nyuuzyou/wikis huggingface dataset:
Vector Space with Standard Affine Structure is Affine Space

== Theorem ==
Let $E$ be a vector space.
Let $\struct {\EE, E, +, -}$ be the standard affine structure on $E$.
Then with this structure, $\EE$ is an affine space.
== Proof ==
We are required to show that:
By definition of the standard affine structure, the addition and subtraction operations are simply those in the vector space $E$.
That is, we want to show that:
By definition the addition operation on a vector space is commutative and associative.
But all three axioms are immediate consequences of commutativity and associativity.
This concludes the proof.
Category:Affine Geometry
Vector Subspace/Examples/Complex Numbers in Quaternion Vector Space

== Example of Vector Subspace ==
The set of complex numbers $\C$ forms a vector subspace of the vector space of quaternions $\H$.
== Proof ==
From Quaternions form Vector Space over Themselves, $\H$ forms a vector space.
From Complex Numbers form Vector Space over Themselves, $\C$ forms a vector space.
From Complex Numbers form Subfield of Quaternions, $\C$ is a subset of $\H$.
Hence the result by definition of vector subspace.
== Sources ==
* 
* 
Category:Examples of Vector Subspaces
Vector Subspace/Examples/Real Numbers in Complex Vector Space

== Example of Vector Subspace ==
The set of real numbers $\R$ forms a vector subspace of the vector space of complex numbers $\C$.
== Proof ==
From Complex Numbers form Vector Space over Themselves, $\C$ forms a vector space.
From Real Numbers form Vector Space, $\R$ forms a vector space.
From Real Numbers form Subfield of Complex Numbers, $\R$ is a subset of $\C$.
Hence the result by definition of vector subspace.
== Sources ==
* 
* 
Category:Examples of Vector Subspaces
Vector Subspace contains Zero Vector

== Theorem ==
Let $V$ be a vector space over a division ring $K$ whose unity is $1_K$.
Let $U \subseteq V$ be a subspace of $V$.
Then $U$ contains the zero vector of $V$.
== Proof ==
Demonstrated during the course of the One-Step Vector Subspace Test.
== Sources ==
* 
* 
Category:Vector Subspaces
Vector Subspace of Algebra over Field Embeds into Unitization as Vector Subspace

== Theorem ==
Let $K$ be a field.
Let $A$ be an algebra over $K$. 
Let $B$ be a vector subspace of $A$. 
Let $A_+$ be the unitization of $A$.
Let:
:$B_0 = \set {\tuple {x, 0_K} : x \in B}$
Then $B_0$ is a vector subspace of $A_+$. 
== Proof ==
Clearly $B_0 \ne \O$. 
From One-Step Vector Subspace Test, it is sufficient to show that for each $u, v \in B_0$ and $\lambda \in K$, we have:
:$u + \lambda v \in B_0$
Let $u, v \in B_0$ and $\lambda \in K$. 
Then there exists $x, y \in A$ such that:
:$u = \tuple {x, 0_K}$
and:
:$v = \tuple {y, 0_K}$
Then by the definition of the unitization, we have:
:$\tuple {x, 0_K} + \lambda \tuple {y, 0_K} = \tuple {x + \lambda y, 0_K} \in B_0$
So by the One-Step Vector Subspace Test, $B_0$ is a vector subspace of $A_+$.
Category:Unitizations of Algebras over Fields
Vector Subspace of Hausdorff Topological Vector Space is Hausdorff Topological Vector Space

== Theorem ==
Let $\struct {K, +_K, \circ_K, \tau_K}$ be a topological field.
Let $\struct {\struct {X, +_X, \circ_X}_K, \tau_X}$ be a Hausdorff topological vector space over $K$.
Let $\struct {Y, +_Y, \circ_Y}_K$ be a vector subspace of $X$. 
where:
:$+_Y : Y \times Y \to Y$ is the restriction of $+_X$ to $Y \times Y$
:$\circ_Y : K \times Y \to Y$ is the restriction of $\circ_X$ to $K \times Y$.
Let $\tau_Y$ be the subspace topology on $Y$ induced by $\tau_X$.
Then $\struct {\struct {Y, +_Y, \circ_Y}_K, \tau_Y}$ is a Hausdorff topological vector space over $K$.
== Proof ==
From Vector Subspace of Topological Vector Space is Topological Vector Space, $\struct {\struct {Y, +_Y, \circ_Y}_K, \tau_Y}$ is a topological vector space over $K$.
Since $\struct {\struct {X, +_X, \circ_X}_K, \tau_X}$ is a Hausdorff topological vector space over $K$, it is Hausdorff.
We are given that $\struct {Y, +_Y, \circ_Y}_K$ is a vector subspace of $X$.
We are also given that $\tau_Y$ is the subspace topology on $Y$ induced by $\tau_X$.
So $\struct {\struct {Y, +_Y, \circ_Y}_K, \tau_Y}$ is a topological subspace of $\struct {\struct {X, +_X, \circ_X}_K, \tau_X}$ over $K$.
By Subspace of Hausdorff Space is Hausdorff, we have that $\struct {\struct {Y, +_Y, \circ_Y}_K, \tau_Y}$ is Hausdorff.
So $\struct {\struct {Y, +_Y, \circ_Y}_K, \tau_Y}$ is a Hausdorff topological vector space over $K$.
Category:Hausdorff Topological Vector Spaces
Vector Subspace of Normed Dual Space is Weak-* Dense iff Separates Points

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a normed vector space over $\GF$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $F \subseteq X^\ast$ be a vector subspace of $X^\ast$.
Then $F$ is everywhere dense in $\struct {X^\ast, w^\ast}$ it separates points.
== Proof ==
=== Necessary Condition ===
Let $F$ be everywhere dense in $\struct {X^\ast, w^\ast}$.
From Weak-* Dense Subset of Normed Dual Space Separates Points, $F$ separates points.
=== Sufficient Condition ===
Let $F$ be a vector subspace separating the points of $X^\ast$.
Suppose that $F$ is not everywhere dense in $\struct {X^\ast, w^\ast}$. 
From Set is Closed iff Equals Topological Closure, we then have that $\map {\cl_{w^\ast} } F \ne X^\ast$. 
Then $\map {\cl_{w^\ast} } F$ is a proper closed vector subspace in $\struct {X^\ast, w^\ast}$. 
From Existence of Non-Zero Continuous Linear Functional vanishing on Proper Closed Subspace of Hausdorff Locally Convex Space, there exists $\Phi \in \struct {X^\ast, w^\ast}^\ast$ such that $\Phi \ne 0$ and: 
:$\map \Phi f = 0$ for $f \in \map {\cl_{w^\ast} } F$.
From Characterization of Continuity of Linear Functional in Weak-* Topology, there exists $x \in X$ such that: 
:$\Phi = x^\wedge$
We then have: 
:$\map {x^\wedge} f = 0$ for $f \in F$.
That is: 
:$\map f x = 0$ for all $f \in F$.
Since $F$ separates points, we obtain $x = 0$, contradicting that $\Phi \ne 0$. 
Hence we have reached a contradiction.
So $F$ is everywhere dense in $\struct {X^\ast, w^\ast}$. 
Category:Weak-* Topologies
Category:Normed Dual Spaces
Vector Subspace of Real Vector Space

== Theorem ==
Let $\R^n$ be a real vector space.
Let $\mathbb W \subseteq \R^n$.
Then $\mathbb W$ is a linear subspace of $\R^n$ :
:$(1): \quad \mathbf 0 \in \mathbb W$, where $\mathbf 0$ is the zero vector with $n$ entries
:$(2): \quad \mathbb W$ is closed under vector addition
:$(3): \quad \mathbb W$ is closed under scalar multiplication.
=== Corollary ===
Criterion $(1)$ may be replaced by:
:$(1'): \quad \mathbb W \ne \O$
that is, that $\mathbb W$ is non-empty.
== Proof ==
=== Proof of Corollary ===
Suppose $\mathbf 0 \in \mathbb W$.
Then $\mathbb W$ contains an element and is non-empty.
Suppose $\mathbb W$ contains an element $\mathbf x \in \R^n$.
Then, by criterion $(3)$:
:$0 \mathbf x \in \mathbb W$
where $0$ is the zero scalar.
But $0 \mathbf x = \mathbf 0$ from Vector Scaled by Zero is Zero Vector, so $\mathbf 0 \in \mathbb W$.
 
== Also see ==
* Two-Step Vector Subspace Test
* Null Space is Subspace
== Sources ==
* 
Category:Linear Algebra
Vector Subspace of Real Vector Space under Chebyshev Metric is Metric Subspace

== Theorem ==
Let $n \in \N$.
Let $A$ be the set of all ordered $n+1$-tuples $\tuple {x_1, x_2, \ldots, x_{n + 1} }$ of real numbers such that $x_{n + 1} = 0$.
Let $d: A \times A \to \R$ be the function defined as:
:$\ds \forall x, y \in A: \map d {x, y} = \max_{i \mathop = 1}^n \set {\size {x_i - y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_{n + 1} }, y = \tuple {y_1, y_2, \ldots, y_{n + 1} }$.
Then $\struct {A, d}$ is a metric subspace of $\struct {\R^{n + 1}, d_\infty}$ where $d_\infty$ is the Chebyshev distance on the real vector space $\R^{n + 1}$.
== Proof ==
The metric given is the Chebyshev distance restricted to the subset $A$ of the real vector space $\R^{n + 1}$.
The result follows from Subspace of Metric Space is Metric Space.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Example $4$
Category:Metric Subspaces
Category:Chebyshev Distance
Vector Subspace of Topological Vector Space is Topological Vector Space

== Theorem ==
Let $K$ be a topological field.
Let $\struct {\struct {X, +_X, \circ_X}_K, \tau_X}$ be a topological vector space over $K$.
Let $Y$ be a vector subspace of $X$. 
Let $\tau_Y$ be the subspace topology on $Y$ induced by $\tau_X$.
Let $+_Y : Y \times Y \to Y$ be the restriction of $+_X$ to $Y \times Y$.
Let $\circ_Y : K \times Y \to Y$ be the restriction of $\circ_X$ to $K \times Y$. 
Then $\struct {\struct {Y, +_Y, \circ_Y}_K, \tau_Y}$ is a topological vector space.
== Proof ==
From the definition of a vector subspace, $\struct {Y, +_Y, \circ_Y}_K$ is a vector space over $K$.
Since $\struct {\struct {X, +_X, \circ_X}_K, \tau_X}$ is a topological vector space, $+_X : X \times X \to X$ and $\circ_X : K \times X \to X$ are continuous.
From Restriction of Continuous Mapping is Continuous, $+_Y : Y \times Y \to Y$ and $\circ_Y : K \times Y \to Y$ are continuous. 
Hence $\struct {\struct {Y, +_Y, \circ_Y}_K, \tau_Y}$ is a topological vector space.
Category:Topological Vector Spaces
Vector Subtraction/Examples/Example 1

== Examples of Vector Subtraction ==
Let:
Then:
:$\mathbf a - \mathbf b = 4 \mathbf i + 7 \mathbf j + 6 \mathbf k$
== Proof ==
By definition:
:$\mathbf a - \mathbf b = \mathbf a + \paren {-\mathbf b}$
Summing $\mathbf a$ and $-\mathbf b$ by components:
== Sources ==
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach: Example $1.1.1$
Category:Examples of Vector Subtraction
Vector Sum/Component Definition/Examples/Example 1

== Examples of Vector Sum: Definition by Components ==
Let:
Then:
:$\mathbf a + \mathbf b = 8 \mathbf i + \mathbf j$
== Proof ==
Summing $\mathbf a$ and $\mathbf b$ by components:
== Sources ==
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach: Example $1.1.1$
Category:Examples of Vector Sum
Vector Sum of Rotated Triangle is Zero

== Theorem ==
Let $T = \triangle ABC$ be embedded in the Cartesian plane $\CC$.
Let the sides of $T$ be directed line segments, that is, vectors.
Let $\CC$ be rotated anticlockwise about the origin by an angle $\theta$.
Let the triangle $T$, as defined in the rotated coordinates, be T'.
Let $\mathbf{0}$ be the zero vector.
Let $+$ be the vector sum.
Then:
:$T' = \mathbf{AB}' + \mathbf{BC}' + \mathbf{CA}' = \mathbf{0}$
== Proof ==
Let $\mathbf{AB}$ have the same magnitude as $AB$ and direction from $A$ to $B$.
Let similar definitions hold for $\mathbf{BC}$ and $\mathbf{CA}$.
By definition of vectors:
:$\mathbf{AB} = \paren { B_x - A_x, B_y - A_y }$
Let $\mathbf{AB}$ in the rotated coordinates be designated $\mathbf{AB}'$.
By definition of vectors:
:$\mathbf{AB}' = \paren { B_x' - A_x', B_y' - A_y' }$
By Equations defining Plane Rotation the $x$-component of $\mathbf{AB}'$ is:
:$\paren { B_x \cos \theta - B_y \sin \theta - A_x \cos \theta + A_y \sin \theta }$
By Vector Magnitude is Invariant Under Rotation the above expression is equal to:
:$\paren { B_x - A_x }$
and the y-component of $\mathbf{AB}'$ is:
:$\paren { B_x \sin \theta + B_y \cos \theta - A_x \sin \theta - A_y \cos \theta }$
By Vector Magnitude is Invariant Under Rotation the above expression is equal to:
:$\paren { B_y - A_y }$
That is, the sides of $T$ are equal to the sides of $T'$:
:$\norm {AB} = \norm {AB'}$
:$\norm {BC} = \norm {BC'}$
:$\norm {CA} = \norm {CA'}$
By Triangle Side-Side-Side Congruence:
:$T \cong T'$
$T'$ consists of the vectors $\mathbf{AB}' + \mathbf{BC}' + \mathbf{CA}'$.
By definition, the terminal point of $\mathbf{CA}$ is the initial point of $\mathbf{AB}$.
It follows that:
$T = \mathbf{AB} + \mathbf{BC} + \mathbf{CA} = \mathbf{0}$
By definition, the terminal point of $\mathbf{CA}'$ is the initial point of $\mathbf{AB}'$.
It follows that:
$T' = \mathbf{AB}' + \mathbf{BC}' + \mathbf{CA}' = \mathbf{0}$
Category:Vectors
Category:Triangles
Vector Sum of System of Parallel or Coplanar Forces may be Single Force or Couple

== Theorem ==
Let $S = \set {\mathbf v_1, \mathbf v_2, \ldots, \mathbf v_n}$ be a system of forces which are either parallel or coplanar
Let $\sum S$ denote the sum of the elements of $S$.
Then $\sum S$ is either a single force or it is a couple.
== Proof ==
== Sources ==
* 
* 
Category:Force
Category:Couples (Mechanics)
Category:Vector Addition
Vector Times Magnitude Same Length As Magnitude Times Vector

== Theorem ==
Let $\mathbf u$ and $\mathbf v$ be two vectors in the vector space $\struct {G, +_G, \circ}_K$ over a division ring $\struct {K, +_K, \times}$ with subfield $\R$ such that $\R \subseteq \map Z K$ with $\map Z K$ the center of $K$
Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be the lengths of $\mathbf u$ and $\mathbf v$ respectively.
Then:
:$\norm {\paren {\norm {\mathbf v} \circ \mathbf u} } = \norm {\paren {\norm {\mathbf u} \circ \mathbf v} }$
== General Proof ==
Let $\mathbf u$ and $\mathbf v$ be two vectors in the vector space $\struct {G, +_G, \circ}_K$.
From the second property of a vector space norm:
:$\norm {\paren {\norm {\mathbf v} \circ \mathbf u } } = \size {\paren {\norm {\mathbf v} } }_K \times \norm {\mathbf u}$
But then since $\norm {\mathbf v}$ is a non-negative real, and the absolute value function is the norm on the reals:
:$\size {\paren {\norm {\mathbf v} } }_K = \norm {\mathbf v}$
So then:
:$\norm {\paren {\norm {\mathbf v} \circ \mathbf u} } = \norm {\mathbf v} \times \norm {\mathbf u}$
and clearly likewise:
:$\norm {\paren {\norm {\mathbf u} \circ \mathbf v} } = \norm {\mathbf v} \times \norm {\mathbf u}$
Since $\norm {\mathbf v}, \norm {\mathbf u} \in \R$ are in the center of $K$, the result follows.
== Proof for Vectors in Euclidean Space ==
Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_n}$ and $\mathbf v = \tuple {v_1, v_2, \ldots, v_n}$ be elements of the real vector space $\R^n$ under the Euclidean norm.
Note that:
:$\norm {\mathbf v} \circ \mathbf u = \tuple {\norm {\mathbf v} \times u_1, \norm {\mathbf v} \times u_2, \ldots, \norm {\mathbf v} \times u_n}$
Similarly:
:$\norm {\paren {\norm {\mathbf u} \circ \mathbf v} } = \norm {\mathbf v} \times \norm {\mathbf u}$
and since $\times$ is commutative, the result follows.
Category:Vector Algebra
Vector as Sum of Orthogonal Base Vectors

== Theorem ==
Let $\mathbf v$ be a vector quantity in ordinary $3$-space.
Let $\mathbf i, \mathbf j, \mathbf k$ be orthonormal base vectors.
Then:
:$\mathbf v = \paren {\mathbf v \cdot \mathbf i} \mathbf i + \paren {\mathbf v \cdot \mathbf j} \mathbf j + \paren {\mathbf v \cdot \mathbf k} \mathbf k$
== Proof ==
== Sources ==
* : Chapter One: $\S 8$. Orthonormal Sets of Functions
Category:Orthonormal Sets
Category:Vector Algebra
Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $L$ be a linearly independent subset of $X$ such that: 
:$U = \map \span L \ne X$
Let: 
:$x \in X \setminus U$
Then $L \cup \set x$ is linearly independent. 
== Proof ==
Take $x_1, \ldots, x_n \in L$ and take $\alpha_1, \ldots, \alpha_n, \alpha_{n + 1} \in K$ such that: 
:$\ds \alpha_{n + 1} x + \sum_{k \mathop = 1}^n \alpha_i x_i = 0$
If $\alpha_{n + 1} = 0$, then we have: 
:$\ds \sum_{k \mathop = 1}^n \alpha_i x_i = 0$
and so $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$ from the linearly independence of $U$.
Suppose that $\alpha_{n + 1} \ne 0$, then we would have: 
:$\ds x = -\alpha^{-1}_{n + 1} \sum_{k \mathop = 1}^n \alpha_i x_i$
This would mean that $x \in U$, (from Linear Span is Linear Subspace) contradicting that $x \in X \setminus U$.
So we have $\alpha_{n + 1} = 0$, and hence $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$.
So $L \cup \set x$ is linearly independent.
Category:Linear Independence
Vectorialization of Affine Space is Vector Space

== Theorem ==
Let $\EE$ be an affine space over a field $K$ with difference space $E$.
Let $\RR = \tuple {p_0, e_1, \ldots, e_n}$ be an affine frame in $\EE$.
Let $\struct {\EE, +, \cdot}$ be the vectorialization of $\EE$.
Then $\struct {\EE, +, \cdot}$ is a vector space.
== Proof ==
By the definition of the vectorialization of an affine space, the mapping $\Theta_\RR : K^n \to \EE$ defined by:
:$\ds \map {\Theta_\RR} {\lambda_1, \ldots, \lambda_n} = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$
is a bijection from $K^n$ to $\EE$.
Therefore, by Homomorphic Image of Vector Space, it suffices to prove that $\Theta_\RR$ is a linear transformation.
By General Linear Group is Group:
:$\Theta_\RR$ is a linear transformation its inverse ${\Theta_\RR}^{-1}$ is a linear transformation.
Therefore, it suffices to show that:
:$\forall p, q \in \EE, \mu \in K: \map { {\Theta_\RR}^{-1} } {\mu \cdot p + q} = \mu \cdot \map { {\Theta_\RR}^{-1} } p + \map { {\Theta_\RR}^{-1} } g$
Thus:
This is the required identity.
Category:Affine Geometry
Vectorization of Product of Three Matrices

== Theorem ==
Let $R$ be a ring.
Let $A, B, C$ be matrices over $R$ such that the matrix product $ABC$ is defined.
Then $\map {\operatorname {vec} }{ABC} = \paren {C^\intercal \otimes A} \cdot \map {\operatorname {vec} } B$ where:
:$\operatorname {vec}$ denotes vectorization
:$C^\intercal$ is the transpose of $C$
:$\otimes$ denotes Kronecker product
:$\cdot$ denotes matrix product
== Proof ==
== Also see ==
* Vectorization of Product of Two Matrices
Category:Linear Algebra
Vectors are Coplanar iff Scalar Triple Product equals Zero

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Cartesian $3$-space:
Let $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ denote the scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Then:
:$\mathbf a \cdot \paren {\mathbf b \times \mathbf c} = 0$
 $\mathbf a$, $\mathbf b$ and $\mathbf c$ are coplanar.
== Proof ==
From Magnitude of Scalar Triple Product equals Volume of Parallelepiped Contained by Vectors:
:$\size {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} }$ equals the volume of the parallelepiped contained by $\mathbf a, \mathbf b, \mathbf c$.
The result follows.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $7$. Products of Three Vectors
* : Chapter $\text I$: Vector Algebra: $\S 4$
Category:Scalar Triple Product
Vectors are Equal iff Components are Equal

== Theorem ==
Two vector quantities are equal they have the same components.
== Proof ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Then by Vector Quantity can be Expressed as Sum of 3 Non-Coplanar Vectors, $\mathbf a$ and $\mathbf b$ can be expressed uniquely as components.
So if $\mathbf a$ and $\mathbf b$ then the components of $\mathbf a$ are the same as the components of $\mathbf b$
Suppose $\mathbf a$ and $\mathbf b$ have the same components: $\mathbf x$, $\mathbf y$ and $\mathbf z$.
Then by definition:
:$\mathbf a = \mathbf x + \mathbf y + \mathbf z$
and also:
:$\mathbf b = \mathbf x + \mathbf y + \mathbf z$
and trivially:
:$\mathbf a = \mathbf b$
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Components of a Vector: $6$. Resolution of a vector
Category:Vectors
Category:Equality
Vectors are Left Cancellable

== Theorem ==
Let $\struct {\mathbf V, +, \circ}$ be a vector space over $\GF$, as defined by the vector space axioms.
Then every $\mathbf v \in \struct {\mathbf V, +}$ is left cancellable:
:$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf c + \mathbf a = \mathbf c + \mathbf b \implies \mathbf a = \mathbf b$
== Proof ==
Utilizing the vector space axioms:
Category:Vector Algebra
Vectors are Right Cancellable

== Theorem ==
Let $\struct {\mathbf V, +, \circ}$ be a vector space over $\GF$, as defined by the vector space axioms.
Then every $\mathbf v \in \struct {\mathbf V, +}$ is right cancellable:
:$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf a + \mathbf c = \mathbf b + \mathbf c \implies \mathbf a = \mathbf b$
== Proof ==
Utilizing the vector space axioms:
== Also see ==
* Vectors are Left Cancellable
Category:Vector Algebra
Vectors from Sum and Difference

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Let $\mathbf c = \mathbf a + \mathbf b$ and $\mathbf d = \mathbf a - \mathbf b$ be given.
Then:
== Proof ==
== Sources ==
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach: Exercise $1.1.1$
Category:Vector Addition
Category:Vector Subtraction
Vectors in Three Dimensional Space with Cross Product forms Lie Algebra

== Theorem ==
Let $S$ be the set of vectors in $3$ dimensional Euclidean space.
Let $\times$ denote the vector cross product on $S$.
Then $\struct {S, \times}$ is a Lie algebra.
== Proof ==
By definition of Lie algebra, it suffices to prove two properties:
:$(1): \forall a \in S: a \times a = 0$
:$(2): \forall a, b, c \in S: a \times \paren {b \times c} + b \times \paren {c \times a} + c \times \paren {a \times b} = 0$
=== Proof of $(1)$ ===
Cross Product of Vector with Itself is Zero
=== Proof of $(2)$ ===
Vector Cross Product satisfies Jacobi Identity
Both properties hold, and the result follows.
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Lie Algebras
Category:Vector Cross Product
Velocity Vector in Polar Coordinates

== Theorem ==
Consider a particle $p$ moving in the plane.
Let the position of $p$ at time $t$ be given in polar coordinates as $\left\langle{r, \theta}\right\rangle$.
Then the velocity $\mathbf v$ of $p$ can be expressed as:
:$\mathbf v = r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r$
where:
:$\mathbf u_r$ is the unit vector in the direction of the radial coordinate of $p$
:$\mathbf u_\theta$ is the unit vector in the direction of the angular coordinate of $p$
== Proof ==
Let the radius vector $\mathbf r$ from the origin to $p$ be expressed as:
:$(1): \quad \mathbf r = r \mathbf u_r$
:
From Derivatives of Unit Vectors in Polar Coordinates:
The velocity of $p$ is by definition the rate of change in its position:
== Sources ==
* : $\S 3.21$: Newton's Law of Gravitation: $(4)$
* 
* 
Category:Polar Coordinates
Velocity from Integration of Equation of Motion

== Theorem ==
Let $P$ be a particle of constant mass $m$ moving under a force $\mathbf F$ as a function of time $t$ according to the equation of motion:
:$m \dfrac {\d^2 \mathbf r} {\d t^2} = \map {\mathbf F} {\mathbf r}$
where $\mathbf r$ is the position vector of $P$.
Let the velocity of $P$ at time $t = 0$ be $\mathbf v_0$.
Then the velocity of $P$ at time $t$ is given by:
:$\mathbf v = \mathbf v_0 + \dfrac {\d \mathbf r} {\d t}$
== Proof ==
== Sources ==
* 
* 
Category:Velocity
Category:Equations of Motion
Velocity of Bead on Brachistochrone

== Theorem ==
Consider a wire bent into the shape of an arc of a cycloid $C$ and inverted so that its cusps are uppermost and on the same horizontal line.
Let $C$ be defined by Equation of Cycloid embedded in a cartesian plane:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Let a bead $B$ be released from some point on the wire.
Let $B$ slide without friction under the influence of a constant gravitational field exerting an acceleration $g$.
Let $s_0$ be the arc length along the cycloid.
Let $s$ be the arc length along the cycloid at any subsequent point in time.
Then the speed $v$ of $B$ relative to $C$ is defined by the equation:
:$4 a v^2 = g \paren {2 s s_0 - s^2}$
== Proof ==
By Brachistochrone is Cycloid, $C$ is a brachistochrone.
== Sources ==
* : $\S 2.11$: Problem $4$
Category:Cycloids
Velocity of Harmonic Wave is Wavelength times Frequency

== Theorem ==
Let $\phi$ be a harmonic wave expressed as:
:$\forall x, t \in \R: \map \phi {x, t} = a \map \cos {\omega \paren {x - c t} }$
where $c$ is the velocity of $\phi$.
Then:
:$c = \nu \lambda$
where:
:$\nu$ is the frequency of $\phi$
:$\lambda$ is the wavelength of $\phi$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 3$: $(7)$
Category:Velocity of Harmonic Wave is Wavelength times Frequency
Category:Velocity of Periodic Wave is Wavelength times Frequency
Category:Harmonic Waves