nyuuzyou/wikis huggingface dataset:
Inverse of Order Isomorphism is Order Isomorphism

== Theorem ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi$ be a bijection from $\struct {S, \preceq_1}$ to $\struct {T, \preceq_2}$.
Then:
:$\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$
is an order isomorphism :
:$\phi^{-1}: \struct {T, \preceq_2} \to \struct {S, \preceq_1}$
is also an order isomorphism.
== Proof ==
Follows directly from the definition of order isomorphism.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an order isomorphism.
Then:
:$\forall x, y, \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
and:
:$\forall \map \phi x, \map \phi y \in T: \map \phi x \preceq_2 \map \phi y \implies \map {\phi^{-1} } {\map \phi x} \preceq_1 \map {\phi^{-1} } {\map \phi y}$
That is:
:$\forall \map \phi x, \map \phi y \in T: \map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.1: \ 2^\circ$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.6$
* : $\S 6.30$
Category:Order Isomorphisms
Inverse of Ordered Ring Isomorphism is Ordered Ring Isomorphism

== Theorem ==
Let $\struct {S, +, \circ, \preceq}$ and $\struct {T, \oplus, *, \preccurlyeq}$ be ordered rings.
Let $\phi: S \to T$ be an ordered ring isomorphism.
Then $\phi^{-1}: T \to S$ is also an ordered ring isomorphism.
== Proof ==
By definition, $\phi$ is a bijection.
By Bijection iff Inverse is Bijection, $\phi^{-1}$ is also a bijection.
By definition, an ordered ring isomorphism from $\phi: \struct {S, +, \circ, \preceq} \to \struct {T, \oplus, *, \preccurlyeq}$ is:
:an order isomorphism from the ordered set $\struct {S, \preceq}$ to the ordered set $\struct {T, \preccurlyeq}$
:a group isomorphism from the group $\struct {S, +}$ to the group $\struct {T, \oplus}$
:a semigroup isomorphism from the semigroup $\struct {S, \circ}$ to the semigroup $\struct {T, *}$.
From Inverse of Order Isomorphism is Order Isomorphism, $\phi^{-1}: \struct {T, \preccurlyeq} \to \struct {S, \preceq}$ is an order isomorphism.
From Inverse of Algebraic Structure Isomorphism is Isomorphism:
:$\phi^{-1}: \struct {T, \oplus} \to \struct {S, +}$ is a group isomorphism
:$\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is a semigroup isomorphism.
From Isomorphism of Abelian Groups, $\phi^{-1}: \struct {T, \oplus} \to \struct {S, +}$ preserves the commutativity of $S$ and $T$.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
Category:Ordered Rings
Inverse of Ordered Semigroup Isomorphism is Isomorphism

== Theorem ==
Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a mapping.
Then:
:$\phi$ is an ordered semigroup isomorphism
:
:$\phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$ is also an ordered semigroup isomorphism.
== Proof ==
=== Sufficient Condition ===
Let $\phi$ be an ordered semigroup isomorphism.
Then by definition $\phi$ is a bijection.
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.
That is:
:$\exists \phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$
From Inverse of Algebraic Structure Isomorphism is Isomorphism:
:$\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is an isomorphism.
Indeed, from Isomorphism Preserves Semigroups, $\phi^{-1}$ is specifically a semigroup isomorphism.
Then from Inverse of Order Isomorphism is Order Isomorphism we have that:
:$\phi^{-1}: \struct {T, \preccurlyeq} \to \struct {S, \preceq}$ is an order isomorphism.
=== Sufficient Condition ===
Let $\phi^{-1}$ be an ordered semigroup isomorphism.
Then $\paren {\phi^{-1} }^{-1}$ is also an ordered semigroup isomorphism.
From Inverse of Inverse of Bijection:
:$\paren {\phi^{-1} }^{-1} = \phi$
and the result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.1$
Category:Ordered Semigroup Isomorphisms
Category:Inverse Mappings
Inverse of Orthogonal Matrix is Orthogonal

== Theorem ==
Let $\mathbf A$ be an orthogonal matrix.
Then its inverse $\mathbf A^{-1}$ is also orthogonal.
== Proof ==
By definition of orthogonal matrix:
:$\mathbf A^\intercal = \mathbf A^{-1}$
where $\mathbf A^\intercal$ is the transpose of $\mathbf A$.
By Inverse of Inverse of Matrix:
:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$
By Transpose of Transpose of Matrix:
:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$
Thus we have:
:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A = \paren {\mathbf A^\intercal}^\intercal = \paren {\mathbf A^{-1} }^\intercal$
and so by definition $\mathbf A^{-1}$ is orthogonal.
Category:Orthogonal Matrices
Category:Inverse Matrices
Inverse of Pascal's Triangle expressed as Matrix

== Theorem ==
Consider Pascal's triangle expressed as a (square) matrix $\mathbf M$, with the top left element holding $\dbinom 0 0$.
:$\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 & 3 & 3 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 & 4 & 6 & 4 & 1 & 0 & 0 & 0 & 0 & \cdots \\
1 & 5 & 10 & 10 & 5 & 1 & 0 & 0 & 0 & \cdots \\
1 & 6 & 15 & 20 & 15 & 6 & 1 & 0 & 0 & \cdots \\
1 & 7 & 21 & 35 & 35 & 21 & 7 & 1 & 0 & \cdots \\
1 & 8 & 28 & 56 & 70 & 56 & 28 & 8 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}$
The inverse $\mathbf M^{-1}$ of $\mathbf M$ is the same as $\mathbf M$ but with alternate elements negated, starting with the elements below the main diagonal:
:$\begin{pmatrix}
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
-1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
-1 & 3 & -3 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\
 1 & -4 & 6 & -4 & 1 & 0 & 0 & 0 & 0 & \cdots \\
-1 & 5 & -10 & 10 & -5 & 1 & 0 & 0 & 0 & \cdots \\
 1 & -6 & 15 & -20 & 15 & -6 & 1 & 0 & 0 & \cdots \\
-1 & 7 & -21 & 35 & -35 & 21 & -7 & 1 & 0 & \cdots \\
 1 & -8 & 28 & -56 & 70 & -56 & 28 & -8 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}$
== Proof ==
We have from Sum over $k$ of $\dbinom r k$ by $\dbinom {s + k} n$ by $\left({-1}\right)^{r - k}$: Corollary:
By definition of matrix multiplication, this is the element $a_{r n}$ of the matrix formed by multiplying the two matrices above.
As can be seen, this results in the identity matrix .
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $54$
Category:Pascal's Triangle
Inverse of Permutation is Permutation

== Theorem ==
If $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
== Proof ==
Let $f: S \to S$ is a permutation of $S$.
By definition, a permutation is a bijection such that the domain and codomain are the same set.
From Bijection iff Inverse is Bijection, it follows $f^{-1}$ is a bijection.
From the definition of inverse relation, the domain of a relation is the codomain of its inverse and vice versa.
Thus the domain and codomain of $f^{-1}$ are both $S$ and it follows that $f^{-1}$ is a permutation.
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations
* : Appendix: Elementary set and number theory
* : $\S 25.4 \ \text{(ii)}$: Some further results and examples on mappings
Category:Permutations
Category:Inverse Mappings
Inverse of Plane Reflection Matrix

== Theorem ==
Let $\mathbf R$ be the matrix associated with a reflection in the plane.
:$\mathbf R = \begin{bmatrix}
\cos 2\alpha & \sin 2\alpha \\
\sin 2\alpha & -\cos 2\alpha
\end{bmatrix}$
Then its inverse matrix $\mathbf R^{-1}$ is itself.
== Proof ==
Consider $\mathbf R \mathbf R$:
Hence, by the definition of the inverse matrix, $\mathbf R$ is the inverse matrix of $\mathbf R$.
Category:Inverse Matrices
Inverse of Plane Rotation Matrix

== Theorem ==
Let $\mathbf R$ be the matrix associated with a rotation of the plane about the origin through an angle of $\alpha$:
:$\mathbf R = \begin{bmatrix}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{bmatrix}$
Then its inverse matrix $\mathbf R^{-1}$ is:
:$\mathbf R^{-1} = \begin{bmatrix}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{bmatrix}$
== Proof ==
Let:
:$\mathbf A = \begin{bmatrix}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{bmatrix}$
Consider $\mathbf R \mathbf A$:
Now consider $\mathbf A \mathbf R$:
Hence, by the definition of the inverse matrix, $\mathbf A$ is the inverse matrix of $\mathbf R$.
Category:Inverse Matrices
Category:Geometric Rotations
Inverse of Product

== Theorem ==
=== Monoid ===
==== General Result ====
=== Group ===
==== General Result ====
== Also known as ==
When first introducing this concept in a course on group theory, some tutors refer to this theorem as the Socks-Shoes Property.
If one thinks of $a$ as putting on socks, $b$ as putting on shoes, $a^{-1}$ as taking off socks, and $b^{-1}$ as taking off shoes, the theorem demonstrates the order in which one must perform these actions. $a b$ would represent putting on socks followed by shoes. In order to take them off, they must be removed in reverse order, that is, $b^{-1} a^{-1}$.
Less informally it is known as the reversal rule (for group inverses).
== Also see ==
* Inverse of Commuting Pair
Category:Abstract Algebra
Category:Inverse of Product
Category:Inverse Elements
Inverse of Product/Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e$.
Let $a, b \in S$ be invertible for $\circ$, with inverses $a^{-1}, b^{-1}$.
Then $a \circ b$ is invertible for $\circ$, and:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$
=== General Result ===
== Proof ==
Similarly for $\paren {b^{-1} \circ a^{-1} } \circ \paren {a \circ b}$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.4$
* : $\S 3.1$: Monoids
Category:Monoids
Category:Inverse Elements
Category:Inverse of Product
Inverse of Product/Monoid/General Result

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e$.
Let $a_1, a_2, \ldots, a_n \in S$ be invertible for $\circ$, with inverses ${a_1}^{-1}, {a_2}^{-1}, \ldots, {a_n}^{-1}$.
Then $a_1 \circ a_2 \circ \cdots \circ a_n$ is invertible for $\circ$, and:
:$\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$
$\map P 1$ is (trivially) true, as this just says:
:$\paren {a_1}^{-1} = {a_1}^{-1}$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\paren {a_1 \circ a_2}^{-1} = {a_2}^{-1} \circ {a_1}^{-1}$
which has been proved in Inverse of Product in Monoid.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1} = {a_k}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$
Then we need to show:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_{k + 1} }^{-1} = {a_{k + 1} }^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$
Category:Monoids
Category:Inverse Elements
Category:Inverse of Product
Category:Proofs by Induction
Inverse of Product of Subsets of Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq G$.
Then:
:$\paren {X \circ Y}^{-1} = Y^{-1} \circ X^{-1}$
where $X^{-1}$ is the inverse of $X$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{G}$
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6 \ \text{(i)}$
Category:Group Theory
Category:Subset Products
Category:Inverse of Product
Inverse of Product of Subsets of Group/Proof 2

== Theorem ==
== Proof ==
=== Superset ===
We will show that $\forall z \in Y^{-1} \circ X^{-1}: z \in \paren {X \circ Y}^{-1}$, from which:
:$Y^{-1} \circ X^{-1} \subseteq \paren {X \circ Y}^{-1}$
Let $z \in Y^{-1} \circ X^{-1}$.
By the definition of subset product:
:$\exists x' \in X^{-1}, y' \in Y^{-1}: z = y' \circ x'$
Then by Inverse of Group Product:
:$(2)\quad z^{-1} = x'^{-1} \circ y'^{-1}$
By the definition of inverse of subset:
:$x'^{-1} \in X$ and $y'^{-1} \in Y$
By the definition of subset product:
:$x'^{-1} \circ y'^{-1} \in X \circ Y$
Thus by $(2)$:
:$z^{-1} \in X \circ Y$
By the definition of inverse of subset:
:$z \in \paren {X \circ Y}^{-1}$
=== Subset ===
We will show that $\forall z \in \paren {X \circ Y}^{-1}: z \in Y^{-1} \circ X^{-1}$, from which:
:$\paren {X \circ Y}^{-1} \subseteq Y^{-1} \circ X^{-1}$
Let $z \in \paren {X \circ Y}^{-1}$.
By the definition of inverse of subset:
:$z^{-1} \in X \circ Y$
From Inverse of Inverse of Subset of Group:
:$z^{-1} \in \paren {X^{-1} }^{-1} \circ \paren {Y^{-1} }^{-1}$
Thus by the superset proof above:
:$z^{-1} \in \paren {Y^{-1} \circ X^{-1} }^{-1}$
From the definition of inverse of subset:
:$z \in Y^{-1} \circ X^{-1}$
Category:Inverse of Product
Inverse of Proper Orthogonal Matrix is Proper Orthogonal

== Theorem ==
Let $\mathbf A$ be a proper orthogonal matrix.
Then its inverse $\mathbf A^{-1}$ is also proper orthogonal.
== Proof ==
By definition of proper orthogonal matrix:
:$\mathbf A$ is orthogonal
:the determinant of $\mathbf A$ is equal to $1$.
By Inverse of Orthogonal Matrix is Orthogonal, $\mathbf A^{-1}$ is orthogonal.
By Determinant of Inverse Matrix:
:$\det \mathbf A^{-1} = \dfrac 1 {\det \mathbf A} = 1$
Hence, by definition, $\mathbf A^{-1}$ is proper orthogonal.
Category:Orthogonal Matrices
Category:Inverse Matrices
Inverse of Real Square Function on Positive Reals

== Theorem ==
Let $f: \R_{\ge 0} \to R_{\ge 0}$ be the restriction of the real square function to the positive real numbers $\R_{\ge 0}$.
The inverse of $f$ is $f^{-1}: \R_{\ge 0} \times R_{\ge 0}$ defined as:
:$\forall x \in \R_{\ge 0}: \map {f^{-1} } x = \sqrt x$
where $\sqrt x$ is the positive square root of $x$.
== Proof ==
From Restriction of Real Square Mapping to Positive Reals is Bijection, $f$ is a bijection.
By definition of the positive square root:
:$y = \sqrt x \iff x = y^2$
for $x, y \in \R_{\ge 0}$.
Hence the result.
== Sources ==
* : Chapter $3$. Mappings: Exercise $3$
Category:Square Function
Inverse of Reflexive Relation is Reflexive

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is reflexive, then so is $\RR^{-1}$.
== Proof ==
Hence the result by definition of reflexive relation.
== Sources ==
* : $\S 1.19$: Some Important Properties of Relations: Exercise $9$
Category:Reflexive Relations
Category:Inverse Relations
Inverse of Relation Compatible with Operation is Compatible

== Theorem ==
Let $\struct {S, \circ}$ be a closed algebraic structure.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\QQ$ be the inverse relation of $\RR$.
Then $\QQ$ is compatible with $\circ$.
== Proof ==
Let $x, y, z \in S$.
Suppose that $x \mathrel \QQ y$.
Then by the definition of $\QQ$:
:$y \mathrel \RR x$.
Since $\RR$ is compatible with $\circ$:
:$\paren {y \circ z} \mathrel \RR \paren {x \circ z}$
and
:$\paren {z \circ y} \mathrel \RR \paren {z \circ x}$
Thus by the definition of $\QQ$:
:$\paren {x \circ z} \mathrel \QQ \paren {y \circ z}$
and
:$\paren {z \circ x} \mathrel \RR \paren {z \circ y}$
Thus $\QQ$ is compatible with $\circ$.
Category:Compatible Relations
Inverse of Relation Isomorphism is Relation Isomorphism

== Theorem ==
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\phi: \struct {S, \RR_1} \to \struct {T, \RR_2}$ be a bijection.
Then:
:$\phi: \struct {S, \RR_1} \to \struct {T, \RR_2}$
is a relation isomorphism
:
:$\phi^{-1}: \struct {T, \RR_2} \to \struct {S, \RR_1}$
is also a relation isomorphism.
== Proof ==
Follows directly from the definition of relation isomorphism.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(b)}$
Category:Relation Isomorphisms
Inverse of Rescaling Matrix

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $r \in R$ be a unit in $R$.
Let $r \, \mathbf I_n$ be the $n \times n$ rescaling matrix of $r$.
Then:
:$\paren {r \, \mathbf I_n}^{-1} = r^{-1} \, \mathbf I_n$
== Proof ==
By definition, a rescaling matrix is also a diagonal matrix.
Hence Inverse of Diagonal Matrix applies, and since $r$ is a unit, it gives the desired result.
Category:Matrix Algebra
Inverse of Right-Total Relation is Left-Total

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^{-1} \subseteq T \times S$ be the inverse of $\RR$.
Then:
:$\RR$ is right-total $\RR^{-1}$ is left-total.
== Proof ==
=== Sufficient Condition ===
Let $\RR$ be right-total.
Then by definition:
:$\forall t \in T: \exists s \in S: \tuple {s, t} \in \RR$
By definition of the inverse of $\RR$, it follows that:
:$\forall t \in T: \exists s \in S: \tuple {t, s} \in \RR^{-1}$
So by definition $\RR^{-1}$ is left-total.
=== Necessary Condition ===
Let $\RR^{-1}$ is left-total.
Then by definition:
:$\forall t \in T: \exists s \in S: \tuple {t, s} \in \RR^{-1}$
and so:
:$\forall t \in T: \exists s \in S: \tuple {s, t} \in \RR$
So by definition $\RR$ is right-total.
Category:Inverse Relations
Category:Right-Total Relations
Category:Left-Total Relations
Inverse of Similarity Mapping

== Theorem ==
Let $G$ be a vector space over a field $K$.
Let $\beta \in K$ such that $\beta \ne 0$.
Let $s_\beta: G \to G$ be the similarity on $G$ defined as:
:$\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$
Let $\paren {s_\beta}^{-1}$ denote the inverse of $s_\beta$.
Then:
:$\paren {s_\beta}^{-1} = s_{\beta^{-1} }$
where $\beta^{-1}$ is the multiplicative inverse in $K$ of $\beta$.
== Proof ==
From Similarity Mapping is Automorphism, $s_\beta$ is an automorphism of $G$.
Hence $s_\beta$ is an vector space isomorphism from $G$ to $G$ itself.
So by definition $s_\beta$ is a bijection.
Hence the existence of this inverse $\paren {s_\beta}^{-1}$ follows from Bijection iff Left and Right Inverse.
By , we have that there exists a multiplicative inverse $\beta^{-1}$ for $\beta$ such that:
:$\beta \beta^{-1} = 1_K = \beta^{-1} \beta$
where $1_K$ is the multiplicative identity of $K$.
Then:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.3$
Category:Similarity Mappings
Inverse of Small Relation is Small

== Theorem ==
Let $a$ be a small class.
Let $a$ also be a relation.
Then the inverse relation of $a$ is small.
== Proof ==
Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, \tuple {y, x} } : \tuple {x, y} \in a}$
Then $A$ maps $a$ to its inverse.
Thus, the inverse of $a$ is the image of $a$ under $A$.
By Image of Small Class under Mapping is Small, the inverse of $a$ is small.
== Sources ==
* : $\S 6.8 \ (1)$
Category:Zermelo-Fraenkel Class Theory
Inverse of Square Matrix over Field is Unique

== Theorem ==
Let $\Bbb F$ be a field, usually one of the standard number fields $\Q$, $\R$ or $\C$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map \MM n$ denote the matrix space of order $n$ square matrices over $\Bbb F$.
Let $\mathbf B$ be an inverse matrix of $\mathbf A$.
Then $\mathbf B$ is the only inverse matrix of $\mathbf A$.
== Proof ==
Consider the algebraic structure $\struct {\map \MM {m, n}, +, \circ}$, where:
:$+$ denotes matrix entrywise addition
:$\circ$ denotes (conventional) matrix multiplication.
From Ring of Square Matrices over Field is Ring with Unity, $\struct {\map \MM {m, n}, +, \circ}$ is a ring with unity.
Hence a fortiori $\struct {\map \MM {m, n}, +, \circ}$ is a monoid.
The result follows directly from Inverse in Monoid is Unique.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.3$ The inverse of a matrix: Fact $1.1$
Category:Inverse Matrices
Inverse of Star of Element in Unital *-Algebra

== Theorem ==
Let $\struct {A, \ast}$ be a unital $\ast$-algebra.
Let $a \in A$ be invertible.
Then $a^\ast$ is invertible and:
:$\paren {a^\ast}^{-1} = \paren {a^{-1} }^\ast$ 
=== Corollary ===
== Proof ==
We have: 
:$a a^{-1} = a^{-1} a = {\mathbf 1}_A$
From $(C^\ast 3)$ in the definition of an involution, we have:
:$\paren {a^{-1} }^\ast a^\ast = a^\ast \paren {a^{-1} }^\ast = {\mathbf 1}_A^\ast$ 
From Identity Element in Unital *-Algebra is Hermitian, we therefore have:
:$\paren {a^{-1} }^\ast a^\ast = a^\ast \paren {a^{-1} }^\ast = {\mathbf 1}_A$
Hence we have:
:$\paren {a^\ast}^{-1} = \paren {a^{-1} }^\ast$
== Sources ==
* : $2.1$: $C^\ast$-Algebras
Category:*-Algebras
Category:Inverse of Star of Element in Unital *-Algebra
Inverse of Star of Element in Unital *-Algebra/Corollary

== Corollary ==
Let $\struct {A, \ast}$ be a unital $\ast$-algebra.
Let $a \in A$. 
Then $a$ is invertible $a^\ast$ is invertible.
== Proof ==
From Inverse of Star of Element in Unital *-Algebra:
:if $a$ is invertible then $a^\ast$ is invertible.
Hence:
:if $a^\ast$ is invertible then $a^{\ast \ast}$ is invertible.
Since $\ast$ is an involution, we have $a^{\ast \ast} = a$ and hence:
:if $a^\ast$ is invertible then $a$ is invertible.
We conclude:
:$a$ is invertible $a^\ast$ is invertible.
Category:Inverse of Star of Element in Unital *-Algebra
Inverse of Stirling's Triangle expressed as Matrix

== Theorem ==
Consider Stirling's triangle of the first kind (signed) expressed as a (square) matrix $\mathbf A$, with the top left element holding $\map s {0, 0}$.
:$\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 1 & 0 & 0 & 0 & 0 & \cdots \\
0 & -1 & 1 & 0 & 0 & 0 & \cdots \\
0 & 2 & -3 & 1 & 0 & 0 & \cdots \\
0 & -6 & 11 & -6 & 1 & 0 & \cdots \\
0 & 24 & -50 & 35 & -10 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}$
Then consider Stirling's triangle of the second kind expressed as a (square) matrix $\mathbf B$, with the top left element holding $\ds {0 \brace 0}$.
:$\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 1 & 0 & 0 & 0 & 0 & \cdots \\
0 & 1 & 1 & 0 & 0 & 0 & \cdots \\
0 & 1 & 3 & 1 & 0 & 0 & \cdots \\
0 & 1 & 7 & 6 & 1 & 0 & \cdots \\
0 & 1 & 15 & 25 & 10 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}$
Then:
:$\mathbf A = \mathbf B^{-1}$
that is:
:$\mathbf B = \mathbf A^{-1}$
== Proof ==
First note that from Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
From First Inversion Formula for Stirling Numbers:
:$\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
From Second Inversion Formula for Stirling Numbers:
:$\ds \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
By definition of matrix multiplication, the element $a_{r n}$ of the matrix formed by multiplying the two matrices above.
As can be seen, this results in the identity matrix .
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $55$
Category:Stirling Numbers