nyuuzyou/wikis huggingface dataset:
Determinant with Columns Transposed

== Theorem ==
If two columns of a matrix with determinant $D$ are transposed, its determinant becomes $-D$.
== Proof ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $1 \le r < s \le n$.
Let $\mathbf B$ be $\mathbf A$ with columns $r$ and $s$ transposed.
Consider:
:the transpose $\mathbf A^\intercal$ of $\mathbf A$
:the transpose $\mathbf B^\intercal$ of $\mathbf B$.
hence $\mathbf B^\intercal$ is $\mathbf A^\intercal$ with rows $r$ and $s$ transposed.
From Determinant with Rows Transposed:
:$\map \det {\mathbf B^\intercal} = -\map \det {\mathbf A^\intercal}$
From from Determinant of Transpose:
:$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$
:$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$
and the result follows.
== Also see ==
* Determinant with Rows Transposed
== Sources ==
* : $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.10 \ (2)$
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$
* 
* 
* 
Category:Determinants
Determinant with Row Multiplied by Constant

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $\mathbf B$ be the matrix resulting from one row of $\mathbf A$ having been multiplied by a constant $c$.
Then:
:$\map \det {\mathbf B} = c \map \det {\mathbf A}$
That is, multiplying one row of a square matrix by a constant multiplies its determinant by that constant.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Determinant with Column Multiplied by Constant
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercises -- Second Set
* 
* 
Category:Determinants
Category:Determinant with Row Multiplied by Constant
Determinant with Row Multiplied by Constant/Proof 1

== Theorem ==
== Proof ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $e$ be the elementary row operation that multiplies rows $i$ by the scalar$c$.
Let $\mathbf B = \map e {\mathbf A}$.
Let $\mathbf E$ be the elementary row matrix corresponding to $e$.
From Elementary Row Operations as Matrix Multiplications:
:$\mathbf B = \mathbf E \mathbf A$
From Determinant of Elementary Row Matrix: Exchange Rows:
:$\map \det {\mathbf E} = c$
Then:
Hence the result.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.10$
Category:Determinant with Row Multiplied by Constant
Determinant with Row Multiplied by Constant/Proof 2

== Theorem ==
== Proof ==
Let:
:$\mathbf A = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{r 1} & a_{r 2} & \cdots & a_{r n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
\end {bmatrix}$
:$\mathbf B = \begin{bmatrix}
b_{1 1} & b_{1 2} & \ldots & b_{1 n} \\
b_{2 1} & b_{2 2} & \ldots & b_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
b_{r 1} & b_{r 2} & \cdots & b_{r n} \\
 \vdots & \vdots & \ddots & \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n n} \\
\end{bmatrix} = \begin{bmatrix}
a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\
a_{2 1} & a_{2 2} & \ldots & a_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
c a_{r 1} & c a_{r 2} & \cdots & c a_{r n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
\end{bmatrix}$
Then from the definition of the determinant:
The constant $c$ is a factor of all the terms in the $\sum_\lambda$ expression and can be taken outside the summation:
Category:Determinant with Row Multiplied by Constant
Determinant with Rows Transposed

== Theorem ==
If two rows of a matrix with determinant $D$ are transposed, its determinant becomes $-D$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Determinant with Columns Transposed
== Sources ==
* 
* 
* 
Category:Determinants
Category:Determinant with Rows Transposed
Determinant with Rows Transposed/Proof 1

== Theorem ==
== Proof ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $1 \le r < s \le n$.
Let $e$ be the elementary row operation that exchanging rows $r$ and $s$.
Let $\mathbf B = \map e {\mathbf A}$.
Let $\mathbf E$ be the elementary row matrix corresponding to $e$.
From Elementary Row Operations as Matrix Multiplications:
:$\mathbf B = \mathbf E \mathbf A$
From Determinant of Elementary Row Matrix: Exchange Rows:
:$\map \det {\mathbf E} = -1$
Then:
Hence the result.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.9$
Category:Determinant with Rows Transposed
Determinant with Rows Transposed/Proof 2

== Theorem ==
== Proof ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $1 \le r < s \le n$.
Let $\rho$ be the permutation on $\N^*_n$ which transposes $r$ and $s$.
From Parity of K-Cycle, $\map \sgn \rho = -1$.
Let $\mathbf A' = \sqbrk {a'}_n$ be $\mathbf A$ with rows $r$ and $s$ transposed.
By the definition of a determinant:
:$\displaystyle \map \det {\mathbf A'} = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a'_{k \map \lambda k} }$
By Permutation of Determinant Indices:
:$\displaystyle \map \det {\mathbf A'} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }$
We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:
:$\displaystyle \map \det {\mathbf A'} = \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} } = -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$
Hence the result.
Category:Determinant with Rows Transposed
Determinant with Unit Element in Otherwise Zero Column

== Theorem ==
Let $D$ be the determinant:
:$D = \begin{vmatrix}
 1 & b_{12} & \cdots & b_{1n} \\
 0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
 0 & b_{n2} & \cdots & b_{nn}
\end{vmatrix}$
Then:
:$D = \begin{vmatrix}
 b_{22} & \cdots & b_{2n} \\
\vdots & \ddots & \vdots \\
 b_{n2} & \cdots & b_{nn}
\end{vmatrix}$
== Proof ==
We note that:
:$D = \begin{vmatrix}
 1 & b_{12} & \cdots & b_{1n} \\
 0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
 0 & b_{n2} & \cdots & b_{nn}
\end{vmatrix}$
is the transpose of:
:$D^\intercal = \begin{vmatrix}
 1 & 0 & \cdots & 0 \\
 b_{12} & b_{22} & \cdots & b_{n2} \\
\vdots & \vdots & \ddots & \vdots \\
 b_{1n} & b_{2n} & \cdots & b_{nn}
\end{vmatrix}$
From Determinant with Unit Element in Otherwise Zero Row:
:$D^\intercal = \begin{vmatrix}
 b_{22} & \cdots & b_{n2} \\
\vdots & \ddots & \vdots \\
 b_{2n} & \cdots & b_{nn}
\end{vmatrix}$
The result follows by Determinant of Transpose.
== Also see ==
* Determinant with Unit Element in Otherwise Zero Row
Category:Determinants
Determinant with Unit Element in Otherwise Zero Row

== Theorem ==
Let $D$ be the determinant:
:$D = \begin {vmatrix}
 1 & 0 & \cdots & 0 \\
b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n n}
\end {vmatrix}$
Then:
:$D = \begin {vmatrix}
b_{2 2} & \cdots & b_{2 n} \\
 \vdots & \ddots & \vdots \\
b_{n 2} & \cdots & b_{n n}
\end {vmatrix}$
== Proof ==
We refer to the elements of:
:$\begin {vmatrix}
 1 & 0 & \cdots & 0 \\
b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n n}
\end {vmatrix}$
as $\begin {vmatrix} b_{i j} \end {vmatrix}$.
Thus $b_{1 1} = 1, b_{1 2} = 0, \ldots, b_{1 n} = 0$.
Then from the definition of determinant:
Now we note:
So only those permutations on $\N^*_n$ such that $\map \lambda 1 = 1$ contribute towards the final summation.
Thus we have:
:$\ds D = \sum_\mu \map \sgn \mu b_{2 \map \mu 2} \cdots b_{n \map \mu n}$
where $\mu$ is the collection of all permutations on $\N^*_n$ which fix $1$.
Hence the result.
== Also see ==
* Determinant with Unit Element in Otherwise Zero Column
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercises -- Second Set
Category:Determinants
Deterministic Time Series/Examples/Arbitrary Cosine Function

== Example of Deterministic Time Series ==
Consider the time series $T$ whose future values are determined by the function:
:$z_t = \map \cos {2 \pi f t}$
for some constant $f$.
Then $T$ is a deterministic time series.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.1$ Time Series and Stochastic Processes: Deterministic and statistical time series
Category:Deterministic Time Series
Dewey Decimal Classification for Number Theory

== Observation ==
The Dewey decimal classification for number theory used to be $512 \cdotp 81$.
Thus we have:
Unfortunately, since $1967$ it was changed to $512 \cdotp 73$, which needs some ingenious manipulation to demonstrate its numerological significance.
== Sources ==
* 
* : $512 \cdotp 73$
* : $512 \cdotp 73$
Category:Numerology
Dewey Decimal Classification for Numerology

== Observation ==
The Dewey decimal classification for numerology is $133 \cdotp 335$.
If you reverse it and add it to itself, you get:
:$666 \cdotp 666$
Spoo-kee ...
== Sources ==
* 
* : $133 \cdotp 335$
* : $133 \cdotp 335$
Category:Numerology
Diaconescu-Goodman-Myhill Theorem

== Theorem ==
The axiom of choice implies the law of excluded middle.
== Proof ==
Let $\mathbb B = \set {0, 1}$. 
Let $p$ be a proposition.
Let the following two sets be defined:
:$A = \set {x \in \mathbb B: x = 0 \lor p}$
:$B = \set {x \in \mathbb B: x = 1 \lor p}$
where $\lor$ denotes the disjunction operator.
We have that:
:$0 \in A$
and:
:$1 \in B$
so both $A$ and $B$ are non-empty
Then the set:
:$X = \set {A, B}$
is a set of non-empty sets.
By the axiom of choice, there exists a choice function:
:$f: X \to \mathbb B$
since $\ds \bigcup X = \mathbb B$.
There are four cases:
:$(1): \quad \map f A = \map f B = 0$
This means that $0 \in B$.
But for that to happen, $\paren {0 = 1} \vee p$ must be true.
So by Disjunctive Syllogism, $p$ is true.
:$(2): \quad \map f A = \map f B = 1$
This means that $1 \in A$.
Arguing similarly to case $(1)$, it follows that $p$ is true in this case also.
:$(3): \quad \map f A = 1 \ne \map f B = 0$
This means that $A \ne B$ (or otherwise $f$ would pick the same element).
But if $p$ is true, that means:
:$A = B = \mathbb B$
which is a contradiction.
Therefore in this case:
:$\neg p$
:$(4): \quad \map f A = 0 \ne \map f B = 1$
Using the same reasoning as in case $(3)$, it is seen that in this case:
:$\neg p$
So by Proof by Cases:
:$\paren {p \vee \neg p}$
That is the Law of Excluded Middle.
== Also known as ==
The Diaconescu-Goodman-Myhill Theorem is also known as Diaconescu's Theorem‎ and the Goodman-Myhill Theorem.
It is ever 's endeavour to attest a theorem to as many contributors as appropriate.
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Set Theory
Category:Axiom of Choice
Diaconescu-Goodman-Myhill Theorem/Historical Note

== Historical Note on Diaconescu-Goodman-Myhill Theorem ==
The proof of the Diaconescu-Goodman-Myhill Theorem was first published in $1975$ by .
It was later independently rediscovered by and and published in $1978$.
However, the first appearance of the result itself was in 's $1967$ work , where he set it as an exercise, without including a solution.
Category:Historical Notes
Diagonal Complement Relation Compatible with Group Operation

== Theorem ==
Let $\left({G, \circ}\right)$ be a group.
Let $\Delta_G$ be the diagonal relation on $G$.
Then $\Delta_G^c = \complement_{G \times G} \Delta_G$ is a relation compatible with $\circ$.
In other words, $\ne$ is a relation compatible with $\circ$.
== Proof ==
By Diagonal Relation is Universally Compatible, $\Delta_G$ is compatible with $\circ$.
By Complement of Relation Compatible with Group is Compatible , $\Delta_G^c$ is also compatible with $\circ$.
Category:Compatible Relations
Diagonal Lemma

== Theorem ==
Let $T$ be the set of theorems of some theory in the language of arithmetic which contains minimal arithmetic.
For any formula $\map B y$ in the language of arithmetic, there is a sentence $G$ such that
:$T \vdash G \leftrightarrow \map B {\hat G}$
where $\hat G$ is the Gödel number of $G$ (more accurately, it is the term in the language of arithmetic obtained by applying the function symbol $s$ to $0$ this many times).
== Proof ==
There is a primitive recursive function $\mathrm {diag}$ which is defined by:
:$\map {\mathrm {diag} } n = \widehat {\map A {\hat A} }$
where:
:$\map A x$ is the formula such that $\hat A = n$.
:the $\hat {}$ sign denotes the Gödel number of the contained formula (and we are not being formal about distinguishing between integers and symbols in the language).
Informally, $\mathrm {diag}$ takes a Gödel number, decodes it to a formula, plugs in the Gödel number for that formula in place of a free variable, and encodes this new formula back to a new Gödel number.
Since $T$ contains $Q$, by Recursive Sets are Definable in Arithmetic applied to the graph of $\mathrm {diag}$, we have that there is some formula $\map {\mathrm {Diag} } {x, y}$ which defines the graph of $\mathrm {diag}$ in $T$.
That is:
:$\map {\mathrm {diag} } n = m$ $T \vdash \map {\mathrm {Diag} } {n, m}$
Let $\map A x$ be the formula:
:$\exists y \paren {\map {\mathrm {Diag} } {x, y} \land \map B y}$
Let $G$ be $\map A {\hat A}$.
We then have $T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$, by checking the definitions.
Let $T' = T \cup \set G$.
Then:
:$T' \vdash \map A {\hat A}$
Hence:
:$T' \vdash \exists y \paren {\map {\mathrm {Diag} } {\hat A, y} \land \map B y}$
But since $\hat G$ is the only number such that $T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$, this gives us:
:$T' \vdash \map B {\hat G}$
Thus:
:$T \vdash G \rightarrow \map B {\hat G}$
Let $T' = T \cup \set {\map B {\hat G} }$.
Again, we have:
:$T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$
so this gives us:
:$T' \vdash \map {\mathrm {Diag} } {\hat A, \hat G} \land \map B {\hat G}$
and hence:
:$T' \vdash \exists y \paren {\map {\mathrm {Diag} } {\hat A, y} \land \map B y}$
But this is the same thing as:
:$T' \vdash G$
Thus:
:$T \vdash \map B {\hat G} \rightarrow G$
Thus $G$ is as claimed.
== Sources ==
* 
Category:Model Theory for Predicate Logic
Diagonal Matrix is Symmetric

== Theorem ==
Let $D$ be a diagonal matrix.
Then $D$ is symmetric.
== Proof ==
By definition of diagonal matrix:
:$\forall j, k: j \ne k \implies a_{jk} = 0 = a_{kj}$
So by definition of transpose of $D$:
:$D = D^\intercal$
where $D^\intercal$ denotes the transpose.
Hence the result, by definition of symmetric matrix.
Category:Symmetric Matrices
Category:Diagonal Matrices
Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation

== Theorem ==
Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\Bbb K$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed $2$-sequence space.
Let $\mathbf x = \tuple {a_1, a_2, a_3, \ldots} \in \ell^2$.
Suppose $\Lambda : \ell^2 \to \ell^2$ is a diagonal operator such that:
:$\Lambda \tuple {a_1, a_2, a_3, \ldots} = \tuple {\lambda_1 a_1, \lambda_2 a_2, \lambda_3 a_3, \ldots}$
Then $\Lambda \in \map {CL} {\ell^2}$.
== Proof ==
=== Linearity ===
Let $\sequence {a_n}_{n \mathop \in \N_{> 0} }, \sequence {b_n}_{n \mathop \in \N_{> 0} } \in \ell^2$.
By linearity of transformations:
:$\Lambda \in \map \LL {\ell^2}$
=== Continuity ===
For $\tuple {a_n}_{n \mathop \in \N_{> 0} }$:
By Continuity of Linear Transformation between Normed Vector Spaces:
:$\Lambda \in \map C {\ell^2}$.
By definition:
:$\Lambda \in \map {CL} {\ell^2}$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Linear Transformations
Category:Continuous Linear Transformations
Category:Continuous Mappings
Diagonal Relation is Equivalence

== Theorem ==
The diagonal relation $\Delta_S$ on a set $S$ is always an equivalence in $S$.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
=== Symmetric ===
=== Transitive ===
== Examples ==
== Sources ==
* : $\S 7$: Relations
* : $\S 2.2$. Equivalence relations: Example $29$
* : Chapter $\text{I}$: Sets and Functions: Relations
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers
* : $\S 6$. Indexed families; partitions; equivalence relations: Example $6.7$
* : Appendix $\text{A}.3$: Equivalence Relations: $(1)$
* : $\S 16$: Equivalence relations
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations
* : Chapter $2$: 'And you do addition?': $\S 2.3$: Relations: Example $2.3.3$
Category:Diagonal Relation is Equivalence
Category:Diagonal Relation
Category:Examples of Equivalence Relations
Category:Equality
Diagonal Relation is Equivalence/Examples/Integers

== Examples of Use of Diagonal Relation is Equivalence ==
Let $\Z$ denote the set of integers.
Let $\RR$ denote the relation on $\Z$ defined as:
:$\forall x, y \in \Z: x \mathrel \RR y \iff x = y$
Then $\RR$ is an equivalence relation such that the equivalence classes are singletons.
== Proof ==
This is an instance of Diagonal Relation is Equivalence.
The result follows from Equivalence Classes of Diagonal Relation.
== Sources ==
* : Chapter $1$: Preliminaries: Exercises: $1.6 \ \text {a)}$
Category:Examples of Use of Diagonal Relation is Equivalence
Diagonal Relation is Equivalence/Examples/Numbers

== Examples of Use of Diagonal Relation is Equivalence ==
Let $\SS$ denote the set of numbers.
Let $\RR$ denote the relation on $S$ defined as:
:$\forall x, y \in S: x \mathrel \RR y \iff x = y$
Then $\RR$ is an equivalence relation such that the equivalence classes are singletons.
== Proof ==
This is an instance of Diagonal Relation is Equivalence.
The result follows from Equivalence Classes of Diagonal Relation.
== Sources ==
* 
* 
Category:Examples of Use of Diagonal Relation is Equivalence
Diagonal Relation is Left Identity

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Then:
:$\Delta_T \circ \RR = \RR$
where $\Delta_T$ is the diagonal relation on $T$, and $\circ$ signifies composition of relations.
== Proof ==
We use the definition of relation equality, as follows:
=== Equality of Domains ===
The domains of $\RR$ and $\Delta_T \circ \RR$ are equal from Domain of Composite Relation:
:$\Dom {\Delta_T \circ \RR} = \Dom \RR$
=== Equality of Codomains ===
From Codomain of Composite Relation, the codomains of $\Delta_T \circ \RR$ and $\Delta_T$ are both equal to $T$.
But from the definition of the diagonal relation, the codomain of $\Delta_T$ is $\Dom {\Delta_T} = T$.
=== Equality of Relations ===
The composite of $\RR$ and $\Delta_T$ is defined as:
:$\Delta_T \circ \RR = \set {\tuple {x, z} \in S \times T: \exists y \in T: \tuple {x, y} \in \RR \land \tuple {y, z} \in \Delta_T}$
But by definition of the diagonal relation on $T$, we have that:
:$\tuple {y, z} \in \Delta_T \implies y = z$
Hence:
:$\Delta_T \circ \RR = \set {\tuple {x, y} \in S \times T: \exists y \in T: \tuple {x, y} \in \RR \land \tuple {y, y} \in \Delta_T}$
But as $\forall y \in T: \tuple {y, y} \in \Delta_T$, this means:
:$\Delta_T \circ \RR = \set {\tuple {x, y} \in S \times T: \tuple {x, y} \in \RR}$
That is:
:$\Delta_T \circ \RR = \RR$
Hence the result.
== Also see ==
* Diagonal Relation is Right Identity
* Identity Mapping is Left Identity
* Identity Mapping is Right Identity
== Sources ==
* : Chapter $0$: Relations
* : $\S 10$: Inverses and Composites
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(d)}$
Category:Diagonal Relation
Category:Identity Mappings
Diagonal Relation is Many-to-One

== Theorem ==
The diagonal relation is many-to-one.
That is:
:$\forall x \in \Dom {\Delta_S}: \tuple {x, y_1} \in \Delta_S \land \tuple {x, y_2} \in \Delta_S \implies y_1 = y_2$
where $\Delta_S$ is the diagonal relation on a set $S$.
== Proof ==
Let $S$ be a set and let $\Delta_S$ be the diagonal relation on $S$.
Let $\tuple {x, y_1} \in \Delta_S \land \tuple {x, y_2} \in \Delta_S$.
From the definition of the diagonal relation:
:$\tuple {x, y_1} = \tuple {x, x}$
:$\tuple {x, y_2} = \tuple {x, x}$
and so $y_1 = y_2$.
== Also see ==
* Definition:Diagonal Mapping
== Sources ==
* : $\S 4$. Relations; functional relations; mappings: Example $4.5$
Category:Diagonal Relation
Category:Many-to-One Relations
Diagonal Relation is Ordering and Equivalence

== Theorem ==
Let $\struct {S, \Delta_S}$ be a relational structure where $\Delta_S$ is the diagonal relation, defined as:
:$\forall x, y \in S: \paren {x, y} \in \Delta_S \iff x = y$
Then $\Delta_S$ is the only relation on $S$ which is both an equivalence and an ordering.
== Proof ==
From Trivial Ordering is Universally Compatible and Diagonal Relation is Equivalence we know that the diagonal relation possesses these properties.
We now need to show that it is the only relation on $S$ which possesses these properties.
Both an equivalence and an ordering are reflexive and transitive.
Also:
:An equivalence relation is symmetric
:An ordering is antisymmetric.
The result follows from Relation is Symmetric and Antisymmetric iff Coreflexive.
== Sources ==
* : $\S 7$: Exercise $1$
Category:Examples of Equivalence Relations
Category:Order Theory
Diagonal Relation is Reflexive (Class Theory)

== Theorem ==
Let $V$ be a basic universe.
Let $\Delta_V$ denote the diagonal relation on $V$:
:$\Delta_V = \set {\tuple {x, x}: x \in V}$
$\Delta_V$ is a reflexive relation.
== Proof ==
So $\Delta_V$ is reflexive.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering
Category:Diagonal Relation
Category:Examples of Reflexive Relations
Diagonal Relation is Right Identity

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Then:
:$\RR \circ \Delta_S = \RR$
where $\Delta_S$ is the diagonal relation on $S$, and $\circ$ signifies composition of relations.
== Proof ==
We use the definition of relation equality, as follows:
=== Equality of Codomains ===
The codomains of $\RR$ and $\RR \circ \Delta_S$ are both equal to $T$ from Codomain of Composite Relation.
=== Equality of Domains ===
The domains of $\RR$ and $\RR \circ \Delta_S$ are also easily shown to be equal.
From Domain of Composite Relation:
:$\Dom {\RR \circ \Delta_S} = \Dom {\Delta_S}$
But from the definition of the diagonal relation:
:$\Dom {\Delta_S} = \Img {\Delta_S} = S$
=== Equality of Relations ===
The composite of $\Delta_S$ and $\RR$ is defined as:
:$\RR \circ \Delta_S = \set {\tuple {x, z} \in S \times T: \exists y \in S: \tuple {x, y} \in \Delta_S \land \tuple {y, z} \in \RR}$
But by definition of the diagonal relation on $S$, we have that:
:$\tuple {x, y} \in \Delta_S \implies x = y$
Hence:
:$\RR \circ \Delta_S = \set {\tuple {y, z} \in S \times T: \exists y \in S: \tuple {y, y} \in \Delta_S \land \tuple {y, z} \in \RR}$
But as $\forall y \in S: \tuple {y, y} \in \Delta_S$, this means:
:$\RR \circ \Delta_S = \set {\tuple {y, z} \in S \times T: \tuple {y, z} \in \RR}$
That is:
:$\RR \circ \Delta_S = \RR$
Hence the result.
== Also see ==
* Diagonal Relation is Left Identity
* Identity Mapping is Right Identity
* Identity Mapping is Left Identity
== Sources ==
* : Chapter $0$: Relations
* : $\S 10$: Inverses and Composites
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(d)}$
Category:Diagonal Relation
Category:Identity Mappings
Diagonal Relation is Serial

== Theorem ==
Let $S$ be a set.
Let $\Delta_S$ be the diagonal relation on $S$.
Then $\Delta_S$ is a serial relation.
== Proof ==
By Diagonal Relation is Equivalence it follows that $\Delta_S$ is reflexive.
The result follows from Reflexive Relation is Serial.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $2 \ \text{(a)}$
Category:Serial Relations