nyuuzyou/wikis huggingface dataset:
Inverse of Straight Line Not Through Inversion Center is Circle Through Inversion Center

== Theorem ==
Let $C$ be a circle on center $O$ in the plane.
Let $T : X \to Y$ be an inversive transformation with $C$ as the inversion circle. 
Then $O$ is the inversion center.
Let $L$ be an arbitrary straight line not containing $O$.
Let $P$ be an arbitrary point on $L$.
Let $P'$ be the image of $P$ under $T$.
Then $P'$ lies on a circle through $O$.
== Proof ==
By Inverse of Circle Through Inversion Center is Straight Line Not Through Inversion Center:
:the image under $T$ of a circle through $C$ is a straight line not through $C$.
By definition of inversive transformation, $T$ is an involution.
Thus, the image under $T$ of a straight line not through $C$ is a circle through $C$.
== Sources ==
* : Chapter $\text{III}$ / $\text{II}$ Section $4$: "Geometrical Transformations. Inversion."
Category:Inversive Transformations
Inverse of Strictly Decreasing Concave Real Function is Concave

== Theorem ==
Let $f$ be a real function which is concave on the open interval $I$.
Let $J = f \left[{I}\right]$.
If $f$ be strictly decreasing on $I$, then $f^{-1}$ is concave on $J$.
== Proof ==
Let:
:$X = f \left({x}\right) \in J$
:$Y = f \left({y}\right) \in J$.
From the definition of concave:
: $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \ge \alpha f \left({x}\right) + \beta f \left({y}\right)$
Let $f$ be strictly decreasing on $I$.
From Inverse of Strictly Monotone Function, it follows that $f^{-1}$ is strictly decreasing on $J$.
Thus:
: $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \ge f^{-1} \left({\alpha X + \beta Y}\right)$
Hence $f^{-1}$ is concave on $J$.
== Also see ==
* Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave
* Inverse of Strictly Decreasing Convex Real Function is Convex
* Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex
* Inverse of Strictly Increasing Convex Real Function is Concave
* Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave
* Inverse of Strictly Increasing Concave Real Function is Convex
* Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex
Category:Concave Real Functions
Inverse of Strictly Decreasing Convex Real Function is Convex

== Theorem ==
Let $f$ be a real function which is convex on the open interval $I$.
Let $J = f \left[{I}\right]$.
If $f$ be strictly decreasing on $I$, then $f^{-1}$ is convex on $J$.
== Proof ==
Let:
:$X = f \left({x}\right) \in J$
:$Y = f \left({y}\right) \in J$.
From the definition of convex:
: $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \le \alpha f \left({x}\right) + \beta f \left({y}\right)$
Let $f$ be strictly decreasing on $I$.
Then from Inverse of Strictly Monotone Function it follows that $f^{-1}$ is strictly decreasing on $J$.
Thus:
: $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \le f^{-1} \left({\alpha X + \beta Y}\right)$
Hence $f^{-1}$ is convex on $J$.
== Also see ==
* Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex
* Inverse of Strictly Decreasing Concave Real Function is Concave
* Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave
* Inverse of Strictly Increasing Convex Real Function is Concave
* Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave
* Inverse of Strictly Increasing Concave Real Function is Convex
* Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex
== Sources ==
* : $\S 12.21 \ (3)$
Category:Convex Real Functions
Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave

== Theorem ==
Let $f$ be a real function which is strictly concave on the open interval $I$.
Let $J = f \sqbrk I$.
If $f$ be strictly decreasing on $I$, then $f^{-1}$ is strictly concave on $J$.
== Proof ==
Let:
:$X = \map f x \in J$
:$Y = \map f y \in J$.
From the definition of strictly concave:
:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} > \alpha \map f x + \beta \map f y$
Let $f$ be strictly decreasing on $I$.
From Inverse of Strictly Monotone Function, it follows that $f^{-1}$ is strictly decreasing on $J$.
Thus:
:$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y > \map {f^{-1} } {\alpha X + \beta Y}$
Hence $f^{-1}$ is strictly concave on $J$.
== Also see ==
* Inverse of Strictly Decreasing Concave Real Function is Concave
* Inverse of Strictly Decreasing Convex Real Function is Convex
* Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex
* Inverse of Strictly Increasing Convex Real Function is Concave
* Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave
* Inverse of Strictly Increasing Concave Real Function is Convex
* Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex
Category:Concave Real Functions
Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex

== Theorem ==
Let $f$ be a real function which is strictly convex on the open interval $I$.
Let $J = f \sqbrk I$.
If $f$ be strictly decreasing on $I$, then $f^{-1}$ is strictly convex on $J$.
== Proof ==
Let:
:$X = \map f x \in J$
:$Y = \map f y \in J$.
From the definition of strictly convex:
:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \map f x + \beta \map f y$
Let $f$ be strictly decreasing on $I$.
Then from Inverse of Strictly Monotone Function it follows that $f^{-1}$ is strictly decreasing on $J$.
Thus:
:$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y < \map {f^{-1} } {\alpha X + \beta Y}$
Hence $f^{-1}$ is strictly convex on $J$.
== Also see ==
* Inverse of Strictly Decreasing Convex Real Function is Convex
* Inverse of Strictly Decreasing Concave Real Function is Concave
* Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave
* Inverse of Strictly Increasing Convex Real Function is Concave
* Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave
* Inverse of Strictly Increasing Concave Real Function is Convex
* Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex
Category:Convex Real Functions
Inverse of Strictly Increasing Concave Real Function is Convex

== Theorem ==
Let $f$ be a real function which is concave on the open interval $I$.
Let $J = f \sqbrk I$.
If $f$ is strictly increasing on $I$, then $f^{-1}$ is convex on $J$.
== Proof ==
Let $X = \map f x \in J, Y = \map f y \in J$.
From the definition of concave:
:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \ge \alpha \map f x + \beta \map f y$
Let $f$ be strictly increasing on $I$.
From Inverse of Strictly Monotone Function, it follows that $f^{-1}$ is strictly increasing on $J$.
Thus:
:$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y \ge \map {f^{-1} } {\alpha X + \beta Y}$
Hence $f^{-1}$ is convex on $J$.
== Also see ==
* Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex
* Inverse of Strictly Increasing Convex Real Function is Concave
* Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave
* Inverse of Strictly Decreasing Concave Real Function is Concave
* Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave
* Inverse of Strictly Decreasing Convex Real Function is Convex
* Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex
Category:Convex Real Functions
Category:Concave Real Functions
Category:Strictly Increasing Real Functions
Inverse of Strictly Increasing Convex Real Function is Concave

== Theorem ==
Let $f$ be a real function which is convex on the open interval $I$.
Let $J = f \sqbrk I$.
If $f$ be strictly increasing on $I$, then $f^{-1}$ is concave on $J$.
== Proof ==
Let:
:$X = \map f x \in J$
:$Y = \map f y \in J$.
From the definition of convex:
:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \le \alpha \map f x + \beta \map f y$
Let $f$ be strictly increasing on $I$.
From Inverse of Strictly Monotone Function it follows that, $f^{-1}$ is strictly increasing on $J$.
Thus:
:$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y \le \map {f^{-1} } {\alpha X + \beta Y}$
Hence $f^{-1}$ is concave on $J$.
== Also see ==
* Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave
* Inverse of Strictly Increasing Concave Real Function is Convex
* Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex
* Inverse of Strictly Decreasing Convex Real Function is Convex
* Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex
* Inverse of Strictly Decreasing Concave Real Function is Concave
* Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave
== Sources ==
* : $\S 12.21 \ (3)$
Category:Convex Real Functions
Category:Concave Real Functions
Category:Strictly Increasing Real Functions
Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex

== Theorem ==
Let $f$ be a real function which is strictly concave on the open interval $I$.
Let $J = f \sqbrk I$.
If $f$ be strictly increasing on $I$, then $f^{-1}$ is strictly convex on $J$.
== Proof ==
Let $X = \map f x \in J, Y = \map f y \in J$.
From the definition of Strictly concave:
:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} > \alpha \map f x + \beta \map f y$
Let $f$ be strictly increasing on $I$.
From Inverse of Strictly Monotone Function, it follows that $f^{-1}$ is strictly increasing on $J$.
Thus:
:$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y > \map {f^{-1} } {\alpha X + \beta Y}$
Hence $f^{-1}$ is strictly convex on $J$.
== Also see ==
* Inverse of Strictly Increasing Concave Real Function is Convex
* Inverse of Strictly Increasing Convex Real Function is Concave
* Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave
* Inverse of Strictly Decreasing Concave Real Function is Concave
* Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave
* Inverse of Strictly Decreasing Convex Real Function is Convex
* Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex
Category:Strictly Increasing Real Functions
Category:Convex Real Functions
Category:Concave Real Functions
Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave

== Theorem ==
Let $f$ be a real function which is strictly convex on the open interval $I$.
Let $J = f \sqbrk I$.
If $f$ be strictly increasing on $I$, then $f^{-1}$ is strictly concave on $J$.
== Proof ==
Let:
:$X = \map f x \in J$
:$Y = \map f y \in J$
From the definition of strictly convex:
:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \, \map f x + \beta \, \map f y$
Let $f$ be strictly increasing on $I$.
From Inverse of Strictly Monotone Function it follows that, $f^{-1}$ is strictly increasing on $J$.
Thus:
:$\alpha \, \map {f^{-1} } X + \beta \, \map {f^{-1} } Y = \alpha x + \beta y < \map {f^{-1} } {\alpha X + \beta Y}$
Hence $f^{-1}$ is strictly concave on $J$.
== Also see ==
* Inverse of Strictly Increasing Convex Real Function is Concave
* Inverse of Strictly Increasing Concave Real Function is Convex
* Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex
* Inverse of Strictly Decreasing Convex Real Function is Convex
* Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex
* Inverse of Strictly Decreasing Concave Real Function is Concave
* Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave
Category:Strictly Increasing Real Functions
Category:Convex Real Functions
Category:Concave Real Functions
Inverse of Strictly Monotone Continuous Real Function is Strictly Monotone and Continuous

== Theorem ==
Let $f$ be a continuous real function which is defined on the closed interval $I := \closedint a b$.
Let $f$ be strictly monotone on $I$.
Then $f$ has an inverse function $f^{-1}$ which is continuous and strictly monotone on $f \sqbrk I$.
== Proof ==
The function $f$ is a bijection from Strictly Monotone Real Function is Bijective.
From Inverse of Strictly Monotone Function, $f^{-1}$ is strictly monotone on on $f \sqbrk I$.
From Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone, $f^{-1}$ is a continuous real function.
Hence the result.
== Examples ==
== Sources ==
* 
* 
Category:Inverse of Strictly Monotone Continuous Real Function is Strictly Monotone and Continuous
Category:Strictly Monotone Real Functions
Category:Continuous Real Functions
Category:Inverse Mappings
Inverse of Strictly Monotone Continuous Real Function is Strictly Monotone and Continuous/Examples/Arbitrary Example 1

== Example of Use of Inverse of Strictly Monotone Continuous Real Function is Strictly Monotone and Continuous ==
Consider the real function:
:$\forall x \in \closedint 0 1: \map f x = y = 2 x + 3$
This has an inverse:
:$\map {f^{-1} } y = x = \dfrac {y - 3} 2$
on the closed interval $\closedint 3 5$
Hence we can say:
:$f: x \mapsto 2 x + 3$ on $\closedint 0 1$
and:
:$f^{-1}: x \mapsto \dfrac {x - 3} 2$ on $\closedint 3 5$
== Sources ==
* 
* 
Category:Examples of Inverse of Strictly Monotone Continuous Real Function is Strictly Monotone and Continuous
Inverse of Strictly Monotone Function

== Theorem ==
Let $f$ be a real function which is defined on $I \subseteq \R$.
Let $f$ be strictly monotone on $I$.
Let the image of $f$ be $J$. 
Then $f$ always has an inverse function $f^{-1}$ and:
:if $f$ is strictly increasing then so is $f^{-1}$
:if $f$ is strictly decreasing then so is $f^{-1}$.
== Proof ==
The function $f$ is a bijection from Strictly Monotone Real Function is Bijective.
Hence from Bijection iff Inverse is Bijection, $f^{-1}$ always exists and is also a bijection.
From the definition of strictly increasing:
:$x < y \iff \map f x < \map f y$
Hence:
:$\map {f^{-1} } x < \map {f^{-1} } y \iff \map {f^{-1} } {\map f x} < \map {f^{-1} } {\map f y}$
and so:
:$\map {f^{-1} } x < \map {f^{-1} } y \iff x < y$
Similarly, from the definition of strictly decreasing:
:$x < y \iff \map f x > \map f y$
Hence:
:$\map {f^{-1} } x < \map {f^{-1} } y \iff \map {f^{-1} } {\map f x} > \map {f^{-1} } {\map f y}$
and so:
:$\map {f^{-1} } x < \map {f^{-1} } y \iff x > y$
== Sources ==
* : $\S 12.9$
Category:Strictly Monotone Real Functions
Category:Inverse Mappings
Inverse of Subgroup

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Then:
:$H^{-1} = H$
where $H^{-1}$ is the inverse of $H$.
== Proof ==
As $H$ is a subgroup of $G$:
:$\forall h \in H: h^{-1} \in H$
The result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.10 \ \text {(a)}$
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6 \ \text{(ii)}$
Category:Subgroups
Category:Subset Products
Inverse of Subset/Group/Examples/Subset of Reals under Multiplication

== Example of Inverse of Subset of Group ==
Let $\struct {\R, \times}$ be the multiplicative group of (non-zero) real numbers.
Let $S = \set {-1, 2}$.
Then the inverse $S^{-1}$ of $S$ is:
:$S^{-1} = \set {-1, \dfrac 1 2}$
== Proof ==
Taking each element of $S$:
== Sources ==
* : Chapter $\text{II}$: Groups: Subgroups
Category:Examples of Inverses of Subsets
Inverse of Subset of Relation is Subset of Inverse

== Theorem ==
Let $S$ and $T$ be sets
Let $\RR_1 = S \times T$ be a relation on $S \times T$.
Let $\RR_2 \subseteq \RR_1$.
Then:
:$\RR_2^{-1} \subseteq \RR_1^{-1}$
where $\RR_1^{-1}$ denotes the inverse of $\RR_1$.
== Proof ==
The result follows by definition of subset.
Category:Subsets
Category:Inverse Relations
Inverse of Supremum in Ordered Group is Infimum of Inverses

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.
Let $x, y \in G$.
Then:
:$\set {x, y}$ admits a supremum in $G$
:
:$\set {x^{-1}, y^{-1} }$ admits an infimum in $G$
in which case:
:$\paren {\sup \set {x, y} }^{-1} = \inf \set {x^{-1}, y^{-1} }$
== Proof ==
Let:
Recall from Inverse of Group Inverse:
From Inverse of Infimum in Ordered Group is Supremum of Inverses:
Then:
:$\set {a, b}$ admits an infimum in $G$
:
:$\set {a^{-1}, b^{-1} }$ admits a supremum in $G$
in which case:
:$\paren {\inf \set {a, b} }^{-1} = \sup \set {a^{-1}, b^{-1} }$
Substituting back for $a$ and $b$:
:$\set {x^{-1}, y^{-1} }$ admits an infimum in $G$
:
:$\set {x, y}$ admits a supremum in $G$
in which case:
:$\paren {\inf \set {x^{-1}, x^{-1} } }^{-1} = \sup \set {x, y}$
Hence from : 
:$\inf \set {x^{-1}, x^{-1} } = \paren {\sup \set {x, y} }^{-1}$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.10 \ \text {(d)}$
Category:Ordered Groups
Category:Suprema
Category:Infima
Inverse of Surjection is Relation both Left-Total and Right-Total

== Theorem ==
Let $f$ be an surjective mapping.
Then its inverse $f^{-1}$ is a relation which is both left-total and right-total.
== Proof ==
We are given that $f$ is a surjective mapping.
By Inverse of Mapping is Right-Total Relation, $f^{-1}$ is a right-total relation.
By definition of surjection, $f$ is itself a right-total relation.
From Inverse of Right-Total Relation is Left-Total, $f^{-1}$ is a left-total relation.
Hence the result.
Category:Inverse Relations
Category:Surjections
Category:Inverse Mappings
Category:Right-Total Relations
Category:Left-Total Relations
Inverse of Symmetric Relation is Symmetric

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is symmetric, then so is $\RR^{-1}$.
== Proof ==
Let $\RR$ be symmetric.
Then from Relation equals Inverse iff Symmetric it follows that $\RR^{-1}$ is also symmetric.
Category:Symmetric Relations
Category:Inverse Relations
Inverse of Transitive Relation is Transitive

== Theorem ==
Let $\RR$ be a relation on a set $S$.
Let $\RR$ be transitive.
Then its inverse $\RR^{-1}$ is also transitive.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.19$: Some Important Properties of Relations: Exercise $7$
Category:Transitive Relations
Category:Inverse Relations
Category:Inverse of Transitive Relation is Transitive
Inverse of Transitive Relation is Transitive/Proof 1

== Theorem ==
== Proof ==
Let $\RR$ be transitive.
Then:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$
Thus:
:$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \in \RR^{-1}$
and so $\RR^{-1}$ is transitive.
Category:Inverse of Transitive Relation is Transitive
Inverse of Transitive Relation is Transitive/Proof 2

== Theorem ==
== Proof ==
Let $\RR$ be transitive.
Thus by definition:
:$\RR \circ \RR \subseteq \RR$
Thus:
Hence the result by definition of transitive relation.
== Sources ==
* : Chapter $0$: Relations
Category:Inverse of Transitive Relation is Transitive
Inverse of Transpose of Matrix is Transpose of Inverse

== Theorem ==
Let $\mathbf A$ be a matrix over a field.
Let $\mathbf A^\intercal$ denote the transpose of $\mathbf A$.
Let $\mathbf A$ be an nonsingular matrix.
Then $\mathbf A^\intercal$ is also nonsingular and:
:$\paren {\mathbf A^\intercal}^{-1} = \paren {\mathbf A^{-1} }^\intercal$
where $\mathbf A^{-1}$ denotes the inverse of $\mathbf A$.
== Proof ==
We have:
Hence $\paren {\mathbf A^{-1} }^\intercal$ is an inverse of $\mathbf A^\intercal$.
From Inverse of Square Matrix over Field is Unique:
:$\paren {\mathbf A^{-1} }^\intercal = \paren {\mathbf A^\intercal}^{-1}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.4$ The transpose of a matrix: Proposition $1.3 \ \text {(b)}$
Category:Transposes of Matrices
Category:Inverse Matrices
Inverse of Unit in Centralizer of Ring is in Centralizer

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $S$ be a subset of $R$.
Let $\map {C_R} S$ denote the centralizer of $S$ in $R$
Let $u \in R$ be a unit of $R$.
Then:
:$u \in \map {C_R} S \implies u^{-1} \in \map {C_R} S$
== Proof ==
Let $u \in R$ be a unit of $R$.
Let $u \in \map {C_R} S$.
Then from Commutation with Inverse in Monoid:
:$u^{-1} \in \map {C_R} S$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.5$
Category:Ring Theory
Category:Subrings
Inverse of Vandermonde Matrix

== Theorem ==
Let $V_n$ be the Vandermonde matrix of order $n$ given by:
:$V_n = \begin {bmatrix}
 x_1 & x_2 & \cdots & x_n \\
 x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
 x_1^n & x_2^n & \cdots & x_n^n
\end{bmatrix}$
Then its inverse $V_n^{-1} = \sqbrk b_n$ can be specified as:
:$b_{i j} = \begin {cases}
\paren {-1}^{j - 1} \paren {\dfrac {\ds \sum_{\substack {1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n - j} \mathop \le n \\ m_1, \ldots, m_{n - j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n - j} } } {x_i \ds \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_m - x_i} } } & : 1 \le j < n \\
\qquad \qquad \qquad \dfrac 1 {x_i \ds \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_i - x_m} } & : j = n
\end{cases}$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $40$
Category:Vandermonde Matrices
Category:Inverse of Vandermonde Matrix
Inverse of Vandermonde Matrix/Corollary

== Corollary to Inverse of Vandermonde Matrix ==
Define for variables $\set {y_1,\ldots, y_k}$ elementary symmetric functions:
Let $\set {x_1, \ldots, x_n}$ be a set of distinct values.
Let $W_n$ and $V_n$ be Vandermonde matrices of order $n$:
:$W_n = \begin{bmatrix}
 1 & x_1 & \cdots & x_1^{n-1} \\
 1 & x_2 & \cdots & x_2^{n-1} \\
\vdots & \vdots & \ddots & \vdots \\
 1 & x_1^{n-1} & \cdots & x_n^{n-1} \\
\end{bmatrix}, \quad
V_n = \begin{bmatrix}
 x_1 & x_2 & \cdots & x_n \\
 x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
 x_1^n & x_2^n & \cdots & x_n^n \\
\end{bmatrix}$
Let their matrix inverses be written as
$W_n^{-1} = \begin{bmatrix} b_{ij} \end{bmatrix}$
$V_n^{-1} = \begin{bmatrix} c_{ij} \end{bmatrix}$.
Then:
== Proof ==
The details appear in Inverse of Vandermonde Matrix/Proof 1, same notation.
Category:Inverse of Vandermonde Matrix
Inverse of Vandermonde Matrix/Eisinberg Formula

== Theorem ==
Let:
Let $W_n$ have a matrix inverse $W_n^{-1} = \begin {bmatrix} d_{ij} \end {bmatrix}$. 
Let $a_n = \map {e_0} {x_1, \ldots, x_n} = 1$.
Then:
== Proof ==
=== Lemma 1 ===
Given values $z_1, \ldots, z_{p + 1}$ and $1 \le m \le p$, then:
=== Lemma 2 ===
Let $X = \set {x_1, \ldots, x_n}$ and $\mathbf u = x_j$ for some $j = 1, \ldots, n$.
Then:
Proof of Lemma 2:
Let $S$ denote the of $(3)$.
Let $U = X \setminus \set {\mathbf u}$.
Then:
Proof of the Theorem
 
 
== Sources ==
* 
* : $\S 1.2.3$: Sums and Products: Exercise $40$
* 
* 
Category:Inverse of Vandermonde Matrix