avewright/proofwiki-math huggingface dataset:
Theorem: Element is Member of Base iff Not Loop

Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $\mathscr B$ denote the set of all bases of $M$.
Let $x \in S$.
Then:
:$\exists B \in \mathscr B: x \in B$ {{iff}} $x$ is not a loop

Proof:
=== Necessary Condition ===
Let $B \in \mathscr B$ such that $x \in B$.
From Singleton of Element is Subset:
:$\set x \subseteq B$
By definition of a base:
:$B \in \mathscr I$
From matroid axiom $(\text I 2)$:
:$\set x \in \mathscr I$
Then $\set x$ is not a dependent subset by definition.
It follows that $x$ is not a loop by definition.
{{qed|lemma}}
Theorem: Distinct Elements are Parallel iff Pair forms Circuit

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x, y \in S : x \ne y$.
Then:
:$x$ and $y$ are parallel {{iff}} $\set {x, y}$ is a circuit

Proof:
=== Necessary Condition ===
Let $x$ and $y$ be parallel.
By definition of parallel:
:$\set x$ is independent 
:$\set y$ is independent 
:$\set {x, y}$ is dependent
Let $A \subseteq \set {x, y}$ be dependent.
Thus:
:$A \ne \set x, \set y$
By matroid axiom $(\text I 1)$:
:$\O$ is independent
Thus:
:$A \ne \O$
From Power Set of Doubleton:
:$\powerset {\set {x, y} } = \set {\O, \set x, \set y, \set {x, y} }$
Thus:
:$A = \set {x, y}$
By definition of a minimal set:
:$\set {x, y}$ is a minimal dependent subset
It follows that $\set {x, y}$ is a circuit by definition.
{{qed|lemma}}
Theorem: Parallel Relationship is Transitive

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x, y, z \in S : x \ne y, x \ne z, y \ne z$.
If $x$ is parallel to $y$ and $y$ is parallel to $z$ then $x$ is parallel to $z$.

Proof:
Let $x$ be parallel to $y$ and $y$ be parallel to $z$.
By definition of parallel:
:$\set x$, $\set y$, $\set z$ are independent subsets
:$\set {x, y}$, $\set {y, z}$ are dependent subsets
To show that $x$ is parallel to $z$ it remains to show that:
:$\set {x, z}$ is dependent
{{AimForCont}} $\set {x, z}$ is independent.
By matroid axiom $(\text I 3)$:
:$\exists w \in \set{x, z} \setminus \set y : \set{w, y} \in \mathscr I$
By definition of the doubleton:
:$\set{x, y} \in \mathscr I \lor \set{z, y} \in \mathscr I$
This contradicts the assumption that $\set {x, y}$, $\set {y, z}$ are dependent subsets.
It follows that:
:$\set {x, z}$ is dependent
{{qed}}
Theorem: Distinct Matroid Elements are Parallel iff Each is in Closure of Other

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\sigma: \powerset S \to \powerset S$ denote the closure operator of $M$.
Let $x, y \in S : x \ne y$.
Then $x$ is parallel to $y$ {{iff}}:
:$(1)\quad x$ and $y$ are not loops
:$(2)\quad x \in \map \sigma {\set y}$
:$(3)\quad y \in \map \sigma {\set x}$

Proof:
=== Lemma ===
{{:Distinct Matroid Elements are Parallel iff Each is in Closure of Other/Lemma}}{{qed|lemma}}
Theorem: Closure of Subset Contains Parallel Elements

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\sigma: \powerset S \to \powerset S$ denote the closure operator of $M$.
Let $A \subseteq S$.
Let $x, y \in S$.
If $x \in \map \sigma A$ and $y$ is parallel to $x$ then:
:$y \in \map \sigma A$

Proof:
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
Let $x \in \map \sigma A$
Let $y$ be parallel to $x$.
By the definitions of the closure operator and depends depends:
:$\map \rho A = \map \rho {A \cup \set x}$
and:
:$y \in \map \sigma A$ {{iff}} $\map \rho A = \map \rho {A \cup \set y}$
From Rank Function is Increasing:
:$\map \rho {A \cup \set x} = \map \rho A \le \map \rho {A \cup \set y} \le \map \rho {A \cup \set {x, y} }$
To show $\map \rho A = \map \rho {A \cup \set y}$ it is sufficient to show:
:$\map \rho {A \cup \set {x, y} } = \map \rho {A \cup \set x}$
From Rank Function is Increasing:
:$\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x, y} }$
By the definition of parallel:
:$\set x$ is independent
From Independent Subset is Contained in Maximal Independent Subset:
:$\exists X \in \mathscr I : \set x \subseteq X \subseteq A \cup \set x : \size X = \map \rho {A \cup \set x}$
From Subset Relation is Transitive:
:$X \subseteq A \cup \set {x, y}$
From Independent Subset is Contained in Maximal Independent Subset:
:$\exists Y \in \mathscr I : X \subseteq Y \subseteq A \cup \set {x, y} : \size Y = \map \rho {A \cup \set {x, y} }$
By definition of a subset:
:$x \in Y$
By definition of parallel:
:$\set {x, y}$ is dependent
From Superset of Dependent Set is Dependent:
:$\set {x, y} \nsubseteq Y$
Then:
:$y \notin Y$
So:
:$Y \subseteq A \cup \set x$
By definition of the rank function:
:$\size Y \le \map \rho {A \cup \set x}$
As $\size Y = \map \rho {A \cup \set {x, y} }$ then:
:$\map \rho {A \cup \set {x, y} } \le \map \rho {A \cup \set x}$
As $\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x, y} }$ then:
:$\map \rho {A \cup \set {x, y} } = \map \rho {A \cup \set x}$
The result follows.
{{qed}}
Theorem: Set with Two Parallel Elements is Dependent

Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x, y \in S$.
Let $x, y$ be parallel elements.
If $x, y \in A$ then $A$ is dependent.

Proof:
Let $x, y \in A$.
From Doubleton of Elements is Subset:
:$\set{x, y} \subseteq A$
By the definition of parallel elements:
:$\set {x, y}$ is dependent
From Superset of Dependent Set is Dependent:
:$A$ is dependent
{{qed}}
Theorem: Loop Belongs to Every Flat

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x \in S$.
If $x$ is a loop and $A$ is a flat subset then $x \in A$.

Proof:
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
We proceed by Proof by Contraposition.
That is, it is shown that:
:if $x \notin A$ then either $x$ is not a loop or $A$ is not a flat subset
Let $x \notin A$.
Let $x$ be a loop.
By definition of a loop:
:$\set x$ is a dependent subset.
From Rank Function is Increasing:
:$\map \rho A \le \map \rho {A \cup \set x}$
Let $X \in \mathscr I$ such that $X \subseteq A \cup \set x$.
From Superset of Dependent Set is Dependent:
:$\set x \nsubseteq X$
From Singleton of Element is Subset:
:$x \notin X$
So:
:$X \subseteq A$
By definition of the rank function:
:$\size X \le \map \rho A$
From Max Operation Yields Supremum of Parameters:
:$\map \rho {A \cup \set x} = \max \set{\size X : X \in \mathscr I \land X \subseteq A \cup \set x} \le \map \rho A$
Then:
:$\map \rho A = \map \rho {A \cup \set x}$
If follows that $A$ is not a flat subset by definition.
It has been shown that:
:if $x \notin A$ then either $x$ is not a loop or $A$ is not a flat subset
The theorem holds by the Rule of Transposition.
{{qed}}
Theorem: Parallel Elements Depend on Same Subsets

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x, y \in S$.
Let $x$ be parallel to $y$.
Then:
:$x$ depends on $A$ {{iff}} $y$ depends on $A$

Proof:
This follows directly from Closure of Subset Contains Parallel Elements.
{{qed}}
Theorem: Matroid Contains No Loops iff Empty Set is Flat

Let $M = \struct{S, \mathscr I}$ be a matroid.
Then:
:$M$ contains no loops {{iff}} the empty set is flat.

Proof:
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
Theorem: Absolutely Continuous Real Function is Uniformly Continuous

Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be an absolutely continuous real function.
Then $f$ is uniformly continuous.

Proof:
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of pairwise disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
Consider specifically the case $n = 1$.
Suppose that $a \le x \le y \le b$ and:
:$0 \le y - x < \delta$
so that: 
:$\size {x - y} < \delta$
From the absolute continuity of $f$ we then have:
:$\size {\map f y - \map f x} < \epsilon$
so:
:$\size {\map f x - \map f y} < \epsilon$
Now suppose that $y < x$ and: 
:$\size {x - y} < \delta$
then: 
:$0 < x - y < \delta$ 
and: 
:$\size {\map f x - \map f y} < \epsilon$ 
So, in fact, for all $x, y \in \closedint a b$ with:
:$\size {x - y} < \delta$
we have:
:$\size {\map f x - \map f y} < \epsilon$
Since $\epsilon$ was arbitrary:
:$f$ is uniformly continuous.
{{qed}}
Category:Absolutely Continuous Real Functions
Category:Uniformly Continuous Real Functions
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Theorem: Closure of Subspace of Normed Vector Space is Subspace

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $Y \subseteq X$ be a subspace of $X$.
Let $Y^-$ be the closure of $Y$.
Then $Y^- \subseteq X$ is also a subspace of $X$.

Proof:
Suppose $y \in Y^-$.
Then there is a sequence $\ds \sequence {y_n}_{n \mathop \in \N} \in Y$ which converges to $y$.
Suppose $y \in Y$ and $y$ is a limit point.
Then one can define a constant sequence:
:$\sequence {y_n}_{n \mathop \in \N} = y$
Theorem: Closure of Convex Subset in Normed Vector Space is Convex

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $C \subseteq X$ be a convex subset of $X$.
Let $C^-$ be the closure of $C$.
Then $C^- \subseteq X$ is also a convex subset of $X$.

Proof:
Let $x, y \in C^-$.
Suppose $x, y$ are limit points.
Then there are sequences $\sequence {x_n}_{n \mathop \in \N}, \sequence {x_n}_{n \mathop \in \N}$ in $C$, such that:
:$\ds \lim_{n \mathop \to \infty} x_n = x$
:$\ds \lim_{n \mathop \to \infty} x_y = y$
Let $\alpha \in \closedint 0 1$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {1 - \alpha} x + \alpha y
      | r = \paren {1 - \alpha} \lim_{n \mathop \to \infty} x_n + \alpha \lim_{n \mathop \to \infty} y_n
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\paren {1 - \alpha} x_n + \alpha y_n}
}}
{{end-eqn}}
Since $C$ is convex:
:$\ds \forall n \in \N : \paren {1 - \alpha} x_n + \alpha y_n \in C \subseteq C^-$
$C^-$ is closed and contains its limit points.
Hence:
:$\paren {1 - \alpha} x + \alpha y \in C^-$
{{qed}}
Theorem: Field is Principal Ideal Domain

Let $F$ be a field.
Then $F$ is a principal ideal domain.

Proof:
Let $F$ be a field.
Let $I \subset F$ be a non-null ideal of $F$.
Let $a \in I$ be non-zero.
Since $F$ is a field, $a^{-1}$ exists.
We have that $1 = a^{-1} \cdot a \in I$.
Since $1 \in I$, for every element $b \in F$:
:$b = b \cdot 1 \in I$
we have that $I = F = \ideal 1$ if $I \ne \set 0$.
Thus the only ideals of $F$ are $\ideal 0 = \set 0$ and $\ideal 1 = F$, which are both principal ideals.
Hence $F$ is a principal ideal domain.
{{qed}}
Category:Principal Ideal Domains
Category:Field Theory
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Theorem: Complex Vector Space is Vector Space

Let $\C$ denote the set of complex numbers.
Then the complex vector space $\C^n$ is a vector space.

Proof:
=== Construction of Complex Vector Space ===
From the definition, a vector space is a unitary module whose scalar ring is a field.
In order to call attention to the precise scope of the operators, let complex addition and complex multiplication be expressed as $+_\C$ and $\times_\C$ respectively.
Then we can express the field of complex numbers as $\struct {\C, +_\C, \times_\C}$.
From Complex Numbers under Addition form Group, $\struct {\C, +}$ is a group.
Again, in order to call attention to the precise scope of the operator, let complex addition be expressed on $\struct {\C, +}$ as $+_G$.
That is, the group under consideration is $\struct {\C, +_G}$.
Consider the cartesian product:
:$\ds \C^n = \prod_{i \mathop = 1}^n \struct {\C, +_G} = \underbrace {\struct {\C, +_G} \times \cdots \times \struct {\C, +_G} }_{n \text{ copies} }$
Let:
:$\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$
:$\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$
be arbitrary elements of $\C^n$.
Let $\lambda$ be an arbitrary element of $\C$.
Let $+$ be the binary operation defined on $\C^n$ as:
:$\mathbf a + \mathbf b = \tuple {a_1 +_G b_1, a_2 +_G b_2, \ldots, a_n +_G b_n}$
Also let $\cdot$ be the binary operation defined on $\C \times \C^n$ as:
:$\lambda \cdot \mathbf a = \tuple {\lambda \times_\C a_1, \lambda \times_\C a_2, \ldots, \lambda \times_\C a_n}$
In this context, $\lambda \times_\C a_j$ is defined as complex multiplication, as is appropriate (both $\lambda$ and $a_j$ are complex numbers).
With this set of definitions, the structure $\struct {\C^n, +, \cdot}$ is a vector space, as is shown in Proof of Complex Vector Space below.
Theorem: Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space

Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $D \subseteq X$ be a subset of $X$.
Let $D^-$ be the closure of $D$.
Then $D$ is dense {{iff}} $D^- = X$.

Proof:
=== Necessary Condition ===
{{:Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition}}{{qed|lemma}}
Theorem: Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition

Let $\struct {X, \norm {\, \cdot \,} }$ is a normed vector space.
Let $D \subseteq X$ be a subset of $X$.
Let $D^-$ be the closure of $D$.
Let $D$ be dense.
Then:
:$D^- = X$

Proof:
Let $x \in X \setminus D$.
Suppose $D$ is dense in $X$.
Then:
:$\forall n \in N : \exists d_n \in D : d_n \in \map {B_{\frac 1 n}} x$
where $\ds \map {B_{\frac 1 n}} x$ is an open ball.
Let $\sequence {d_n}_{n \mathop \in \N}$ be a sequence in $D$.
Then:
:$\forall n \in \N : \norm {x - d_n} < \frac 1 n$
Hence, $x$ is a limit point of $D$.
In other words, $x \in D^-$.
We have just shown that:
:$x \in X \setminus D \implies x \in D^-$
Hence:
:$X \setminus D \subseteq D^-$.
By definition of closure:
:$D \subseteq D^-$
Therefore:
{{begin-eqn}}
{{eqn | l = X
      | r = D \cup \paren {X \setminus D}
}}
{{eqn | o = \subseteq
      | r = D^-
}}
{{eqn | o = \subseteq
      | r = X
}}
{{end-eqn}}
Thus:
:$X = D^-$.
Theorem: Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient Condition

Let $\struct {X, \norm {\, \cdot \,} }$ is a normed vector space.
Let $D \subseteq X$ be a subset of $X$.
Let $D^-$ be the closure of $D$.
Let $D^- = X$.
Then $D$ is dense.

Proof:
Let $X = D^-$.
We have to show, that for every $x \in X$ there is an open ball with an element from $D^-$.
We have that $X = D \cup \paren {X \setminus D}$.
Suppose $x \in X \setminus D$.
Then $x \in D^- \setminus D$.
Hence, $x$ is a limit point of $D$.
Therefore, there is a sequence $\sequence {d_n}_{n \mathop \in \N}$ in $D$ which converges to $x$.
Thus:
:$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \norm {x - d_N} < \epsilon$
In other words:
:$\ds d_N \in D \implies d_N \in \map {B_\epsilon} x$
Therefore:
:$d_N \in D \cap \map {B_\epsilon} x$.
Suppose $x \in D$.
Let $\epsilon > 0$.
Then $x \in \map {B_\epsilon} x \cap D$.
From both parts and definition we conclude that $D^-$ is dense in $X$.
Theorem: Modulus of Limit/Normed Vector Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\sequence {x_n}$ be a convergent sequence in $R$ to the limit $x$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = x$.
Then:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n} = \norm x$

Proof:
By the Reverse Triangle Inequality:
:$\cmod {\norm {x_n} - \norm x} \le \norm {x_n - x}$
Hence by the Squeeze Theorem and Convergent Sequence Minus Limit:
:$\norm {x_n} \to \norm x$
as $n \to \infty$.
{{Qed}}
Theorem: Matrix Entrywise Addition forms Abelian Group

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $\struct {R, +, \circ}$.
Then $\struct {\map {\MM_R} {m, n}, +}$, where $+$ is matrix entrywise addition, is a group.

Proof:
We have by definition that matrix entrywise addition is a specific instance of a Hadamard product.
By definition of a ring, the structure $\struct {R, +}$ is a group.
As $\struct {R, +}$ is a fortiori a monoid, it follows from Matrix Space Semigroup under Hadamard Product that $\struct {\map {\MM_R} {m, n}, +}$ is also a monoid.
As $\struct {R, +}$ is a group, it follows from Negative Matrix is Inverse for Matrix Entrywise Addition that all elements of $\struct {\map {\MM_R} {m, n}, +}$ have an inverse element.
From Matrix Entrywise Addition is Commutative it follows that $\struct {\map {\MM_R} {m, n}, +}$ is an Abelian group.
The result follows.
{{Qed}}
Theorem: Negative Matrix is Inverse for Matrix Entrywise Addition over Ring

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $\struct {R, +, \circ}$.
Let $\mathbf A$ be an element of $\map {\MM_R} {m, n}$.
Let $-\mathbf A$ be the negative of $\mathbf A$.
Then $-\mathbf A$ is the inverse for the operation $+$, where $+$ is matrix entrywise addition.

Proof:
Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A + \paren {-\mathbf A}
      | r = \sqbrk a_{m n} + \paren {-\sqbrk a_{m n} }
      | c = Definition of $\mathbf A$
}}
{{eqn | r = \sqbrk a_{m n} + \sqbrk {-a}_{m n}
      | c = {{Defof|Negative Matrix over Ring}}
}}
{{eqn | r = \sqbrk {a + \paren {-a} }_{m n}
      | c = {{Defof|Matrix Entrywise Addition over Ring}}
}}
{{eqn | r = \sqbrk {0_R}_{m n}
      | c = {{Defof|Ring Negative}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf A + \paren {-\mathbf A}
      | r = \mathbf 0_R
      | c = {{Defof|Zero Matrix over Ring}}
}}
{{end-eqn}}
The result follows from Zero Matrix is Identity for Matrix Entrywise Addition over Ring.
{{qed}}
Theorem: Zero Matrix is Identity for Matrix Entrywise Addition over Ring

Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.
Let $\mathbf 0_R = \sqbrk {0_R}_{m n}$ be the zero matrix of $\map {\MM_R} {m, n}$.
Then $\mathbf 0_R$ is the identity element for matrix entrywise addition.

Proof:
Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A + \mathbf 0_R
      | r = \sqbrk a_{m n} + \sqbrk {0_R}_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
}}
{{eqn | r = \sqbrk {a + 0_R}_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk a_{m n}
      | c = {{Ring-axiom|A3}} is $0_R$
}}
{{eqn | ll= \leadsto
      | l = \mathbf A + \mathbf 0_R
      | r = \mathbf A
      | c = {{Defof|Zero Matrix over Ring}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \mathbf 0_R + \mathbf A
      | r = \sqbrk {0_R}_{m n} + \sqbrk a_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
}}
{{eqn | r = \sqbrk {0_R + a}_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk a_{m n}
      | c = {{Ring-axiom|A3}} is $0_R$
}}
{{eqn | ll= \leadsto
      | l = \mathbf 0_R + \mathbf A
      | r = \mathbf A
      | c = {{Defof|Zero Matrix over Ring}}
}}
{{end-eqn}}
{{qed}}
Theorem: Zero Matrix is Identity for Matrix Entrywise Addition over Ring

Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.
Let $\mathbf 0_R = \sqbrk {0_R}_{m n}$ be the zero matrix of $\map {\MM_R} {m, n}$.
Then $\mathbf 0_R$ is the identity element for matrix entrywise addition.

Proof:
By definition, matrix entrywise addition is the '''Hadamard product''' with respect to ring addition.
We have from {{Ring-axiom|A3}} that the identity element of ring addition is the ring zero $0_R$.
The result then follows directly from Zero Matrix is Identity for Hadamard Product.
{{qed}}
Theorem: Properties of Matrix Entrywise Addition over Ring

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $S$ over an algebraic structure $\struct {R, +, \circ}$.
Let $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$.
Let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation of matrix entrywise addition satisfies the following properties:
:$+$ is closed on $\map {\MM_R} {m, n}$
:$+$ is associative on $\map {\MM_R} {m, n}$
:$+$ is commutative on $\map {\MM_R} {m, n}$.

Proof:
=== Matrix Entrywise Addition over Ring is Closed ===
{{:Matrix Entrywise Addition over Ring is Closed}}
Theorem: Matrix Entrywise Addition is Associative

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is associative on $\map \MM {m, n}$.
That is:
:$\paren {\mathbf A + \mathbf B} + \mathbf C = \mathbf A + \paren {\mathbf B + \mathbf C}$
for all $\mathbf A$, $\mathbf B$ and $\mathbf C$ in $\map \MM {m, n}$.

Proof:
From:
:Integers form Ring
:Rational Numbers form Ring
:Real Numbers form Ring
:Complex Numbers form Ring
the standard number systems $\Z$, $\Q$, $\R$ and $\C$ are rings.
Hence we can apply Matrix Entrywise Addition over Ring is Associative.
{{qed|lemma}}
The above cannot be applied to the natural numbers $\N$, as they do not form a ring.
However, from Natural Numbers under Addition form Commutative Monoid, the algebraic structure $\struct {\N, +}$ is a commutative monoid.
By definition, matrix entrywise addition is the '''Hadamard product''' with respect to addition of numbers.
The result follows from Associativity of Hadamard Product.
{{qed}}
Theorem: Matrix Entrywise Addition is Associative

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is associative on $\map \MM {m, n}$.
That is:
:$\paren {\mathbf A + \mathbf B} + \mathbf C = \mathbf A + \paren {\mathbf B + \mathbf C}$
for all $\mathbf A$, $\mathbf B$ and $\mathbf C$ in $\map \MM {m, n}$.

Proof:
Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be matrices whose order is $m \times n$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf A + \mathbf B} + \mathbf C
      | r = \paren {\sqbrk a_{m n} + \sqbrk b_{m n} } + \sqbrk c_{m n}
      | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$
}}
{{eqn | r = \sqbrk {a + b}_{m n} + \sqbrk c_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk {\paren {a + b} + c}_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk {a + \paren {b + c} }_{m n}
      | c = Associative Law of Addition
}}
{{eqn | r = \sqbrk a_{m n} + \sqbrk {b + c}_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk a_{m n} + \paren {\sqbrk b_{m n} + \sqbrk c_{m n} }
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \mathbf A + \paren {\mathbf B + \mathbf C}
      | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$
}}
{{end-eqn}}
{{qed}}
Theorem: Matrix Entrywise Addition over Ring is Commutative

Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.
For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is commutative on $\map {\MM_R} {m, n}$.
That is:
:$\mathbf A + \mathbf B = \mathbf B + \mathbf A$
for all $\mathbf A$ and $\mathbf B$ in $\map {\MM_R} {m, n}$.

Proof:
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of the $m \times n$ matrix space over $R$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A + \mathbf B
      | r = \sqbrk a_{m n} + \sqbrk b_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf B$
}}
{{eqn | r = \sqbrk {a + b}_{m n}
      | c = {{Defof|Matrix Entrywise Addition over Ring}}
}}
{{eqn | r = \sqbrk {b + a}_{m n}
      | c = {{Ring-axiom|A2}}
}}
{{eqn | r = \sqbrk b_{m n} + \sqbrk a_{m n}
      | c = {{Defof|Matrix Entrywise Addition over Ring}}
}}
{{eqn | r = \mathbf B + \mathbf A
      | c = Definition of $\mathbf A$ and $\mathbf B$
}}
{{end-eqn}}
{{qed}}