x^2$
contradicting that:
:$\alpha^2 + \beta^2 + 2 \beta t \le \norm x^2$ for each $t \in \R$. 
Now suppose that $\beta < 0$.
Take:
:$\ds t < -\frac 1 {2 \beta} \norm x^2$
so that:
:$\alpha^2 + \beta^2 + 2 \beta t > \alpha^2 + \beta^2 + \norm x^2 \ge \norm x^2$
similarly. 
We again get a contradiction to the fact that:
:$\alpha^2 + \beta^2 + 2 \beta t \le \norm x^2$ for each $t \in \R$. 
Hence we have a contradiction to our assumption that $\beta \ne 0$. 
Hence we have $\beta = 0$. 
Hence:
:$\map \phi x = \alpha \in \R$
{{qed}}
Category:Characters (Banach Algebras)
Category:C*-Algebras
Category:Hermitian Elements of *-Algebras
a3fn4niqxd6v3smrb63qw4ei2tsjv1i
Theorem: Character on C*-Algebra is *-Algebra Homomorphism

Let $\tuple {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\phi$ be a character on $A$.
Then $\phi$ is a $\ast$-algebra homomorphism.

Proof:
Let $x \in A$.
From Element of *-Algebra Uniquely Decomposes into Hermitian Elements, there exists Hermitian elements $a, b \in A$ such that:
:$x = a + i b$
We have:
{{begin-eqn}}
{{eqn	| l = \map \phi {x^\ast}
		| r = \map \phi {a^\ast - i b^\ast}
		| c = $(\text C^\ast 2)$, $(\text C^\ast 4)$ in the definition of an involution
}}
{{eqn	| r = \map \phi a - i \map \phi b
		| c = $a, b$ are Hermitian and $\phi$ is linear
}}
{{eqn	| r = \overline {\map \phi a + i \map \phi b}
		| c = Character on Unital C*-Algebra is Real at Hermitian Elements
}}
{{eqn	| r = \overline {\map \phi {a + i b} }
		| c = $\phi$ is linear
}}
{{eqn	| r = \overline {\map \phi x}
}}
{{end-eqn}}
{{qed}}
Category:Characters (Banach Algebras)
Category:C*-Algebras
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Theorem: Spectral Radius of Normal Element of C*-Algebra Equal to Norm

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $r_A$ be the spectral radius in $A$.
Let $x \in A$ be normal.
Then $\map {r_A} x = \norm x$.

Proof:
We first take $x$ Hermitian.
From the $\text C^\ast$ identity, we have:
:$\norm {x^2} = \norm {x x^\ast} = \norm x^2$
for each $x \in A$.