nyuuzyou/wikis huggingface dataset:
Change of Basis Matrix under Linear Transformation/Converse/Corollary

== Corollary to Change of Basis Matrix under Linear Transformation: Converse ==
Let $R$ be a commutative ring with unity.
Let $G$ be a free unitary $R$-module of finite dimensions $n$.
Let $\sequence {a_n}$ be an ordered basis of $G$.
Let $\mathbf A$ and $\mathbf B$ be square matrices of order $n$ over $R$.
Let there exist an nonsingular matrix $\mathbf P$ of order $n$ such that:
:$\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$
Then there exist:
:a linear operator $u$ on $G$
:an ordered basis $\sequence { {a_n}'}$ of $G$
such that:
:$\mathbf A = \sqbrk {u; \sequence {a_n} }$
:$\mathbf B = \sqbrk {u; \sequence { {a_n}'} }$
where $\sqbrk {u; \sequence {a_n} }$ denotes the matrix of $u$ relative to $\sequence {a_n}$.
== Proof ==
Follows directly from Change of Basis Matrix under Linear Transformation: Converse.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.5$: Corollary
Category:Change of Basis Matrix under Linear Transformation
Change of Basis Matrix under Linear Transformation/Corollary

== Corollary to Change of Basis Matrix under Linear Transformation ==
Let $R$ be a commutative ring with unity.
Let $G$ be a free unitary $R$-module of finite dimension $n$.
Let $\sequence {a_n}$ and $\sequence { {a_n}'}$ be ordered bases of $G$.
Let $u: G \to G$ be a linear operator on $G$.
Let $\sqbrk {u; \sequence {a_n} }$ denote the matrix of $u$ relative to $\sequence {a_n}$.
Let:
:$\mathbf A = \sqbrk {u; \sequence {a_n} }$
:$\mathbf B = \sqbrk {u; \sequence { {a_n}'} }$
Then:
:$\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$
where $\mathbf P$ is the matrix corresponding to the change of basis from $\sequence {a_n}$ to $\sequence { {a_n}'}$.
== Proof ==
This is an instance of Change of Basis Matrix under Linear Transformation.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.4$: Corollary
Category:Change of Basis Matrix under Linear Transformation
Change of Coordinate Vector Under Change of Basis

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n>0$.
Let $\BB$ and $\CC$ be bases of $M$.
Let $\mathbf M_{\BB, \CC}$ be the change of basis matrix from $\BB$ to $\CC$.
Let $m \in M$.
Let $\sqbrk m_\BB$ and $\sqbrk m_\CC$ be its coordinate vectors relative to $\BB$ and $\CC$ respectively.
Then $\sqbrk m_\BB = \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC$.
== Proof ==
Intuitively, when we consider $\BB$ and $\CC$ as row vectors, this is because:
:$\CC = \BB \cdot \mathbf M_{\BB, \CC}$ and:
:$\BB \cdot \sqbrk m_\BB = \CC \cdot \sqbrk m_\CC$ imply:
:$\BB \cdot \sqbrk m_\BB = \BB \cdot \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC$.
Because $\BB$ is a basis, this implies $\sqbrk m_\BB = \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC$.
This can be formalized by giving $R \times M$ the structure of a ring. Alternatively, this can be verified directly, which boils down to re-proving that that matrix multiplication is associative.
== Also see ==
* Change of Coordinate Vectors Under Linear Mapping, an analogous result for linear transformations, of which this is a special case
* Change of Basis Matrix under Linear Transformation
Category:Change of Basis
Change of Coordinate Vectors Under Linear Transformation

== Theorem ==
Let $R$ be a ring with unity.
Let $M, N$ be free $R$-modules of finite dimension $m, n > 0$ respectively.
Let $\AA, \BB$ be ordered bases of $M$ and $N$ respectively.
Let $f: M \to N$ be a linear transformation.
Let $\mathbf M_{f, \BB, \AA}$ be its matrix relative to $\AA$ and $\BB$.
Then for all $m \in M$:
:$\sqbrk {\map f m}_\BB = \mathbf M_{f, \BB, \AA} \cdot \sqbrk m_\AA$
where $\sqbrk {\, \cdot \,}_-$ denotes the coordinate vector with respect to a basis.
== Proof ==
Both sides are linear in $m$ and they coincide on the elements of $\AA$ by definition of $\mathbf M_{f, \BB, \AA}$.
So they are equal for all $m \in M$.
== Also see ==
* Change of Coordinate Vector Under Change of Basis, an analogous result for change of basis
Category:Linear Transformations
Change of Index Variable of Product

== Theorem ==
:$\ds \prod_{\map R i} a_i = \prod_{\map R j} a_j$
where $\ds \prod_{\map R i} a_i$ denotes the product over $a_i$ for all $i$ that satisfy the propositional function $\map R i$.
== Proof ==
Let $S = \set {i \in \Z: \map R i}$.
Let $T = \set {j \in \Z: \map R j}$.
Let $i \in S$.
Then $\map R i$.
Let $j = i$.
By Leibniz's Law, $\map R j$.
Thus $i \in T$.
By definition of subset:
:$S \subseteq T$
Similarly, let $j \in T$.
Then $\map R j$.
Let $i = j$.
By Leibniz's Law, $\map R i$.
Thus $j \in S$.
By definition of subset:
:$T \subseteq S$
Thus by definition of set equality:
:$S = T$
Thus:
== Sources ==
* : $\S 1.2.3$: Sums and Products
Category:Continued Products
Change of Index Variable of Summation

== Theorem ==
:$\ds \sum_{\map R i} a_i = \sum_{\map R j} a_j$
where $\ds \sum_{\map R i} a_i$ denotes the summation over $a_i$ for all $i$ that satisfy the propositional function $\map R i$.
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: $(5)$
Category:Summations
Change of Index Variable of Supremum

== Theorem ==
Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\map R i$ be a propositional functions of $i \in I$.
Let $\ds \sup_{\map R i} a_i$ be the indexed supremum on $\family {a_i}$.
Then:
:$\ds \sup_{\map R i} a_i = \sup_{\map R j} a_j$
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $35$
Category:Suprema
Change of Lead in Prime Number Race 4n+1 vs. 4n-1

== Theorem ==
Consider the prime number race between $4 n + 1$ and $4 n - 1$.
While the prime numbers of the form $4 n - 1$ appear usually to be in the majority, the lead changes from one to the other an infinite number of times.
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $26,861$
* : $26,861$
Category:Prime Number Races
Change of Measures Formula for Integrals

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$ such that: 
:$\nu$ is absolutely continuous with respect to $\mu$.
Let $g$ be a Radon-Nikodym derivative of $\nu$ with respect to $\mu$.
Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.
Then: 
:$\ds \int f \rd \nu = \int \paren {f \cdot g} \rd \mu$
where:
:$f \cdot g$ is the pointwise product of $f$ and $g$
:$\ds \int \cdot \rd \nu$ denotes the integral of a positive $\Sigma$-measurable function with respect to $\nu$.
=== Corollary ===
== Proof ==
First consider $f$ a positive simple function. 
From Simple Function has Standard Representation, there exists:
:a finite sequence $a_0, \ldots, a_n$ of non-negative real numbers
:a partition $E_0, E_1, \ldots, E_n$ of $X$ into $\Sigma$-measurable sets
such that: 
:$\ds f = \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then:
Now suppose that $f$ is a general positive $\Sigma$-measurable function.
From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for each $x \in X$.
For each $n \in \N$, we have:
:$\ds \int f_n \rd \nu = \int \paren {f_n \cdot g} \rd \mu$
From the Monotone Convergence Theorem (Measure Theory), we have: 
:$\ds \lim_{n \mathop \to \infty} \int f_n \rd \nu = \int f \rd \nu$
Note that $\sequence {f_n \cdot g}_{n \mathop \in \N}$ is an increasing sequence with $f_n \cdot g \to f \cdot g$.
So, from the Monotone Convergence Theorem (Measure Theory), we have:
:$\ds \lim_{n \mathop \to \infty} \int \paren {f_n \cdot g} \rd \mu = \int \paren {f \cdot g} \rd \mu$
So, we have: 
:$\ds \int f \rd \nu = \int \paren {f \cdot g} \rd \mu$
Category:Radon-Nikodym Derivatives
Category:Change of Measures Formula for Integrals
Change of Measures Formula for Integrals/Corollary

== Corollary ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$ such that: 
:$\nu$ is absolutely continuous with respect to $\mu$.
Let $g$ be a Radon-Nikodym derivative of $\nu$ with respect to $\mu$.
Let $f : X \to \overline \R$ be a $\nu$-integrable function.
Then $f \cdot g$ is $\mu$-integrable with: 
:$\ds \int f \rd \nu = \int \paren {f \cdot g} \rd \mu$
where:
:$f \cdot g$ is the pointwise product of $f$ and $g$
:$\ds \int \cdot \rd \nu$ denotes the integral of a $\nu$-integrable function with respect to $\nu$.
== Proof ==
From Pointwise Product of Measurable Functions is Measurable, we have: 
:$f \cdot g$ is $\Sigma$-measurable.
We show that: 
:$\ds \int \paren {f \cdot g}^+ \rd \mu < \infty$
and:
 
:$\ds \int \paren {f \cdot g}^- \rd \mu < \infty$ 
where $\paren {f \cdot g}^+$ and $\paren {f \cdot g}^-$ denote the positive part and negative part of $f \cdot g$ respectively. 
We have:
and: 
So $f \cdot g$ is $\mu$-integrable.
Then: 
Category:Change of Measures Formula for Integrals
Change of Variables Theorem (Multivariable Calculus)

== Theorem ==
Let $\mathbf F: U \to V$ be a diffeomorphism between open subsets of $\R^2$.
Let $D^* \subset U$ and $D = \map {\mathbf F} {D^*} \subset V$ be bounded subsets.
Let $f: D \to \R$ be a bounded function.
Then
:$\ds \iint_D \map f {x, y} \rd x \rd y = \iint_{D^*} \map f {\map {\mathbf F} {u, v} } \size {\map {\det D \mathbf F} {u, v} } \rd u \rd v$
where $\map {\det D \mathbf F} {u, v}$ is the Jacobian.
The equality means that the left-hand integral exists if and only if the right-hand integral does and that, if so, the two integrals are equal.
== Proof ==
== Sources ==
* : $\S$ 1.1032: Change of Variable in an Integral
* 
Category:Integral Calculus
Change of Variables in Indexed Summation

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a, b, c, d$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f: \closedint a b \to \mathbb A$ be a mapping.
Let $g: \closedint c d \to \closedint a b$ be a bijection.
Then we have an equality of indexed summations:
:$\ds \sum_{i \mathop = a}^b \map f i = \sum_{i \mathop = c}^d \map f {\map g i}$
== Outline of Proof ==
We use Indexed Summation over Translated Interval to translate $\closedint c d$ to $\closedint a b$.
This reduces the problem to Indexed Summation does not Change under Permutation.
== Proof ==
Because $g : \closedint c d \to \closedint a b$ is a bijection, these sets are equivalent.
By Cardinality of Integer Interval, $\closedint a b$ has cardinality $b - a + 1$.
Thus:
:$b - a + 1 = d - c + 1$
Thus
:$c - a = d - b$
By Indexed Summation over Translated Interval:
:$\ds \sum_{i \mathop = c}^d \map f {\map g i} = \sum_{i \mathop = a}^b \map f {\map g {i + c - a} }$
By Translation of Integer Interval is Bijection, the mapping $T : \closedint a b \to \closedint c d$ defined as:
:$\map T k = k + c - a$
is a bijection.
By Composite of Bijections is Bijection, $g \circ T$ is a permutation of $\closedint a b$.
We have:
== Also see ==
* Change of Variables in Summation over Finite Set
Category:Summations
Change of Variables in Summation over Finite Set

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ and $T$ be finite sets.
Let $f: S \to \mathbb A$ be a mapping.
Let $g: T \to S$ be a bijection.
Then we have an equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S} \map f s = \sum_{t \mathop \in T} \map f {\map g t}$
== Proof ==
Let $n$ be the cardinality of $S$ and $T$.
Let $\N_{<n}$ be an initial segment of the natural numbers.
Let $h : \N_{<n} \to T$ be a bijection.
By definition of summation:
:$\ds \sum_{t \mathop \in T} \map f {\map g t} = \sum_{i \mathop = 0}^{n - 1} \map f {\map g {\map h i} }$
By Composite of Bijections is Bijection, the composition $g \circ h : \N_{<n} \to S$ is a bijection.
By definition of summation:
:$\ds \sum_{s \mathop\in S} \map f s = \ds \sum_{i \mathop = 0}^{n - 1} \map f {\map g {\map h i} }$
== Also see ==
* Change of Variables in Indexed Summation
* Finite Summation does not Change under Permutation
Category:Summations
Chapman-Kolmogorov Equation

== Theorem ==
Let $X$ be a homogeneous Markov chain with $n$-step transition probability matrix:
:$\mathbf P^{\paren n} = \sqbrk { {p_{j k} }^{\paren n} }_{j, k \mathop \in S}$
where:
:${p_{j k} }^{\paren n} = \condprob {X_n = k} {X_0 = j} $ is the $n$-step transition probability.
Then:
:$\mathbf P^{\paren {n + m} } = \mathbf P^{\paren n} \mathbf P^{\paren m}$
or equivalently:
:$\ds {p_{i j} }^{\paren {n + m} } = \sum_{k \mathop \in S} {p_{i k} }^{\paren n} {p_{k j} }^{\paren m}$
== Proof ==
We consider the conditional probability on the :
Category:Markov Chains
Character of Representations over C are Algebraic Integers

== Theorem ==
Let $G$ be a finite group.
Let $\chi$ be the character of any $\C \left[{G}\right]$-module $\left({V, \rho}\right)$.
Then for all $g \in G$, it follows that $\chi \left({g}\right)$ is an algebraic integer.
== Proof ==
By the definition of character:
:$\chi \left({g}\right) = \operatorname{Tr} \left({\rho_g}\right)$
where:
: $\rho \in \hom \left({\C \left[{G}\right], \operatorname{Aut} \left({V}\right)}\right): \vec {e_g} \mapsto \rho_g$
by definition.
Fix an arbitrary $g \in G$.
Let $\left\vert{g}\right\vert$ denote the order of $g$.
The trace $\operatorname{Tr} \left({\rho_g}\right)$ of $\rho_g$ is defined as the sum of the eigenvalues of $\rho_g$.
From Eigenvalues of G-Representation are Roots of Unity, we have that any eigenvalue $\lambda$ of $\rho_g$ is a root of unity whose order is $\left\vert{g}\right\vert$.
We have that $\lambda$ satisfies the monic polynomial $x^{\left\vert{g}\right\vert} - 1$
Hence we have that $\lambda$ is an algebraic integer.
From Ring of Algebraic Integers, we have that the sum of the eigenvalues is also an algebraic integer.
Thus $\chi \left({g}\right)$ is an algebraic integer.
Category:Module Theory
Category:Group Theory
Category:Complex Analysis
Character on Banach Algebra is Continuous

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.
Let $\phi : A \to \C$ be a character on $A$.
Then $\phi$ is continuous and further:
:$\norm \phi_{A^\ast} \le 1$
=== Corollary ===
== Proof ==
Let $\map G A$ be the group of units of $A$.
Suppose first that $\struct {A, \norm {\, \cdot \,} }$ is unital.
We show that:
:$\cmod {\map \phi x} \le \norm x$ for each $x \in A$.
From Continuity of Linear Functionals, we will then have that $\phi$ is continuous with $\norm \phi_{A^\ast} \le 1$.
 that $x \in A$ is such that:
:$\cmod {\map \phi x} > \norm x$ 
Since then $\map \phi x \ne 0$, we have $x \ne {\mathbf 0}_A$.
From , we have:
:$\ds \norm {\frac x {\map \phi x} } < 1$
From Element of Unital Banach Algebra Close to Identity is Invertible, we have that:
:$\ds {\mathbf 1}_A - \frac x {\map \phi x} \in \map G A$
So there exists $z \in A$ such that:
:$\ds z \paren { {\mathbf 1}_A - \frac x {\map \phi x} } = {\mathbf 1}_A$
From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have $\map \phi { {\mathbf 1}_A} = 1$. 
So, we have:
This is a contradiction since $A \ne \set { {\mathbf 0}_A}$.
Hence we have:
:$\norm {\map \phi x} \le \norm x$ for all $x \in A$.
So we have that $\phi$ is continuous with:
:$\norm \phi_{A^\ast} \le 1$ 
Now suppose that $\struct {A, \norm {\, \cdot \,} }$ is not unital.
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.
Define $\phi_+ : A_+ \to \C$ by: 
:$\map {\phi_+} {\tuple {x, \lambda} } = \map \phi x + \lambda$
for each $\tuple {x, \lambda} \in A_+$.
From Character on Non-Unital Banach Algebra induces Character on Unitization, $\phi_+$ is a character on $A_+$.
Further:
:$\map {\phi_+} {\tuple {x, 0} } = \map \phi x$ for each $x \in A$.
From the unital case, we have that $\phi_+$ is continuous with $\norm {\phi_+}_{A_+^\ast} \le 1$.
Then, we have:
So $\phi$ is continuous with $\norm \phi_{A^\ast} \le 1$.
Category:Characters (Banach Algebras)
Category:Character on Banach Algebra is Continuous
Character on Banach Algebra is Continuous/Corollary

== Corollary ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.
Let $\phi : A \to \C$ be a character on $A$.
Then:
:$\norm \phi_{A^\ast} = 1$
== Proof ==
From Character on Banach Algebra is Continuous, we have that $\phi$ is continuous with $\norm \phi_{A^\ast} \le 1$. 
So, we have:
:$\ds \sup_{x \in A, \, \norm x = 1} \cmod {\map \phi x} \le 1$
We have that $\norm { {\mathbf 1}_A} = 1$ and $\map \phi { {\mathbf 1}_A} = 1$, so we have:
:$\ds \sup_{x \in A, \, \norm x = 1} \cmod {\map \phi x} = 1$
So $\norm \phi_{A^\ast} = 1$.
Category:Character on Banach Algebra is Continuous
Character on Banach Algebra is Surjective

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $\phi : A \to \C$ be a character on $A$. 
Then $\phi$ is surjective.
== Proof 1 ==
== Proof 2 ==
Category:Characters (Banach Algebras)
Category:Character on Banach Algebra is Surjective
Character on Banach Algebra is Surjective/Proof 1

== Theorem ==
== Proof ==
From Image of Submodule under Linear Transformation is Submodule, $\phi \sqbrk A$ is a vector subspace of $\C$. 
From Dimension of Proper Subspace is Less Than its Superspace, we have:
:$\dim \phi \sqbrk A \le \dim \C = 1$
and so we either have $\phi \sqbrk A = \set 0$ or $\phi \sqbrk A = \C$.
Since $\phi \ne 0$ by the definition of a character, we have $\phi \sqbrk A = \C$.
Category:Character on Banach Algebra is Surjective
Character on Banach Algebra is Surjective/Proof 2

== Theorem ==
== Proof ==
As $\phi$ is non-zero, there exists an $x_0 \in A$ such that:
:$\map \phi {x_0} \in \C \setminus \set 0$
Thus, for each $a \in \C$:
and:
Category:Character on Banach Algebra is Surjective
Character on C*-Algebra is *-Algebra Homomorphism

== Theorem ==
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:
Category:Characters (Banach Algebras)
Category:C*-Algebras
Character on Non-Unital Banach Algebra induces Character on Unitization

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$ that is not unital.
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.
Let $\phi$ be a character on $A$.
Define: 
:$\map {\phi_+} {\tuple {x, \lambda} } = \map \phi x + \lambda$
for each $\tuple {x, \lambda} \in A_+$. 
Then $\phi_+$ is a character on $A_+$.
== Proof ==
Let $\tuple {x, \lambda}, \tuple {y, \mu} \in A_+$ and $t \in \C$.
We have: 
and so $\phi_+$ is linear.
To show that $\phi_+$ is a character, it remains to show that:
:$\map {\phi_+} {\tuple {x, \lambda} \tuple {y, \mu} } = \map {\phi_+} {\tuple {x, \lambda} } \map {\phi_+} {\tuple {y, \lambda} }$
We have:
So $\phi_+$ is a character.
Category:Characters (Banach Algebras)
Character on Unital Banach Algebra is Uniquely Identified by Kernel

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$. 
Let $\phi, \psi : A \to \C$ be characters on $A$ such that:
:$\ker \phi = \ker \psi$
Then $\phi = \psi$.
== Proof ==
From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have $\map \phi { {\mathbf 1}_A} = 1$ and $\map \psi { {\mathbf 1}_A} = 1$. 
Let $x \in A$.
We have:
So we have $x - \map \phi x {\mathbf 1}_A \in \ker \phi$.
Since $\ker \phi = \ker \psi$, we have $x - \map \phi x {\mathbf 1}_A \in \ker \psi$.
So, we have:
:$\map \psi {x - \map \phi x {\mathbf 1}_A} = 0$
We therefore obtain that:
So we have $\map \phi x = \map \psi x$.
Since $x \in A$ was arbitrary, we obtain $\phi = \psi$.
Category:Characters (Banach Algebras)
Character on Unital Banach Algebra is Unital Algebra Homomorphism

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.
Let $\phi : A \to \C$ be a character on $A$. 
Then $\phi$ is a unital algebra homomorphism.
== Proof ==
By the definition of a character, $\phi$ is a non-zero algebra homomorphism.
We only need to verify that:
:$\map \phi { {\mathbf 1}_A} = 1$
We have:
:$\map \phi { {\mathbf 1}_A} = \map \phi { {\mathbf 1}_A^2} = \paren {\map \phi { {\mathbf 1}_A} }^2$
So, we have:
:$\map \phi { {\mathbf 1}_A} \in \set {0, 1}$
Note that for all $x \in A$, we have:
:$\map \phi x = \map \phi { {\mathbf 1}_A} \map \phi x$
Hence if we had $\map \phi { {\mathbf 1}_A} = 0$, we would have $\phi = 0$.
Since $\phi \ne 0$, we must therefore have $\map \phi { {\mathbf 1}_A} = 1$.
So $\phi$ is a unital algebra homomorphism.
== Sources ==
* : $4.10$: The continuity of characters
Category:Characters (Banach Algebras)
Character on Unital C*-Algebra has Modulus One at Unitary Elements

== Theorem ==
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 unitary.
Then:
:$\cmod {\map \phi x} = 1$
== Proof ==
Let ${\mathbf 1}_A$ be the identity element of $A$. 
Since $x$ is unitary, we have:
:$x x^\ast = {\mathbf 1}_A$
Hence:
:$\map \phi {x x^\ast} = \map \phi { {\mathbf 1}_A}$
From Character on C*-Algebra is *-Algebra Homomorphism, we have:
:$\map \phi {x x^\ast} = \map \phi x \map \phi {x^\ast} = \map \phi x \overline {\map \phi x}$
From Product of Complex Number with Conjugate, we therefore have:
:$\map \phi {x x^\ast} = \cmod {\map \phi x}^2$ 
From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have:
:$\map \phi { {\mathbf 1}_A} = 1$ 
So we have:
:$\cmod {\map \phi x}^2 = 1$
giving:
:$\cmod {\map \phi x} = 1$ 
Category:Characters (Banach Algebras)
Category:C*-Algebras
Category:Unitary Elements of Unital *-Algebras
Character on Unital C*-Algebra is Real at Hermitian Elements

== Theorem ==
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: 
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 $t \in \R$. 
 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:
:$\map \phi x = \alpha \in \R$
Category:Characters (Banach Algebras)
Category:C*-Algebras
Category:Hermitian Elements of *-Algebras
Characterisation of Cauchy Sequence in Non-Archimedean Norm

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with non-Archimedean norm $\norm {\, \cdot \,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Then:
:$\sequence {x_n}$ is a Cauchy sequence
:
:$\ds \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n} = 0$
=== Corollary ===
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : $\S 3.2$: Completions: Lemma $3.2.2$
Category:Normed Division Rings
Category:Cauchy Sequences
Category:Non-Archimedean Norms
Category:Cauchy Sequences in Normed Division Rings
Category:Characterisation of Cauchy Sequence in Non-Archimedean Norm