nyuuzyou/wikis huggingface dataset:
Condition for Isomorphism between Structures Induced by Permutations

== Theorem ==
Let $S$ be a set.
Let $\oplus$ and $\otimes$ be closed operations on $S$ such that both $\oplus$ and $\otimes$ have the same identity.
Let $\sigma$ and $\tau$ be permutations on $S$.
Let $\oplus_\sigma$ and $\otimes_\tau$ be the operations on $S$ induced on $\oplus$ by $\sigma$ and on $\otimes$ by $\tau$ respectively:
:$\forall x, y \in S: x \oplus_\sigma y := \map \sigma {x \oplus y}$
:$\forall x, y \in S: x \otimes_\tau y := \map \tau {x \otimes y}$
Let $f: S \to S$ be a mapping.
Then:
:$f$ is an isomorphism from $\struct {S, \oplus_\sigma}$ to $\struct {S, \otimes_\tau}$
:
:$f$ is an isomorphism from $\struct {S, \oplus}$ to $\struct {S, \otimes}$ satisfying the condition:
:::$f \circ \sigma = \tau \circ f$
:where $\circ$ denotes composition of mappings.
== Proof ==
Recall that:
:an isomorphism is a bijection which is a homomorphism
:a permutation is a bijection from a set to itself.
Hence on both sides of the double implication:
:$f$ is a permutation on $S$
:both $f \circ \sigma$ and $\tau \circ f$ are permutations on $S$.
So bijectivity of all relevant mappings can be taken for granted throughout the following.
=== Necessary Condition ===
Let $f$ be an isomorphism from $\struct {S, \oplus}$ to $\struct {S, \otimes}$ satisfying the condition:
:$f \circ \sigma = \tau \circ f$
We have:
demonstrating that $f$ is a homomorphism from $\struct {S, \oplus_\sigma}$ to $\struct {S, \otimes_\tau}$.
As $f$ is a bijection, it follows by definition that $f$ is an isomorphism from $\struct {S, \oplus_\sigma}$ to $\struct {S, \otimes_\tau}$.
=== Sufficient Condition ===
Let $f: \struct {S, \oplus_\sigma} \to \struct {S, \otimes_\tau}$ be an isomorphism.
Let $e \in S$ be the identity for both $\oplus$ and $\otimes$, .
We have:
Thus $f$ is a homomorphism, and thus an isomorphism, from $\struct {S, \oplus}$ to $\struct {S, \otimes}$.
Now we have:
In particular, this holds for $y = e$, so:
Hence by Equality of Mappings:
:$\tau \circ f = f \circ \sigma$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.9 \ \text {(b)}$
Category:Operations Induced by Permutations
Category:Isomorphisms (Abstract Algebra)
Condition for Limits of Computable Real Sequences to be Computable

== Theorem ==
Let $\sequence {x_k}_{k \in \N}$ be a computable real sequence.
Let $\sequence {y_n}_{n \in \N}$ be a real sequence.
If there exists total recursive function $\phi : \N^2 \to \N$ such that:
:$\forall n, p \in \N: \size {x_{\map \phi {n, p}} - y_n} < \dfrac 1 {p + 1}$
then:
:$\sequence {y_n}_{n \in \N}$
is a computable real sequence.
== Proof ==
By Corollary to Computable Real Sequence iff Limits of Computable Rational Sequences, there exists a computable rational sequence $\sequence {a_N}$ such that, for all $k, p \in \N$:
:$\size {a_{\map \pi {k, p} } - x_k} < \dfrac 1 {p + 1}$
where $\pi$ is the Cantor pairing function.
Let $\psi : \N^2 \to \N$ be defined as:
:$\map \psi {n, p} = \map \pi {\map \phi {n, 2 p + 1}, 2 p + 1}$
which is total recursive by:
* Cantor Pairing Function is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
By Corollary to Computable Subsequence of Computable Rational Sequence is Computable, there exists a computable rational sequence $\sequence {b_N}$ such that, for all $n, p \in \N$:
:$b_{\map \pi {n, p}} = a_{\map \psi {n, p}}$
We have, for all $n, p \in \N$:
The result then follows from Corollary to Computable Real Sequence iff Limits of Computable Rational Sequences.
Category:Computability Theory
Condition for Linear Dependence of Linear Functionals in terms of Kernel

== Theorem ==
Let $V$ be a vector space over a field $\GF$.
Let $f, f_1, \ldots, f_n: V \to \GF$ be linear functionals.
Suppose that:
:$\ds \bigcap_{i \mathop = 1}^n \ker f_i \subseteq \ker f$
where $\ker f$ denotes the kernel of $f$.
Then there exist $\alpha_1, \ldots, \alpha_n \in \GF$ such that:
:$\ds \forall v \in V: \map f v = \sum_{i \mathop = 1}^n \alpha_i \map {f_i} v$
That is: 
:$f \in \span \set {f_1, \ldots, f_n}$
== Proof 1 ==
 
== Proof 2 ==
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra: Proposition $1.4$
Category:Condition for Linear Dependence of Linear Functionals in terms of Kernel
Category:Linear Functionals
Condition for Linear Dependence of Linear Functionals in terms of Kernel/Proof 1

== Theorem ==
== Proof ==
For $i = 1, \ldots, n$, let $w_i$ be such that:
:$w_i \not\in \ker f_i$
:$w_i \in \ker f_j, j \ne i$
Suppose first that the $w_i$ all exist.
Let $v_i = \dfrac 1 {\map {f_i} {w_i} } w_i$.
Then since $f_i$ is linear:
:$\map {f_i} {v_i} = 1$
Furthermore for $j \ne i$, $\map {f_j} {v_i} = 0$.
Now let $v \in V$ be arbitrary, and define:
:$\ds w = v - \sum_{i \mathop = 1}^n \map {f_i} v v_i$
Then:
Since $j$ was arbitrary, it follows from the premise on $f$ that:
:$\map f w = 0$
That is to say:
:$\ds 0 = \map f v - \sum_{i \mathop = 1}^n \map {f_i} v \map f {v_i}$
Setting $\alpha_i = \map f {v_i}$ we find:
:$\ds \map f v = \sum_{i \mathop = 1}^n \alpha_i \map {f_i} v$
Now the case that such $w_i$ does not exist for a particular $i$.
Then:
:$\ds \bigcap_{i \mathop = 1}^n \ker f_i = \bigcap_{j \mathop \ne i} \ker f_j$
and $\set{ f_j: j \ne i}$ fulfils the condition of the result.
The result for $f_1, \ldots, f_n$ follows by setting $\alpha_i = 0$.
Category:Condition for Linear Dependence of Linear Functionals in terms of Kernel
Condition for Linear Dependence of Linear Functionals in terms of Kernel/Proof 2

== Theorem ==
== Proof ==
Define $T : X \to {\GF}^n$ by: 
:$\map T x = \paren {\map {g_1} x, \map {g_2} x, \ldots, \map {g_n} x}$
for each $x \in X$.
We show that $T$ is linear. 
Let $x, y \in X$ and $\alpha, \beta \in \GF$. 
We have: 
so $T$ is linear.
Now, we have $\map T x = 0$ $\map {g_i} x = 0$ for each $x \in X$. 
That is, :
:$\ds x \in \bigcap_{i \mathop = 1}^n \ker g_i$
so that:
:$\ds \ker T = \bigcap_{i \mathop = 1}^n \ker g_i \subseteq \ker f$
Now, suppose that $x, y \in X$ have $T x = T y$.
Then by the linearity of $T$, we have: 
:$\map T {x - y} = 0$
so $x - y \subseteq \ker f$.
So $\map f {x - y} = 0$, and hence $\map f x = \map f y$, for these $x, y \in X$.
We can therefore define a map $\widetilde f : \Img T \to \GF$ by: 
:$\map {\widetilde f} {\map T x} = \map f x$ 
for each $x \in \Img T$.
From Image of Vector Subspace under Linear Transformation is Vector Subspace, $\Img T$ is a vector subspace of ${\GF}^n$.
Let $\set {e_1, \ldots, e_n}$ be the standard basis for ${\GF}^n$.
Let $\set {v_1, \ldots, v_k}$ be a basis of $\Img T$.
From Linearly Independent Set is Contained in some Basis, $\set {v_1, \ldots, v_k}$ is contained in some basis $\set {v_1, \ldots, v_n}$ of ${\GF}^n$. 
Define the mapping $h : {\GF}^n \to \GF$ by taking: 
:$\ds \map h {v_i} = \map {\widetilde f} {v_i}$ for $1 \le i \le k$
and:
:$\ds \map h {v_i} = 0$ for $k < i \le n$
and taking: 
:$\ds \map h {\sum_{i \mathop = 1}^n \alpha_i v_i} = \sum_{i \mathop = 1}^n \alpha_i \map h {v_i}$ 
for scalars $\alpha_1, \alpha_2, \ldots, \alpha_n \in \GF$.
By construction, $h$ is a linear functional.
We then have, for each $x \in X$: 
with $\map h {e_i} \in \GF$.
So we have: 
:$f \in \span \set {g_1, \ldots, g_n}$
Category:Condition for Linear Dependence of Linear Functionals in terms of Kernel
Condition for Linear Divisor of Polynomial

== Theorem ==
Let $\map P x$ be a polynomial in $x$.
Let $a$ be a constant.
Then $x - a$ is a divisor of $\map P x$ $a$ is a root of $P$.
== Proof ==
From the Little Bézout Theorem, the remainder of $\map P x$ when divided by $x - a$ is equal to $\map P a$.
=== Sufficient Condition ===
Let $x - a$ be a divisor of $\map P x$.
From the Little Bézout Theorem, the remainder of $\map P x$ when divided by $x - a$ is equal to $\map P a$.
By definition of divisor, the remainder of $\map P x$ when divided by $x - a$ equals $0$.
That is:
:$\map P a = 0$
It follows by definition that $a$ is a root of $P$.
=== Necessary Condition ===
Let $a$ be a root of $P$.
From the Little Bézout Theorem, and by Polynomial Long Division, we have:
:$\map P x = \paren {x - a} \map Q x + \map P a$
where $\map Q x$ is a polynomial in $x$ of degree one less than $\map P x$.
By definition of root of polynomial:
:$\map P a = 0$
So we have:
:$\map P x = \paren {x - a} \map Q x$
and the result follows by definition of divisor.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 37$. Roots of Polynomials: Theorem $70$
* 
* 
Category:Polynomial Theory
Condition for Linear Operation on Complex Numbers to be of Finite Order

== Theorem ==
Let $A$ be the operation on the complex numbers $\C$ defined as:
:$\map A x = \alpha x + \beta$
Then $A$ is of finite order greater than $1$ $\alpha$ is a root of unity other than $1$.
== Proof ==
== Sources ==
* : Chapter $\text {I}$: The Group Concept: Examples: $(3)$
Category:Roots of Unity
Condition for Linear Transformation

== Theorem ==
Let $G$ be a unitary $R$-module.
Let $H$ be an $R$-module.
Let $\phi: G \to H$ be a mapping.
Then $\phi$ is a linear transformation :
:$\forall x, y \in G: \forall \lambda, \mu \in R: \map \phi {\lambda x + \mu y} = \lambda \map \phi x + \mu \map \phi y$
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
Let $\phi$ be such that the condition is satisfied.
Let $\lambda = \mu = 1_R$.
Then:
:$\map \phi {x + y} = \map \phi x + \map \phi y$
Now let $\mu = 0_R$.
Then:
:$\map \phi {\lambda x} = \lambda \map \phi x$
Thus the conditions are fulfilled for $\phi$ to be a homomorphism, that is, a linear transformation.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Condition for Lines to be Conjugate

== Theorem ==
Let $\CC$ be a circle of radius $r$ whose center is at the origin of a Cartesian plane.
Let $\PP$ and $\QQ$ be conjugate lines with respect to $\CC$:
Then:
:$l_1 l_2 + m_1 m_2 = \dfrac {n_1 n_2} {r^2}$
== Proof ==
By definition of conjugate lines, $\PP$ and $\QQ$ are the polars of points $P$ and $Q$ respectively, such that $P$ lies on $\QQ$ and $Q$ lies on $\PP$.
From Coordinates of Pole of Given Polar, $P$ is given by:
:$P = \tuple {-\dfrac {l_1} {n_1} r^2, -\dfrac {m_1} {n_1} r^2}$
We have that $P$ lies on $\QQ$.
Substituting $x = -\dfrac {l_1} {n_1} r^2$ and $y = -\dfrac {m_1} {n_1} r^2$ in the equation of $\QQ$, we obtain:
from which the result follows.
== Sources ==
* : Chapter $\text {III}$. The Circle: $8$. Reciprocal property of pole and polar
Category:Conjugate Lines
Condition for Lipschitz Condition to be Satisfied

== Theorem ==
Let $\phi$ be a real function.
Then $\phi$ satisfies the Lipschitz condition on a closed real interval $\closedint a b$ if:
:$\forall y \in \closedint a b: \exists A \in \R: \size {\map {\phi'} y} \le A$
== Proof ==
Integrating both sides of $\size {\map {\phi'} y} \le A$ gives us:
 
On the interval $\closedint a b$ it follows that $\size {\map \phi y}$ is bounded by the greater of $A a$ and $A b$.
Hence the result.
Category:Calculus
Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure with $1$ operation.
Let $\struct {A, \circ {\restriction_A} }$ and $\struct {B, \circ {\restriction_B} }$ be closed algebraic substructures of $\struct {S, \circ}$, where $\circ {\restriction_A}$ and $\circ {\restriction_B}$ are the operations induced by the restrictions of $\circ$ to $A$ and $B$ respectively.
Let the mapping $\phi: A \times B \to S$ be defined as:
:$\forall \tuple {a, b} \in A \times B: \map \phi {a, b} = a \circ b$
where $A \times B$ denotes the Cartesian product of $A$ and $B$.
Then:
:$\phi$ is a bijection
:
:for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.
== Proof ==
First we establish that from Set of Finite Subsets under Induced Operation is Closed:
:$A \times B \subseteq S$
Thus:
:$\forall \tuple {a, b} \in A \times B: \exists s \in S: s = a \circ b = \map \phi {a, b}$
Thus $\phi$ is indeed a mapping.
=== Sufficient Condition ===
Let $\phi$ be a bijection.
As $\phi$ is a bijection, it is surjective, so:
:$S \subseteq A \times B$
Thus:
:$S = A \times B$
That is, every element of $S$ is an element of $A \times B$.
Thus, every element of $S$ can be expressed as a product of the form $a \circ b$ such that $\tuple {a, b} \in A \times B$ in at least one way.
As $\phi$ is a bijection, it is injective.
Let $\map \phi {a_1, b_1} = \map \phi {a_2, b_2}$.
Then by definition of injective:
:$\tuple {a_1, b_1} = \tuple {a_2, b_2}$
and thus:
:$a_1 = a_2, b_1 = b_2$
From the definition of $\phi$, this means:
:$a_1 \circ b_1 = a_2 \circ b_2$
Thus, every element of $S$ that can be expressed as a product of the form $a \circ b$ such that $\tuple {a, b} \in A \times B$ can be thus expressed uniquely.
In summary:
:$\forall s \in S: \exists_1 \tuple {a, b} \in A \times B: s = a \circ b$
where $\exists_1$ denotes unique existence.
=== Necessary Condition ===
Let it be the case that:
:for all $s \in S$: there exists a unique $\tuple {a, b} \in A \times B$ such that $a \circ b = s$.
Hence by definition of $\phi$:
:$\forall s \in S: \exists \tuple {a, b} \in A \times B: s = \map \phi {a, b}$
That is, $\phi$ is surjective.
Now let $\tuple {a, b}$ and $\tuple {c, d}$ in $A \times B$ such that:
:$\map \phi {a, b} = \map \phi {c, d}$
That is:
:$\exists s \in S: s = a \circ b = c \circ d$
But such a representation of $s$ is unique.
That is:
:$\tuple {a, b} = \tuple {c, d}$
and so $\phi$ is by definition an injection.
Thus $\phi$ is both a surjection and an injection.
Hence the result by definition of bijection.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces
Category:Internal Direct Products
Condition for Mapping between Structures to be Homomorphism

== Theorem ==
Let $\struct {A, \odot}$ and $\struct {B, \circledast}$ be magmas.
Let $\struct {A \times B, \otimes}$ be the external direct product of $\struct {A, \odot}$ and $\struct {B, \circledast}$.
Let $\phi: A \to B$ be a mapping.
Let $\phi$ be considered as a subset of the Cartesian product $A \times B$.
Then:
:$\phi$ is a homomorphism
:
:the algebraic structure $\struct {\phi, \otimes_\phi}$ is a submagma of $\struct {A \times B, \otimes}$.
== Proof ==
Let $\phi$ be a homomorphism
Let $\tuple {a, b}, \tuple {c, d} \in A \times B$ such that:
Then:
That is:
:$\struct {\phi, \otimes_\phi}$ is a closed subset of $\struct {A \times B, \otimes}$
which means the same thing as:
:$\struct {\phi, \otimes_\phi}$ is a submagma of $\struct {A \times B, \otimes}$.
The argument reverses, so:
:if $\struct {\phi, \otimes_\phi}$ is a submagma of $\struct {A \times B, \otimes}$
then:
:$\phi$ is a homomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.15$
Category:Homomorphisms (Abstract Algebra)
Category:External Direct Products
Condition for Mapping from Quotient Set to be Injection

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR$ be an equivalence relation on $S$.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $S / \RR$ be the quotient set of $S$ induced by $\RR$.
Let $q_\RR: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Let the mapping $\phi: S / \RR \to T$ defined as:
:$\phi \circ q_\RR = f$
be well-defined.
Then:
:$\phi$ is an injection
:
:$\forall x, y \in S: \tuple {x, y} \in \RR \iff \map f x = \map f y$
== Proof ==
From Condition for Mapping from Quotient Set to be Well-Defined, $\phi$ is well-defined :
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
By definition of injection, $\phi$ is injective :
:$\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \eqclass x \RR = \eqclass y \RR$
that is:
:$\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \tuple {x, y} \in \RR$
By definition of $\phi$:
:$\map \phi {\eqclass x \RR} = \map f x$
and so:
:$\map f x = \map f y \implies \tuple {x, y} \in \RR$
Hence the result.
== Motivation ==
== Also see ==
* Definition:Well-Defined Mapping
* Condition for Mapping from Quotient Set to be Well-Defined
* Mapping from Quotient Set when Defined is Unique
* Condition for Mapping from Quotient Set to be Surjection
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.5$
Category:Quotient Mappings
Category:Quotient Sets
Category:Injections
Condition for Mapping from Quotient Set to be Surjection

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR$ be an equivalence relation on $S$.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $S / \RR$ be the quotient set of $S$ induced by $\RR$.
Let $q_\RR: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Let the mapping $\phi: S / \RR \to T$ defined as:
:$\phi \circ q_\RR = f$
be well-defined.
Then:
:$\phi$ is a surjection
:
:$f$ is a surjection.
== Proof ==
We are given that:
:$\phi \circ q_\RR = f$
is well-defined.
Note that from Quotient Mapping is Surjection, $q_\RR$ is a surjection.
=== Sufficient Condition ===
Let $\phi$ be a surjection.
Then from Composite of Surjections is Surjection:
:$\phi \circ q_\RR$ is a surjection.
=== Necessary Condition ===
Let $f = \phi \circ q_\RR$ be a surjection.
Then from Surjection if Composite is Surjection:
:$\phi$ is a surjection.
Hence the result.
== Motivation ==
== Also see ==
* Definition:Well-Defined Mapping
* Condition for Mapping from Quotient Set to be Well-Defined
* Mapping from Quotient Set when Defined is Unique
* Condition for Mapping from Quotient Set to be Injection
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.5$
Category:Quotient Mappings
Category:Quotient Sets
Category:Surjections
Condition for Mapping from Quotient Set to be Well-Defined

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR$ be an equivalence relation on $S$.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $S / \RR$ be the quotient set of $S$ induced by $\RR$.
Let $q_\RR: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Then:
:there exists a mapping $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$
:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
::$\begin {xy} \xymatrix@L + 2mu@ + 1em {
 S \ar[r]^*{f}
 \ar[d]_*{q_\RR}
& T \\
S / \RR \ar@{-->}[ur]_*{\phi} } \end {xy}$
=== Mapping from Quotient Set when Defined is Unique ===
=== Condition for Mapping from Quotient Set to be Injection ===
=== Condition for Mapping from Quotient Set to be Surjection ===
== Motivation ==
== Proof ==
From Condition for Composite Mapping on Left, we have:
:$\exists \phi: S / \RR \to T$ such that $\phi$ is a mapping and $\phi \circ q_\RR = f$
:
:$\forall x, y \in S: \map {q_\RR} x = \map {q_\RR} y \implies \map f x = \map f y$
But by definition of the quotient mapping induced by $\RR$:
:$\map {q_\RR} x = \map {q_\RR} y \iff \tuple {x, y} \in \RR$
Hence the result.
== Also see ==
* Definition:Well-Defined Mapping
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Problem $\text{CC}$: Factoring through Quotients
* : Chapter $1$: Equivalence Relations: $\S 19$
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.5$
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra
Category:Quotient Mappings
Category:Quotient Sets
Condition for Mapping from Quotient Vector Space to be Well-Defined

== Theorem ==
Let $V, W$ be vector spaces.
Let $T: V \to W$ be a linear transformation.
Let $M$ be a subspace of $V$.
Let $V / M$ be the quotient vector space of $V$ by $M$.
Let $Q_M: V \to V / M$ be the associated quotient mapping.
Then:
:there exists a linear transformation $L: V / M \to W$ such that $L \circ Q_M = T$
:
:$M \subseteq \ker T$
::$\begin {xy} \xymatrix@L + 2mu@ + 1em {V \ar[r]^*{T} \ar[d]_*{Q_M} & W \\ V / M \ar@{-->}[ur]_*{L} } \end {xy}$
== Proof ==
By Condition for Mapping from Quotient Set to be Well-Defined, it follows that:
:$L: V / M \to W$ exists
:$\forall v, v' \in V: v + M = v' + M \implies T v = T v'$
Now $v + M = v' + M$ $v - v' \in M$.
We are given that $T$ is a linear transformation, so:
:$T v = T v' \iff T \paren{ v - v' } = 0$
In particular then, with $v' = 0$ it follows that:
:$L: V / M \to W$ exists
:$\forall v \in V: v \in M \implies T v = 0$
and the latter states that $M \subseteq \ker T$.
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra: Proposition $1.3$
Category:Quotient Vector Spaces
Condition for Membership is Right Compatible with Ordinal Exponentiation

== Theorem ==
Let $x, y, z$ be ordinals.
Let $z$ be the successor of some ordinal $w$.
Then:
:$x < y \iff x^z < y^z$
== Proof ==
=== Sufficient Condition ===
Suppose $x < y$.
By Subset is Right Compatible with Ordinal Exponentiation:
:$x^w \le y^w$
Then:
Thus the sufficient condition is satisfied.
=== Necessary Condition ===
Suppose $x^z < y^z$.
From Subset is Right Compatible with Ordinal Exponentiation:
:$y \le x \implies y^z \le x^z$
This contradicts:
:$x^z < y^z$
so:
:$y \nleq x$
By Ordinal Membership is Trichotomy:
:$x < y$
== Sources ==
* : $\S 8.36$
Category:Ordinal Arithmetic
Condition for Membership of Equivalence Class

== Theorem ==
Let $\RR$ be an equivalence relation on a set $S$.
Let $\eqclass x \RR$ denote the $\RR$-equivalence class of $x$.
Then:
:$\forall y \in S: y \in \eqclass x \RR \iff \tuple {x, y} \in \RR$
== Proof ==
From the definition of an equivalence class:
:$\eqclass x \RR = \set {y \in S: \tuple {x, y} \in \RR}$
Let $y \in S$ such that $y \in \eqclass x \RR$.
Then by definition $\tuple {x, y} \in \RR$.
Similarly, let $\tuple {x, y} \in \RR$.
Again by definition, $y \in \eqclass x \RR$.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.3 \ (1)$
Category:Equivalence Classes
Condition for Method of Least Squares to match Maximum Likelihood Estimation

== Theorem ==
Let $X_n$ and $Y_n$ be sets of $n$ paired observations of random variables $X$ and $Y$.
For the paired observations $\tuple {x_i, y_i}$:
:let $x_i$ be error-free
:let the errors in $y_i$ be identical and normally distributed with expectation $0$.
Then the method of least squares yields a result equivalent to the maximum likelihood estimator.
== Proof ==
== Sources ==
* 
* 
Category:Method of Least Squares (Statistics)
Category:Maximum Likelihood Estimation
Condition for Nonzero Eigenvalue of Compact Operator

== Theorem ==
Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.
Let $T \in \map {B_0} H$ be a compact operator.
Let $\lambda \in \Bbb F, \lambda \ne 0$ be a nonzero scalar.
Suppose that the following holds:
:$\inf \set {\norm {\paren {T - \lambda I} h}_H: \norm h_H = 1} = 0$
Then $\lambda \in \map {\sigma_p} T$, that is, $\lambda$ is an eigenvalue for $T$.
=== Corollary ===
== Proof ==
By hypothesis, there is a sequence of unit vectors $\sequence{h_n}$ such that $\cmod{(T-\lambda) h_n} \to 0$.
Since $T$ is compact, by definition 2 there is a $f\in H$ and a subsequence $\sequence{h_{n_k}}$ such that $T h_{n_k}\to f$. 
But $h_{n_k}=\lambda^{-1}\left[(\lambda-T) h_{n_k}+T h_{n_k}\right] \to\lambda^{-1} f$.
So $1=\norm{\lambda^{-1} f}=\cmod\lambda^{-1}\norm{f}$ and $f \neq 0$.
Also, it must be that $T h_{n_k} \to \lambda^{-1} T f$.
Since $T h_{n_k} \to f, f=\lambda^{-1} T f$, or $T f=\lambda f$.
That is, $f \in \ker(T-\lambda)$ and $f \neq 0$, so $\lambda \in \sigma_p(T)$.
== Sources ==
* $II.4.14, 15$
Category:Linear Transformations on Hilbert Spaces
Condition for Nonzero Eigenvalue of Compact Operator/Corollary

== Corollary to Condition for Nonzero Eigenvalue of Compact Operator ==
Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.
Let $T \in \map {B_0} H$ be a compact operator.
Let $\lambda \in \Bbb F, \lambda \ne 0$ be a nonzero scalar.
Suppose $\lambda \notin \map {\sigma_p} T$ and $\bar \lambda \notin \map {\sigma_p} {T^*}$.
Then $T - \lambda I$ is invertible.
Furthermore, $\paren {T - \lambda I}^{-1}$ is a bounded linear operator.
== Proof ==
Category:Linear Transformations on Hilbert Spaces