hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === When $L' = L$, the claim is trivial. Let $L' \ne L$ be described by the equation: :$\alpha'_1 x + \alpha'_2 y = \beta'$ Without loss of generality, let $\alpha'_1 \ne 0$ (the case $\alpha'_2 \ne 0$ is similar). Then for $\tuple {x, y} \in L'$ to hold, one needs: {{begin-eqn}} {{eqn | l = \alpha'_1 x + \alpha'_2 y       | r = \beta' }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \frac {-\alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1} }} {{end-eqn}} For $L'$ to be parallel to $L$, it is required that then $\tuple {x, y} \notin L$, that is: {{begin-eqn}} {{eqn | l = \alpha_1 x + \alpha_2 y       | o = \ne       | r = \beta }} {{eqn | ll= \leadstoandfrom       | l = \alpha_1 \paren {\frac {- \alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1} } + \alpha_2 y       | o = \ne       | r = \beta }} {{eqn | ll= \leadstoandfrom       | l = \paren {\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} } y + \alpha_1 \frac {\beta'} {\alpha'_1}       | o = \ne       | r = \beta }} {{eqn | ll= \leadstoandfrom       | l = \paren {\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} } y       | o = \ne       | r = \beta - \alpha_1 \frac {\beta'} {\alpha'_1} }} {{end-eqn}} It follows that necessarily $\beta - \alpha_1 \frac {\beta'} {\alpha'_1} \ne 0$, or taking $y = 0$ would yield equality. The only remaining way to obtain the desired inequality for all $y$ is that: :$\alpha_2 - \alpha_1 \dfrac {\alpha'_2} {\alpha'_1} = 0$ One observes that now $\alpha_1 = 0 \implies \alpha_2 = 0$. However, as $L: \alpha_1 x + \alpha_2 y = \beta$ is a straight line in $\R^2$, it cannot be that $\alpha_1 = \alpha_2 = 0$. So $\alpha_1 \ne 0$, and one finds: :$\alpha'_2 = \dfrac {\alpha'_1} {\alpha_1} \alpha_2$ Hence obtain: {{begin-eqn}} {{eqn | l = \alpha'_1 x + \alpha'_2 y       | r = \beta' }} {{eqn | ll= \leadstoandfrom       | l = \frac {\alpha'_1} {\alpha_1} \paren {\alpha_1 x + \alpha_2 y}       | r = \beta' }} {{eqn | ll= \leadstoandfrom       | l = \alpha_1 x + \alpha_2 y       | r = \beta' \frac {\alpha_1} {\alpha'_1} }} {{end-eqn}} That is, $L'$ is described by an equation of the required form. {{qed|lemma}} {{proofread}} === Sufficient Condition === Let $L' \ne L$ be a straight line given by the equation: :$\alpha_1 x + \alpha_2 y = \beta'$ Suppose we have a point $\mathbf x = \tuple {x, y} \in L \cap L'$. Then, as $\mathbf x \in L$, it also satisfies: :$\alpha_1 x + \alpha_2 y = \beta$ It follows that $\beta = \beta'$, so $L = L'$. This contradiction shows that $L \cap L' = \O$, that is, $L$ and $L'$ are parallel. The remaining case is when $L' = L$.  By definition, $L$ is parallel to itself.  The result follows. {{qed}}
=== Necessary Condition === {{ProofWanted}} === Sufficient Condition === Let $P' \ne P$ be a plane given by the equation: :$\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma'$ Suppose we have a point $\mathbf x = \left({x_1, x_2, x_3}\right) \in P \cap P'$. Then, as $\mathbf x \in P$, it also satisfies: :$\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ It follows that $\gamma = \gamma'$, so $P = P'$. This contradiction shows that $P \cap P' = \varnothing$, i.e., $P$ and $P'$ are parallel. The remaining case is when $P' = P$. By definition, $P$ is parallel to itself.  The result follows. {{qed}}
$\struct {S, \cdot}$ is a semigroup and is therefore closed and associative. As $\struct {S, \cdot}$ is closed, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Closure of Hadamard Product. As $\struct {S, \cdot}$ is associative, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Associativity of Hadamard Product. Thus if $\struct {S, \cdot}$ is a semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$. If $\struct {S, \cdot}$ is commutative, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Commutativity of Hadamard Product. Thus if $\struct {S, \cdot}$ is a commutative semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$. Let $\struct {S, \cdot}$ be a monoid, with identity $e$. Then from Zero Matrix is Identity for Hadamard Product, $\struct {\map {\MM_S} {m, n}, \circ}$ also has an identity and is therefore also a monoid. {{Qed}} Category:Hadamard Product 5bd25x7uey0z2g4kzye6z130c57p57z
As $\struct {G, \cdot}$, being a group, is a monoid, it follows from Matrix Space Semigroup under Hadamard Product that $\struct {\map {\MM_G} {m, n}, \circ}$ is also a monoid. As $\struct {G, \cdot}$ is a group, it follows from Negative Matrix is Inverse for Hadamard Product that all elements of $\struct {\map {\MM_G} {m, n}, \circ}$ have an inverse. The result follows. {{Qed}}
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{p q}$ be matrices. From inspection of the subscripts, we can see that both $\paren {\mathbf A \mathbf B} \mathbf C$ and $\mathbf A \paren {\mathbf B \mathbf C}$ are defined: $\mathbf A$ has $n$ columns and $\mathbf B$ has $n$ rows, while $\mathbf B$ has $p$ columns and $\mathbf C$ has $p$ rows. Consider $\paren {\mathbf A \mathbf B} \mathbf C$. Let $\mathbf R = \sqbrk r_{m p} = \mathbf A \mathbf B, \mathbf S = \sqbrk s_{m q} = \mathbf A \paren {\mathbf B \mathbf C}$. Then: {{begin-eqn}}      {{eqn | l = s_{i j}       | r = \sum_{k \mathop = 1}^p r_{i k} \circ c_{k j}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | l = r_{i k}       | r = \sum_{l \mathop = 1}^n a_{i l} \circ b_{l k}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | ll= \leadsto       | l = s_{i j}       | r = \sum_{k \mathop = 1}^p \paren {\sum_{l \mathop = 1}^n a_{i l} \circ b_{l k} } \circ c_{k j} }} {{eqn | r = \sum_{k \mathop = 1}^p \sum_{l \mathop = 1}^n \paren {a_{i l} \circ b_{l k} } \circ c_{k j}       | c = {{Ring-axiom|D}} }} {{end-eqn}} Now consider $\mathbf A \paren {\mathbf B \mathbf C}$. Let $\mathbf R = \sqbrk r_{n q} = \mathbf B \mathbf C, \mathbf S = \sqbrk s_{m q} = \mathbf A \paren {\mathbf B \mathbf C}$. Then: {{begin-eqn}} {{eqn | l = s_{i j}       | r = \sum_{l \mathop = 1}^n a_{i l} \circ r_{l j}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | l = r_{l j}       | r = \sum_{k \mathop = 1}^p b_{l k} \circ c_{k j}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | ll= \leadsto       | l = s_{i j}       | r = \sum_{l \mathop = 1}^n a_{i l} \circ \paren {\sum_{k \mathop = 1}^p b_{l k} \circ c_{k j} } }} {{eqn | r = \sum_{l \mathop = 1}^n \sum_{k \mathop = 1}^p a_{i l} \circ \paren {b_{l k} \circ c_{k j} }       | c = {{Ring-axiom|D}} }} {{end-eqn}} Using {{Ring-axiom|M1}}: :$\displaystyle s_{i j} = \sum_{k \mathop = 1}^p \sum_{l \mathop = 1}^n \paren {a_{i l} \circ b_{l k} } \circ c_{k j} = \sum_{l \mathop = 1}^n \sum_{k \mathop = 1}^p a_{i l} \circ \paren {b_{l k} \circ c_{k j} } = s'_{i j}$ It is concluded that: :$\paren {\mathbf A \mathbf B} \mathbf C = \mathbf A \paren {\mathbf B \mathbf C}$ {{qed}}
The proof that $M$ is an isomorphism is straightforward. {{stub|The proof that $M$ is an isomorphism.}} The relation: : $\left[{v \circ u; \left \langle {c_m} \right \rangle, \left \langle {a_p} \right \rangle}\right] = \left[{v; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right] \left[{u; \left \langle {b_n} \right \rangle, \left \langle {a_p} \right \rangle}\right]$ follows from Relative Matrix of Composition of Linear Mappings. {{qed}} {{Proofread}}
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{n p}$ be matrices over a ring $\struct {R, +, \circ}$. Consider $\mathbf A \paren {\mathbf B + \mathbf C}$. Let $\mathbf R = \sqbrk r_{n p} = \mathbf B + \mathbf C, \mathbf S = \sqbrk s_{m p} = \mathbf A \paren {\mathbf B + \mathbf C}$. Let $\mathbf G = \sqbrk g_{m p} = \mathbf A \mathbf B, \mathbf H = \sqbrk h_{m p} = \mathbf A \mathbf C$. Then: {{begin-eqn}} {{eqn | l = s_{i j}       | r = \sum_{k \mathop = 1}^n a_{i k} \circ r_{k j}       | c =  }} {{eqn | l = r_{k j}       | r = b_{k j} + c_{k j}       | c =  }} {{eqn | ll= \leadsto       | l = s_{i j}       | r = \sum_{k \mathop = 1}^n a_{i k} \circ \paren {b_{k j} + c_{k j} }       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j} + \sum_{k \mathop = 1}^n a_{i k} \circ c_{k j}       | c =  }} {{eqn | r = g_{i j} + h_{i j}       | c =  }} {{end-eqn}} Thus: :$\mathbf A \paren {\mathbf B + \mathbf C} = \paren {\mathbf A \mathbf B} + \paren {\mathbf A \mathbf C}$ A similar construction shows that: :$\paren {\mathbf B + \mathbf C} \mathbf A = \paren {\mathbf B \mathbf A} + \paren {\mathbf C \mathbf A}$ {{Qed}}
In Unit Matrix is Identity for Matrix Multiplication, it is demonstrated that: :$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$  Hence the result, by definition of identity element {{qed}}
From the definition of matrix multiplication, the product of two matrices is another matrix. The order of an $m \times n$ multiplied by an $n \times p$ matrix is $m \times p$. The entries of that product matrix are elements of the ring over which the matrix is formed. Thus an $n \times n$ matrix over $R$ multiplied by an $n \times n$ matrix over $R$ gives another $n \times n$ matrix over $R$. Hence the result. {{qed}} Category:Conventional Matrix Multiplication Category:Algebraic Closure 9qwfqajxemqtzbxb0nsxfztyliq2lvf
: Matrix Multiplication is Closed. : Matrix Multiplication is Associative. : The Unit Matrix is Unity of Ring of Square Matrices. {{qed}}
From Ring of Square Matrices over Ring with Unity we have that $\struct {\map {\MM_R} n, +, \times}$ is a ring with unity. However, Matrix Multiplication is not Commutative. Hence $\struct {\map {\MM_R} n, +, \times}$ is not a commutative ring for $n \ge 2$. For $n = 1$ we have that: {{begin-eqn}} {{eqn | lo= \forall \mathbf A, \mathbf B \in \map {\MM_R} 1:       | l = \mathbf A \mathbf B       | r = a_{11} b_{11}       | c = where $\mathbf A = \begin {pmatrix} a_11 \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} b_11 \end {pmatrix}$ }} {{eqn | r = b_{11} a_{11}       | c = as $R$ is a commutative ring }} {{eqn | r = \mathbf {B A}       | c =  }} {{end-eqn}} Thus, for $n = 1$, $\struct {\map {\MM_R} n, +, \times}$ ''is'' a commutative ring. {{qed}}
From Product of Change of Basis Matrices and Change of Basis Matrix Between Equal Bases: * $\left[{I_M; \mathcal A, \mathcal B}\right] \left[{I_M; \mathcal B, \mathcal A}\right] = \left[{I_M; \mathcal A, \mathcal A}\right] = I_n$ * $\left[{I_M; \mathcal B, \mathcal A}\right] \left[{I_M; \mathcal A, \mathcal B}\right] = \left[{I_M; \mathcal B , \mathcal B}\right] = I_n$ Hence the result. {{qed}}
From Change of Basis is Invertible, if $\left \langle {b_n} \right \rangle$ is an ordered basis of $G$ then $\mathbf P$ is invertible. Now let $\mathbf P$ be invertible. Then by the corollary to Linear Transformations Isomorphic to Matrix Space, there is an automorphism $u$ of $G$ which satisfies $\mathbf P = \left[{u; \left \langle {a_n} \right \rangle}\right]$. Therefore, as $\forall j \in \left[{1 \,.\,.\, n}\right]: b_j = u \left({a_j}\right)$, it follows that $\left \langle {b_n} \right \rangle$ is also an ordered basis of $G$. {{Qed}}
We have $u = I_H \circ u \circ I_G$ and $\mathbf Q^{-1} = \left[{I_H; \left \langle {{b_m}'} \right \rangle, \left \langle {b_m} \right \rangle}\right]$. Thus by Linear Transformations Isomorphic to Matrix Space: {{begin-eqn}} {{eqn | l = \mathbf Q^{-1} \mathbf A \mathbf P       | r = \left[{I_H; \left \langle { {b_m}'} \right \rangle, \left \langle {b_m} \right \rangle}\right] \left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right] \left[{I_G; \left \langle {a_n} \right \rangle, \left \langle { {a_n}'} \right \rangle}\right]       | c =  }} {{eqn | r = \left[{I_H \circ u \circ I_G; \left \langle { {b_m}'} \right \rangle, \left \langle { {a_n}'} \right \rangle}\right]       | c =  }} {{eqn | r = \mathbf B       | c =  }} {{end-eqn}} {{qed}}
Checking in turn each of the critera for equivalence: === Reflexive === $\mathbf A = \mathbf{I_m}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all $m \times n$ matrices $\mathbf A$. Thus reflexivity holds. {{qed|lemma}} === Symmetric === Let $\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$. As $\mathbf P$ and $\mathbf Q$ are both invertible, we have: {{begin-eqn}} {{eqn | l=\mathbf Q \mathbf B \mathbf P^{-1}       | r=\mathbf Q \mathbf Q^{-1} \mathbf A \mathbf P \mathbf P^{-1}       | c= }} {{eqn | r=\mathbf{I_m} \mathbf A \mathbf{I_n}       | c= }} {{eqn | r=\mathbf A       | c= }} {{end-eqn}} Thus symmetry holds. {{qed|lemma}} === Transitive === Let $\mathbf B = \mathbf Q_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf Q_2^{-1} \mathbf B \mathbf P_2$. Then $\mathbf C = \mathbf Q_2^{-1} \mathbf Q_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$. Transitivity follows from the definition of invertible matrix, that the product of two invertible matrices is itself invertible. {{qed|lemma}} Hence the result by definition of equivalence relation. {{qed}}
Checking in turn each of the critera for equivalence: === Reflexive === $\mathbf A = \mathbf{I_n}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all order $n$ square matrices $\mathbf A$. So matrix similarity is reflexive. {{qed|lemma}} === Symmetric === Let $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$. As $\mathbf P$ is invertible, we have: {{begin-eqn}} {{eqn | l = \mathbf P \mathbf B \mathbf P^{-1}       | r = \mathbf P \mathbf P^{-1} \mathbf A \mathbf P \mathbf P^{-1}       | c =  }} {{eqn | r = \mathbf{I_n} \mathbf A \mathbf{I_n}       | c =  }} {{eqn | r = \mathbf A       | c =  }} {{end-eqn}} So matrix similarity is symmetric. {{qed|lemma}} === Transitive === Let $\mathbf B = \mathbf P_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf P_2^{-1} \mathbf B \mathbf P_2$. Then $\mathbf C = \mathbf P_2^{-1} \mathbf P_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$. The result follows from the definition of invertible matrix, that the product of two invertible matrices is itself invertible. So matrix similarity is transitive. {{qed|lemma}} So, by definition, matrix similarity is an equivalence relation. {{qed}}
If $\mathbf A \sim \mathbf B$ then $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$. Let $\mathbf Q = \mathbf P$. Then $\mathbf A$ are equivalent to $\mathbf B$, as: :$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$ {{qed}}
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$ such that $\mathbf A \equiv \mathbf B$. Let $S$ and $T$ be vector spaces of dimensions $n$ and $m$ over $K$. Let $\mathbf A$ be the matrix of a linear transformation $u: S \to T$ relative to the ordered bases $\sequence {a_n}$ of $S$ and $\sequence {b_m}$ of $T$. Let $\psi: K^m \to T$ be the isomorphism defined as: :$\displaystyle \map \psi {\sequence {\lambda_m} } = \sum_{k \mathop = 1}^m \lambda_k b_k$ Then $\psi$ takes the $j$th column of $\mathbf A$ into $\map u {a_j}$. Hence it takes the subspace of $K^m$ generated by the columns of $\mathbf A$ onto the codomain of $u$. Thus $\map \rho {\mathbf A} = \map \rho u$, and equivalent matrices over a field have the same rank. Now let $\map {\mathcal L} {K^n, K^m}$ be the set of all linear transformations from $K^n$ to $K^m$. Let $u, v \in \map {\mathcal L} {K^n, K^m}$ such that $\mathbf A$ and $\mathbf B$ are respectively the matrices of $u$ and $v$ relative to the standard ordered bases of $K^n$ and $K^m$. Let $r = \map \phi {\mathbf A}$. By Linear Transformation from Ordered Basis less Kernel, there exist ordered bases $\sequence {a_n}, \sequence {a'_n}$ of $K^n$ such that: :$\sequence {\map u {a_r} }$ and $\sequence {\map v {a'_r} }$ are ordered bases of $\map u {K^n}$ and $\map v {K^n}$ respectively and such that: :$\set {a_k: k \in \closedint {r + 1} n}$ and $\set {a'_k: k \in \closedint {r + 1} n}$ are respectively bases of the kernels of $u$ and $v$. Thus, by Results concerning Generators and Bases of Vector Spaces there exist ordered bases $\sequence {b_m}$ and $\sequence {b'_m}$ of $K^m$ such that $\forall k \in \closedint 1 r$: {{begin-eqn}} {{eqn | l = b_k       | r = \map u {a_k}       | c =  }} {{eqn | l = b'_k       | r = \map v {a'_k}       | c =  }} {{end-eqn}} Let $z$ be the automorphism of $K^n$ which satisfies $\forall k \in \closedint 1 n: \map z {a'_k} = a_k$. Let $w$ be the automorphism of $K^m$ which satisfies $\forall k \in \closedint 1 m: \map w {b'_k} = b_k$. Then:: $\map {\paren {w^{-1} \circ u \circ z} } {a'_k} = \begin{cases} \map {w^{-1} } {b_k} = \map v {a_k} & : k \in \closedint 1 r \\ 0 = \map v {a_k} & : k \in \closedint 1 {r + 1} n \end{cases}$ So $w^{-1} \circ u \circ z = v$. Now let $\mathbf P$ be the matrix of $z$ relative to the standard ordered bases of $K^n$, and let $\mathbf Q$ be the matrix of $w$ relative to the standard ordered basis of $K^m$. Then $\mathbf P$ and $\mathbf Q$ are invertible and: :$\mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$ and thus: :$\mathbf A \equiv \mathbf B$ {{Qed}} {{Proofread}}
Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{n p}$ Let $\mathbf A \mathbf B = \sqbrk c_{m p}$. Then from the definition of matrix product: :$\displaystyle \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$ So, let $\paren {\mathbf A \mathbf B}^\intercal = \sqbrk r_{p m}$. The dimensions are correct, because $\mathbf A \mathbf B$ is an $m \times p$ matrix, thus making $\paren {\mathbf A \mathbf B}^\intercal$ a $p \times m$ matrix. Thus: :$\displaystyle \forall j \in \closedint 1 p, i \in \closedint 1 m: r_{j i} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$ Now, let $\mathbf B^\intercal \mathbf A^\intercal = \sqbrk s_{p m}$ Again, the dimensions are correct because $\mathbf B^\intercal$ is a $p \times n$ matrix and $\mathbf A^\intercal$ is an $n \times m$ matrix. Thus: :$\displaystyle \forall j \in \closedint 1 p, i \in \closedint 1 m: s_{j i} = \sum_{k \mathop = 1}^n b_{k j} \circ a_{i k}$ As the underlying structure of $\mathbf A$ and $\mathbf B$ is a commutative ring, then $a_{i k} \circ b_{k j} = b_{k j} \circ a_{i k}$. Note the order of the indices in the term in the summation sign on the {{RHS}} of the above. They are reverse what they would normally be because we are multiplying the transposes together. Thus it can be seen that $r_{j i} = s_{j i}$ and the result follows. {{qed}}
Let $u: K^n \to K^m$ be the linear transformation such that $\mathbf A$ is the matrix of $u$ relative to the standard ordered bases of $K^n$ and $K^m$. Let $\rho \left({\mathbf A}\right)$ be the rank of $\mathbf A$. Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$. Similar notations on $u$ denote the rank and transpose of $u$. We have $\rho \left({\mathbf A}\right) = \rho \left({u}\right)$ and $\rho \left({\mathbf A^\intercal}\right) = \rho \left({u^\intercal}\right)$, but $\rho \left({u^\intercal}\right) = \rho \left({u}\right)$ from Rank and Nullity of Transpose. {{finish}}
Taking the group axioms in turn: === Group Axiom $\text G 0$: Closure === The matrix product of two $n \times n$ matrices is another $n \times n$ matrix. The matrix product of two invertible matrices is another invertible matrix. Thus $\GL {n, K}$ is closed. {{qed|lemma}} === Group Axiom $\text G 1$: Associativity === Matrix Multiplication is Associative. {{qed|lemma}} === Group Axiom $\text G 2$: Identity === From Unit Matrix is Unity of Ring of Square Matrices, the unit matrix serves as the identity of $\GL {n, K}$. {{qed|lemma}} === Group Axiom $\text G 3$: Inverses === From the definition of invertible matrix, the inverse of any invertible matrix $\mathbf A$ is $\mathbf A^{-1}$. {{qed}}
We can see the truth of this by writing them out in full. :$\displaystyle \sum_{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$ can be written as: {{begin-eqn}} {{eqn | l = \alpha_{1 1} x_1 + \alpha_{1 2} x_2 + \ldots + \alpha_{1 n} x_n       | r = \beta_1       | c =  }} {{eqn | l = \alpha_{2 1} x_1 + \alpha_{2 2} x_2 + \ldots + \alpha_{2 n} x_n       | r = \beta_2       | c = }} {{eqn | o = \vdots }} {{eqn | l = \alpha_{m 1} x_1 + \alpha_{m 2} x_2 + \ldots + \alpha_{m n} x_n       | r = \beta_m       | c =  }} {{end-eqn}} while $\sqbrk \alpha_{m n} \sqbrk x_{n 1} = \sqbrk \beta_{m 1}$ can be written as: :$\begin {bmatrix} \alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\ \alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\       \vdots &       \vdots & \ddots &       \vdots \\ \alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n} \end {bmatrix} \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} = \begin {bmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_m \end {bmatrix}$ So the question: :Find a solution to the following system of $m$ simultaneous linear equations in $n$ variables is equivalent to: :Given the following element $\mathbf A \in \map {\MM_K} {m, n}$ and $\mathbf b \in \map {\MM_K} {m, 1}$, find the set of all $\mathbf x \in \map {\MM_K} {n, 1}$ such that $\mathbf A \mathbf x = \mathbf b$ where $\map {\MM_K} {m, n}$ is the $m \times n$ matrix space over $S$. {{qed}}
From the definition of an infinite cyclic group, we have: :$G = \gen a = \set {a^k: k \in \Z}$ Let us define the mapping: :$\phi: \Z \to G: \map \phi k = a^k$. We now show that $\phi$ is an isomorphism. From Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism, $\phi$ is a homomorphism. Now we show that $\phi$ is a surjection. As $G$ is cyclic, every element of $G$ is apower of $a$ for some $a \in G$ such that $G = \gen a$. Thus: :$\forall x \in G: \exists k \in \Z: x = a^k$ By the definition of $\phi$: :$\map \phi k = a^k = x$ Thus $\phi$ is surjective. Now we show that $\phi$ is an injection. This follows directly from Powers of Infinite Order Element, where: :$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$ Thus $\phi$ is an injective, surjective homomorphism, thus: : $G \cong \struct {\Z, +}$ as required. {{Qed}}
The fact that $H \le G$ follows from the definition of subgroup generator. By Infinite Cyclic Group is Isomorphic to Integers: :$G \cong \struct {\Z, +}$ Now we show that $H$ is of infinite order. Suppose $\exists h \in H, h \ne e: \exists r \in \Z, r > 0: h^r = e$. But: :$h \in H \implies \exists s \in \Z, s > 0: h = g^s$ where $g = a^k$. Thus: :$e = h^r = \paren {g^s}^r = \paren {\paren {a^k}^s}^r = a^{k s r}$ and thus $a$ is of finite order. This would mean that $G$ was also of finite order. So $H$ must be of infinite order. From Subgroup of Cyclic Group is Cyclic, as $G$ is cyclic, then $H$ must also be cyclic. From Infinite Cyclic Group is Isomorphic to Integers: :$H \cong \struct {\Z, +}$ Therefore, as $G \cong \struct {\Z, +}$: :$H \cong G$ {{qed}}
Let $C_n = \gen {a: a^n = e_{C_n} }$, that is, let $a$ be a generator of $C_n$. Let us define $\phi: \struct {\Z, +} \to C_n$ such that: :$\forall k \in \Z: \map \phi k = a^k$ Then from the First Isomorphism Theorem: :$\Img \phi = C_n = \struct {\Z, +} / \map \ker \phi$ We now need to show that $\map \ker \phi = n \Z$. We have: :$\map \ker \phi = \set {k \in \Z: a^k = e_{C_n} }$ Let $x \in \map \ker \phi$. Then $a^x = e_{C_n}$ and thus: :$n \divides x$ The result follows. {{qed}}
As $\struct {R, +, \circ}$ is a ring, it follows that $\struct {R, +}$ is an abelian group. Thus by Subgroup of Abelian Group is Normal, all subgroups of $\struct {R, +, \circ}$ are normal. So from the definition of quotient group, it follows directly that $X +_\mathcal P Y$ in $\powerset R$ is also the sum in the quotient ring $R / J$. The set $\ideal 5$ of all integral multiples of $5$ is a principal ideal of the ring $\Z$. {{finish}} In the ring $\Z / \ideal 5$ we have: :$\ideal 5 \circ \ideal 5 = \ideal 5$ However, in $\powerset \Z$, we have $\ideal 5 \circ_\mathcal P \ideal 5 = \ideal {25}$. {{finish|Plenty work needed here.}}