nyuuzyou/wikis huggingface dataset:
Vector Field is Expressible as Gradient of Scalar Field iff Conservative

== Theorem ==
Let $R$ be a region of space.
Let $\mathbf V$ be a vector field acting over $R$.
Then $\mathbf V$ can be expressed as the gradient of some scalar field $F$ $\mathbf V$ is a conservative vector field.
== Proof ==
Let $\mathbf V_F$ be a vector field which is the gradient of some scalar field $F$:
:$\mathbf V_F = \grad F = \nabla F$
:
Let $A$ and $B$ be two points in $R$.
Let $\text {Path $1$}$ be an arbitrary path from $A$ to $B$ lying entirely in $R$.
At the point $P$, let $\d \mathbf l$ be a small element of $\text {Path $1$}$.
Let $\mathbf V_F$ make an angle $\theta$ with $\d \mathbf l$.
Then at $P$:
:$V_F \cos \theta \d l = \mathbf V_F \cdot \d \mathbf l$
where $V_F$ and $\d l$ are the magnitudes of $\mathbf V_F$ and $\d \mathbf l$ respectively.
Let $\mathbf r$ be the position vector of the point $P$ as it passes from $A$ to $B$.
Then $\d \mathbf l$ is the same as $\d \mathbf r$, and so we can write:
Hence the contour integral of $\mathbf V_F$ from $A$ to $B$ is:
Since only the end values feature in this expression, it follows that the actual route through $R$ taken by $\text {Path $1$}$ is immaterial.
That is, the value of $\ds \int_A^B \mathbf V_F \cdot \d \mathbf l$ is independent of the actual path from $A$ to $B$ along which the contour integral is taken.
Let $\text {Path $2$}$ now be an arbitrary path from $B$ back to $A$, so that $\text {Path $1$}$ and $\text {Path $2$}$ together make a closed loop.
Since the limits of integration are reversed for $\text {Path $2$}$, we have:
:$\ds \int_B^A \mathbf V_F \cdot \d \mathbf l = F_A - F_B$
Hence we have:
:$\ds \oint \paren {\grad F} \cdot \d \mathbf l = 0$
That is, $\mathbf V_F$ is a conservative vector field.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $2 a$. The Operation $\nabla S$: $(4.5)$
Category:Gradient Operator
Category:Conservative Vector Fields
Vector Inverse is Negative Vector

== Theorem ==
Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms.
Then:
:$\forall \mathbf v \in \mathbf V: -\mathbf v = -1_F \circ \mathbf v$
== Proof ==
so $-1_F \circ \mathbf v$ is an additive inverse of $\mathbf v$.
From Additive Inverse in Vector Space is Unique:
:$-1_F \circ \mathbf v = -\mathbf v$
== Sources ==
* : $1$. Preliminary ideas; sets, vector spaces: $1.2$. Vector Spaces
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Theorem $64 \ \text{(v)}$
Category:Vector Algebra
Vector Magnitude is Invariant Under Rotation

== Theorem ==
Let $\mathbf v$ be an arbitrary vector in the Cartesian plane $\CC$.
Let the coordinate system then be rotated in the anticlockwise direction by an arbitrary angle $\theta$.
Then:
the magnitude of $\mathbf v$ is unchanged in the new coordinate system.
== Proof 1 ==
== Proof 2 ==
Category:Vector Magnitude is Invariant Under Rotation
Category:Geometric Rotations
Category:Vector Length
Vector Magnitude is Invariant Under Rotation/Proof 1

== Theorem ==
== Proof ==
Let $P = \tuple {x_1, y_1}$ be the initial point of $\mathbf v$.
Let $Q = \tuple {x_2, y_2}$ be the terminal point of $\mathbf v$.
Then:
Then we have:
where $\size {\mathbf v}$ denotes the magnitude of $\mathbf v$.
The square of $\size {\mathbf v}$ is:
By Equations defining Plane Rotation:
:$x' = x \cos \theta - y \sin \theta$
:$y' = x \sin \theta + y \cos \theta$
Let the vector in the rotated coordinates be $\mathbf v'$.
The $x$-component of $\mathbf v'$ is:
:$X' = x_2 \cos \theta - y_2 \sin \theta - x_1 \cos \theta + y_1 \sin \theta$
The $y$-component of $\mathbf v'$ is:
:$Y' = x_2 \sin \theta + y_2 \cos \theta - x_1 \sin \theta - y_1 \cos \theta$
We compute the square of $X'$ and $Y'$ separately:
There are $16$ terms so we group them on four lines as a bookkeeping device:
Then:
Group the terms on separate lines:
All the terms with $\sin \theta \cos \theta$ cancel.
We have six pairs of terms remaining.
By Sum of Squares of Sine and Cosine:
:$\sin^2 \theta + \cos^2 \theta = 1$
So:
Compare with the results for $\mathbf v$:
:$\size {\mathbf v'}^2 = \size {\mathbf v}^2$
Since magnitude is positive, when we take the square root:
:$\size {\mathbf v'} = \size {\mathbf v}$
Category:Vector Magnitude is Invariant Under Rotation
Vector Product is Zero only if Factor is Zero

== Theorem ==
Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms.
Then:
:$\forall \lambda \in F: \forall \mathbf v \in \mathbf V: \lambda \circ \mathbf v = \bszero \implies \paren {\lambda = 0_F \lor \mathbf v = \mathbf 0}$
where $\bszero \in \mathbf V$ is the zero vector.
== Proof ==
 that:
:$\exists \lambda \in F: \exists \mathbf v \in \mathbf V: \lambda \circ \mathbf v = \bszero \land \lambda \ne 0_F \land \mathbf v \ne \bszero$
which is the negation of the exposition of the theorem.
Utilizing the vector space axioms:
 
which contradicts the assumption that $\mathbf v \ne \mathbf 0$.
== Also see ==
* Zero Vector Scaled is Zero Vector
* Vector Scaled by Zero is Zero Vector
* Zero Vector Space Product iff Factor is Zero
Category:Vector Algebra
Vector Quantity as Scalar Product of Unit Vector Quantity

== Theorem ==
Let $\mathbf a$ be a vector quantity.
Then:
:$\mathbf a = \size {\mathbf a} \mathbf {\hat a}$
where:
:$\size {\mathbf a}$ denotes the magnitude of $\mathbf a$
:$\mathbf {\hat a}$ denotes the unit vector in the direction $\mathbf a$.
== Proof ==
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $2$. Graphical Representation of Vectors
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and Subtraction of Vectors: $5$. Multiplication by a number
Category:Scalar Multiplication
Category:Unit Vectors
Vector Quantity can be Expressed as Sum of 3 Non-Coplanar Vectors

== Theorem ==
Let $\mathbf r$ be a vector quantity embedded in space.
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be non-coplanar.
Then $\mathbf r$ can be expressed uniquely as the resultant of $3$ vector quantities which are each parallel to one of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
== Proof ==
Let $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ be unit vectors in the directions of $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.
Let $O$ be a point in space.
Take $\vec {OP} := \mathbf r$.
With $OP$ as its space diagonal, construct a parallelepiped with edges $OA$, $OB$ and $OC$ parallel to $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ respectively.
Only one such parallelepiped can be so constructed.
Let $x$, $y$ and $z$ be the length of the edges $OA$, $OB$ and $OC$ respectively.
Then:
Thus $\mathbf r$ is the resultant of the $3$ components $x \mathbf {\hat a}$, $y \mathbf {\hat b}$ and $z \mathbf {\hat c}$ which, by construction, are parallel to $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.
The fact that only one parallelepiped can be constructed in the above proves uniqueness.
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Components of a Vector: $6$. Resolution of a vector
* : Volume One: Chapter $1$ Vectors in Classical Physics: $1.2$ The Resolution of a Vector into Components
Category:Vectors
Vector Scaled by Zero is Zero Vector

== Theorem ==
Let $F$ be a field whose zero is $0_F$.
Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms.
Then:
:$\forall \mathbf v \in \mathbf V: 0_F \circ \mathbf v = \bszero$
== Proof ==
== Also see ==
* Zero Vector Scaled is Zero Vector
* Vector Product is Zero only if Factor is Zero
* Zero Vector Space Product iff Factor is Zero
* Vector Inverse is Negative Vector
== Sources ==
* : $1$. Preliminary ideas; sets, vector spaces: $1.2$. Vector Spaces
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Theorem $64 \ \text{(iii)}$
Category:Vector Algebra
Vector Space/Examples/Arrows through Point in 3D Space

== Example of Vector Space ==
The set $\mathbf V$ of all arrows through a given point in ordinary $3$-dimensional Euclidean space forms a vector space whose scalar field is the set of real numbers $\R$.
Thus $\mathbf V$ is itself a vector space of $3$ dimensions.
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $61$
* : Chapter $7$: Vector Spaces: $\S 34$. Dimension: Example $65$
Category:Examples of Vector Spaces
Vector Space has Basis

== Theorem ==
Let $K$ be a division ring.
Let $V$ be a vector space over $K$.
Then $V$ has a basis.
== Proof ==
The result follows from Vector Space has Basis between Linearly Independent Set and Spanning Set.
It suffices to find a linearly independent subset $L \subseteq V$ that is contained in a spanning set $S \subseteq V$.
By Empty Set is Linearly Independent, $L$ can be taken to be the empty set.
Or if $V$ is nonzero, by Singleton is Linearly Independent, $L$ can be taken to be any singleton of $V$.
$S$ can be taken to be $V$, since $V$ trivially spans itself.
Therefore, $L$ and $S$ exist and $L \subseteq S$ so $V$ has a basis $B$ with $L \subseteq B \subseteq S$.
== Also see ==
* Finitely Generated Vector Space has Finite Basis
Category:Bases of Vector Spaces
Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set

== Theorem ==
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 $H \subseteq B \subseteq F$.
== Outline of Proof ==
Of all the sets between $H$ and $F$ that generate $G$, we choose one with the fewest elements, using the Well-Ordering Principle.
Using the minimality, we show that this generator is linearly independent.
== Proof ==
Let $\mathbb S$ be the set of all $S \subseteq G$ such that $S$ is a generator for $G$ and that $H \subseteq S \subseteq F$.
Because $F \in \mathbb S$, it follows that $\mathbb S \ne \O$.
Because $F$ is finite, then so is every element of $\mathbb S$.
Let $R = \set {r \in \Z: r = \card S \in \mathbb S}$.
That is, $R$ is the set of all the integers which are the number of elements in generators for $G$ that are subsets of $F$.
Let $n$ be the smallest element of $R$.
Let $B$ be an element of $\mathbb S$ such that $\card B = n$.
We note that as $H$ is a linearly independent set, it does not contain $0$ by Subset of Module Containing Identity is Linearly Dependent.
Then $0 \notin B$, or $B \setminus \set 0$ would be a generator for $G$ with $n - 1$ elements.
This would contradict the definition of $n$.
Let $m = \card H$.
Let $\sequence {a_n}$ be a sequence of distinct vectors such that $H = \set {a_1, \ldots, a_m}$ and $B = \set {a_1, \ldots, a_n}$.
Suppose $B$ were linearly dependent.
By Linearly Dependent Sequence of Vector Space, there would exist $p \in \closedint 2 n$ and scalars $\mu_1, \ldots, \mu_{p - 1}$ such that $\ds a_p = \sum_{k \mathop = 1}^{p - 1} \mu_k a_k$.
This shows that $\set {a_1, \ldots, a_p}$ is linearly dependent.
If $p \le m$, we would have $\set {a_1, \ldots, a_p} \subseteq H$.
But by Superset of Linearly Dependent Set is Linearly Dependent, $H$ would be linearly dependent, which is a contradiction.
Thus $p > m$ and therefore $B' = B \setminus \set {a_p}$ would contain $H$.
Now if $\ds x = \sum_{k \mathop = 1}^n \lambda_k a_k$, then:
:$\ds x = \sum_{k \mathop = 1}^{p - 1} \paren {\lambda_k + \lambda_p \mu_k} a_k + \sum_{k \mathop = p + 1}^n \lambda_k a_k$
Hence $B'$ would be a generator for $G$ containing $n - 1$ elements, which contradicts the definition of $n$.
Thus $B$ must be linearly independent and hence is a basis.
== Also see ==
* Vector Space has Basis between Linearly Independent Set and Spanning Set, for a proof without the hypothesis that $G$ is finitely generated
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.7$
Category:Bases of Vector Spaces
Category:Generators of Vector Spaces
Vector Space has Basis between Linearly Independent Set and Spanning Set

== Theorem ==
Let $V$ be a vector space over a field $F$.
Let $L$ be a linearly independent subset of $V$.
Let $S$ be a set that spans $V$.
Suppose that $L \subseteq S$.
Then $V$ has a basis $B$ such that $L \subseteq B \subseteq S$.
=== Corollary ===
== Outline of Proof ==
We use Zorn's Lemma to construct a maximal linearly independent subset.
== Proof ==
Let $\mathscr I$ be the set of linearly independent subsets of $S$ that contain $L$, ordered by inclusion.
Note that $L \in \mathscr I$, so $\mathscr I \ne \O$.
Let $\mathscr C$ be a chain in $\mathscr I$.
Let $C = \bigcup \mathscr C$.
 that $C$ is linearly dependent.
Then there exist $v_1, v_2, \ldots, v_n \in C$ and $r_1, r_2, \ldots, r_n \in F$ such that $r_1 \ne 0$:
:$\ds \sum_{k \mathop = 1}^n r_k v_k = 0$
Then there are $C_1, C_2, \ldots, C_n \in \mathscr C$ such that $v_k \in C_k$ for each $k \in \set {1, 2, \ldots, n}$.
Since $\mathscr C$ is a chain, $C_1 \cup C_2 \cup \cdots \cup C_n$ must equal $C_k$ for some $k \in \set {1, 2, \ldots, n}$.
But then $C_k \in \mathscr C$ and $C_k$ is linearly dependent, which is a contradiction.
Thus $C$ is linearly independent.
By Zorn's Lemma, $\mathscr I$ has a maximal element $M$ (one that is not contained in any other element).
Since $M \in \mathscr I$, $M$ is linearly independent.
All that remains is to show that $M$ spans $V$.
 there exists a $v \in V \setminus \map \span M$.
Then, since $S$ spans $V$, there must be an element $s$ of $S$ such that $s \notin \map \span M$.
Then $M \cup \set s$ is linearly independent.
Thus $M \cup \set s \supsetneq M$, contradicting the maximality of $M$.
Thus $M$ is a linearly independent subset of $V$ that spans $V$.
Therefore, by definition, $M$ is a basis for $V$.
== Also see ==
* Existence of Vector Space Bases implies Axiom of Choice
* Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, for the finitely generated case
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra: Proposition $1.2$
Category:Generators of Vector Spaces
Category:Bases of Vector Spaces
Vector Space has Unique Additive Identity/Proof 1

== Theorem ==
== Proof ==
Let $V$ be a vector space.
Let $0$ and $0'$ both be additive identities for $V$.
Because $0$ is an additive identity:
:$0' = 0' + 0$
From commutativity of vector space addition:
:$0' + 0 = 0 + 0'$
Because $0'$ is an additive identity:
:$0 + 0 = 0$
Hence:
:$0' = 0$
Thus $V$ has a unique additive identity.
== Sources ==
* , p. 15
Category:Vector Space has Unique Additive Identity
Vector Space obtains Norm Structure through Linear Isomorphism

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be vector spaces over $\GF$.
Let $\norm {\, \cdot \,}_Y$ be a norm on $Y$.
Let $\phi : X \to Y$ be a linear isomorphism.
Define $\norm {\, \cdot \,}_X : X \to \R$ by:
:$\norm x_X = \norm {\map \phi x}_Y$
for each $x \in X$.
Then $\norm {\, \cdot \,}_X$ is a norm on $X$.
== Proof ==
First since $\norm y_Y \in \hointr 0 \infty$ for each $y \in Y$, we have $\norm {\map \phi x}_Y \in \hointr 0 \infty$ for each $x \in X$. 
=== Proof of ===
Suppose that $x \in X$ has:
:$\norm x_X = 0$
Then:
:$\norm {\map \phi x}_Y = 0$
Then from for $\norm {\, \cdot \,}_Y$, we have that:
:$\map \phi x = {\mathbf 0}_Y$
Since $\phi$ is a linear isomorphism, by Linear Transformation Maps Zero Vector to Zero Vector, we have $x = {\mathbf 0}_X$.
=== Proof of ===
Let $\lambda \in \GF$ and $x \in X$.
We have:
=== Proof of ===
Let $u, v \in X$.
We have:
 
Category:Normed Vector Spaces
Vector Space of All Mappings is Vector Space

== Theorem ==
Let $\struct {K, +, \circ}$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the vector space of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_K$ is a $K$-vector space.
== Proof ==
Follows directly from Module of All Mappings is Module and the definition of vector space.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.4$
Category:Examples of Vector Spaces
Vector Space of Continuous on Closed Interval Real Functions is not Finite Dimensional

== Theorem ==
Let $I := \closedint 0 1$ be a closed real interval.
Let $\struct {\map C I, +, \, \cdot \,}_\R$ be the continuous on closed interval real function vector space.
Then $\struct {\map C I, +, \, \cdot \,}_\R$ is not finite dimensional.
== Proof ==
=== Monomials are linearly independent ===
Let $d \in \N_{>0}$.
Consider the set of real monomials of the following form:
:$\map {x_n} t = t^n$
where $n \in \N_{>0}$ and $n \le d$.
 the set of $x_n$ is not linearly independent.
Then:
:$\forall n \in \N_{>0}: n \le d: \exists \alpha_n \in \R: \neg \forall n: \alpha_n \ne 0$
and:
:$\ds \sum_{k \mathop = 1}^d \alpha_k t^k = 0$
Let $m \in \N_{>0}: m \le d$ be the smallest index such that $\alpha_m \ne 0$.
Then:
:$\ds \forall t \in \closedint 0 1: \sum_{k \mathop = m}^d \alpha_k t^k = 0$
or
:$\ds \forall t \in \hointl 0 1: \sum_{k \mathop = m}^d \alpha_k t^{k - d} = 0$
Note that:
:$\ds \forall n \in \N_{>0}: \exists t \in \hointl 0 1: t = \frac 1 n$
Thus:
:$\ds \forall n \in \N_{>0}: \sum_{k \mathop = m}^d \frac {\alpha_k} {n^{d - k} } = 0$
Passing the limit $n \to \infty$ gives us $\alpha_d = 0$.
This is a contradiction.
Hence, the set of $x_n$ is linearly independent.
=== $\struct {\map C I, +, \, \cdot \,}_\R$ is not finite dimensional ===
 $\struct {\map C I, +, \, \cdot \,}_\R$ is finite dimensional and has the dimension $d$. 
Any independent set of cardinality $d$ in a $d$-dimensional vector space is a basis for this vector space.
Then the set of monomials $x_n$ with $n \in \N_{>0}$ and $n \le d$ is a basis for $\struct {\map C I, +, \, \cdot \,}_\R$.
The constant function $\map x t = 1$ belongs to $\map C I$.
Then:
:$\ds \forall n \in \N_{>0}: n \le d: \exists \beta_n \in \R: 1 = \sum_{k \mathop = 1}^d \beta_k \map {x_k} t$
Let $t = 0$.
Then $1 = 0$.
This is a contradiction.
Hence, $\struct {\map C I, +, \, \cdot \,}_\R$ is not finite dimensional.
== Sources ==
* : Chapter $1.1$: Normed and Banach spaces. Vector Spaces
Category:Vector Spaces
Category:Continuous Real Functions
Vector Space of Sequences with Finite Support is Vector Space

== Theorem ==
Let $\struct {K, +_K, \circ_K}$ be a division ring.
Let $\struct{V, +, \cdot}$ be the vector space of sequences with finite support in $K$.
Then $V$ is a vector space over $K$.
== Proof ==
Consider $V$ as a subset of the vector space of all mappings from $\N$ to $K$.
Let us apply the One-Step Vector Subspace Test.
Thus, let $\sequence{a_n}, \sequence{b_n} \in V$ be sequences over $K$ with finite support, and let $\lambda \in K$.
Then we need to show that $\sequence{ a_n + \lambda b_n }$ has finite support.
It is immediate that if:
:$a_n = b_n = 0$
then:
:$a_n + \lambda b_n = 0$ 
Therefore:
:$\set{ n \in \N: a_n + \lambda b_n \ne 0} \subseteq \set{ n \in \N: a_n \ne 0 } \cup \set{ n \in \N: b_n \ne 0 }$
By Union of Finite Sets is Finite, it follows that $\sequence{ a_n + \lambda b_n }$ has finite support.
The result follows by the One-Step Vector Subspace Test.
Category:Examples of Vector Spaces
Vector Space on Cartesian Product is Vector Space

== Theorem ==
Let $\struct {K, +, \circ}$ be a division ring.
Let $n \in \N_{>0}$.
Let $\struct {K^n, +, \times}_K$ be the $K$-vector space $K^n$.
Then $\struct {K^n, +, \times}_K$ is a $K$-vector space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.1$
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $62$
Category:Examples of Vector Spaces
Category:Vector Space on Cartesian Product
Vector Space on Field Extension is Vector Space

== Theorem ==
Let $\struct {K, +, \times}$ be a field.
Let $L / K$ be a field extension over $K$.
Let $\struct {L, +, \times}_K$ be the a vector space of $L$ over $K$.
Then $\struct {L, +, \times}_K$ is a vector space.
== Proof ==
By definition, $L / K$ is a field extension over $K$.
Thus, by definition, $K$ is a subfield of $L$.
Thus, also by definition, $K$ is a division subring of $L$.
The result follows by Vector Space over Division Subring is Vector Space.
Category:Examples of Vector Spaces
Vector Space over Division Subring/Examples/Real Numbers in Complex Numbers

== Example of Vector Space over Division Subring ==
Consider the field of complex numbers $\struct {\C, +, \times}$, which is a ring with unity whose unity is $1$.
Consider the field of real numbers $\struct {\R, +, \times}$, which is a division subring of $\struct {\C, +, \circ}$ such that $1 \in \R$.
Then $\struct {\C, +, \times_\R}_\R$ is an $\R$-vector space, where $\times_\R$ is the restriction of $\times$ to $\R \times \C$.
$\struct {\C, +, \times_\R}_\R$ is of dimension $2$.
The set $\set {1 + 0 i, 0 + i}$ forms a basis of $\struct {\C, +, \times_\R}_\R$, as do any two complex numbers which are not real multiples of each other.
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $64$
* : Chapter $7$: Vector Spaces: $\S 34$. Dimension: Example $68$
Category:Examples of Vector Spaces
Vector Space over Division Subring is Vector Space

== Theorem ==
Let $\struct {L, +_L, \times_L}$ be a division ring.
Let $K$ be a division subring of $\struct {L, +_L, \times_L}$.
Let $\struct {G, +_G, \circ}_L$ be a $L$-vector space.
Let $\circ_K$ be the restriction of $\circ$ to $K \times G$.
Hence let $\struct {G, +_G, \circ_K}_K$ be the vector space induced by $K$.
Then $\struct {G, +_G, \circ_k}_k$ is indeed a $K$-vector space.
=== Special Case ===
== Proof ==
A vector space over a division ring $D$ is by definition a unitary module over $D$. 
$S$ is a division ring by assumption.
$\struct {R, +, \circ_S}_S$ is a unitary module by Subring Module is Module/Special Case.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.3$
Category:Examples of Vector Spaces
Vector Space over Division Subring is Vector Space/Special Case

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose unity is $1_R$.
Let $S$ be a division subring of $R$, such that $1_R \in S$.
The vector space $\struct {R, +, \circ_S}_S$ over $\circ_S$ is a $S$-vector space.
== Proof ==
A vector space over a division ring $D$ is by definition a unitary module over $D$. 
$S$ is a division ring by assumption.
$\struct {R, +, \circ_S}_S$ is a unitary module by Subring Module is Module/Special Case.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.2$
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $64$
Category:Examples of Vector Spaces
Vector Space over an Infinite Field is not equal to the Union of Proper Subspaces

== Theorem ==
Let $U_1, U_2, \dots, U_n$ be proper subspaces of a vector space $V$ over an infinite field $F$, 
then $V$ is not equal to the union of $U_1, U_2, \dots, U_n$.
== Proof ==
 $V = V_1 \cup \cdots \cup V_n$.
We can assume that $n \geq 2$, and that $m$ is minimal $n$ with this property.
Choose and fix some $y \in V \setminus V_1$.
Let $x \in V_1$.
Since $F$ is infinite field, we can choose a subset $S \subset F$ of size $m+1$.
For each $\alpha \in S$, we can find some
:$\map i \alpha \in \set{1, \cdots, m}$
such that $x+\alpha y \in V_{\map i \alpha}$ because $V = V_1 \cup \cdots \cup V_m$.
The function $i: S \to \set{1, \cdots, m}$ cannot be injective.
So we can find $\alpha \neq \beta$ in $S$ such that $x + \alpha y$ and $x + \beta y$ both lie in $V_{\map i \alpha}$.
But then
:$\ds y = \frac{\paren{x + \alpha y} - \paren{x + \beta y}}{\alpha - \beta} \in V_{\map i \alpha}$.
Since $y \notin V_1$ by assumption, we conclude that $\map i \alpha > 1$.
Then $x = \paren{x + \alpha y} - \alpha y \in V_{\map i \alpha} \subseteq V_2 \cup \cdots \cup V_m$ for every $x \in V_1$.
Hence
:$V_1 \subseteq V_2 \cup \cdots \cup V_m$
This implies that $V = V_2 \cup \cdots \cup V_m$, which contradicts the minimality of $m$.
== Source ==
* : Exercise $1.17$
Category:Vector Subspaces