nyuuzyou/wikis huggingface dataset:
Set of Integer Multiples/Examples/Set of Even Numbers

== Example of Set of Integer Multiples ==
The set of even integers is defined as:
:$2 \Z := \set {x \in \Z: 2 \divides x}$
That is:
:$2 \Z := \set {\ldots, -6, -4, -2, 0, 2, 4, 6, \ldots}$
== Sources ==
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 4$
Category:Examples of Sets of Integer Multiples
Set of Integer Multiples is Integral Ideal

== Theorem ==
Let $m \in \Z$ be an integer.
Let $m \Z$ denote the set of integer multiples of $m$.
Then $m \Z$ is an integral ideal.
== Proof ==
First note that $m \times 0 \in m \Z$ whatever $m$ may be.
Thus $m \Z \ne \O$.
Let $a, b \in m \Z$.
Then:
and:
Let $r \in \Z$.
Then:
Thus the conditions for $m \Z$ to be an integral ideal are fulfilled.
Hence the result.
== Sources ==
* : $\text {2-1}$ Euclid's Division Lemma: Exercise $4$
Category:Integral Ideals
Category:Sets of Integer Multiples
Set of Integer Multiples of GCD

== Theorem ==
Let $m, n \in \Z$.
Let $m \Z$ denote the set of integer multiples of $m$
Then:
:$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$
where $\gcd$ denotes greatest common divisor.
== Proof ==
Let $x \in m \Z \cup n \Z$.
Then either:
:$m \divides x$
or:
:$n \divides x$
In both cases:
:$\gcd \set {m, n} \divides x$
and so:
:$x \in \gcd \set {m, n} \Z$
Hence by definition of subset:
:$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $11$
Category:Sets of Integer Multiples
Category:Greatest Common Divisor
Set of Integers Bounded Above by Integer has Greatest Element

== Theorem ==
Let $\Z$ be the set of integers.
Let $\le$ be the ordering on the integers.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded above in $\struct {\Z, \le}$.
Then $S$ has a greatest element.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Set of Integers Bounded Above has Greatest Element
* Set of Integers Bounded Below by Integer has Smallest Element
* Well-Ordering Principle
== Sources ==
* : Introduction $\S 5$: The system of integers: Exercise $1$
* : Exercise $2.3: \ 4$
* : $\S 10.2$: The well-ordering principle
Category:Number Theory
Category:Integers
Category:Set of Integers Bounded Above has Greatest Element
Set of Integers Bounded Above by Integer has Greatest Element/Proof 1

== Theorem ==
== Proof ==
$S$ is bounded above, so $\exists M \in \Z: \forall s \in S: s \le M$.
Hence $\forall s \in S: 0 \le M - s$.
Thus the set $T = \set {M - s: s \in S} \subseteq \N$.
The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$.
Hence:
:$\paren {\forall s \in S: b_T \le M - s} \land \paren {\exists g_S \in S: b_T = M - g_S}$
So:
		 
So $g_S$ is the greatest element of $S$.
== Sources ==
* : $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (3)$
Category:Set of Integers Bounded Above has Greatest Element
Set of Integers Bounded Above by Integer has Greatest Element/Proof 2

== Theorem ==
== Proof ==
Since $S$ is bounded above, $\exists M \in \Z: \forall s \in S: s \le M$.
Hence we can define the set $S' = \set {-s: s \in S}$.
$S'$ is bounded below by $-M$.
So from Set of Integers Bounded Below by Integer has Smallest Element, $S'$ has a smallest element, $-g_S$, say, where $\forall s \in S: -g_S \le -s$.
Therefore $g_S \in S$ (by definition of $S'$) and $\forall s \in S: s \le g_S$.
So $g_S$ is the greatest element of $S$.
Category:Set of Integers Bounded Above has Greatest Element
Set of Integers Bounded Above by Real Number has Greatest Element

== Theorem ==
Let $\Z$ be the set of integers.
Let $\le$ be the usual ordering on the real numbers $\R$.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded above in $\struct {\R, \le}$.
Then $S$ has a greatest element.
== Proof ==
Let $S$ be bounded above by $x \in \R$.
By the Axiom of Archimedes, there exists an integer $n \ge x$.
Then $S$ is bounded above by $n$.
By Set of Integers Bounded Above by Integer has Greatest Element, $S$ has a greatest element.
== Also see ==
* Set of Integers Bounded Above has Greatest Element
* Set of Integers Bounded Below by Real Number has Smallest Element
Category:Set of Integers Bounded Above has Greatest Element
Set of Integers Bounded Below by Integer has Smallest Element

== Theorem ==
Let $\Z$ be the set of integers.
Let $\le$ be the ordering on the integers.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded below in $\struct {\Z, \le}$.
Then $S$ has a smallest element.
== Proof ==
We have that $S$ is bounded below in $\Z$.
So:
: $\exists m \in \Z: \forall s \in S: m \le s$
Hence:
: $\forall s \in S: 0 \le s - m$
Thus:
:$T = \set {s - m: s \in S} \subseteq \N$
The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$.
Hence:
:$\paren {\forall s \in S: b_T \le s - m} \land \paren {\exists b_S \in S: b_T = b_S - m}$
So:
So $b_S$ is the smallest element of $S$.
== Also see ==
* Set of Integers Bounded Below has Smallest Element
* Set of Integers Bounded Above by Integer has Greatest Element
* Well-Ordering Principle
== Sources ==
* : Introduction $\S 5$: The system of integers: Exercise $1$
* : $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (3)$
* : $\S 10.1$: The well-ordering principle
Category:Set of Integers Bounded Below has Smallest Element
Set of Integers Bounded Below by Real Number has Smallest Element

== Theorem ==
Let $\Z$ be the set of integers.
Let $\le$ be the usual ordering on the real numbers $\R$.
Let $\O \subset S \subseteq \Z$ such that $S$ is bounded below in $\struct {\R, \le}$.
Then $S$ has a smallest element.
== Proof ==
Let $S$ be bounded below by $x \in \R$.
By the Axiom of Archimedes, there exists an integer $n \le x$.
Then $S$ is bounded below by $n$.
By Set of Integers Bounded Below by Integer has Smallest Element, $S$ has a smallest element.
== Also see ==
* Set of Integers Bounded Below has Smallest Element
* Set of Integers Bounded Above by Real Number has Greatest Element
Category:Set of Integers Bounded Below has Smallest Element
Set of Integers can be Well-Ordered

== Theorem ==
The set of integers $\Z$ can be well-ordered with an appropriately chosen ordering.
== Proof ==
Consider the ordering $\preccurlyeq \subseteq \Z \times \Z$ defined as:
:$x \preccurlyeq y \iff \left({\left\vert{x}\right\vert < \left\vert{y}\right\vert}\right) \lor \left({\left\vert{x}\right\vert = \left\vert{y}\right\vert \land x \le y}\right)$
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $15 \ \text{(b)}$
Category:Integers
Category:Well-Orderings
Set of Integers is not Bounded

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
The set $\Z$ of integers is not bounded in $\R$.
== Proof ==
Let $a \in \R$.
Let $K \in \R_{>0}$.
Consider the open $K$-ball $\map {B_K} a$.
By the Axiom of Archimedes there exists $n \in \N$ such that $n > a + K$.
As $\N \subseteq \Z$:
:$\exists n \in \Z: a + K < n$
and so:
:$n \notin \map {B_K} a$
As this applies whatever $a$ and $K$ are, it follows that there is no $\map {B_K} a$ which contains all the integers.
Hence the result, by definition of bounded space.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Real Analysis
Category:Boundedness
Set of Integers is not Compact

== Theorem ==
Let $\Z$ be the set of integers.
Then $\Z$ is not compact.
== Proof ==
Let $\R$ be the real number line considered as an Euclidean space.
From Set of Integers is not Bounded, $\Z$ is not bounded in $\R$.
The result follows by definition of compact.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Compact Spaces
Set of Integers is not Well-Ordered by Usual Ordering

== Theorem ==
The set of integers $\Z$ is not well-ordered under the usual ordering $\le$.
== Proof ==
 $\Z$ is a well-ordered set.
Then by definition, all subsets of $\Z$ has a smallest element.
But take $\Z$ itself.
Suppose $x \in \Z$ is a smallest element.
Then $x - 1 \in \Z$.
But $x - 1 < x$, which contradicts the supposition that $x \in \Z$ is a smallest element.
Hence there can be no such smallest element.
So by Proof by Contradiction, $\Z$ is not well-ordered by $\le$.
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $15 \ \text{(a)}$
Category:Integers
Category:Well-Orderings
Set of Integers under Addition is Isomorphic to Set of Even Integers under Addition

== Theorem ==
Let $\struct {\Z, +}$ be the algebraic structure formed by the set of integers under the operation of addition.
Let $\struct {2 \Z, +}$ be the algebraic structure formed by the set of even integers under the operation of addition.
Then $\struct {\Z, +}$ and $\struct {2 \Z, +}$ are isomorphic.
== Proof ==
Let $f: \Z \to 2 \Z$ be the mapping:
:$\forall n \in \Z: \map f n = 2 n$
From Bijection between Integers and Even Integers, $f$ is a bijection.
Let $m, n \in \Z$.
Then:
Thus $f$ is an isomorphism by definition.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.4$
Category:Integers
Category:Even Integers
Category:Additive Group of Integers
Set of Integers with GCD of 1 are not necessarily Pairwise Coprime

== Theorem ==
Let $S$ be a set of integers such that $S$ has more than $2$ elements:
:$S = \set {s_1, s_2, \ldots, s_n}$
Let:
:$\map \gcd S = 1$
where $\gcd$ denotes the GCD of $S$.
Then it is not necessarily the case that there exist a pair of elements of $S$ which are themselves pairwise coprime:
:$\exists i, j \in \set {1, 2, \ldots, n}: \gcd \set {s_i, s_j} = 1$
== Proof ==
Proof by Counterexample
Let $S = \set {6, 10, 15}$.
We have:
Hence the result.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Greatest Common Divisor
Category:Coprime Integers
Set of Intersections with Superset is Cover

== Theorem ==
Let $S$ be a set.
Let $\CC$ be a cover of $S$.
Let $T \supseteq S$ be a superset of $S$.
Then:
:$\set {C \cap T : C \in \CC}$
is a cover of $S$.
== Proof ==
Let $x \in S$ be arbitrary.
By definition of cover, there is some $C \in \CC$ such that:
:$x \in C$
By definition of superset:
:$x \in T$
Therefore, by definition of intersection:
:$x \in C \cap T$
As $x \in S$ was arbitrary, the result follows.
Category:Covers
Category:Set Intersection
Set of Inverse Positive Integers with Zero is Compact

== Theorem ==
Let $K$ be the set of inverse positive integers with zero:
:$\ds K := \set {1, \frac 1 2, \frac 1 3, \dots} \cup \set 0$
Let $\struct {\R, \size {\, \cdot \,}}$ be the normed vector space of real numbers.
Then $K$ is compact in real numbers.
== Proof ==
We have that $K \subset \closedint 0 1$.
Hence, $K$ is bounded.
Furthermore:
:$\ds \R \setminus K = \openint {-\infty} 0 \cup \paren {\bigcup_{n \mathop = 1}^\infty \openint {\frac 1 {n + 1}} {\frac 1 n}} \cup \openint 1 \infty$
By Union of Open Sets of Normed Vector Space is Open, $\R \setminus K$ is open.
By definition, $K$ is closed.
By Heine-Borel theorem, $K$ is compact.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Integers
Category:Compact Spaces
Set of Invertible Continuous Transformations is Open Subset of Continuous Linear Transformations in Supremum Operator Norm Topology

== Theorem ==
Let $X$ be a Banach space.
Let $\map {CL} X$ be the continuous linear operator space on $X$.
Let $\map {GL} X$ denote the set of all invertible continuous linear operators on $X$.
Then $\map {GL} X \subseteq \map {CL} X$ in the supremum operator norm topology.
== Proof ==
Let $T_0 \in \map {GL} X$.
By definition:
:$T_0^{-1} \in \map {CL} X$.
Let $T \in \map {B_\epsilon} {T_0}$ where $\map {B_\epsilon} x$ is an open ball in $\struct {\map {GL} X, \norm {\, \cdot \,} }$ topology.
By definition:
:$\norm {T - T_0} < \epsilon$
We also have that:
Zero operator is not invertible.
Hence, $T_0^{-1} \ne \mathbf 0$ and $\norm {T_0^{-1} } \ne 0$.
Choose $\epsilon$ such that $\epsilon \norm {T_0^{-1} } < 1$, for example, $\epsilon = \dfrac 1 {2 \norm {T_0^{-1} } }$.
Then:
By Neumann series theorem:
:$I + \paren {T - T_0} \circ T_0^{-1} \in \map {GL} X$
However, $T_0 \in \map {GL} X$ too.
Hence:
We have that $T_0 \in \map {GL} X$ and $I + \paren {T - T_0}\circ T_0^{-1} \in \map {GL} X$.
Also, Composite of Bijections is Bijection and Bijection is Invertible.
Hence, their composition is also in $\map {GL} X$:
:$T = \paren {I + \paren {T - T_0}\circ T_0^{-1} } \circ T_0 \in \map {GL} X$
$T$ was arbitrary.
So we have the implication that:
:$\forall T \in \text{topology of } \struct {\map {GL} X, \norm {\, \cdot \,} } : \paren {T \in \map {GL} X} \implies \paren {T \in \map {CL} X}$
or in other words:
:$\map {GL} X \subseteq \map {CL} X$
in the supremum operator norm topology.
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Continuous Linear Transformations
Category:Inverse Mappings
Category:Banach Spaces
Category:Topology
Set of Invertible Mappings forms Symmetric Group

== Theorem ==
Let $S$ be a set.
Let $\GG$ be the set of all invertible mappings from $S$ to $S$.
Then $\struct {\GG, \circ}$ is the symmetric group on $S$.
== Proof ==
Let $\struct {S^S, \circ}$ be the algebraic structure formed from the set of all mappings from $S$ to itself.
From Set of all Self-Maps under Composition forms Monoid, $\struct {S^S, \circ}$ is a monoid.
By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
By Bijection iff Inverse is Bijection, it follows that all the invertible elements of $S^S$ are exactly the permutations on $S$.
The result follows from Invertible Elements of Monoid form Subgroup of Cancellable Elements.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
* 
* 
Category:Symmetric Groups
Category:Inverse Mappings
Set of Isolated Points of Metric Space is Disjoint from Limit Points

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H'$ be the set of limit points of $H$.
Let $H^i$ be the set of isolated points of $H$.
Then:
:$H' \cap H^i = \O$
== Proof ==
Let $a \in H_i$.
Then by definition of isolated point:
:$\exists \epsilon \in \R_{>0}: \set {x \in H: \map d {x, a} < \epsilon} = \set a$
But by :
:$\map d {a, a} = 0$
and so:
:$\set {x \in H: 0 < \map d {x, a} < \epsilon} = \O$
So by definition $a$ is not a limit point of $H$.
That is:
:$a \notin H'$
or:
:$a \in \relcomp A {H'}$
It follows from Intersection with Complement is Empty iff Subset that:
:$H' \cap H^i = \O$
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $6$
Category:Limit Points
Category:Isolated Points
Set of Isometries in Complex Plane under Composition forms Group

== Theorem ==
Let $S$ be the set of all bijective complex functions $f: \C \to \C$ which preserve distance when embedded in the complex plane.
That is:
:$\size {\map f a - \map f b} = \size {a - b}$
where $\size z$ denotes the complex modulus of $z \in \C$.
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.
== Proof ==
From Complex Plane is Metric Space, $\C$ can be treated as a metric space.
We have that $f$ is:
:a distance-preserving mapping.
:a bijection.
Hence $f$ is an isometry on $\C$.
Taking the group axioms in turn:
=== ===
Let $f$ and $g$ be isometries on $\C$.
We have:
Thus $g \circ f$ is an isometry, and so $\struct {S, \circ}$ is closed.
=== ===
By Composition of Mappings is Associative, $\circ$ is an associative operation.
=== ===
The identity mapping is the identity element of $\struct {S, \circ}$.
=== ===
We have that $f$ is an isometry.
Hence $f$ is also a bijection.
Hence for $f \in S$, we have that $f^{-1}$ is also an isometry.
Thus every element of $f$ has an inverse $f^{-1}$.
All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{B iii}$
Category:Complex Numbers
Category:Isometries (Euclidean Geometry)
Category:Examples of Groups
Set of Isometries in Euclidean Space under Composition forms Group

== Theorem ==
Let $\struct {\R^n, d}$ be a real Euclidean space of $n$ dimensions.
Let $S$ be the set of all mappings $f: \R^n \to \R^n$ which preserve distance:
That is:
:$\map d {\map f a, \map f b} = \map d {a, b}$
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.
== Proof ==
From Euclidean Metric on Real Vector Space is Metric, $\R^n$ is a metric space.
Hence it is seen that a complex function $f: \C \to \C$ which preserves distance is in fact an isometry on $\C$.
Taking the group axioms in turn:
=== ===
Let $f$ and $g$ be isometries on $\C$.
We have:
Thus $g \circ f$ is an isometry, and so $\struct {S, \circ}$ is closed.
=== ===
By Composition of Mappings is Associative, $\circ$ is an associative operation.
=== ===
The identity mapping is the identity element of $\struct {S, \circ}$.
=== ===
By definition, an isometry is a bijection.
For $f \in S$, we have that $f^{-1}$ is also an isometry.
Thus every element of $f$ has an inverse $f^{-1}$.
All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.
== Sources ==
* : Euclidean, Affine, and Differentiable Structure on $R^n$: $\text {1-1}$: Euclidean $n$-Space, Linear $n$-Space, and Affine $n$-Space
Category:Euclidean Metric
Category:Isometries (Euclidean Geometry)
Category:Examples of Groups
Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal

== Theorem ==
Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $a_1, a_2, \dotsc, a_n$ be non-zero elements of $D$.
Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$
Then for some $x \in D$:
:$J = \ideal x$
where $\ideal x$ denotes the principal ideal generated by $x$.
== Proof ==
Let the unity of $D$ be $1_D$.
By definition of principal ideal:
:$\ds \ideal a = \set {\sum_{i \mathop = 1}^n r_i \circ a \circ s_i: n \in \N, r_i, s_i \in D}$
Let $x, y \in J$.
By definition of linear combination:
and:
Thus:
Then we have:
Thus by the Test for Ideal, $J$ is an ideal of $D$.
As $D$ is a principal ideal domain, it follows that $J$ is a principal ideal.
Thus by definition of principal ideal:
:$J = \ideal x$
for some $x \in D$.
== Sources ==
* : $\S 62.4$ Factorization in an integral domain
Category:Principal Ideal Domains
Set of Linear Subspaces is Closed under Intersection

== Theorem ==
Let $\struct {V, +, \circ}_K$ be a $K$-vector space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed family of subspaces of $V$.
Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a subspace of $V$.
== Proof ==
It needs to be demonstrated that $M$ is:
:$(1): \quad$ a closed algebraic structure under $+$
:$(2): \quad$ closed for scalar product $\circ$.
So let $a, b \in M$.
By definition of intersection, $a, b \in M_i$ for all $i \in I$.
As the $M_i$ are subspaces of $V$, $a + b \in M_i$ for all $i \in I$.
That is, by definition of intersection, $a + b \in M$.
It follows that $M$ is closed under $+$.
Now let $\lambda \in K, a \in M$.
By definition of intersection, $a \in M_i$ for all $i \in I$.
As the $M_i$ are subspaces of $V$, $\lambda \circ a \in M_i$ for all $i \in I$.
That is, by definition of intersection, $\lambda \circ a \in M$.
It follows that $M$ is closed under $\circ$.
Hence the result, by definition of subspace.
Category:Vector Subspaces
Set of Linear Transformations is Isomorphic to Matrix Space

== Theorem ==
Let $R$ be a ring with unity.
Let $F$, $G$ and $H$ be free $R$-modules of finite dimension $p, n, m > 0$ respectively.
Let $\sequence {a_p}$, $\sequence {b_n}$ and $\sequence {c_m}$ be ordered bases 
Let $\map {\LL_R} {G, H}$ denote the set of all linear transformations from $G$ to $H$.
Let $\map {\MM_R} {m, n}$ be the $m \times n$ matrix space over $R$.
Let $\sqbrk {u; \sequence {c_m}, \sequence {b_n} }$ be the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$.
Let $M: \map {\LL_R} {G, H} \to \map {\MM_R} {m, n}$ be defined as:
:$\forall u \in \map {\LL_R} {G, H}: \map M u = \sqbrk {u; \sequence {c_m}, \sequence {b_n} }$
Then $M$ is a module isomorphism.
=== Corollary ===
== Proof ==
Let $u, v \in \map {\LL_R} {G, H}$ such that:
:$\map M u = \map M v$
We have that the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$ is defined as the $m \times n$ matrix $\sqbrk \alpha_{m n}$ where:
:$\ds \forall \tuple {i, j} \in \closedint 1 m \times \closedint 1 n: \map u {b_j} = \sum_{i \mathop = 1}^m \alpha_{i j} \circ c_i$
and it is seen that $\map M u$ and $\map M v$ are the same object.
That is:
:$\map M u = \map M v \implies u = v$
and $M$ is seen to be injective.
== Motivation ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.1$
Category:Linear Transformations
Category:Matrix Algebra
Category:Set of Linear Transformations is Isomorphic to Matrix Space
Set of Linear Transformations is Isomorphic to Matrix Space/Corollary

== Corollary to Set of Linear Transformations is Isomorphic to Matrix Space ==
Let $R$ be a commutative ring with unity.
Let $M: \struct {\map {\LL_R} G, +, \circ} \to \struct {\map {\MM_R} n, +, \times}$ be defined as:
:$\forall u \in \map {\LL_R} G: \map M u = \sqbrk {u; \sequence {a_n} }$
Then $M$ is an isomorphism.
== Proof ==
Follows directly from Set of Linear Transformations is Isomorphic to Matrix Space.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.2$
Category:Set of Linear Transformations is Isomorphic to Matrix Space
Set of Linear Transformations is Isomorphic to Matrix Space/Motivation

== Motivation for Set of Linear Transformations is Isomorphic to Matrix Space ==
What Set of Linear Transformations is Isomorphic to Matrix Space tells us is two things:
# That the relative matrix of a linear transformation can be considered to be the same thing as the transformation itself
# To determine the relative matrix for the composite of two linear transformations, what you do is multiply the relative matrices of those linear transformations.
Thus one has a means of direct arithmetical manipulation of linear transformations, thereby transforming geometry into algebra.
In fact, matrix multiplication was purposely defined (some would say designed) so as to produce exactly this result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Set of Linear Transformations is Isomorphic to Matrix Space
Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings

== Theorem ==
Let $R$ be a commutative ring.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Then $\map {\LL_R} {G, H}$ is a submodule of the $R$-module $H^G$.
=== Commutative and Unitary Ring ===
== Proof ==
By definition of abelian group, the center of a commutative ring $R$ is $R$ itself.
So, we need to show that every $\phi \in \map {\LL_R} {G, H}$ fulfils the conditions to be a linear transformation:
:$(1): \quad \forall x, y \in G: \map {\paren {\lambda \circ \phi} } {x +_G y} = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y$
:$(2): \quad \forall x \in G: \forall \mu \in R: \map {\paren {\lambda \circ \phi} } {\mu \circ x} = \mu \circ \map {\paren {\lambda \circ \phi} } x$
for all $\phi \in \map {\LL_R} {G, H}$.
From Linear Transformation from Center of Scalar Ring, these conditions are indeed fulfulled.
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations