nyuuzyou/wikis huggingface dataset:
Necessary and Sufficient Condition for Integral Parametric Functional to be Independent of Parametric Representation

== Theorem ==
Let $x: \R \to \R$ and $y: \R \to \R$ be real functions.
Let $J \sqbrk {x, y}$ be a functional of the form:
:$\ds J \sqbrk {x, y} = \int_{t_0}^{t_1} \map \Phi {t, x, y, \dot x, \dot y} \rd t$
where $\dot y$ denotes the derivative of $y$ $t$:
:$\dot y = \dfrac {\d y} {\d t}$
Then $J \sqbrk {x, y}$ depends only on the curve in the $x y$-plane defined by the parametric equations:
:$x = \map x t$, $y = \map y t$ 
and not on the choice of the parametric representation of the curve the integrand $\Phi$ does not involve $t$ explicitly 
and is a positive-homogeneous function of degree $1$ in $\dot x$ and $\dot y$.
== Proof ==
In the functional:
:$\ds \JJ \sqbrk y = \int_{x_0}^{x_1} \map F {x, y, y'} \rd x$
let the argument $y$ stand for a curve which is given in a parametric form. 
In other words, let the curve be described by the vector $\tuple {\map y t,\map x t}$ rather than $\paren {\map y x, x}$.
Then the functional can be rewritten as:
The function on the does not involve $t$ explicitly.
Suppose that it is positive-homogeneous of degree 1 in $\map {\dot x} t$ and $\map {\dot y} t$:
:$\map \Phi {x, y, \lambda \dot x, \lambda \dot y} = \lambda \map \Phi {x, y, \dot x, \dot y}$ for all $\lambda > 0$.
Now we will show that the value of such functional depends only on the curve in the $x y$-plane defined by the parametric equations $x = \map x t$ and $y = \map y t$, and not on the functions $\map x t$, $\map y t$ themselves. 
Let a new parameter $\tau$ be chosen such that $t = \map t \tau$, where $\dfrac {\d t} {\d \tau} > 0$.
Let the interval $\closedint {t_0} {t_1}$ be mapped onto $\closedint {\tau_0} {\tau_1}$.
Since $\Phi$ is positive-homogeneous of degree $1$ in $\dot x$ and $\dot y$, it follows that:
== Sources ==
* : $\S 2.10$: Variational Problems in Parametric Form
Category:Calculus of Variations
Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Dependent on N Functions

== Theorem ==
Let $K$ be a (real) functional, such that:
:$\ds K \sqbrk {\mathbf h} = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x$
where:
:$\mathbf h$ is an $N$-dimensional vector
:$\mathbf Q$ is a $N \times N$ matrix
:$\mathbf P$ is a $N\times N$ symmetric positive definite matrix.
Let $\closedint a b$ be such that it does not contain a point conjugate to $a$.
Then:
:$\forall \mathbf h: \map {\mathbf h} a = \map {\mathbf h} b = 0: K \sqbrk {\mathbf h} > 0$
 the above holds.
== Proof ==
=== Necessary Condition ===
Let $\mathbf W$ be an arbitrary differentiable symmetric matrix.
Then
Suppose, $\mathbf W$ is such that:
:$\mathbf Q + \mathbf W' = \mathbf W \mathbf P^{-1} \mathbf W$
Then:
Note that:
:$\mathbf P^{1/2} \mathbf h' + \mathbf P^{-1/2} \mathbf W \mathbf h \ne 0$
unless:
:$\forall x\in\closedint a b: \map {\mathbf h} x = 0$
However, this contradicts the absence of conjugate points.
Hence:
:$K > 0$
=== Sufficient Condition ===
Consider the following functional:
:$\ds \int_a^b \paren {K t + \mathbf h' \paren {1 - t} \mathbf h'} \rd x$
The corresponding Euler's equations are:
:$-\map {\dfrac \d {\d x} } {t \mathbf P \mathbf h' + \paren {1 - t} \mathbf h'} + t \mathbf Q \mathbf h = 0$
Suppose the interval $\closedint a b$ contains a point $\tilde a$ conjugate to $a$.
Then the determinant $\size {h_{ij} }$ vanishes.
Therefore there exists a linear combination of $h_i$ not identically equal to zero such that $\map {\mathbf h} {\tilde a} = 0$.
Furthermore, since the Euler's equations are continuous $t$, so is the solution of this equation.
 $\tilde a = b$.
By lemma, $K$ vanishes.
This contradicts the positive definiteness of $K$.
Therefore, $\tilde a \ne b$.
Thus, for $t = 1$ the conjugate point may only reside in $\openint a b$.
== Sources ==
* : $\S 5.29$: Generalization to n Unknown Functions
Category:Calculus of Variations
Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Lemma 1

== Theorem ==
Let the function $\map h x$ satisfy the equation:
:$-\map {\dfrac \d {\d x} } {P h'} + Q h = 0$
Let $\map h x$ have the boundary conditions:
:$\map h a = \map h b = 0$
Then:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x = 0$
== Proof ==
== Sources ==
: $\S 5.26$: Analysis of the Quadratic Functional $ \int_a^b \paren {P h'^2 + Q h^2} \rd x$
Category:Calculus of Variations
Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Lemma 2

== Theorem ==
Let $\map h x : \closedint a b \to \R$ be continuously differentiable $\forall x \in \closedint a b$.
Suppose the function $\map h x$ satisfies the equation:
:$-\map {\dfrac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h = 0$
subject to the boundary conditions:
:$\map h {a, t} = \map h {b, t} = 0$
Then:
:$\ds \int_a^b \paren {\paren {P h'^2 + Q h^2} t + \paren {1 - t} h'^2} \rd x = 0$
== Proof ==
== Sources ==
: $\S 5.26$: Analysis of the Quadratic Functional $\int_a^b \paren {P h'^2 + Q h^2} \rd x$
Category:Calculus of Variations
Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space/Corollary 1

== Theorem ==
Let $\mathbb K = \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\mathbb K$.
Let $CL$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\struct {\map {CL} {X, \mathbb K}, \norm {\, \cdot \,} }$ is a Banach space.
== Proof ==
:$X' := \map {CL} {X, \mathbb K}$ is the dual space of $X$
So this theorem is the same as Normed Dual Space is Banach Space.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. When is $\map {CL} {X, Y}$ complete?
Category:Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space
Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space/Corollary 2

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space.
Let $CL$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\struct {\map {CL} {X, X}, \norm{\, \cdot \,}}$ is a Banach space.
== Proof ==
Take $Y = X$ in Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. When is $\map {CL} {X, Y}$ complete?
Category:Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space
Negated Restricted Universal Quantifier

== Theorem ==
Let $x$ and $A$ be sets.
Let $\map P x$ be a propositional function.
:$\neg \forall x \in A : \map P x \iff \exists x \in A : \neg \map P x $
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
Category:Universal Quantifier
Negated Upper Index of Binomial Coefficient

== Theorem ==
Let $r \in \R, k \in \Z$.
Then:
:$\dbinom r k = \paren {-1}^k \dbinom {k - r - 1} k$
where $\dbinom r k$ is a binomial coefficient.
=== Corollary 1 ===
=== Corollary 2 ===
=== Complex Numbers ===
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.3$: Binomial Coefficients
* : $\S 1.2.6$: Binomial Coefficients: $\text{G}$: $(17)$
Category:Binomial Coefficients
Category:Negated Upper Index of Binomial Coefficient
Negated Upper Index of Binomial Coefficient/Complex Numbers

== Theorem ==
For all $z, w \in \C$ such that it is not the case that $z$ is a negative integer and $t, w$ integers:
:$\dbinom z w = \dfrac {\map \sin {\pi \paren {w - z - 1} } } {\map \sin {\pi z} } \dbinom {w - z - 1} w$
where $\dbinom z w$ is a binomial coefficient.
== Proof ==
By definition of Binomial Coefficient:
:$\dbinom z w = \ds \lim_{\zeta \mathop \to z} \lim_{\tau \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\omega + 1} \map \Gamma {\zeta - \omega + 1} }$
Euler's Reflection Formula gives:
:$\forall z \notin \Z: \map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Thus:
and:
Hence:
Now we have:
and:
Thus:
and the result follows.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $42$ (Solution)
Category:Negated Upper Index of Binomial Coefficient
Negated Upper Index of Binomial Coefficient/Corollary 1

== Corollary to Negated Upper Index of Binomial Coefficient ==
Let $r \in \R, k \in \Z$.
Then:
:$\dbinom {-r} k = \paren {-1}^k \dbinom {r + k - 1} k$
where $\dbinom {-r} k$ is a binomial coefficient.
== Proof 1 ==
== Proof 2 ==
Category:Binomial Coefficients
Category:Negated Upper Index of Binomial Coefficient
Negated Upper Index of Binomial Coefficient/Corollary 2

== Corollary to Negated Upper Index of Binomial Coefficient ==
Let $n, m \in \Z$.
Then:
:$\dbinom n m = \paren {-1}^{n - m} \dbinom {-\paren {m + 1} } {n - m}$
where $\dbinom n m$ is a binomial coefficient.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(19)$
Category:Binomial Coefficients
Category:Negated Upper Index of Binomial Coefficient
Negated Upper Index of Gaussian Binomial Coefficient

== Theorem ==
Let $q \in \R_{\ne 1}, r \in \R, k \in \Z$.
Then:
:$\dbinom r k_q = \paren {-1}^k \dbinom {k - r - 1} k_q q^{k r - k \paren {k - 1} / 2}$
where $\dbinom r k_q$ is a binomial coefficient.
== Proof ==
First note that:
Then:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $58$
Category:Gaussian Binomial Coefficients
Category:Negated Upper Index of Binomial Coefficient
Negation implies Negation of Conjunction/Case 1

== Theorem ==
:$\neg p \implies \neg \paren {p \land q}$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T43}$
Category:Negation implies Negation of Conjunction
Negation implies Negation of Conjunction/Case 2

== Theorem ==
:$\neg q \implies \neg \left({p \land q}\right)$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T44}$
Category:Negation implies Negation of Conjunction
Negation of Conditional implies Negation of Consequent

== Theorem ==
:$\vdash \neg \paren {p \implies q} \implies \neg q$
== Proof ==
== Sources ==
* : $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T22}$
Category:Conditional
Category:Logical Negation
Negation of Finite State Machine

== Theorem ==
Let $F = \tuple {S, A, I, \Sigma, T}$ be a finite state machine.
Then there exists a finite state machine $F'$ such that $\map L {F'} = \map \complement {\map L F}$.
== Proof ==
Construct $F' = \tuple {S, S \setminus A, I, \Sigma, T}$.
The state transitions are identical to those of $F$.
However, when the input word is exhausted, the final state is an accepting state it is not an accepting state in $F$, as required.
Category:Abstract Machines
Negation of Integer is Primitive Recursive

== Theorem ==
Let $f^- : \N \to \N$ be defined as:
:$\map {f^-} {x_\Z} = \paren {-x}_\Z$
where:
:$k_\Z$ denotes the code number for the integer $k$
Then $f^-$ is a primitive recursive function.
== Proof ==
Define $f^-$ as:
$\quad\map {f^-} {x_\Z} = \begin{cases}
\map {c^-} {\size x} & : x > 0 \\
\map {c^+} {\size x} & : \text{otherwise}
\end{cases}$
where:
:$\map {c^-} x = \paren {-x}_\Z$
:$\map {c^+} x = x_\Z$
The function is primitive recursive by:
* Definition by Cases is Primitive Recursive/Corollary
* Set of Strictly Positive Integers is Primitive Recursive
* Code Number for Non-Negative Integer is Primitive Recursive
* Code Number for Non-Positive Integer is Primitive Recursive
For $x > 0$, we have:
:$x = \size x$
Thus:
:$\map {f^-} x = \paren {-x}_\Z$
by definition of $c^-$, as required.
For $x = 0$, we have:
:$x = 0 = -\size x$
Thus:
:$\map {f^-} 0 = 0_\Z = \paren {-0}_\Z$
by definition of $c^+$, as required.
For $x < 0$, we have:
:$x = -\size x$
Thus:
:$\map {f^-} x = x_\Z = \paren {-\paren {-x}}_\Z$
by definition of $c^+$, as required.
Category:Primitive Recursive Functions
Negation of Ordering of Natural Numbers is Provable

== Theorem ==
Let $x, y \in \N$ be natural numbers.
Suppose $x \ge y$.
Let $\sqbrk a$ denote the unary representation of $a \in \N$.
Then:
:$\neg \paren {\sqbrk x < \sqbrk y}$
is a theorem of minimal arithmetic.
== Proof ==
Fix $x \in \N$.
Proceed by induction on $y$.
=== Basis for the Induction ===
Suppose $y = 0$.
The result:
:$\neg \paren {\sqbrk x < 0}$
follows from Axiom $\text M 7$
=== Induction Hypothesis ===
Suppose that the result holds for $y \in \N$.
=== Induction Step ===
By Trichotomy Law for Natural Numbers, one of:
:$x < y$
:$x = y$
:$x > y$
holds.
Suppose $x \le y$.
Then $x < \map s y$.
Then the conditions for the result fail, and so it follows from False Statement implies Every Statement.
Otherwise, suppose $x > y$.
Then, by hypothesis:
:$\neg \paren {\sqbrk x < \sqbrk y}$
is a theorem.
Additionally, by Inequality of Natural Numbers is Provable:
:$\neg \paren {\sqbrk x = \sqbrk y}$
is a theorem.
By Rule of Conjunction and Conjunction of Negations:
:$\neg \paren {\sqbrk x < \sqbrk y \lor \sqbrk x = \sqbrk y}$
The result follows from Axiom $\text M 8$ and Modus Tollendo Tollens.
By the Principle of Mathematical Induction, the result holds for every $y \in \N$.
As $x \in \N$ was arbitrary, the result holds for every $x, y \in \N$.
Category:Ordering on Natural Numbers
Category:Proofs by Induction
Negation of Propositional Function in Two Variables

== Theorem ==
Let $\map P {x, y}$ be a propositional function of two Variables.
Then:
:$\neg \forall x: \exists y: \map P {x, y} \iff \exists x: \forall y: \neg \map P {x, y}$
That is:
:It is not the case that for all $x$ a value of $y$ can be found to satisfy $\map P {x, y}$
means the same thing as:
:There exists at least one value of $x$ such that for all $y$ it is not possible to satisfy $\map P {x, y}$
== Proof ==
== Sources ==
* : $\S 1$: Some mathematical language: Variables and quantifiers
Category:Universal Quantifier
Category:Existential Quantifier
Negative Binomial Distribution Gives Rise to Probability Mass Function

== Theorem ==
Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
=== Type $1$ ===
=== Type $2$ ===
Category:Negative Binomial Distribution Gives Rise to Probability Mass Function
Category:Negative Binomial Distributions
Category:Probability Mass Functions
Negative Binomial Distribution Gives Rise to Probability Mass Function/Type 1

== Theorem ==
Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the Type $1$ negative binomial distribution with parameters $n$ and $p$ ($0 < p < 1$).
Then $X$ gives rise to a probability mass function.
== Proof ==
By definition:
:$\Img X = \set {n, n + 1, n + 2, \ldots}$
:$\map \Pr {X = k} = \dbinom {k - 1} {n - 1} p^n \paren {1 - p}^{k - n}$
Then:
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
== Sources ==
* : $\S 2.2$: Examples
Category:Negative Binomial Distribution Gives Rise to Probability Mass Function
Category:Negative Binomial Distribution (Type 1)
Category:Probability Mass Functions
Negative Binomial Distribution Gives Rise to Probability Mass Function/Type 2

== Theorem ==
Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the type $2$ negative binomial distribution with parameters $n$ and $p$ ($0 < p < 1$).
Then $X$ gives rise to a probability mass function.
== Proof ==
By definition:
:$\Img X = \set {0, 1, 2, \ldots}$
:$\map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k \paren {1 - p}^n$
Then:
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
Category:Negative Binomial Distribution (Type 2)
Negative Binomial Distribution as Generalized Geometric Distribution/Type 1

== Theorem ==
The type $1$ negative binomial distribution is a generalization of the shifted geometric distribution:
Let $\sequence {Y_i}$ be a Bernoulli process with parameter $p$.
Let $\FF$ be the experiment which consists of:
:Perform the Bernoulli trial $Y_i$ as many times as it takes to achieve $n$ successes, and then stop.
Let $k$ be the number of Bernoulli trials that need to be taken in order to achieve up to (and including) the $n$th success.
Let $\FF'$ be the experiment which consists of:
:Perform the Bernoulli trial $Y_i$ until one success is achieved, and then stop.
Then $k$ is modelled by the experiment:
:Perform experiment $\FF'$ until $n$ failures occur, and then stop.
== Proof ==
Consider the experiment $\FF$ as described.
By definition, $\FF$ is modelled by a negative binomial distribution with parameters $n$ and $p$:
:$\ds \forall k \in \Z, k \ge n: \map \Pr {Y = k} = \binom {k - 1} {n - 1} q^{k - n} p^n$
where $q = 1 - p$.
Now consider the experiment $\FF'$ as described.
By Bernoulli Process as Shifted Geometric Distribution, $\FF'$ is modelled by a geometric distribution with parameter $p$:
:$\forall k \in \Z, k \ge 1: \map \Pr {Y = k} = q^{k - 1} p$
where $q = 1 - p$.
Category:Negative Binomial Distribution as Generalized Geometric Distribution
Category:Negative Binomial Distribution (Type 1)
Category:Geometric Distribution
Negative Binomial Distribution as Generalized Geometric Distribution/Type 2

== Theorem ==
The type $2$ negative binomial distribution is a generalization of the geometric distribution:
Let $\sequence {X_i}$ be a Bernoulli process with parameter $p$.
Let $\EE$ be the experiment which consists of:
:Perform the Bernoulli trial $X_i$ until $n$ failures occur, and then stop.
Let $k$ be the number of successes before before $n$ failures have been encountered.
Let $\EE'$ be the experiment which consists of:
:Perform the Bernoulli trial $X_i$ until one failure occurs, and then stop.
Then $k$ is modelled by the experiment:
:Perform experiment $\EE'$ until $n$ failures occur, and then stop.
== Proof ==
Consider the experiment $\EE$ as described.
By definition, $\EE$ is modelled by a type $2$ negative binomial distribution with parameters $n$ and $p$:
:$\forall k \in \Z, k \ge 0: \map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k q^n $
where $q = 1 - p$.
Now consider the experiment $\EE'$ as described.
By Bernoulli Process as Geometric Distribution, $\EE'$ is modelled by a geometric distribution with parameter $p$:
:$\forall k \in \Z, k \ge 0: \map \Pr {X = k} = p^k q$
where $q = 1 - p$.
== Sources ==
* 
* 
Category:Negative Binomial Distribution as Generalized Geometric Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Geometric Distribution
Negative Infinity is Minimal

== Theorem ==
Let $\struct {\overline \R, \le}$ be the extended real numbers with the usual ordering.
Then $-\infty$ is a minimal element of $\overline \R$.
== Proof ==
By the definition of the usual ordering on the extended real numbers:
:${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$
Suppose $x \in \overline \R$ and $x \le -\infty$.
That is:
:$\tuple {x, -\infty} \in {\le}$
By the definition of union, $\tuple {x, -\infty}$ must lie in one of the three sets whose union forms $\le$.
Since ${\le_\R} \subseteq \R \times \R$ and $-\infty \notin \R$:
:$\tuple {x, -\infty} \notin {\le_\R}$
Since $+\infty \ne -\infty$ by the definition of the extended real numbers:
:$\tuple {x, -\infty} \notin \set {\tuple {x, +\infty}: x \in \overline \R}$
Therefore:
:$\tuple {x, -\infty} \in \set {\tuple {-\infty, x}: x \in \overline \R}$
and we conclude that $x = -\infty$.
That is, $-\infty$ is a minimal element of $\overline \R$.
== Also see ==
* Positive Infinity is Maximal
Category:Order Theory
Category:Extended Real Numbers
Negative Infinity is Smallest Element

== Theorem ==
Let $\left({\overline \R, \le}\right)$ be the extended real numbers with their usual ordering.
Then $-\infty$ is the smallest element of $\overline \R$.
== Proof ==
We have, by definition of the usual ordering on $\overline \R$:
:$\forall x \in \overline \R: -\infty \le x$
That is, $-\infty$ is the smallest element of $\overline \R$.
== Also see ==
* Positive Infinity is Greatest Element
Category:Order Theory
Category:Extended Real Numbers
Negative Linear Transformation

== Theorem ==
Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be abelian groups.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}$ and $\struct {H, +_H, \circ_H}$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\phi: G \to H$ be a linear transformation.
Let $-\phi$ be the pointwise negative of $\phi$:
:$\forall \phi \in \map {\LL_R} {G, H}: \forall x \in G: \map {\paren {-\phi} } x = -\paren {\map \phi x}$
Then $-\phi: G \to H$ is also a linear transformation.
== Proof ==
From the definition of a module, the group $\struct {H, +_H}$ is abelian.
Therefore we can apply Inverse Mapping in Induced Structure of Homomorphism to Abelian Group to show that $-\phi: G \to H$ is a homomorphism.
Let $\lambda \in R, x \in G$.
Then:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Negative Matrix is Inverse for Hadamard Product

== Theorem ==
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 $\struct {G, \cdot}$.
Let $\mathbf A$ be an element of $\map {\MM_G} {m, n}$.
Let $-\mathbf A$ be the negative of $\mathbf A$.
Then $-\mathbf A$ is the inverse for the operation $\circ$, where $\circ$ is the Hadamard product.
== Proof ==
Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_G} {m, n}$.
Then:
The result follows from Zero Matrix is Identity for Hadamard Product.
== Also see ==
* Negative Matrix is Inverse for Matrix Entrywise Addition
Category:Negative Matrices
Category:Hadamard Product
Negative Matrix is Inverse for Matrix Entrywise Addition

== Theorem ==
Let $\Bbb F$ denote one of the standard number systems.
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over $\Bbb F$.
Let $\mathbf A$ be an element of $\map \MM {m, n}$.
Let $-\mathbf A$ be the negative of $\mathbf A$.
Then $-\mathbf A$ is the inverse for the operation $+$, where $+$ is matrix entrywise addition.
== Proof ==
Let $\mathbf A = \sqbrk a_{m n} \in \map \MM {m, n}$.
Then:
The result follows from Zero Matrix is Identity for Matrix Entrywise Addition.
== Also see ==
* Negative Matrix is Inverse for Matrix Entrywise Addition over Ring
* Negative Matrix is Inverse for Hadamard Product
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $7$
Category:Negative Matrices
Negative Matrix is Inverse for Matrix Entrywise Addition over Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $\struct {R, +, \circ}$.
Let $\mathbf A$ be an element of $\map {\MM_R} {m, n}$.
Let $-\mathbf A$ be the negative of $\mathbf A$.
Then $-\mathbf A$ is the inverse for the operation $+$, where $+$ is matrix entrywise addition.
== Proof ==
Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Then:
The result follows from Zero Matrix is Identity for Matrix Entrywise Addition over Ring.
== Also see ==
* Negative Matrix is Inverse for Hadamard Product
Category:Negative Matrices
Negative Number is Congruent to Modulus minus Number

== Theorem ==
:$\forall m, n \in \Z: -m \equiv n - m \pmod n$
where $\bmod n$ denotes congruence modulo $n$.
== Proof ==
Let $-m = r + k n$.
Then $-m + n = r + \paren {k + 1} n$
and the result follows directly by definition.
== Also see ==
* Wilson's Theorem, where this is used:
:$-1 \equiv \paren {p - 1}! \pmod p \iff \text {$p$ is prime}$
Category:Modulo Arithmetic