nyuuzyou/wikis huggingface dataset:
Kernel of Linear Transformation is Null Space of Matrix Representation

== Theorem ==
Let $V$ and $W$ be finite dimensional vector spaces.
Let $\phi: V \to W$ be a linear transformation from $V$ to $W$.
Let $\tuple {e_1, \ldots, e_n}$ and $\tuple {f_1, \ldots, f_m}$ be ordered bases of $V$ and $W$ respectively.
Let $A$ be the matrix of $\phi$ in these bases.
Define $f: V \to \R^n$ by:
:$\ds \sum_{i \mathop = 1}^n a_i e_i \mapsto \tuple {a_1, \ldots, a_n}$
and $g : W \to \R^m$ by:
:$\ds \sum_{i \mathop = 1}^m b_i f_i \mapsto \tuple {b_1, \ldots, b_m}$
Let $\map N A = \set {x \in \R^n: A x = 0}$ be the null space of $A$.
Let $\map \ker \phi = \set {x \in V: \phi x = 0}$ be the kernel of $\phi$.
Then:
:$f \sqbrk {\map \ker \phi} = \map N A$
and
:$f^{-1} \sqbrk {\map N A} = \map \ker \phi$
where $f \sqbrk X$ denotes the image set of a subset $X$ of the domain of $f$.
== Proof ==
By the definition of the matrix $A$:
:$A \circ f = g \circ \phi$
Therefore if $x \in \map \ker \phi$ we have:
:$A \map f x = \map g {\map \phi x} = \map g 0 = 0$
This shows that:
:$f \sqbrk {\map \ker \phi} \subseteq \map N A$
Now let $x = \tuple {x_1, \ldots, x_n} \in \map N A$.
Let $y = x_1 e_1 + \cdots + x_n e_n \in V$.
We have:
:$\map {g \circ \phi} y = \map {A \circ f} y = \map A {x_1, \ldots, x_n}^T = 0$
so:
:$\map \phi y = 0 f_1 + \cdots + 0 f_m = 0$
This shows that $y \in \map \ker \phi$.
Since $x = \map f y$, we have shown that:
:$\map N A \subseteq \map f {\map \ker \phi}$
Therefore:
:$\map f {\map \ker \phi} = \map N A$
We deduce immediately from the definitions that:
:$\map \ker \phi \subseteq f^{-1} \sqbrk {\map N A}$
Now suppose that $x \in f^{-1} \sqbrk {\map N A}$.
Then:
:$\map f x \in \map N A$
Therefore:
:$\map {g \circ \phi} x = \map {A \circ f} x = 0$
so:
:$\map \phi x = 0 f_1 + \cdots + 0 f_m = 0$
This shows that:
:$f^{-1} \sqbrk {\map N A} \subseteq \map \ker \phi$
as required.
Category:Kernels of Linear Transformations
Category:Null Spaces
Kernel of Linear Transformation is Orthocomplement of Image of Adjoint

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $\map \BB {\HH, \KK}$ denote the set of bounded linear transformations from $\HH$ to $\KK$.
Let $A \in \map \BB {\HH, \KK}$ be a bounded linear transformation.
Then $\ker A = \paren {\Img {A^*} }^\perp$, where:
:$A^*$ denotes the adjoint of $A$
:$\ker A$ is the kernel of $A$
:$\Img {A^*}$ is the image of $A^*$
:$\perp$ signifies orthocomplementation
=== Corollary ===
== Proof ==
Let $x \in \HH$ be arbitrary.
First fix $x \in \map \ker A$ and let $y \in \Img {A^\ast}$ be arbitrary.
We then have: 
:$y = A^\ast u$
for some $u \in \KK$. 
We have: 
:$\innerprod x y_\HH = \innerprod x {A^\ast u}_\HH$
By the definition of the adjoint, we have: 
:$\innerprod x y_\HH = \innerprod {A x} u_\KK$
But $x \in \map \ker A$, so:
:$\innerprod {A x} u_\KK = 0$ 
So: 
:$\innerprod x y_\HH = 0$ for any $y \in \Img {A^\ast}$.
So:
:$x \in \paren {\Img {A^\ast} }^\bot$
So from the definition of set inclusion, we have: 
:$\map \ker A \subseteq \paren {\Img {A^\ast} }^\bot$
Now let $x \in \paren {\Img {A^\ast} }^\bot$. 
Then for each $y \in \Img {A^\ast}$ we have: 
:$\innerprod x y_\HH = 0$
For each $y \in \Img {A^\ast}$ we can write $y = A^\ast u$ for some $u \in \KK$. 
Then we have: 
:$\innerprod x {A^\ast u}_\HH = 0$
for each $u \in \KK$, and so:
:$\innerprod {A x} u_\KK = 0$
for each $u \in \KK$ from the definition of the adjoint. 
In particular, we are free to pick $u = A x$.
Then:
:$\innerprod {A x} {A x}_\KK = 0$
So $A x = 0$ from the definition of the inner product.
So:
:$x \in \ker A$
So, from the definition of set inclusion, we have: 
:$\paren {\Img {A^\ast} }^\bot \subseteq \ker A$
Hence by definition of set equality:
:$\ker A = \paren {\Img {A^*} }^\perp$
== Sources ==
* : $\text {II}.2.19$
Category:Adjoints
Category:Linear Transformations on Hilbert Spaces
Kernel of Linear Transformation is Orthocomplement of Image of Adjoint/Corollary

== Corollary to Kernel of Linear Transformation is Orthocomplement of Image of Adjoint ==
Let $\HH$ be a Hilbert space.
Let $\map \BB \HH$ denote the set of normal operators on $\HH$.
Let $A \in \map \BB \HH$ be a normal operator.
Then:
:$\ker A = \paren {\Img A}^\perp$
where:
:$A^*$ denotes the adjoint of $A$
:$\ker A$ is the kernel of $A$
:$\Img A$ is the image of $A$
:$\perp$ signifies orthocomplementation
== Proof ==
From Kernel of Linear Transformation is Orthocomplement of Image of Adjoint, we have: 
:$\ker A = \paren {\Img {A^\ast} }^\perp$
From Kernel of Normal Operator is Kernel of Adjoint, we then have: 
:$\ker A = \ker A^\ast$
and so: 
:$\ker A^\ast = \paren {\Img {A^\ast} }^\perp$
Substituting $A^\ast$ for $A$ we obtain:
:$\ker A^{\ast \ast} = \paren {\Img {A^{\ast \ast} } }^\perp$
From Adjoint is Involutive, we can conclude:
:$\ker A = \paren {\Img A}^\perp$
Category:Adjoints
Category:Linear Transformations on Hilbert Spaces
Kernel of Linear Transformation is Submodule

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}_R$ and $\struct {H, +_H, \circ_H}_R$ be $R$-modules.
Let $\phi: G \to H$ be a linear transformation.
Let $\map \ker \phi$ denote the kernel of $\phi$.
Then $\map \ker \phi$ is a submodule of $G$.
== Proof ==
By definition, the kernel of $\phi$ is defined as:
:$\map \ker \phi := \phi^{-1} \sqbrk {\set {e_H} }$
where $e_H$ is the identity of $\struct {H, +_H}$.
where $\phi^{-1} \sqbrk S$ denotes the preimage of $S$ under $\phi$.
From Null Module is Module:
:$\struct {\set {e_H}, +_H, \circ_H}_R$ is a module
where $\struct {\set {e_H}, +_H, \circ_H}_R$ is the null module.
We have that $\struct {\set {e_H}, +_H, \circ_H}_R$ is a module which is a subset of $H$.
It follows that $\struct {\set {e_H}, +_H, \circ_H}_R$ is a submodule of $H$.
The result follows from Preimage of Submodule under Linear Transformation is Submodule.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.2$
Category:Linear Transformations
Kernel of Magma Homomorphism is Submagma

== Theorem ==
Let $\struct {S, *}$ be a magma.
Let $\struct {T, \circ}$ be an algebraic structure with an identity $e$.
Let $\phi: S \to T$ be a homomorphism.
Then the kernel of $\phi$ is a submagma of $\struct {S, *}$.
That is:
:$\struct {\map {\phi^{-1} } e, *}$ is a submagma of $\struct {S, *}$
where $\map {\phi^{-1} } e$ denote the preimage of $e$.
== Proof ==
Let $x, y \in \map {\phi^{-1} } e$.
By the definition of a magma, $S$ is closed under $*$.
That is:
:$\forall x, y \in S: x * y \in S$
Hence:
:$x * y \in \Dom \phi$
It is to be shown that:
:$x * y \in \map {\phi^{-1} } e$
Thus:
Hence the result.
Category:Abstract Algebra
Kernel of Multiple Function on Ring with Characteristic Zero is Trivial

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $g_a: \Z \to R$ be the mapping from the integers into $R$ defined as:
:$\forall n \in \Z:\forall a \in R: \map {g_a} n = n \cdot a$
where $\cdot$ denotes the multiple operation.
Let $a \in R$ such that $a$ is not a zero divisor of $R$.
Let the characteristic of $R$ be $0$.
Then:
:$\map \ker {g_a} = \set {0_R}$
where $\ker$ denotes the kernel of $g_a$.
That is:
:$n \cdot a = 0_R$
:
:$n = 0$
== Proof ==
For $n = 0$, we trivially have $n \cdot a = 0_R$.
 $\exists n \ne 0: n \cdot a = 0_R$.
Then:
This contradicts our assertion that the characteristic of $R$ is $0$.
Hence by Proof by Contradiction there can be no such $n \ne 0$ such that $n \cdot a = 0_R$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.8 \ 2^\circ$
Category:Homomorphism from Integers into Ring with Unity
Kernel of Non-Zero Divisor Multiple Function is Primary Ideal of Characteristic

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the characteristic of $R$ be $p$.
Let $g_a: \Z \to R$ be the mapping from the integers into $R$ defined as:
:$\forall n \in \Z:\forall a \in R: \map {g_a} n = n \cdot a$
where $\cdot$ denotes the multiple operation.
Let $a \in R$ such that $a$ is not a zero divisor of $R$.
Then:
:$\map \ker {g_a} = \ideal p$
where:
:$\map \ker {g_a}$ is the kernel of $g_a$
:$\ideal p$ is the principal ideal of $\Z$ generated by $p$.
== Proof ==
From Principal Ideal of Characteristic of Ring is Subset of Kernel of Multiple Function we have:
:$\ideal p \subseteq \map \ker {g_a}$
for all $a \in R$.
It remains to be shown that for all $a \in R$ such that $a$ is not a zero divisor of $R$:
:$\map \ker {g_a} \subseteq \ideal p$
So:
Hence the result by definition of subset.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.8 \ 2^\circ$
Category:Homomorphism from Integers into Ring with Unity
Kernel of Normal Operator is Kernel of Adjoint

== Theorem ==
Let $H$ be a Hilbert space.
Let $A \in \map B H$ be a normal operator.
Then:
:$\ker A = \ker A^*$
where:
:$\ker$ denotes kernel
:$A^*$ denotes the adjoint of $A$.
== Proof ==
Let $x \in H$ be arbitrary.
Then:
Hence, by definition of set equality:
:$\ker A = \ker A^*$
Category:Adjoints
Category:Linear Transformations on Hilbert Spaces
Kernel of Orthogonal Projection

== Theorem ==
Let $H$ be a Hilbert space.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Then: 
:$\ker P_K = K^\bot$
where:
:$\ker P_K$ denotes the kernel of $P_K$
:$K^\bot$ denotes the orthocomplement of $K$.
== Proof ==
We first prove that: 
:$\ker P_K \subseteq K^\bot$
Let $h \in \ker P_K$.
Then: 
:$\map {P_K} h = 0$
From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, ​we have: 
:$h - \map {P_K} h \in K^\bot$
That is: 
:$h \in K^\bot$
So:
:$\ker P_K \subseteq K^\bot$
We now prove that: 
:$K^\bot \subseteq \ker P_K$
Let $h \in K^\bot$. 
From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, we have that: 
:$\map {P_K} h$ is the unique point such that $h - \map {P_K} h \in K^\bot$ 
So, since $h - 0 \in K^\bot$, we have $\map {P_K} h = 0$.
So, we have $h \in \ker P_K$, and we obtain: 
:$K^\bot = \ker P_K$
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text {(d)}$
Category:Orthogonal Projections
Kernel of Projection in Plane between Lines passing through Origin

== Theorem ==
Let $M$ and $N$ be distinct lines in the plane both of which pass through the origin $O$.
Let $\pr_{M, N}$ be the projection on $M$ along $N$:
:$\forall x \in \R^2: \map {\pr_{M, N} } x =$ the intersection of $M$ with the line through $x$ parallel to $N$.
Then $N$ is the kernel of $\pr_{M, N}$.
== Proof ==
Let $\LL$ be the straight line through $x$ which is parallel to $N$.
Let $\map {\pr_{M, N} } x = \tuple {0, 0}$.
By definition, $\map {\pr_{M, N} } x$ is the intersection of $M$ with $\LL$.
However, as $\map {\pr_{M, N} } x = \tuple {0, 0}$, it follows that $\LL$ is coincident with $N$.
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.5$
Category:Geometric Projections
Kernel of Quotient Mapping

== Theorem ==
Let $V$ be a vector space.
Let $M$ be a subspace of $V$.
Let $Q: V \to V / M$ be the quotient mapping.
Then:
:$\ker Q = M$
where $\ker Q$ is the kernel of $Q$.
== Proof ==
For $v \in V$, we have that:
:$v \in \ker Q$
:$v + M = 0 + M$
That is, :
:$v \in M$
Hence:
:$\ker Q = M$
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra
Category:Quotient Vector Spaces
Category:Quotient Mappings
Kernel of Ring Epimorphism is Ideal

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Then:
=== Kernel of Ring Homomorphism is Ideal ===
=== Quotient Ring of Kernel of Ring Epimorphism ===
=== Ring Epimorphism with Trivial Kernel is Isomorphism ===
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 1^\circ$
Category:Ring Epimorphisms
Category:Quotient Rings
Category:Ideal Theory
Kernel of Ring Homomorphism is Ideal

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
The kernel of $\phi$ is an ideal of $R_1$.
== Proof ==
By Kernel of Ring Homomorphism is Subring, $\map \ker \phi$ is a subring of $R_1$.
Let $s \in \map \ker \phi$, so $\map \phi s = 0_{R_2}$.
Suppose $x \in R_1$. Then:
and similarly for $\map \phi {s \circ_1 x}$.
The result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 1^\circ$
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Theorem $46 \ \text{(ii)}$
* : $\S 2.2$: Homomorphisms: Lemma $2.6 \ \text{(i)}$
* : $\S 58.5$ Ideals
Category:Ring Homomorphisms
Category:Ideal Theory
Kernel of Ring Homomorphism is Subring

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then the kernel of $\phi$ is a subring of $R_1$.
== Proof ==
From Ring Homomorphism of Addition is Group Homomorphism and Kernel of Group Homomorphism is Subgroup:
:$\struct {\map \ker \phi, +_1} \le \struct {R_1, +_1}$
where $\le$ denotes subgroup.
Let $x, y \in \map \ker \phi$.
Thus $x \circ_1 y \in \map \ker \phi$.
Thus the conditions for Subring Test are fulfilled, and $\map \ker \phi$ is a subring of $R_1$.
== Sources ==
* : $\S 57.3$ Ring homomorphisms: $\text{(ii)}$
Category:Ring Homomorphisms
Category:Subrings
Kernel of Transpose of Linear Transformation is Annihilator of Image

== Theorem ==
Let $G$ and $H$ be $n$-dimensional vector spaces over a field.
Let $\map \LL {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $u \in \map \LL {G, H}$.
Let $u^\intercal$ be the transpose of $u$.
Then:
:$\map \ker {u^\intercal}$ is the annihilator of the image of $u$
where $\map \ker {u^\intercal}$ denotes the kernel of $u^\intercal$.
== Proof ==
From the definitions of:
:the transpose $u^\intercal$
:the annihilator $\paren {\map u G}^\circ$
it follows that:
:$\map {u^\intercal} y = 0 \iff y \in \paren {\map u G}^\circ$
Thus:
:$\map \ker {u^\intercal} = \paren {\map u G}^\circ$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.12$
Category:Linear Algebra
Khinchin's Law

== Theorem ==
Let $P$ be a population.
Let $P$ have mean $\mu$ and finite variance. 
Let $\sequence {X_n}_{n \mathop \ge 1}$ be a sequence of random variables forming a random sample from $P$.
Let: 
:$\ds {\overline X}_n = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
:${\overline X}_n \xrightarrow p \mu$
where $\xrightarrow p$ denotes convergence in probability.
 
== Proof ==
Let $\sigma$ be the standard deviation of $P$.
By the definition of convergence in probability, we aim to show that: 
:$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} = 1$
for all real $\epsilon > 0$.
Let $\epsilon > 0$ be a real number.
By Variance of Sample Mean:
:$\var {{\overline X}_n} = \dfrac {\sigma^2} n$
By the Bienaymé-Chebyshev Inequality, we have for real $k > 0$: 
:$\map \Pr {\size { {\overline X}_n - \mu} \ge \dfrac {k \sigma} {\sqrt n}} \le \dfrac 1 {k^2}$
As $\sigma > 0$ and $n > 0$, we can set: 
:$k = \dfrac {\sqrt n} {\sigma} \epsilon$
This gives: 
:$\map \Pr {\size {{\overline X}_n - \mu} \ge \epsilon} \le \dfrac {\sigma^2} {n \epsilon^2}$
We therefore have: 
So: 
:$1 - \dfrac {\sigma^2} {n \epsilon^2} \le \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} \le 1$
We have: 
:$\ds \lim_{n \mathop \to \infty} \paren {1 - \dfrac {\sigma^2} {n \epsilon^2} } = 1$
and: 
:$\ds \lim_{n \mathop \to \infty} 1 = 1$
So by the Squeeze Theorem: 
:$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} = 1$
for all real $\epsilon > 0$.
== Also known as ==
== Also see ==
* Bernoulli's Theorem, also known as the Law of Large Numbers
* Kolmogorov's Law, also known as the Strong Law of Large Numbers
== Sources ==
* 
* 
* 
Category:Khinchin's Law
Category:Laws of Large Numbers
Category:Probability Theory
Killing Form of Orthogonal Lie Algebra

== Theorem ==
Let $\mathbb K \in \set {\C, \R}$.
Let $n$ be a positive integer.
Let $\map {\mathfrak {so}_n} {\mathbb K}$ be the Lie algebra of the special orthogonal group $\map {\operatorname {SO_n} } {\mathbb K}$.
Then its Killing form is $B: \tuple {X, Y} \mapsto \paren {n - 2} \map \tr {X Y}$.
== Proof ==
=== Lemma ===
Let $R$ be a ring with unity.
Let $n$ be a positive integer.
Let $E_{ij}$ denote the matrix with only zeroes except a $1$ at the $\tuple {i, j}$th position.
Then for all $X, Y \in R^{n \times n}$:
:$\ds \sum_{1 \mathop \le i \mathop < j \le n} \map \tr {\paren {\map X {E_{ij} - E_{ji} } Y}^t \paren {E_{ij} - E_{ji} } } = \map \tr Y \map \tr X - \map \tr {Y^t X}$
== Also see ==
* Killing Form of Symplectic Lie Algebra
Category:Lie Algebras
Killing Form of Symplectic Lie Algebra

== Theorem ==
Let $\mathbb K \in \set {\C, \R}$.
Let $n$ be a positive integer.
Let $\map {\mathfrak {sp}_{2 n} } {\mathbb K}$ be the Lie algebra of the symplectic group $\map {\operatorname {Sp} } {2 n, \mathbb K}$.
Then its Killing form is $B: \tuple {X, Y} \mapsto \paren {2 n + 2} \map \tr {X Y}$.
== Proof ==
=== Lemma ===
Let $R$ be a ring with unity.
Let $n$ be a positive integer.
Let $E_{ij}$ denote the matrix with only zeroes except a $1$ at the $\tuple {i, j}$th position.
Let $X, Y \in R^{2 n \times 2 n}$.
Let $X = \begin {pmatrix} X_{11} & X_{12} \\ X_{21} & X_{22} \end{pmatrix}$ and $Y = \begin{pmatrix} Y_{11} & Y_{12} \\ Y_{21} & Y_{22}\end{pmatrix}$
Then:
:$\ds \sum_{i, j \mathop = 1}^{2 n} \map \tr {\paren {X E_{i j} Y}^t E_{i j} } = \map \tr Y \map \tr X$
:$\ds \sum_{i, j \mathop = 1}^n \map \tr {\paren {X E_{ij} Y}^t E_{j + n, i + n} } = \map \tr {Y_{1 2}^t X_{2 1} }$
=== Proof ===
Use Trace of Alternating Product of Matrices and Almost Zero Matrices.
----
Use Definition:Frobenius Inner Product and Trace in Terms of Orthonormal Basis and the fact that the $\sequence {E_{i j} }_{i \mathop \le n, j \mathop \ge n + 1}, \sequence {E_{i j} }_{i \mathop \ge n + 1, j \mathop \le n}, \sequence {E_{i j} - E_{j + n, i + n} } / \sqrt 2$ are an orthonormal basis of $\mathfrak {sp}_{2 n}$.
== Also see ==
* Killing Form of Orthogonal Lie Algebra
Category:Lie Algebras
Kinetic Energy/Examples/1 kilogram at 1 metre per second

== Example of Kinetic Energy ==
A body of mass $1 \ \mathrm {kg}$ moving at $1 \ \mathrm m \, \mathrm s^{-1}$ has a kinetic energy of $\dfrac 1 2$ a joule.
== Proof ==
== Sources ==
* : Chapter $1$: Electrical Quantities: Definitions and Laws: Definitions: Energy
Category:Examples of Kinetic Energy
Kinetic Energy of Body at Constant Angular Speed

== Theorem ==
Let $B$ be a body rotating at an angular speed $\omega$ about some axis of rotation $R$.
Let $I$ denote the moment of inertia of $B$ about $R$.
Then the kinetic energy $T$ of $B$ brought about by this rotation is given by:
:$T = \dfrac {I \omega^2} 2$
== Proof ==
== Sources ==
* 
* 
* 
* 
Category:Kinetic Energy
Kinetic Energy of Body at Constant Speed

== Theorem ==
Let $B$ be a body of mass $m$ moving at a speed of $v$.
Let $v$ be considerably less than the speed of light.
Then the kinetic energy $T$ of $B$ is given by:
:$T \approx \dfrac {m v^2} 2$
== Proof ==
== Sources ==
* 
* 
Category:Kinetic Energy
Kinetic Energy of Classical Particle

== Theorem ==
Let $\MM$ be an $n$-dimensional Euclidean manifold.
Let $P$ be a particle with an inertial mass $m_i$.
Let $t$ be the time variable of $P$.
Suppose the position of $P$ is a real differentiable $n$-dimensional vector-valued mapping $\mathbf x = \map {\mathbf x} t$.
Then the kinetic energy of a classical particle $P$ is:
:$T = \dfrac {m_i} 2 \paren {\dfrac {\d \mathbf x} {\d t} }^2$
where $\paren {\dfrac {\d \mathbf x} {\d t} }^2$ is the dot product of the vector $\dfrac {\d \mathbf x} {\d t}$ with itself.
== Proof ==
== Sources ==
* : $\S 4.21$: The Principle of Least Action
Category:Lagrangian Mechanics
Category:Kinetic Energy
Kinetic Energy of Motion

== Physical Law ==
Let $B$ be a body of mass $m$ moving with speed $v$ such that $v$ is very much less than the speed of light.
Then its kinetic energy $K$ is given by:
:$K = \dfrac {m v^2} 2$
== Proof ==
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
Category:Physics
Category:Mechanics
Kleene Closure is Free Monoid

== Theorem ==
Let $S$ be a set.
Let $S^*$ be its Kleene closure, and let $i: S \to S^*$ be the insertion of generators.
Then $\struct {S^*, i}$ is a free monoid over $S$.
== Proof ==
By Kleene Closure is Monoid, $S^*$ is a monoid.
It remains to verify the universal property.
=== Existence ===
Let $\struct {N, \circ}$ be a monoid with identity $e_N$, and let $\card N$ be its underlying set.
Let $f: S \to \card N$ be a mapping.
Define $\bar f: S^* \to N$ by:
:$\map {\bar f} - := e_N$
:$\map {\bar f} {\sequence {s_1, \ldots, s_n} } := \map f {s_1} \circ \cdots \circ \map f {s_n}$
The first condition is precisely saying that $\bar f$ preserves identity elements, and by:
it follows that $\bar f$ is a monoid homomorphism.
It also satisfies:
:$\map {\bar f \circ i} s = \map {\bar f} {\sequence s} = \map f s$
and hence meets the set requirements.
=== Uniqueness ===
It remains to verify the uniqueness of $\bar f$.
So let $g: S^* \to N \in \mathbf{Mon}_1$ be a monoid homomorphism such that for all $s \in S$:
$\map {g \circ i} s = \map f s$
Then for any $\sequence {s_1, \ldots, s_n} \in S^*$, compute:
Hence $\struct {S^*, i}$ is a free monoid for $S$.
== Sources ==
* : $\S 1.7$: Proposition $1.9$
Category:Abstract Algebra
Kleene Closure is Monoid

== Theorem ==
Let $S$ be a set, and let $S^*$ be its Kleene closure.
Let $*$ denote concatenation of ordered tuples.
Then $\struct {S^*, *}$ is a monoid.
== Proof ==
First, to prove that $\struct {S^*, *}$ is a semigroup.
That is, to prove $*$ is associative.
Let $s, s', s \in S^*$ be sequences of lengths $n, n', n$, respectively.
Then:
Hence, by Equality of Mappings:
:$s * \paren {s' * s} = \paren {s * s'} * s$
that is, $*$ is associative.
Now, to prove $\struct {S^*, *}$ has an identity $e$.
It follows immediately from the length of a concatenation that $e$ must have length $0$.
That is, the only choice for $e$ is the empty sequence.
Now, for any $s \in S^*$:
:$\map {e * s} i = \begin {cases} \map e i & \text {if } 1 \le i \le 0 \\ \map s i & \text {if } 0 < i < 0 + n \end {cases}$
from which we see that $e * s = s$.
Also:
:$\map {s * e} i = \begin {cases} \map s i & \text {if } 1 \le i \le n \\ \map e i & \text {if } n < i < n + 0 \end {cases}$
which shows $s * e = s$.
So indeed the empty sequence is an identity element of $\struct {S^*, *}$.
Hence $\struct {S^*, *}$ is a monoid.
== Sources ==
* : $\S 1.7$
Category:Sequences
Klein Four-Group/Cayley Table

== Cayley Table for Klein $4$-Group ==
The Klein $4$-group can be described completely by showing its Cayley table:
$\quad \begin {array} {c|cccc}
& e & a & b & c \\
\hline
e & e & a & b & c \\
a & a & e & c & b \\
b & b & c & e & a \\
c & c & b & a & e \\
\end {array}$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem
* : $\S 1.5$: Example $15$
* : Chapter $2$: The Definition of Group Structure: $\S 26 \iota$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $6$
* : $\S 44.5$ Some consequences of Lagrange's Theorem
* 
Category:Klein Four-Group
Category:Examples of Cayley Tables
Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic

== Theorem ==
The Klein $4$-group $K_4$ and the cyclic group of order $4$ $C_4$ are not isomorphic.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Group Isomorphisms/Order 4
Category:Klein Four-Group
Category:Cyclic Group of Order 4
Category:Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic