nyuuzyou/wikis huggingface dataset:
Inverse of Vandermonde Matrix/Proof 2

== Proof ==
=== Definition 1 ===
:$V_n = \begin{bmatrix}
 x_1 & \cdots & x_n \\
 x_1^2 & \cdots & x_n^2 \\
 \vdots & \ddots & \vdots \\
 x_1^{n} & \cdots & x_n^{n} \\
\end{bmatrix} \quad$ 
:$V = \begin{bmatrix}
 1 & \cdots & 1 \\
 x_1 & \cdots & x_n \\
 \vdots & \ddots & \vdots \\
 x_1^{n-1} & \cdots & x_n^{n-1} \\
\end{bmatrix} \quad$ 
:$D = 
\begin{bmatrix}
 x_1 & \cdots & 0 \\
 \vdots & \ddots & \vdots \\
 0 & \cdots & x_n \\
\end{bmatrix},
\quad
P = \begin{bmatrix}
 \map {p_1} {x_1} & \cdots & 0 \\
 \vdots & \ddots & \vdots \\
 0 & \cdots & \map {p_n} {x_n} \\
 \end{bmatrix} \quad$ Definition:Diagonal Matrix
:$E = \begin{bmatrix}
 E_{11} & \cdots & E_{1n} \\
 \vdots & \ddots & \vdots \\
 E_{n1} & \cdots & E_{nn} \\
 \end{bmatrix} \quad$ Matrix of symmetric functions
:where for $\mathbf {1 \mathop \le i, j \mathop \le n}$:
=== Lemma 1 ===
=== Lemma 2 ===
:$E V = P$
:$V^{-1} = P^{-1} E$ provided $\set {x_1, \ldots, x_n}$ is a set of distinct values.
:$V_n^{-1} = D^{-1} V^{-1}$ provided $\set {x_1, \ldots, x_n}$ is a set of distinct values, all nonzero.
Proof of Lemma 2
Matrix multiply establishes $E V = P$, provided:
:$(1): \quad \sum_{k \mathop = 1}^n E_{i k} x_j^{k - 1} = \begin{cases} 0 & i \ne j \\ \map {p_i} {x_i} & i = j \end{cases}$
 
Polynomial $\map {p_i} {x}$ is zero for $x \in \set {x_1,\ldots,x_n} \setminus \set {x_i}$.
 
Then (1) is equivalent to
:$(2): \quad \sum_{k \mathop = 1}^n \paren {-1}^{n - k} \map {e_{n - k} } {\set {x_1, \ldots, x_n} \setminus \set {x_i} } x_j^{k - 1} = \map {p_i} {x_j}$
Apply Viète's Formulas to degree $n - 1$ monic polynomial $\map {p_i} {\mathbf u}$:
:$(3): \quad \sum_{k \mathop = 1}^n \paren {-1}^{n - k} \map {e_{n - k} } {set {x_1, \ldots, x_n} \setminus \set {x_i} } {\mathbf u}^{k - 1} = \map {p_i} {\mathbf u}$
Substitute $\mathbf u = x_j$ into (3), proving (2) holds.
Then (2) and (1) hold, proving $E V = P$.
Assume $\set {x_1, \ldots, x_n}$ is a set of distinct values.
Then $\map \det P$ is nonzero.
By Matrix is Nonsingular iff Determinant has Multiplicative Inverse, $P$ has an inverse $P^{-1}$.
Multiply $E V = P$ by $P^{-1}$, then: 
:$V^{-1} = P^{-1} E\quad$ Left or Right Inverse of Matrix is Inverse
Similarly, $D^{-1}$ exists provided $\set {x_1, \ldots, x_n}$ is a set of nonzero values.
Then $V_n^{-1} = D^{-1} V^{-1}$ by Lemma 1. 
Proof of the Theorem
Assume $\set {x_1, \ldots, x_n}$ is a set of distinct values.
Let $d_{i j}$ denote the entries in $V^{-1}$. Then:
Then:
 
Assume $\set {x_1, \ldots, x_n}$ is a set of distinct values, all nonzero.
Let $b_{i j}$ denote the entries in $V_n^{-1}$.
Then:
Factor $\paren {-1}^{n - 1}$ from the denominator of (4) to agree with Knuth (1997). 
== Sources ==
* 
* 
* 
Category:Inverse of Vandermonde Matrix
Inverse of a Line Through Circle Center is a Line

== Theorem ==
Let $C$ be a circle in the plane on center $O$.
Let the radius of $C$ be $r$.
Let $L$ be an arbitrary ray through $O$.
Let $P$ be an otherwise arbitrary point, but lying on $L$.
Let $T : X \to Y$ be an inversive transformation with $C$ as the inversion circle. 
Then $O$ is the inversion center.
Then the image of $P$ under $T$ also lies on $L$.
== Proof ==
Let $P'$ be the image of $P$ under $T$, with $P \ne O$.
By the definition of inversion:
:$P'$ lies on $L$
We deal with $O$ separately. By the definition of ray:
:$O$ lies on $L$
:
A point has no length, it has no magnitude.
So we cannot compute $\dfrac {r^2} {OO}$ to find $O'$, the image of $O$.
But by the definition of inversion, the image of $O$ is the point at infinity.
By the definition the point at infinity lies on every infinite half-line through $O$.
Hence $O$ lies on $L$.
Therefore the image under $T$ of every point on $L$ lies on $L$.
Category:Inversive Geometry
Inverse reverses Preorder of Positive Invertible Elements of Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering on $A$.
Let $a, b \in A$ be positive and invertible such that:
:$a \le_A b$
Then we have:
:$0 \le_A b^{-1} \le_A a^{-1}$
== Proof ==
First suppose that $c \in A$ satisfies:
:$c \ge_A {\mathbf 1}_A$
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:$c \le_A \norm c {\mathbf 1}_A$ 
From Bound on Inverse in Canonical Preordering of Unital C*-Algebra we have:
:$\norm c^{-1} {\mathbf 1}_A \le_A c^{-1} \le_A {\mathbf 1}_A$
From Spectrum of Inverse of Element of Unital Algebra, we have:
:$\map {\sigma_A} {b^{-1} } = \set {z^{-1} : z \in \map {\sigma_A} b}$
Since $b$ is positive, we have $\map {\sigma_A} b \subseteq \openint 0 \infty$.
Hence $\map {\sigma_A} {b^{-1} } \subseteq \openint 0 \infty$.
So ${\mathbf 0}_A \le_A b^{-1}$.
Let $x \in A$ be the positive square root of $a$. 
That is, $a = x^2$ and $x$ is positive.
Since $x$ commutes with itself, we have that $x$ is invertible from Product of Commuting Elements in Monoid is Unit iff Each Element is Unit.
From Conjugation in C*-Algebra preserves Positivity: Corollary, we have:
:$x^{-1} a x^{-1} \le_A x^{-1} b x^{-1}$ 
We have:
:$x^{-1} a x^{-1} = x^{-1} x^2 x^{-1} = {\mathbf 1}_A$
Hence we have:
:${\mathbf 1}_A \le_A x^{-1} b x^{-1}$
From Bound on Inverse in Canonical Preordering of Unital C*-Algebra, we have that:
:$\paren {x^{-1} b x^{-1} }^{-1} \le_A {\mathbf 1}_A$
From Inverse of Product, we have:
:$\paren {x^{-1} b x^{-1} }^{-1} = x b^{-1} x$ 
Hence:
:$x b^{-1} x \le_A {\mathbf 1}_A$
Hence from Conjugation in C*-Algebra preserves Positivity: Corollary, we have:
:$b^{-1} \le_A x^{-1} x^{-1}$
From Inverse of Product, we have $x^{-1} x^{-1} = \paren {x^2}^{-1} = a^{-1}$
Hence we have:
:$b^{-1} \le_A a^{-1}$
== Sources ==
* : $2.2$: Positive Elements of $C^\ast$-Algebras
Category:Positive Elements of C*-Algebras
Inverses in Group Direct Product

== Theorem ==
Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $g^{-1}$ be an inverse of $g \in \struct {G, \circ_1}$.
Let $h^{-1}$ be an inverse of $h \in \struct {H, \circ_2}$.
Then $\tuple {g^{-1}, h^{-1} }$ is the inverse of $\tuple {g, h} \in \struct {G \times H, \circ}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Group Direct Products
Category:Inverses in Group Direct Product
Inverses in Group Direct Product/Proof 1

== Theorem ==
== Proof ==
Let $e_G$ be the identity for $\struct {G, \circ_1}$
Let $e_H$ be the identity for $\struct {H, \circ_2}$.
Then:
So the inverse of $\tuple {g, h}$ is $\tuple {g^{-1}, h^{-1} }$.
Category:Inverses in Group Direct Product
Inverses in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then for each $h \in H$, the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
== Proof ==
Let $h \in H$.
Let:
:$h'$ be the inverse of $h$ in $H$
:$h^{-1}$ be the inverse of $h$ in $G$.
From Identity of Subgroup:
:$h \circ h' = e$
From Inverse in Group is Unique, it follows that $h' = h^{-1}$.
== Sources ==
* : $\S 36.2 \ \text{(ii)}$: Subgroups
Category:Subgroups
Inverses of Elements Related by Compatible Relation

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\RR$ be a relation compatible with $\circ$.
Then:
:$\forall x, y \in G: x \mathrel \RR y \iff y^{-1} \mathrel \RR x^{-1}$
=== Corollary ===
== Proof ==
Let $e$ be the group identity of $G$.
By Relation Compatible with Group Operation is Strongly Compatible: Corollary:
:$(1): \quad x \mathrel \RR y \iff e \mathrel \RR y \circ x^{-1}$
By Relation Compatible with Group Operation is Strongly Compatible: Corollary, also:
:$(2): \quad y^{-1} \mathrel \RR x^{-1} \iff e \mathrel \RR \paren {y^{-1} }^{-1} \circ x^{-1}$
By Inverse of Group Inverse:
:$\paren {y^{-1} }^{-1} = y$
Thus, we can rewrite $(2)$ as:
:$(3): \quad y^{-1} \mathrel \RR x^{-1} \iff e \mathrel \RR y \circ x^{-1}$
Now note that the of $(3)$ is the same as the in $(1)$.
We conclude that:
:$x \mathrel \RR y \iff y^{-1} \mathrel \RR x$
Category:Relations Compatible with Group Operation
Inverses of Elements Related by Compatible Relation/Corollary

== Corollary to Inverses of Elements Related by Compatible Relation ==
Let $\struct {G, \circ}$ be a group with identity $e$.
Let $\RR$ be a relation compatible with $\circ$.
Then the following hold:
:$\forall x, y \in G:$
::$x \mathrel \RR e \iff e \mathrel \RR x^{-1}$
::$e \mathrel \RR x \iff x^{-1} \mathrel \RR e$
== Proof ==
From Inverse of Identity Element is Itself:
:$e^{-1} = e$
From Inverses of Elements Related by Compatible Relation:
Substituting $e$ for $y$ gives:
:$x \mathrel \RR e \iff e \mathrel \RR x^{-1}$
Substituting $e$ for $x$ and $x$ for $y$ gives:
:$e \mathrel \RR x \iff x^{-1} \mathrel \RR e$
Category:Relations Compatible with Group Operation
Inverses of Right-Total and Left-Total Relations

== 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:
=== Inverse of Right-Total Relation is Left-Total ===
=== Inverse of Left-Total Relation is Right-Total ===
Category:Inverse Relations
Category:Right-Total Relations
Category:Left-Total Relations
Inversion Mapping Reverses Ordering in Ordered Group

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.
Let $x, y \in G$.
Let $\prec$ be the reflexive reduction of $\preceq$.
Then the following equivalences hold:
=== Corollary ===
== Proof ==
By the definition of an ordered group, $\preceq$ is a relation compatible with $\circ$.
Thus by Inverses of Elements Related by Compatible Relation, we obtain the first result:
:$x \preccurlyeq y \iff y^{-1} \preccurlyeq x^{-1}$
By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is also compatible with $\circ$.
Thus by again Inverses of Elements Related by Compatible Relation, we obtain the second result:
:$x \prec y \iff y^{-1} \prec x^{-1}$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.3$
Category:Ordered Groups
Category:Inversion Mappings
Category:Inversion Mapping Reverses Ordering in Ordered Group
Inversion Mapping Reverses Ordering in Ordered Group/Corollary

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $x \in G$.
Then the following equivalences hold:
== Proof 1 ==
== Proof 2 ==
Category:Inversion Mapping Reverses Ordering in Ordered Group
Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 1

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $x \in G$.
Then the following equivalences hold:
== Proof ==
By Inversion Mapping Reverses Ordering in Ordered Group:
Since $e^{-1} = e$, the theorem holds.
Category:Inversion Mapping Reverses Ordering in Ordered Group
Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 2

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $x \in G$.
Then the following equivalences hold:
== Proof ==
By the definition of an ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.
Thus by Inverses of Elements Related by Compatible Relation: Corollary:
By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is also compatible with $\circ$.
Thus again by Inverses of Elements Related by Compatible Relation: Corollary:
Category:Inversion Mapping Reverses Ordering in Ordered Group
Inversion Mapping is Automorphism iff Group is Abelian

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$, defined as:
:$\forall g \in G: \map \iota g = g^{-1}$
Then $\iota$ is an automorphism $G$ is abelian.
== Proof ==
From Inversion Mapping is Permutation, $\iota$ is a permutation.
It remains to be shown that $\iota$ has the morphism property $G$ is abelian.
=== Sufficient Condition ===
Suppose $\iota$ is an automorphism.
Then:
Thus from Inverse of Commuting Pair, $x$ commutes with $y$.
This holds for all $x, y \in G$.
So $\struct {G, \circ}$ is abelian by definition.
=== Necessary Condition ===
Let $\struct {G, \circ}$ be abelian.
Thus $\iota$ has the morphism property and is therefore an automorphism.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: Examples: $(9)$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \gamma$
:::except this source requests only that the morphism property is demonstrated, and not the bijectivity.
* : $\S 1.2$
:::except this source proves only the necessary condition. 
Category:Inversion Mappings
Category:Abelian Groups
Category:Group Automorphisms
Inversion Mapping is Involution

== Theorem ==
Let $G$ be a group, and let $\iota: G \to G$ be the inversion mapping.
Then $\iota$ is an involution.
That is:
:$\forall g \in G: \map \iota {\map \iota g} = g$
== Proof ==
Let $g \in G$.
Then:
which establishes the result.
Category:Inversion Mappings
Inversion Mapping is Isomorphism from Ordered Abelian Group to its Dual

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered Abelian group.
Let $\struct {G, \circ, \succcurlyeq}$ be the dual of $\struct {G, \circ, \preccurlyeq}$.
Let $\phi: \struct {G, \circ, \preccurlyeq} \to \struct {G, \circ, \succcurlyeq}$ be the inversion mapping from $\struct {G, \circ, \preccurlyeq}$ to $\struct {G, \circ, \succcurlyeq}$ defined as:
:$\forall x \in G: \map \phi x = x^{-1}$
where $x^{-1}$ denotes the inverse of $x$ in $\struct {G, \circ}$.
Then $\phi$ is an isomorphism.
== Proof ==
$\struct {G, \circ, \preccurlyeq}$ is an ordered semigroup.
Hence from Dual of Ordered Semigroup is Ordered Semigroup, $\struct {G, \circ, \succcurlyeq}$ is also an ordered semigroup.
That is, $\succcurlyeq$ is compatible with $\circ$.
From Inversion Mapping is Permutation, the inversion mapping is a bijection.
From Inversion Mapping is Automorphism iff Group is Abelian, $\phi: \struct {G, \circ} \to \struct {G, \circ}$ is an automorphism and so an isomorphism.
It remains to be shown that $\phi: \struct {G, \preccurlyeq} \to \struct {G, \succcurlyeq}$ is an order isomorphism.
We note that from Inversion Mapping is Involution that:
:$\phi = \phi^{-1}$
Let $x, y \in G$ be arbitrary.
We have:
Then we have:
Hence we have shown that $\phi$ is an order-preserving mapping in both directions.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.3$
Category:Ordered Groups
Category:Abelian Groups
Category:Dual Orderings
Category:Inversion Mappings
Category:Group Isomorphisms
Category:Order Isomorphisms
Inversion Mapping is Isomorphism to Opposite Group

== Definition ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be its opposite group.
That is:
:$\forall g, h \in G: g \circ h = h * g$
Let $\iota: G \to G$ be the inversion mapping for $\struct {G, \circ}$.
Then $\iota: \struct {G, \circ} \to \struct {G, *}$ is a group isomorphism.
== Proof ==
By Inversion Mapping is Involution, $\iota$ is an involution.
By Involution is Permutation, $\iota$ is a permutation and hence by definition a bijection.
It remains to show that $\iota$ is a group homomorphism.
Let $g, h \in G$.
Then:
Hence $\iota$ is a group homomorphism.
By definition it follows that $\iota$ is also a group isomorphism.
Category:Opposite Groups
Category:Inversion Mappings
Inversion Mapping is Mapping

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$.
Then $\iota$ is indeed a mapping.
== Proof ==
To show that $\iota$ is a mapping, it is sufficient to show that:
:$\map \iota a \ne \map \iota b \implies a \ne b$:
Category:Inversion Mappings
Inversion Mapping is Permutation

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$.
Then $\iota$ is a permutation on $G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions
Category:Inversion Mapping is Permutation
Category:Inversion Mappings
Category:Permutations
Inversion Mapping is Permutation/Proof 1

== Theorem ==
== Proof ==
The inversion mapping on $G$ is the mapping $\iota: G \to G$ defined by:
:$\forall g \in G: \map \iota g = g^{-1}$
where $g^{-1}$ is the inverse or $g$.
By Inversion Mapping is Involution, $\iota$ is an involution:
:$\forall g \in G: \map \iota {\map \iota g} = g$
The result follows from Involution is Permutation.
Category:Inversion Mapping is Permutation
Inversion Mapping is Permutation/Proof 2

== Theorem ==
== Proof ==
=== Proof of Surjection ===
Let $a \in G$.
By definition of $\iota$:
:$\iota(a^{-1}) = \left({a^{-1}}\right)^{-1}$
By Inverse of Inverse:
:$\left({a^{-1}}\right)^{-1} = a$
Hence $a$ has a preimage.
Since $a$ was arbitrary, $\iota$ is a surjection.
=== Proof of Injection ===
Suppose for some $a, b \in G$ that:
:$\iota \left({a}\right) = \iota \left({b}\right)$
Then by the definition of $\iota$:
:$a^{-1} = b^{-1}$
It follows from Inverse in Group is Unique that:
:$a = b$
Hence $\iota$ is an injection.
Hence by definition $\iota$ is a bijection.
A bijection from a set to itself is by definition a permutation.
Category:Inversion Mapping is Permutation
Inversion Mapping on Ordered Group is Dual Order-Isomorphism

== Theorem ==
Let $\struct {G, \circ, \preceq}$ be an ordered group.
Let $\iota: G \to G$ be the inversion mapping, defined by $\map \phi x = x^{-1}$.
Then $\iota$ is a dual order-isomorphism.
== Proof ==
By Inversion Mapping is Involution and Involution is Permutation, $\iota$ is a permutation and so by definition bijective.
Let $x, y \in G$ such that $x \prec y$.
Then $y^{-1} \prec x^{-1}$ by Inversion Mapping Reverses Ordering in Ordered Group.
Thus $\map \iota y \prec \map \iota x$.
Since this holds for all $x$ and $y$ with $x \prec y$, $\iota$ is strictly decreasing.
If $\map \iota x \prec \map \iota y$, then $\map \iota {\map \iota y} \prec \map \iota {\map \iota x}$ by the above.
Thus by Inverse of Group Inverse: $y \prec x$.
Therefore, $\iota$ reverses ordering in both directions, and is thus a dual isomorphism.
Category:Ordered Groups
Category:Inversion Mappings
Inversion Mapping on Topological Group is Homeomorphism

== Theorem ==
Let $T = \struct {G, \circ, \tau}$ be a topological group.
Let $\phi: G \to G$ be the inversion mapping of $T$.
Then $\phi$ is a homeomorphism.
== Proof ==
From the definition of topological group, $\phi$ is continuous.
By Inversion Mapping is Involution, $\phi$ is an involution.
By Continuous Involution is Homeomorphism, $\phi$ is a homeomorphism.
Category:Inversion Mappings
Category:Homeomorphisms (Topological Spaces)
Category:Topological Groups
Inversive Transformation is Conformal Transformation

== Theorem ==
Let $\CC$ be a circle embedded in a Cartesian plane $\EE$ whose center $O$ is at the origin $\tuple {0, 0}$ and whose radius is $r$.
Let $f$ be the inversive transformation of $\EE$ $\CC$.
Then $f$ is a conformal transformation.
== Proof ==
== Sources ==
* 
* 
Category:Inversive Transformations
Invertibility of Arithmetic Functions

== Theorem ==
Let $f: \N \to \C$ be an arithmetic function.
Then $f$ has a Dirichlet inverse :
:$\map f 1 \ne 0$
== Proof ==
Let $*$ denote Dirichlet convolution.
Let $\varepsilon$ denote the identity arithmetic function.
=== Sufficient Condition ===
Let $f$ have a Dirichlet inverse $g$:
:$f * g = \varepsilon$
Then:
Thus $\map f 1 \ne 0$.
=== Necessary Condition ===
Suppose that $\map f 1 \ne 0$.
We want to find an arithmetic function $g$ such that:
:$(1): \quad \map {\paren {f * g} } 1 = 1$
:$(2): \quad \map {\paren {f * g} } n = 0$ for all $n > 1$
We have:
:$\map {\paren {f * g} } 1 = \map f 1 \map g 1$
So we have no choice but to define:
:$\map g 1 = \paren {\map f 1}^{-1}$
and condition $(1)$ is satisfied.
Condition $(2)$ can be written as:
That is:
:$\map g n = -\dfrac 1 {\map f 1} \ds \sum_{\substack {d \mathop \divides n \\ d \mathop > 1} } \map f d \map g {\frac n d}$
We can recursively define $g$ by this formula, starting with $\map g 1$ as above.
By definition, $g$ then satisfies:
:$g * f = \varepsilon$
== Also see ==
* Units of Ring of Arithmetic Functions
Category:Arithmetic Functions
Invertibility of Identity Minus Operator

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a Banach space over $\GF$. 
Let $T : X \to X$ be a bounded linear operator such that: 
:$\norm T_{\map \BB X} < 1$
where $\norm {\, \cdot \,}_{\map \BB X}$ denotes the norm of a bounded linear operator. 
Then $I - T$ is invertible as a bounded linear operator.
In particular: 
:$\ds \paren {I - T}^{-1} = \sum_{n \mathop = 0}^\infty T^n$
 
=== Corollary ===
== Proof ==
For each $n \in \N$, define: 
:$\ds S_n = \sum_{k \mathop = 0}^n T^k$
We argue first that $\sequence {S_n}_{n \mathop \in \N}$ is convergent. 
Since $X$ is a Banach space, it is enough to show that $\sequence {S_n}_{n \mathop \in \N}$ is Cauchy. 
Let $\epsilon > 0$. 
Let $m, n \in \N$ with $n > m$. 
Then we have: 
From Sum of Infinite Geometric Sequence, we have: 
:$\ds \sum_{k \mathop = 0}^\infty \norm T_{\map \BB X}^k$
since $\norm T_{\map \BB X} < 1$. 
So: 
:$\ds \sequence {\sum_{k \mathop = 0}^n \norm T_{\map \BB X}^k}_{n \mathop \in \N}$ is Cauchy.
So we can pick $N \in \N$ such that for $n > m \ge N$ we have: 
:$\ds \sum_{k \mathop = m + 1}^n \norm T_{\map \BB X}^k < \epsilon$ 
So for $n > m \ge N$ we have: 
:$\norm {S_n - S_m}_{\map \BB X} < \epsilon$
Hence from we have: 
:$\norm {S_n - S_m}_{\map \BB X} < \epsilon$
for all $n, m \in \N$ with $n, m \ge N$. 
So $\sequence {S_n}_{n \mathop \in \N}$ is Cauchy and hence convergent in $\map \BB X$. 
Denote: 
:$\ds S = \lim_{n \mathop \to \infty} S_n = \sum_{k \mathop = 0}^\infty T^k$ 
To finish we show that $\paren {I - T} S = S \paren {I - T} = I$. 
Now for each $n \in \N$, we have: 
and similarly: 
We now show that $I - T^{k + 1} \to I$ in $\map \BB X$. 
We have: 
On the other hand by Convergence of Product in Normed Algebra, we have: 
:$S_n \paren {I - T} \to S \paren {I - T}$
and: 
:$\paren {I - T} S_n \to \paren {I - T} S$
in $\map \BB X$, while: 
:$S_n \paren {I - T} = \paren {I - T} S_n = I - T^{k + 1} \to I$ 
So we have: 
:$S \paren {I - T} = \paren {I - T} S = I$
Category:Bounded Linear Operators
Category:Invertibility of Identity Minus Operator
Invertibility of Identity Minus Operator/Corollary

== Corollary to Invertibility of Identity Minus Operator ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a Banach space over $\GF$. 
Let $T : X \to X$ be a invertible bounded linear operator. 
Let $S : X \to X$ be a bounded linear operator such that: 
:$\norm S_{\map \BB X} \norm {T^{-1} }_{\map \BB X} < 1$
Then $T + S : X \to X$ is an invertible bounded linear operator.
== Proof ==
We have: 
:$\norm {-S T^{-1} }_{\map \BB X} \le \norm S_{\map \BB X} \norm {T^{-1} }_{\map \BB X} < 1$
from , Norm on Bounded Linear Transformation is Submultiplicative.
From Invertibility of Identity Minus Operator, we have: 
:$I - \paren {-S T^{-1} } = I + S T^{-1}$ is an invertible bounded linear operator.
So from Composition of Bounded Linear Transformations is Bounded, we have that: 
:$\paren {I + S T^{-1} } T = T + S$ is a invertible bounded linear operator.
Category:Invertibility of Identity Minus Operator
Invertibility of Identity Transformation Plus Product of Two Continuous Linear Transformations

== Theorem ==
Let $\struct {X, \norm{, \cdot ,} }$ be the normed vector space.
Let $I : X \to X$ be the identity mapping.
Let $\map {CL} X := \map {CL} {X, X}$ be the continuous linear transformation space on $X$.
Suppose $I + A \circ B$ is invertible, where $\circ$ denotes the composition of mappings.
Then $I + B \circ A$ is invertible, with the inverse given by:
:$\paren {I + B \circ A}^{-1} = I - B \circ \paren {I + A \circ B}^{-1} \circ A$
== Proof ==
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Continuous Linear Transformations
Category:Inverse Mappings