hcju/proofwikiqa huggingface dataset:
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}}
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}}
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}}
=== Matrix Entrywise Addition over Ring is Closed === {{:Matrix Entrywise Addition over Ring is Closed}} === Matrix Entrywise Addition over Ring is Associative === {{:Matrix Entrywise Addition over Ring is Associative}} === Matrix Entrywise Addition over Ring is Commutative === {{:Matrix Entrywise Addition over Ring is Commutative}} Category:Matrix Entrywise Addition ofmyntkjczgnuup663ekqa4e6fcfv7v
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}}
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}}
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}}
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}}
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}}
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}}
Let $\mathbf A = \sqbrk a_{m n}$ be matrices. Let $\mathbf 0 \mathbf A$ 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 \mathbf 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       | lo= \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 \mathbf A       | r = \sqbrk 0_{r n}       | c =  }} {{end-eqn}} Hence $\mathbf 0 \mathbf A$ is the Zero Matrix of order $r \times n$. Let $\mathbf A \mathbf 0$ 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 \mathbf 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       | lo= \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 \mathbf A       | r = \sqbrk 0_{m s}       | c =  }} {{end-eqn}} Hence $\mathbf A \mathbf 0$ is the Zero Matrix of order $m \times s$. {{qed|lemma}} If $\mathbf 0$ is of order $n \times m$,then both $\mathbf A \mathbf 0$ and $\mathbf 0 \mathbf A$ are defined, and: {{begin-eqn}} {{eqn | l = \mathbf A \mathbf 0       | r = \sqbrk 0_{m m} }} {{eqn | l = \mathbf 0 \mathbf A       | r = \sqbrk 0_{n n} }} {{end-eqn}} {{qed}}
=== Lemma: Left Identity === {{:Unit Matrix is Identity for Matrix Multiplication/Left}}{{qed|lemma}} === Lemma: Right Identity === {{:Unit Matrix is Identity for Matrix Multiplication/Right}}{{qed|lemma}} Thus: $\mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$ Hence, by definition, $\mathbf I_n$ is an identity element for (conventional) matrix multiplication over $\map {\MM_R} n$. That $\mathbf I_n$ is ''the'' identity element follows from Identity is Unique. {{qed}}
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 | ll= \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 csco6xzhtl62zukbf5e63iq3i03fg0u
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 | ll= \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 8pmjc10owrj8op15gsw82jetlcmq0k4
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 $1.15$.}}
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}}
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}}
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}}
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 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 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}}
=== Necessary Condition === {{:Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition}}{{qed}}
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}}
From Max 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 and Min Operations dlvc1y82jhfq4bosca2cw83mjb3m8al
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
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}}
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 swum7srrfe0166pzo0sd3i1yzcw3b9d
From Empty Set is Subset of All Sets: :$\O \in \powerset {\set x}$ Let $A \in \powerset {\set x}$ such that $A \ne \O$ That is: {{begin-eqn}} {{eqn | r = A \subseteq \set x \land A \ne \O       | o =       | c =  }} {{eqn | ll= \leadsto       | r = A \subseteq \set x \land \exists y : y \in A       | o =       | c = {{Defof|Empty Set}} }} {{eqn | ll= \leadsto       | r = A \subseteq \set x \land \exists y : y \in A \land y \in \set x       | o =       | c = {{Defof|Subset}} }} {{eqn | ll= \leadsto       | r = A \subseteq \set x \land \exists y : y \in A \land y = x       | o =       | c = {{Defof|Singleton}} }} {{eqn | ll= \leadsto       | r = A \subseteq \set x \land x \in A       | o =       | c = }} {{eqn | ll= \leadsto       | r = A \subseteq \set x \land \set x \subseteq A       | o =       | c = Singleton of Element is Subset }} {{eqn | ll= \leadsto       | r = A = \set x       | o =       | c = {{Defof|Set Equality}} }} {{end-eqn}} So a subset of $\set x$ is either $\O$ or $\set x$. {{qed}} Category:Power Set Category:Singletons 0z1pqmxvgh8yira067g3l8s1prfj25s
The following algorithm generates a sequence of elementary row operations which convert $\mathbf A$ to $\mathbf B$. Let $\mathbf A' = \sqbrk {a'}_{m n}$ denote the state of $\mathbf A$ after having processed the latest step. After each step, an implicit step can be included that requires that the form of $\mathbf A'$ is inspected to see if it is in the form $\mathbf B$, and if so, terminating the algorithm, but this is not essential. :$(1): \quad$ Are all elements in the first column of $\mathbf A$ equal to $0$? :::If so, there is nothing to do, and the required row operation is the unit matrix $\mathbf I_m$. :::Otherwise, move on to step $(2)$. :$(2): \quad$ Is element $a_{1 1}$ equal to $0$? :::If so: ::::$\text (a): \quad$ find the smallest $k$ such that row $k$ of $\mathbf A$ such that $a_{k 1} \ne 0$ ::::$\text (b): \quad$ use the elementary row operation $r_1 \leftrightarrow r_k$ which will result $a'_{1 1} = a_{k 1}$ and $a'_{k 1} = 0$. :Move on to step $(3)$. :$(3): \quad$ Is element $a'_{1 1}$ equal to $1$? :::If so, use the elementary row operation $r_1 \to \lambda r_1$ where $\lambda = \dfrac 1 {a'_{1 1} }$, which will result $a'_{1 1} = 1$. :Move on to step $4$ :$(4): \quad$ For each row $j$ from $2$ to $m$, do the following: :::Is $a_{j 1} \ne 0$? ::::If so, use the elementary row operation $r_j \leftrightarrow r_j + \mu r_1$, where $\mu = -\dfrac {a'_{j 1} } {a'{1 1} }$, which will result in $a'_{j 1} = 0$. This will result in an $m \times n$ matrix in the required form. Exercising the above algorithm will have generated a sequence of elementary row operations $e_1, e_2, \ldots, e_t$. For each $e_k$ we create the elementary row matrix $\mathbf E_k$. We then assemble the matrix product: :$\mathbf R := \mathbf E_t \mathbf E_{t - 1} \mathbf E_{t - 2} \dotsm \mathbf E_2 \mathbf E_1$ From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, $\mathbf R$ is the resulting $m \times m$ matrix corresponding to the row operation which is used to convert $\mathbf A$ to $\mathbf B$. {{qed}}
For ongoing brevity in this proof, the term '''non-unity echelon matrix''' will be used to refer to this variant echelon matrix in which the leading coefficients are not necessarily equal to $1$. === $(1)$ implies $(2)$ === Let $\mathbf A$ be an non-unity echelon matrix by definition $1$. Then, apart from zero rows, each row starts with strictly more zeroes than the one before it. {{AimForCont}} there exist adjacent rows in $\mathbf A$ of the form: :$\begin {pmatrix} 0 & 0 & \cdots & 0 & x_1 & x_2 & \cdots \\ 0 & 0 & \cdots & 0 & y_1 & y_2 & \cdots \\ \end {pmatrix}$ where $y_1 \ne 0$. If $x_1 \ne 0$, then the $2$nd of these rows starts with the same number of zeroes as the row before it. If $x_1 = 0$, then the $2$nd of these rows starts with fewer zeroes as the row before it. In both cases, this contradicts our definition of a non-unity echelon matrix. That is, there there exist no adjacent rows in $\mathbf A$ of the form: :$\begin {pmatrix} 0 & 0 & \cdots & 0 & x_1 & x_2 & \cdots \\ 0 & 0 & \cdots & 0 & y_1 & y_2 & \cdots \\ \end {pmatrix}$ where $y_1 \ne 0$. That is $\mathbf A$ is a non-unity echelon matrix by definition $2$. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\mathbf A$ be a non-unity echelon matrix by definition $2$. Then by definition there exist no adjacent rows in $\mathbf A$ of the form: :$\begin {pmatrix} 0 & 0 & \cdots & 0 & x_1 & x_2 & \cdots \\ 0 & 0 & \cdots & 0 & y_1 & y_2 & \cdots \\ \end {pmatrix}$ where $y_1 \ne 0$. Suppose row $k$ does not start with a sequence of zeroes. Then it cannot start with strictly more zeroes than the previous row unless $k = 1$, in which case there is no previous row. Hence criterion $1$ of definition $1$ is satisfied. Suppose row $k$, where $k > 1$, is not a zero row. Let its leading coefficient be in column $r$. Then the leading coefficient of row $k - 1$ must be in column $s$ where $s < r$. That is, row $k$ starts with strictly more zeroes than row $k - 1$. Hence criterion $2$ of definition $1$ is satisfied. Suppose row $k$ is a zero row such that $k < m$. Then by definition row $k + 1$ cannot have a leading coefficient. So row $k + 1$ and all rows following must be zero rows. Hence criterion $3$ of definition $1$ is satisfied. Thus $\mathbf A$ is a non-unity echelon matrix by definition $1$. {{qed}}