hcju/proofwikips huggingface dataset:
Direct Product of Vector Spaces is Vector Space
Let $K$ be a field. Let $V_1, V_2, \ldots, V_n$ be $K$-vector spaces. Let $\struct {V, + , \circ}_K$ be their direct product. Then $\struct {V, + , \circ}_K$ is a $K$-vector space.
Definition:Vector Space on Cartesian Product
Let $\struct {K, +, \circ}$ be a division ring. Let $n \in \N_{>0}$. Let $+: K^n \times K^n \to K^n$ be defined as: :$\tuple {\alpha_1, \ldots, \alpha_n} + \tuple {\beta_1, \ldots, \beta_n} = \tuple {\alpha_1 +_R \beta_1, \ldots, \alpha_n +_R \beta_n}$ Let $\times: K \times K^n \to K^n$ be defined as: :$\lambda \times \tuple {\alpha_1, \ldots, \alpha_n} = \tuple {\lambda \times_R \alpha_1, \ldots, \lambda \times_R \alpha_n}$ Then $\struct {K^n, +, \times}_K$ is '''the $K$-vector space $K^n$'''.
Definition:Vector Space over Division Subring
Let $\struct {R, +, \circ}$ be a ring with unity whose unity is $1_R$. Let $S$ be a division subring of $R$, such that $1_R \in S$. Then $\struct {R, +, \circ_S}_S$, where $\circ_S$ is the restriction of $\circ$ to $S \times R$, is the '''vector space on $R$ over the division subring $S$'''.
Division Ring is Vector Space over Prime Subfield
Let $\struct {K, +, \times}$ be a division ring. Let $\struct {S, +, \times}$ be the prime subfield of $K$ Then $\struct {K, +, \times_S}_S$ is an $S$-vector space, where $\times_S$ is the restriction of $\times$ to $S \times K$.
Vector Space over Subring
Let $K$ be a division subring of the division ring $\struct {L, +_L, \times_L}$. Let $\struct {G, +_G, \circ}_L$ be a $L$-vector space. Then $\struct {G, +_G, \circ_K}_K$ is a $K$-vector space, where $\circ_K$ is the restriction of $\circ$ to $K \times G$. The $K$-vector space $\struct {G, +_G, \circ_K}_K$ is called the '''$K$-vector space obtained from $\struct {L, +_L, \times_L}$ by restricting scalar multiplication'''.
Subspaces of Dimension 2 Real Vector Space
Take the $\R$-vector space $\left({\R^2, +, \times}\right)_\R$. Let $S$ be a subspace of $\left({\R^2, +, \times}\right)_\R$. Then $S$ is one of: : $(1): \quad \left({\R^2, +, \times}\right)_\R$ : $(2): \quad \left\{{0}\right\}$ : $(3): \quad$ A line through the origin.
Subspace of Real Continuous Functions
Let $\mathbb J = \set {x \in \R: a \le x \le b}$ be a closed interval of the real number line $\R$. Let $\map \CC {\mathbb J}$ be the set of all continuous real functions on $\mathbb J$. Then $\struct {\map \CC {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
Singleton is Linearly Independent
Let $K$ be a division ring. Let $\struct {G, +_G}$ be a group whose identity is $e$. Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space whose zero is $0_K$. Let $x \in G: x \ne e$. Then $\set x$ is a linearly independent subset of $G$.
Linearly Dependent Sequence of Vector Space
Let $\struct {G, +}$ be a group whose identity is $\mathbf 0$. Let $\struct {G, +, \circ}_K$ be a $K$-vector space. Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct non-zero vectors of $G$. Then $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent : :$\exists p \in \closedint 2 n: a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$
Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set
Let $K$ be a division ring. Let $G$ be a finitely generated $K$-vector space. Let $H$ be a linearly independent subset of $G$. Let $F$ be a finite generator for $G$ such that $H \subseteq F$. Then there is a basis $B$ for $G$ such that $H \subseteq B \subseteq F$.
Bases of Finitely Generated Vector Space have Equal Cardinality
Let $K$ be a division ring. Let $G$ be a finitely generated $K$-vector space. Then any two bases of $G$ are finite and equivalent. {{improve|I still think there's a simpler way of saying "they have the same number of elements" than bringing in all that top-heavy technical set theoretic language of "set equivalence" and "cardinality" and so on. Since we are talking about a finite set, the complexities which arise with regard to transfinites do not arise, so there is no direct need to go into such depth.}}
Linearly Independent Subset also Independent in Generated Subspace
Let $G$ be a finitely generated $K$-vector space. Let $S$ be a linearly independent subset of $G$. Let $M$ be the subspace of $G$ generated by $S$. If $M \ne G$, then $\forall b \in G: b \notin M$, the set $S \cup \set b$ is linearly independent.
Sufficient Conditions for Basis of Finite Dimensional Vector Space
Let $K$ be a division ring. Let $n \ge 0$ be a natural number. Let $E$ be an $n$-dimensional vector space over $K$. Let $B \subseteq E$ be a subset such that $\card B = n$.  : $(1): \quad$ $B$ is a basis of $E$. : $(2): \quad$ $B$ is linearly independent. : $(3): \quad$ $B$ is a generator for $E$.
Dimension of Proper Subspace is Less Than its Superspace
Let $G$ be a vector space whose dimension is $n$. Let $H$ be a subspace of $G$. Then $H$ is finite dimensional, and $\map \dim H \le \map \dim G$. If $H$ is a proper subspace of $G$, then $\map \dim H < \map \dim G$.
Results concerning Generators and Bases of Vector Spaces
Let $E$ be a vector space of $n$ dimensions. Let $G$ be a generator for $E$. Then $G$ has the following properties:
Grassmann's Identity
Let $K$ be a division ring. Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space. Let $M$ and $N$ be finite-dimensional subspaces of $G$. Then the sum $M + N$ and intersection $M \cap N$ are finite-dimensional, and: :$\map \dim {M + N} + \map \dim {M \cap N} = \map \dim M + \map \dim N$
Rank Plus Nullity Theorem
Let $G$ be an $n$-dimensional vector space. Let $H$ be a vector space. Let $\phi: G \to H$ be a linear transformation. Let $\map \rho \phi$ and $\map \nu \phi$ be the rank and nullity respectively of $\phi$. Then the image of $\phi$ is finite-dimensional, and: :$\map \rho \phi + \map \nu \phi = n$ By definition of rank and nullity, it can be seen that this is equivalent to the alternative way of stating this result: :$\map \dim {\Img \phi} + \map \dim {\map \ker \phi} = \map \dim G$ {{wtd|and the theorem is applicable to matrices}}
Linear Transformation of Vector Space Monomorphism
Let $G$ and $H$ be a $K$-vector space. Let $\phi: G \to H$ be a linear transformation. Then $\phi$ is a monomorphism  for every linearly independent sequence $\sequence {a_n}$ of vectors of $G$, $\sequence {\map \phi {a_n} }$ is a linearly independent sequence of vectors of $H$.
Linear Transformation of Vector Space Equivalent Statements
Let $G$ and $H$ be $n$-dimensional vector spaces. Let $\phi: G \to H$ be a linear transformation. Then these statements are equivalent: : $(1): \quad \phi$ is an isomorphism. : $(2): \quad \phi$ is a monomorphism. : $(3): \quad \phi$ is an epimorphism. : $(4): \quad \phi \left({B}\right)$ is a basis of $H$ for every basis $B$ of $G$. : $(5): \quad \phi \left({B}\right)$ is a basis of $H$ for some basis $B$ of $G$.
Results Concerning Annihilator of Vector Subspace
Let $G$ be an $n$-dimensional vector space over a field. Let $J: G \to G^{**}$ be the evaluation isomorphism. Let $G^*$ be the algebraic dual of $G$. Let $G^{**}$ be the algebraic dual of $G^*$. Let $M$ be an $m$-dimensional subspace of $G$. Let $N$ be a $p$-dimensional subspace of $G^*$. Let $M^\circ$ be the annihilator of $M$. Then: :$(1): \quad M^\circ$ is an $\paren {n - m}$-dimensional subspace of $G^*$, and $M^{\circ \circ} = \map J M$ :$(2): \map {\quad J^{-1} } {N^\circ}$ is an $\paren {n - p}$-dimensional subspace of $G$ :$(3): \quad$ The mapping $M \to M^\circ$ is a bijection from the set of all $m$-dimensional subspaces of $G$ onto the set of all $\paren {n - m}$-dimensional subspaces of $G^*$ :$(4): \quad$ Its inverse is the bijection $N \to \map {J^{-1} } {N^\circ}$.
Vector Subspace Dimension One Less
Let $K$ be a field. Let $M$ be a subspace of the $n$-dimensional vector space $K^n$. The following statements are equivalent: :$(1): \quad \map \dim M = n - 1$ :$(2): \quad M$ is the kernel of a nonzero linear form :$(3): \quad$ There exists a sequence $\sequence {\alpha_n} $ of scalars, not all of which are zero, such that: :::$M = \set {\tuple {\lambda_1, \ldots, \lambda_n} \in K^n: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}$ Also, suppose the above hold. Let $\sequence {\beta_n}$ be a sequence of scalars such that: :$M = \set {\tuple {\lambda_1, \ldots, \lambda_n} \in K^n: \beta_1 \lambda_1 + \cdots + \beta_n \lambda_n = 0}$ Then there is a non-zero scalar $\gamma$ such that: :$\forall k \in \closedint 1 n: \beta_k = \gamma \alpha_k$
Rank and Nullity of Transpose
Let $G$ and $H$ be $n$-dimensional vector spaces over a field. Let $\mathcal L \left({G, H}\right)$ be the set of all linear transformations from $G$ to $H$. Let $u \in \mathcal L \left({G, H}\right)$. Let $u^t$ be the transpose of $u$. Then: : $(1): \quad$ $u$ and $u^t$ have the same rank and nullity : $(2): \quad$ $\ker \left({u^t}\right)$ is the annihilator of the image of $u$ : $(3): \quad$ The image of $u^t$ is the annihilator of $\ker \left({u}\right)$.
Linear Operator on the Plane
Let $\phi$ be a linear operator on the real vector space of two dimensions $\R^2$. Then $\phi$ is completely determined by an ordered tuple of $4$ real numbers.
Similarity Mapping of Plane is Linear Operator
Let $G$ be a vector space over a field $K$. Let $\beta \in K$. Then the mapping: :$s_\beta: G \to G$ defined by $\map {s_\beta} {\mathbf x} = \beta \mathbf x$ is a linear operator on $G$. If $\beta \ne 0$ then $s_\beta$ is an automorphism of $G$, and $\paren {s_\beta}^{-1} = s_{\beta^{-1} }$ The linear operators $s_\beta$, where $\beta \ne 0$, are called '''similarities of $G$'''.
Cantor-Dedekind Hypothesis
The points on an infinite straight line are in one-to-one correspondence with the set $\R$ of real numbers. Hence the set of all points on an infinite straight line and $\R$ are equinumerous.
Full Angle measures 2 Pi Radians
One full angle is equal to $2 \pi$ radians. :$2 \pi \approx 6 \cdotp 28318 \, 53071 \, 79586 \, 4769 \ldots$ {{OEIS|A019692}}
Rotation of Plane about Origin is Linear Operator
Let $r_\alpha$ be the plane rotation of the plane about the origin through an angle of $\alpha$. That is, let $r_\alpha: \R^2 \to \R^2$ be the mapping defined as: :$\forall x \in \R^2: \map {r_\alpha} x = \text { the point into which a rotation of } \alpha \text{ carries } x$ Then $r_\alpha$ is a linear operator determined by the ordered sequence: : $\tuple {\cos \alpha  -\sin \alpha, \sin \alpha + \cos \alpha}$
Stretching and Contraction Mappings of Plane are Linear Operators
Let $s_\beta: \R^2 \to \R^2$ be a similarity of $\R^2$. Then $s_{-1}$ is the same as the rotation $r_{\pi}$ of the plane about the origin one half turn. If $\beta \ge 1$, then $s_\beta$ is called a '''stretching''', and if $0 < \beta \le 1$, $s_\beta$ is called a '''contraction'''. If $\beta < 0$, then $s_\beta$ is a stretching or contraction followed by a rotation one half turn. It is also the same as a rotation one half turn followed by a stretching or contraction.
Reflection of Plane in Line through Origin is Linear Operator
Let $M$ be a straight line in the plane passing through the origin. Then the '''reflection''' $s_M$ of $\R^2$ in $M$ is the rotation of the plane in space through one half turn about $M$ as an axis. :$s_M \circ s_M = I_{\R^2}$ and hence: :$s_M = s_M^{-1}$ If $M$ is the $x$-axis then $\map {s_M} {\lambda_1, \lambda_2} = \tuple {\lambda_1, -\lambda_2}$. If $M$ is the $y$-axis then $\map {s_M} {\lambda_1, \lambda_2} = \tuple {-\lambda_1, \lambda_2}$. In general, $s_M$ is a linear operator for every straight line $M$ through the origin.
Projection of Straight Line on Another in Plane
Let $M$ and $N$ be distinct straight lines through the plane through the origin. Let $\operatorname{pr}_{M, N}$ be the projection on $M$ along $N$. $M$ and $N$ are respectively the codomain and kernel of $\operatorname{pr}_{M, N}$. {{explain|As the kernel is a concept defined in relation to a homomorphism, it needs to be clarified what homomorphism is being considered.}} :$\operatorname{pr}_{M, N} \left({x}\right) = x \iff x \in M$ If $M$ is the $x$-axis and $N$ is the $y$-axis, then $\operatorname{pr}_{M, N} \left({\lambda_1, \lambda_2}\right) = \left({\lambda_1, 0}\right)$. If $M$ is the $y$-axis and $N$ is the $x$-axis, then $\operatorname{pr}_{M, N} \left({\lambda_1, \lambda_2}\right) = \left({0, \lambda_2}\right)$. Any such projection is a linear operator.
Condition for Straight Lines in Plane to be Parallel
Let $L: \alpha_1 x + \alpha_2 y = \beta$ be a straight line in $\R^2$. Then the straight line $L'$ is parallel to $L$ iff there is a $\beta' \in \R^2$ such that: :$L' = \set {\tuple {x, y} \in \R^2: \alpha_1 x + \alpha_2 y = \beta'}$
Equation of Plane
A plane $P$ is the set of all $\tuple {x, y, z} \in \R^3$, where: :$\alpha_1 x + \alpha_2 y + \alpha_3 z = \gamma$ where $\alpha_1, \alpha_2, \alpha_3, \gamma \in \R$ are given, and not all of $\alpha_1, \alpha_2, \alpha_3$ are zero.
Planes are Subspaces of Space
The two-dimensional subspaces of $\R^3$ are precisely the homogeneous planes of solid analytic geometry.
Matrix Space Semigroup under Hadamard Product
Let $\map {\MM_S} {m, n}$ be the matrix space over a semigroup $\struct {S, \cdot}$. Then the algebraic structure $\struct {\map {\MM_S} {m, n}, \circ}$, where $\circ$ is the Hadamard product, is also a semigroup. If $\struct {S, \cdot}$ is a commutative semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$. If $\struct {S, \cdot}$ is a monoid then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
Hadamard Product over Group forms Group
Let $\struct {G, \cdot}$ be a group whose identity is $e$. Let $\map {\MM_G} {m, n}$ be a $m \times n$ matrix space over $\struct {G, \cdot}$. Then $\struct {\map {\MM_G} {m, n}, \circ}$, where $\circ$ is Hadamard product, is also a group.
Matrix Multiplication is Associative
Let $R$ be a ring. Matrix multiplication (conventional) is associative.
Linear Transformations Isomorphic to Matrix Space
Let $R$ be a ring with unity. Let $F$, $G$ and $H$ be free $R$-modules of finite dimension $p,n,m>0$ respectively. Let $\left \langle {a_p} \right \rangle$, $\left \langle {b_n} \right \rangle$ and $\left \langle {c_m} \right \rangle$ be ordered bases  Let $\operatorname{Hom} \left({G, H}\right)$ be the set of all linear transformations from $G$ to $H$. Let $\mathcal M_R \left({m, n}\right)$ be the $m \times n$ matrix space over $R$. Let $\left[{u; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right]$ be the matrix of $u$ relative to $\left \langle {b_n} \right \rangle$ and $\left \langle {c_m} \right \rangle$. Let $M: \operatorname{Hom} \left({G, H}\right) \to \mathcal M_R \left({m, n}\right)$ be defined as: :$\forall u \in \operatorname{Hom} \left({G, H}\right): M \left({u}\right) = \left[{u; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right]$ Then $M$ is an isomorphism of modules, and: :$\forall u \in \operatorname{Hom} \left({F, G}\right), v \in \operatorname{Hom} \left({G, H}\right): \left[{v \circ u; \left \langle {c_m} \right \rangle, \left \langle {a_p} \right \rangle}\right] = \left[{v; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right] \left[{u; \left \langle {b_n} \right \rangle, \left \langle {a_p} \right \rangle}\right]$
Matrix Multiplication Distributes over Matrix Addition
Matrix multiplication (conventional) is distributive over matrix entrywise addition.
Unit Matrix is Unity of Ring of Square Matrices
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 $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$. The unit matrix over $R$: :$\mathbf I_n = \begin {pmatrix} 1_R & 0_R & 0_R & \cdots & 0_R \\ 0_R & 1_R & 0_R & \cdots & 0_R \\ 0_R & 0_R & 1_R & \cdots & 0_R \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0_R & 0_R & 0_R & \cdots & 1_R \end {pmatrix}$ is the identity element of $\struct {\map {\MM_R} n, +, \times}$.
Matrix Multiplication is Closed
Let $\struct {R, +, \circ}$ be a ring. Let $\map {\MM_R} n$ be a $n \times n$ matrix space over $R$. Then matrix multiplication (conventional) over $\map {\MM_R} n$ is closed.
Square Matrices over Real Numbers under Multiplication form Monoid
Let $\map {\mathcal M_\R} n$ be a $n \times n$ matrix space over the set of real numbers $\R$. Then the set of all $n \times n$ real matrices $\map {\mathcal M_\R} n$ under matrix multiplication (conventional) forms a monoid.
Ring of Square Matrices over Commutative Ring with Unity
Let $R$ be a commutative ring with unity. Let $n \in \Z_{>0}$ be a (strictly) positive integer. Let $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$. Then $\struct {\map {\MM_R} n, +, \times}$ is a ring with unity. However, for $n \ge 2$, $\struct {\map {\MM_R} n, +, \times}$ is not a commutative ring.
Invertible Matrix corresponds to Automorphism
Let $R$ be a ring with unity. Let $n \in \Z_{>0}$ be a (strictly) positive integer. Let $G$ be an $n$-dimensional $R$-module. Let $\map {\mathcal M_R} n$ be the $n \times n$ matrix space over $R$. Let $\map {\mathcal L_R} G$ be the set of all linear operators on $G$. Then the invertible elements of the ring of square matrices $\struct {\map {\mathcal M_R} n, +, \times}$ correspond directly to automorphisms of $\map {\mathcal L_R} G$.
Change of Basis is Invertible
Let $R$ be a ring with unity. Let $M$ be a free $R$-module of finite dimension $n>0$. Let $\mathcal A$ and $\mathcal B$ be ordered bases of $M$. Let $\mathbf P$ be the change of basis matrix from $\mathcal A$ to $\mathcal B$. Then $\mathbf P$ is invertible, and its inverse $\mathbf P^{-1}$ is the change of basis matrix from $\mathcal B$ to $\mathcal A$.
Matrix Corresponding to Change of Basis under Linear Transformation
Let $R$ be a ring with unity. Let $G$ and $H$ be free $R$-modules of finite dimensions $n,m>0$ respectively. Let $\left \langle {a_n} \right \rangle$ and $\left \langle {{a_n}'} \right \rangle$ be ordered bases of $G$. Let $\left \langle {b_m} \right \rangle$ and $\left \langle {{b_m}'} \right \rangle$ be ordered bases of $H$. Let $u: G \to H$ be a linear transformation, and let $\left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right]$ be the matrix of $u$ relative to $\left \langle {a_n} \right \rangle$ and $\left \langle {b_m} \right \rangle$. Let: : $\mathbf A = \left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right]$ : $\mathbf B = \left[{u; \left \langle {{b_m}'} \right \rangle, \left \langle {{a_n}'} \right \rangle}\right]$ Then: :$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$ where: : $\mathbf P$ is the matrix corresponding to the change of basis from $\left \langle {a_n} \right \rangle$ to $\left \langle {{a_n}'} \right \rangle$ : $\mathbf Q$ is the matrix corresponding to the change of basis from $\left \langle {b_m} \right \rangle$ to $\left \langle {{b_m}'} \right \rangle$.
Matrix Equivalence is Equivalence Relation
Matrix equivalence is an equivalence relation.
Equivalent Matrices have Equal Rank
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$. Let $\map \phi {\mathbf A}$ denote the rank of $\mathbf A$. Let $\mathbf A \equiv \mathbf B$ denote that $\mathbf A$ and $\mathbf B$ are matrix equivalent. Then: :$\mathbf A \equiv \mathbf B$ : :$\map \phi {\mathbf A} = \map \phi {\mathbf B}$
Transpose of Matrix Product
Let $\mathbf A$ and $\mathbf B$ be matrices over a commutative ring such that $\mathbf A \mathbf B$ is defined. Then $\mathbf B^\intercal \mathbf A^\intercal$ is defined, and: :$\paren {\mathbf A \mathbf B}^\intercal = \mathbf B^\intercal \mathbf A^\intercal$ where $\mathbf X^\intercal$ is the transpose of $\mathbf X$.
General Linear Group is Group
Let $K$ be a field. Let $\GL {n, K}$ be the general linear group of order $n$ over $K$. Then $\GL {n, K}$ is a group.
Transpose of Row Matrix is Column Matrix
Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a row matrix. Then $\mathbf x^\intercal$, the transpose of $\mathbf x$, is a column matrix: :$\begin {bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}^\intercal = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$
Transpose of Transpose of Matrix
Let $\mathbf A$ be a matrix. Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$. Then: :$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$
Solution to Simultaneous Linear Equations
Let $\displaystyle \forall i \in \closedint 1 m: \sum _{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$ be a system of simultaneous linear equations. where all of $\alpha_1, \ldots, a_n, x_1, \ldots x_n, \beta_i, \ldots, \beta_m$ are elements of a field $K$. Then $x = \tuple {x_1, x_2, \ldots, x_n}$ is a solution of this system : :$\sqbrk \alpha_{m n} \sqbrk x_{n 1} = \sqbrk \beta_{m 1}$ where $\sqbrk a_{m n}$ is an $m \times n$ matrix.
Infinite Cyclic Group is Isomorphic to Integers
Let $G$ be an infinite cyclic group. Then $G$ is isomorphic to the additive group of integers: $G \cong \struct {\Z, +}$.
Subgroup of Infinite Cyclic Group is Infinite Cyclic Group
Let $G = \gen a$ be an infinite cyclic group generated by $a$, whose identity is $e$. Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$. Let $H = \gen g$. Then $H \le G$ and $H \cong G$. Thus, all non-trivial subgroups of an infinite cyclic group are themselves infinite cyclic groups. A subgroup of $G = \gen a$ is denoted as follows: :$n G := \gen {a^n}$ This notation is usually used in the context of $\struct {\Z, +}$, where $n \Z$ is (informally) understood as '''the set of integer multiples of $n$'''.
Quotient Group of Infinite Cyclic Group by Subgroup
Let $C_n$ be the cyclic group of order $n$. Then: :$C_n \cong \dfrac {\struct {\Z, +} } {\struct {n \Z, +} } = \dfrac \Z {n \Z}$ where: :$\Z$ is the additive group of integers :$n \Z$ is the additive group of integer multiples :$\Z / n \Z$ is the quotient group of $\Z$ by $n \Z$. Thus, every cyclic group is isomorphic to one of: :$\Z, \dfrac \Z \Z, \dfrac \Z {2 \Z}, \dfrac \Z {3 \Z}, \dfrac \Z {4 \Z}, \ldots$
Ring Operations on Coset Space of Ideal
Let $\struct {R, +, \circ}$ be a ring. Let $\powerset R$ be the power set of $R$. Let $J$ be an ideal of $R$. Let $X$ and $Y$ be cosets of $J$. Let $X +_\mathcal P Y$ be the sum of $X$ and $Y$, where $+_\mathcal P$ is the operation induced on $\powerset R$ by $+$. Similarly, let $X \circ_\mathcal P Y$ be the product of $X$ and $Y$, where $\circ_\mathcal P$ is the operation induced on $\powerset R$ by $\circ$. Then: * The sum $X +_\mathcal P Y$ in $\powerset R$ is also their sum in the quotient ring $R / J$. * The product $X \circ_\mathcal P Y$ in $\powerset R$ may be a proper subset of their product in $R / J$.
Property of Being an Ideal is not Transitive
Let $J_1$ be an ideal of a ring $R$. Let $J_2$ be an ideal of $J_1$. Then $J_2$ need not necessarily be an ideal of $R$.
Ideals Containing Ideal Form Lattice
Let $J$ be an ideal of a ring $R$. Let $\mathbb L_J$ be the set of all ideal of $R$ which contain $J$. Then the ordered set $\struct {\mathbb L_J, \subseteq}$ is a lattice.
Ideals Containing Ideal Isomorphic to Quotient Ring
Let $J$ be an ideal of a ring $R$. Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$. Let the ordered set $\left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be the set of all ideals of $R / J$. Let the mapping $\Phi_J: \left({\mathbb L_J, \subseteq}\right) \to \left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be defined as: :$\forall a \in \mathbb L_J: \Phi_J \left({a}\right) = q_J \left({a}\right)$ where $q_J: a \to a / J$ is the quotient epimorphism from $a$ to $a / J$ from the definition of quotient ring. Then $\Phi_J$ is an isomorphism.
Ring of Integers is Principal Ideal Domain
The integers $\Z$ form a principal ideal domain.
Principal Ideals of Integers
Let $J$ be a non-zero ideal of $\Z$. Then $J = \ideal b$ where $b$ is the smallest strictly positive integer belonging to $J$.