avewright/proofwiki-math huggingface dataset:
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:
By definition, matrix entrywise addition is the '''Hadamard product''' of $\mathbf A$ and $\mathbf B$ with respect to ring addition.
We have from {{Ring-axiom|A2}} that ring addition is commutative.
The result then follows directly from Commutativity of Hadamard Product.
{{qed}}
Theorem: Zero Matrix is Identity for Matrix Entrywise Addition

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
Let $\mathbf 0 = \sqbrk 0_{m n}$ be the zero matrix of $\map \MM {m, n}$.
Then $\mathbf 0$ 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

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
Let $\mathbf 0 = \sqbrk 0_{m n}$ be the zero matrix of $\map \MM {m, n}$.
Then $\mathbf 0$ 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: Zero Matrix is Identity for Matrix Entrywise Addition

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
Let $\mathbf 0 = \sqbrk 0_{m n}$ be the zero matrix of $\map \MM {m, n}$.
Then $\mathbf 0$ is the identity element for matrix entrywise addition.

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 whose zero is the number $0$ (zero).
Hence we can apply Zero Matrix is Identity for Matrix Entrywise Addition over Ring.
{{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 whose identity is $0$ (zero).
By definition, matrix entrywise addition is the '''Hadamard product''' with respect to addition of numbers.
The result follows from Zero Matrix is Identity for Hadamard Product.
{{qed}}
Theorem: Zero Matrix is Identity for Matrix Entrywise Addition

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
Let $\mathbf 0 = \sqbrk 0_{m n}$ be the zero matrix of $\map \MM {m, n}$.
Then $\mathbf 0$ is the identity element for matrix entrywise addition.

Proof:
Let $\mathbf A = \sqbrk a_{m n} \in \map \MM {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A + \mathbf 0
      | r = \sqbrk a_{m n} + \sqbrk 0_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
}}
{{eqn | r = \sqbrk {a + 0}_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk a_{m n}
      | c = Identity Element of Addition on Numbers
}}
{{eqn | ll= \leadsto
      | l = \mathbf A + \mathbf 0
      | r = \mathbf A
      | c = {{Defof|Zero Matrix}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \mathbf 0 + \mathbf A
      | r = \sqbrk 0_{m n} + \sqbrk a_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
}}
{{eqn | r = \sqbrk {0 + a}_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk a_{m n}
      | c = Identity Element of Addition on Numbers
}}
{{eqn | ll= \leadsto
      | l = \mathbf 0 + \mathbf A
      | r = \mathbf A
      | c = {{Defof|Zero Matrix}}
}}
{{end-eqn}}
{{qed}}
Theorem: Matrix Scalar Product Distributes over Number Addition

Let $\GF$ denote one of the standard number systems.
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over $\GF$.
For $\mathbf A \in \map \MM {m, n}$ and $\lambda \in \GF$, let $\lambda \mathbf A$ be defined as the matrix scalar product of $\lambda$ and $\mathbf A$.
The matrix scalar product is distributive on $\map \MM {m, n}$, in the following sense:
For all $\mathbf A$ in $\map \MM {m, n}$ and $\lambda, \mu \in \GF$:
:$\paren {\lambda + \mu} \mathbf A = \lambda \mathbf A + \mu \mathbf A$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\lambda + \mu} \mathbf A
      | r = \paren {\lambda + \mu} \sqbrk a_{m n}
      | c = Definition of $\mathbf A$
}}
{{eqn | r = \sqbrk {\paren {\lambda + \mu} a}_{m n}
      | c = {{Defof|Matrix Scalar Product}}
}}
{{eqn | r = \sqbrk {\lambda a + \mu a}_{m n}
      | c = Distributive Property
}}
{{eqn | r = \sqbrk {\lambda a}_{m n} + \sqbrk {\mu a}_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \lambda \mathbf A + \mu \mathbf A
      | c = Definition of $\mathbf A$
}}
{{end-eqn}}
{{qed}}
Theorem: Matrix Scalar Product with Zero gives Zero Matrix

Let $\Bbb F$ denote one of the standard number systems.
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over $\Bbb F$.
For $\mathbf A \in \map \MM {m, n}$ and $\lambda$ \in $\Bbb F$, let $\lambda \mathbf A$ be defined as the matrix scalar product of $\lambda$ and $\mathbf A$.
When $\lambda = 0$, we have for all $\mathbf A$ in $\map \MM {m, n}$:
:$0 \mathbf A = \mathbf 0$
where $\mathbf 0$ denotes the zero matrix.

Proof:
{{begin-eqn}}
{{eqn | l = 0 \mathbf A
      | r = 0 \sqbrk a_{m n}
      | c = Definition of $\mathbf A$
}}
{{eqn | r = \sqbrk {0 a}_{m n}
      | c = {{Defof|Matrix Scalar Product}}
}}
{{eqn | r = \sqbrk 0_{m n}
      | c = Zero is Zero Element for Multiplication
}}
{{eqn | r = \mathbf 0
      | c = {{Defof|Zero Matrix}}
}}
{{end-eqn}}
{{qed}}
Theorem: Negative Matrix is Inverse for Matrix Entrywise Addition

Let $\Bbb F$ denote one of the standard number systems.
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over $\Bbb F$.
Let $\mathbf A$ be an element of $\map \MM {m, n}$.
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 $\mathbf A = \sqbrk a_{m n} \in \map \MM {m, n}$.
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}}
}}
{{eqn | r = \sqbrk {a + \paren {-a} }_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk 0_{m n}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \mathbf A + \paren {-\mathbf A}
      | r = \mathbf 0
      | c = {{Defof|Zero Matrix}}
}}
{{end-eqn}}
The result follows from Zero Matrix is Identity for Matrix Entrywise Addition.
{{qed}}
Theorem: Zero Matrix is Zero for Matrix Multiplication

Let $\struct {R, +, \times}$ be a ring.
Let $\mathbf A$ be a matrix over $R$ of order $m \times n$
Let $\mathbf 0$ be a zero matrix whose order is such that either:
:$\mathbf {0 A}$ is defined
or:
:$\mathbf {A 0}$ is defined
or both.
Then:
:$\mathbf {0 A} = \mathbf 0$
or:
:$\mathbf {A 0} = \mathbf 0$
whenever they are defined.
The order of $\mathbf 0$ will be according to the orders of the factor matrices.

Proof:
Let $\mathbf A = \sqbrk a_{m n}$ be matrices.
Let $\mathbf {0 A}$ be defined.
Then $\mathbf 0$ is of order $r \times m$ for $r \in \Z_{>0}$.
Thus we have:
{{begin-eqn}}     
{{eqn | l = \mathbf {0 A}
      | r = \mathbf C
      | c = 
}}
{{eqn | l = \sqbrk 0_{r m} \sqbrk a_{m n}
      | r = \sqbrk c_{r n}
      | c = Definition of $\mathbf 0$ and $\mathbf A$
}}
{{eqn | ll= \leadsto
      | q = \forall i \in \closedint 1 r, j \in \closedint 1 n
      | l = c_{i j}
      | r = \sum_{k \mathop = 1}^m 0_{i k} \times a_{k j}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \sum_{k \mathop = 1}^m 0
      | c = {{Defof|Zero Matrix}}
}}
{{eqn | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \mathbf {0 A}
      | r = \sqbrk 0_{r n}
      | c = 
}}
{{end-eqn}}
Hence $\mathbf {0 A}$ is the Zero Matrix of order $r \times n$.
Let $\mathbf {A 0}$ be defined.
Then $\mathbf 0$ is of order $n \times s$ for $s \in \Z_{>0}$.
Thus we have:
{{begin-eqn}}     
{{eqn | l = \mathbf {A 0}
      | r = \mathbf C
      | c = 
}}
{{eqn | l = \sqbrk a_{m n} \sqbrk 0_{n s}
      | r = \sqbrk c_{m s}
      | c = Definition of $\mathbf A$ and $\mathbf 0$
}}
{{eqn | ll= \leadsto
      | q = \forall i \in \closedint 1 m, j \in \closedint 1 s
      | l = c_{i j}
      | r = \sum_{k \mathop = 1}^n a_{i k} \times 0_{k j}
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n 0
      | c = {{Defof|Zero Matrix}}
}}
{{eqn | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \mathbf {0 A}
      | r = \sqbrk 0_{m s}
      | c = 
}}
{{end-eqn}}
Hence $\mathbf {A 0}$ is the Zero Matrix of order $m \times s$.
{{qed|lemma}}
If $\mathbf 0$ is of order $n \times m$,then both $\mathbf {A 0}$ and $\mathbf {0 A}$ are defined, and:
{{begin-eqn}}
{{eqn | l = \mathbf {A 0}
      | r = \sqbrk 0_{m m}
}}
{{eqn | l = \mathbf {0 A}
      | r = \sqbrk 0_{n n}
}}
{{end-eqn}}
{{qed}}
Theorem: Unit Matrix is Identity for Matrix Multiplication

Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} n$ denote the metric space of square matrices of order $n$ over $R$.
Let $\mathbf I_n$ denote the unit matrix of order $n$:
Then:
:$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$
That is, the unit matrix $\mathbf I_n$ is the identity element for (conventional) matrix multiplication over $\map {\MM_R} n$.

Proof:
=== Lemma: Left Identity ===
{{:Unit Matrix is Identity for Matrix Multiplication/Left}}{{qed|lemma}}
Theorem: Unit Matrix is Identity for Matrix Multiplication/Left

Let $\map {\MM_R} {m, n}$ denote the $m \times n$ metric space over $R$.
Let $I_m$ denote the unit matrix of order $m$.
Then:
:$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf I_m \mathbf A  = \mathbf A$

Proof:
Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Let $\sqbrk b_{m n} = \mathbf I_m \sqbrk a_{m n}$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i \in \closedint 1 m, j \in \closedint 1 n
      | l = b_{i j}
      | r = \sum_{k \mathop = 1}^m \delta_{i k} a_{k j}
      | c = where $\delta_{i k}$ is the Kronecker delta: $\delta_{i k} = 1_R$ when $i = k$ otherwise $0_R$
}}
{{eqn | r = a_{i j}
      | c = 
}}
{{end-eqn}}
Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_m$ is shown to be a left identity.
{{qed}}
Category:Unit Matrix is Identity for Matrix Multiplication
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Theorem: Unit Matrix is Identity for Matrix Multiplication/Right

Let $\map {\MM_R} {m, n}$ denote the $m \times n$ metric space over $R$.
Let $I_n$ denote the unit matrix of order $n$.
Then:
:$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf A \mathbf I_n  = \mathbf A$

Proof:
Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Let $\sqbrk b_{m n} = \sqbrk a_{m n} \mathbf I_n$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i \in \closedint 1 m, j \in \closedint 1 n
      | l = b_{i j}
      | r = \sum_{k \mathop = 1}^n a_{i k} \delta_{k j}
      | c = where $\delta_{k j}$ is the Kronecker delta: $\delta_{k j} = 1_R$ when $k = j$ otherwise $0_R$
}}
{{eqn | r = a_{i j}
      | c = 
}}
{{end-eqn}}
Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_n$ is shown to be a right identity.
{{qed}}
Category:Unit Matrix is Identity for Matrix Multiplication
0i0mtd5k4tifa6k3t1b6d57ylf1rcjq
Theorem: Left and Right Inverses of Square Matrix over Field are Equal

Let $\Bbb F$ be a field, usually one of the standard number fields $\Q$, $\R$ or $\C$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map \MM n$ denote the matrix space of order $n$ square matrices over $\Bbb F$.
Let $\mathbf B$ be a left inverse matrix of $\mathbf A$.
Then $\mathbf B$ is also a right inverse matrix of $\mathbf A$.
Similarly, let $\mathbf B$ be a right inverse matrix of $\mathbf A$.
Then $\mathbf B$ is also a right inverse matrix of $\mathbf A$.

Proof:
Consider the algebraic structure $\struct {\map \MM {m, n}, +, \circ}$, where:
:$+$ denotes matrix entrywise addition
:$\circ$ denotes (conventional) matrix multiplication.
From Ring of Square Matrices over Field is Ring with Unity, $\struct {\map \MM {m, n}, +, \circ}$ is a ring with unity.
Hence a fortiori $\struct {\map \MM {m, n}, +, \circ}$ is a monoid.
The result follows directly from Left Inverse and Right Inverse is Inverse.
{{qed}}
{{mistake|That's not what it actually says. What the above link says is that ''if'' $\mathbf A$ has both a right inverse matrix ''and'' a left inverse matrix, then those are equal and can be called an inverse matrix. It does not say that if $\mathbf B$ is a left inverse matrix then it is automatically a right inverse matrix. I'll sort this out when I get to exercise $1.15$.}}
Theorem: Inverse of Square Matrix over Field is Unique

Let $\Bbb F$ be a field, usually one of the standard number fields $\Q$, $\R$ or $\C$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map \MM n$ denote the matrix space of order $n$ square matrices over $\Bbb F$.
Let $\mathbf B$ be an inverse matrix of $\mathbf A$.
Then $\mathbf B$ is the only inverse matrix of $\mathbf A$.

Proof:
Consider the algebraic structure $\struct {\map \MM {m, n}, +, \circ}$, where:
:$+$ denotes matrix entrywise addition
:$\circ$ denotes (conventional) matrix multiplication.
From Ring of Square Matrices over Field is Ring with Unity, $\struct {\map \MM {m, n}, +, \circ}$ is a ring with unity.
Hence a fortiori $\struct {\map \MM {m, n}, +, \circ}$ is a monoid.
The result follows directly from Inverse in Monoid is Unique.
{{qed}}
Theorem: Rank of Empty Set is Zero

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho : \powerset S \to \Z$ be the rank function of $M$.
Then:
:$\map \rho \O = 0$

Proof:
By matroid axiom $(\text I 1)$:
:$\O$ is independent
From Rank of Independent Subset Equals Cardinality:
:$\map \rho \O = \size \O$
From Cardinality of Empty Set:
:$\size \O = 0$
The result follows.
{{qed}}
Theorem: Rank Function is Increasing

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho: \powerset S \to \Z$ be the rank function of $M$.
Let $A, B \subseteq S$ be subsets of $S$ such that $A \subseteq B$.
Then:
:$\map \rho A \le \map \rho B$

Proof:
Now:
{{begin-eqn}}
{{eqn | l = \map \rho A
      | r = \max \set {\size X : X \subseteq A \land X \in \mathscr I}
      | c = {{Defof|Rank Function (Matroid)|Rank Function}}
}}
{{eqn | r = \max \set {\size X : X \in \powerset A \land X \in \mathscr I}
      | c = {{Defof|Power Set}} of $\O$
}}
{{eqn | r = \max \set {\size X : X \in \powerset A \cap \mathscr I}
      | c = {{Defof|Set Intersection}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \map \rho B
      | r = \max \set {\size X : X \in \powerset B \cap \mathscr I}
}}
{{end-eqn}}
From Power Set of Subset:
:$\powerset A \subseteq \powerset B$
From Set Intersection Preserves Subsets:
:$\powerset A \cap \mathscr I \subseteq \powerset B \cap \mathscr I$
It follows that:
:$\set {\size X : X \in \powerset A \cap \mathscr I} \subseteq \set {\size X : X \in \powerset B \cap \mathscr I}$
From Max of Subfamily of Operands Less or Equal to Max:
:$\max \set {\size X : X \in \powerset B \cap \mathscr I} \le \max \set {\size X : X \in \powerset B \cap \mathscr I}$
{{qed}}
Theorem: Bounds for Rank of Subset

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho: \powerset S \to \Z$ be the rank function of $M$.
Let $A \subseteq S$ be subset of $S$.
Then:
:$0 \le \map \rho A \le \size A$

Proof:
By definition of the rank function:
{{begin-eqn}}
{{eqn | l = \map \rho A
      | r = \max \set {\size X : X \subseteq A \land X \in \mathscr I}
}}
{{end-eqn}}
From Cardinality of Subset of Finite Set:
:$\forall X \subseteq A : \size X \le \size A$
In particular:
:$\forall X \subseteq A : X \in \mathscr I$ then $\size X \le \size A$
From Max Operation Yields Supremum of Operands:
:$\max \set {\size X : X \subseteq A \land X \in \mathscr I} \le \size A$
From Empty Set is Subset of All Sets:
:$\O \subseteq A$
From Cardinality of Empty Set:
:$\size \O = 0$
By matroid axiom $(\text I 1)$:
:$\O \in \mathscr I$
From Max Operation Yields Supremum of Operands:
:$0 \le \max \set {\size X : X \subseteq A \land X \in \mathscr I}$
It follows that:
:$0 \le \map \rho A \le \size A$
{{qed}}
Theorem: Inverse of Transpose of Matrix is Transpose of Inverse

Let $\mathbf A$ be a matrix over a field.
Let $\mathbf A^\intercal$ denote the transpose of $\mathbf A$.
Let $\mathbf A$ be an nonsingular matrix.
Then $\mathbf A^\intercal$ is also nonsingular and:
:$\paren {\mathbf A^\intercal}^{-1} = \paren {\mathbf A^{-1} }^\intercal$
where $\mathbf A^{-1}$ denotes the inverse of $\mathbf A$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf A^{-1} }^\intercal \mathbf A^\intercal
      | r = \paren {\mathbf A \mathbf A^{-1} }^\intercal
      | c = Transpose of Matrix Product
}}
{{eqn | r = \mathbf I^\intercal
      | c = {{Defof|Inverse Matrix}}: $\mathbf I$ denotes Unit Matrix
}}
{{eqn | r = \mathbf I
      | c = {{Defof|Unit Matrix}}
}}
{{end-eqn}}
Hence $\paren {\mathbf A^{-1} }^\intercal$ is an inverse of $\mathbf A^\intercal$.
From Inverse of Square Matrix over Field is Unique:
:$\paren {\mathbf A^{-1} }^\intercal = \paren {\mathbf A^\intercal}^{-1}$
{{qed}}
Theorem: Max Operation Yields Supremum of Parameters/General Case

Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
Then:
:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$

Proof:
We will prove the result by induction on the number of operands $n$.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
Theorem: Max of Subfamily of Operands Less or Equal to Max

Let $\struct {S, \preceq}$ be a totally ordered set.
Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
Let $\set{k_1, k_2, \dotsc, k_m} \subseteq \set{1, 2, \dotsc, n}$
Then:
:$\max \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m}} \preceq \max \set {x_1, x_2, \dotsc, x_n}$
where:
:$\max$ denotes the max operation

Proof:
From Max Operation Yields Supremum of Operands:
:$\max \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m}} = \sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} }$
and
:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
Since $\set {k_1, k_2, \dotsc, k_m} \subseteq \set {1, 2, \dotsc, n}$ then:
:$\set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \subseteq \set {x_1, x_2, \dotsc, x_n}$
From Supremum of Subset:
:$\sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \preceq \sup \set {x_1, x_2, \dotsc, x_n}$
{{qed}}
Category:Max Operation
jcfapxzwy2tcjvkldcbmq2zbmu05zf9
Theorem: Elementary Matrix corresponding to Elementary Row Operation/Scale Row

Let $e$ be the elementary row operation acting on $\mathbf I$ as:
{{begin-axiom}}
{{axiom | n = \text {ERO} 1
        | t = For some $\lambda \in K_{\ne 0}$, multiply row $k$ of $\mathbf I$ by $\lambda$
        | m = r_k \to \lambda r_k
}}
{{end-axiom}}

Proof:
By definition of the unit matrix:
:$I_{a b} = \delta_{a b}$
where:
:$I_{a b}$ denotes the element of $\mathbf I$ whose indices are $\tuple {a, b}$.
By definition, $\mathbf E$ is the square matrix of order $m$ formed by applying $e$ to the unit matrix $\mathbf I$.
That is, all elements of row $k$ of $\mathbf I$ are to be multiplied by $\lambda$.
By definition of unit matrix, all elements of row $k$ are $0$ except for element $I_{k k}$, which is $1$.
Thus in $\mathbf E$:
:$E_{k k} = \lambda \cdot 1 = \lambda$
The elements in all the other rows of $\mathbf E$ are the same as the corresponding elements of $\mathbf I$.
Hence the result.
{{qed}}
Category:Elementary Matrix corresponding to Elementary Row Operation
6u2jxz2ubtu2iv2m2jq3xwayfy66vm4
Theorem: Elementary Matrix corresponding to Elementary Row Operation/Scale Row and Add

Let $e$ be the elementary row operation acting on $\mathbf I$ as:
{{begin-axiom}}
{{axiom | n = \text {ERO} 2
        | t = For some $\lambda \in K$, add $\lambda$ times row $j$ to row $i$
        | m = r_i \to r_i + \lambda r_j
}}
{{end-axiom}}

Proof:
By definition of the unit matrix:
:$I_{a b} = \delta_{a b}$
where:
:$I_{a b}$ denotes the element of $\mathbf I$ whose indices are $\tuple {a, b}$.
By definition, $\mathbf E$ is the square matrix of order $m$ formed by applying $e$ to the unit matrix $\mathbf I$.
That is, all elements of row $i$ of $\mathbf I$ are to have the corresponding elements of row $j$ added to them after the latter have been multiplied by $\lambda$.
By definition of unit matrix:
:all elements of row $i$ are $0$ except for element $I_{i i}$, which is $1$.
:all elements of row $j$ are $0$ except for element $I_{j j}$, which is $1$.
Thus in $\mathbf E$:
:where $a \ne i$, $E_{a b} = \delta_{a b}$
:where $a = i$:
::$E_{a b} = \delta_{a b}$ where $b \ne j$
::$E_{a b} = \delta_{a b} + \lambda \cdot 1$ where $b = j$
That is:
:$E_{a b} = \delta_{a b}$ for all elements of $\mathbf E$
except where $a = i$ and $b = j$, at which element:
:$E_{a b} = \delta_{a b} + \lambda$
That is:
:$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$
Hence the result.
{{qed}}
Theorem: Elementary Matrix corresponding to Elementary Row Operation/Exchange Rows

Let $e$ be the elementary row operation acting on $\mathbf I$ as:
{{begin-axiom}}
{{axiom | n = \text {ERO} 3
        | t = Interchange rows $i$ and $j$
        | m = r_i \leftrightarrow r_j
}}
{{end-axiom}}

Proof:
By definition of the unit matrix:
:$I_{a b} = \delta_{a b}$
where:
:$I_{a b}$ denotes the element of $\mathbf I$ whose indices are $\tuple {a, b}$.
By definition, $\mathbf E$ is the square matrix of order $m$ formed by applying $e$ to the unit matrix $\mathbf I$.
That is, all elements of row $i$ of $\mathbf I$ are to be exchanged with the corresponding elements of row $j$.
By definition of unit matrix:
:all elements of row $i$ are $0$ except for element $I_{i i}$, which is $1$.
:all elements of row $j$ are $0$ except for element $I_{j j}$, which is $1$.
Thus in $\mathbf E$:
:where $a \ne i$ and $a \ne j$, $E_{a b} = \delta_{a b}$ (all rows except $i$ and $j$ are unchanged)
:where $a = i$, $E_{a b} = \delta_{j b}$ (the contents of row $j$)
:where $a = j$, $E_{a b} = \delta_{i b}$ (the contents of row $i$)
That is:
:$E_{a b} = \begin {cases} \delta_{a b} & : \text {if $a \ne i$ and $a \ne j$} \\ \delta_{j b} & : \text {if $a = i$} \\ \delta_{i b} & : \text {if $a = j$} \end {cases}$
Hence the result.
{{qed}}
Category:Elementary Matrix corresponding to Elementary Row Operation
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