avewright/proofwiki-math huggingface dataset:
Theorem: Power Set of Singleton

Let $x$ be an object.
Then the power set of the singleton $\set x$ is:
:$\powerset {\set x} = \set {\O, \set x}$

Proof:
From Empty Set is Subset of All Sets:
:$\O \in \powerset {\set x}$
Let $A \in \powerset {\set x}$ such that $A \ne \O$
That is:
{{begin-eqn}}
{{eqn | r = A \subseteq \set x \land A \ne \O
      | o =
      | c = 
}}
{{eqn | ll= \leadsto
      | r = A \subseteq \set x \land \exists y : y \in A
      | o =
      | c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadsto
      | r = A \subseteq \set x \land \exists y : y \in A \land y \in \set x
      | o =
      | c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
      | r = A \subseteq \set x \land \exists y : y \in A \land y = x
      | o =
      | c = {{Defof|Singleton}}
}}
{{eqn | ll= \leadsto
      | r = A \subseteq \set x \land x \in A
      | o =
      | c =
}}
{{eqn | ll= \leadsto
      | r = A \subseteq \set x \land \set x \subseteq A
      | o =
      | c = Singleton of Element is Subset
}}
{{eqn | ll= \leadsto
      | r = A = \set x
      | o =
      | c = {{Defof|Set Equality}}
}}
{{end-eqn}}
So a subset of $\set x$ is either $\O$ or $\set x$.
{{qed}}
Category:Power Set
Category:Singletons
0z1pqmxvgh8yira067g3l8s1prfj25s
Theorem: Row Operation to Clear First Column of Matrix

Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over a field $K$.
Then there exists a row operation to convert $\mathbf A$ into another $m \times n$ matrix $\mathbf B = \sqbrk b_{m n}$ with the following properties:
:$(1): \quad$ Except possibly for element $b_{1 1}$, all the elements of column $1$ are $0$
:$(2): \quad$ If $b_{1 1} \ne 0$, then $b_{1 1} = 1$.
This process is referred to as '''clearing the first column'''.

Proof:
The following algorithm generates a sequence of elementary row operations which convert $\mathbf A$ to $\mathbf B$.
Let $\mathbf A' = \sqbrk {a'}_{m n}$ denote the state of $\mathbf A$ after having processed the latest step.
After each step, an implicit step can be included that requires that the form of $\mathbf A'$ is inspected to see if it is in the form $\mathbf B$, and if so, terminating the algorithm, but this is not essential.
:$(1): \quad$ Are all elements in the first column of $\mathbf A$ equal to $0$?
:::If so, there is nothing to do, and the required row operation is the unit matrix $\mathbf I_m$.
:::Otherwise, move on to step $(2)$.
:$(2): \quad$ Is element $a_{1 1}$ equal to $0$?
:::If so:
::::$\text (a): \quad$ find the smallest $k$ such that row $k$ of $\mathbf A$ such that $a_{k 1} \ne 0$
::::$\text (b): \quad$ use the elementary row operation $r_1 \leftrightarrow r_k$ which will result $a'_{1 1} = a_{k 1}$ and $a'_{k 1} = 0$.
:Move on to step $(3)$.
:$(3): \quad$ Is element $a'_{1 1}$ equal to $1$?
:::If so, use the elementary row operation $r_1 \to \lambda r_1$ where $\lambda = \dfrac 1 {a'_{1 1} }$, which will result $a'_{1 1} = 1$.
:Move on to step $4$
:$(4): \quad$ For each row $j$ from $2$ to $m$, do the following:
:::Is $a_{j 1} \ne 0$?
::::If so, use the elementary row operation $r_j \leftrightarrow r_j + \mu r_1$, where $\mu = -\dfrac {a'_{j 1} } {a'{1 1} }$, which will result in $a'_{j 1} = 0$.
This will result in an $m \times n$ matrix in the required form.
Exercising the above algorithm will have generated a sequence of elementary row operations $e_1, e_2, \ldots, e_t$.
For each $e_k$ we create the elementary row matrix $\mathbf E_k$.
We then assemble the matrix product:
:$\mathbf R := \mathbf E_t \mathbf E_{t - 1} \mathbf E_{t - 2} \dotsm \mathbf E_2 \mathbf E_1$
From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, $\mathbf R$ is the resulting $m \times m$ matrix corresponding to the row operation which is used to convert $\mathbf A$ to $\mathbf B$.
{{qed}}
Theorem: Matrix is Row Equivalent to Echelon Matrix

Let $\mathbf A = \sqbrk a_{m n}$ be a matrix of order $m \times n$ over a field $F$.
Then $A$ is row equivalent to an echelon matrix of order $m \times n$.

Proof:
Using the operation Row Operation to Clear First Column of Matrix, $\mathbf A$ is converted to $\mathbf B$, which will be in the form:
:<nowiki>$\begin{bmatrix}
0      & \cdots &      0 &      1 & b_{1, j + 1} & \cdots & b_{1 n} \\
0      & \cdots &      0 &      0 & b_{2, j + 1} & \cdots & b_{2 n} \\
\vdots & \ddots & \vdots & \vdots &       \vdots & \ddots &  \vdots \\
0      & \cdots &      0 &      0 & b_{m, j + 1} & \cdots & b_{m n} \\
\end{bmatrix}$</nowiki>
If some zero rows have appeared, do some further elementary row operations, that is row interchanges, to put them at the bottom.
We then address our attention to the submatrix:
:<nowiki>$\begin{bmatrix}
b_{2, j + 1} & b_{2, j + 2} & \cdots & b_{2 n} \\
b_{3, j + 1} & b_{3, j + 2} & \cdots & b_{3 n} \\
      \vdots &       \vdots & \ddots &  \vdots \\
b_{m, j + 1} & b_{m, j + 2} & \cdots & b_{m n} \\
\end{bmatrix}$</nowiki>
and perform the same operation on that.
This results in the submatrix being transformed into the form:
:<nowiki>$\begin{bmatrix}
     1 & c_{2, j + 2} & \cdots & c_{2 n} \\
     0 & c_{3, j + 2} & \cdots & c_{3 n} \\
\vdots &       \vdots & \ddots &  \vdots \\
     0 & c_{m, j + 2} & \cdots & c_{m n} \\
\end{bmatrix}$</nowiki>
Again, we process the submatrix:
:<nowiki>$\begin{bmatrix}
c_{3, j + 2} & \cdots & c_{3 n} \\
      \vdots & \ddots &  \vdots \\
c_{m, j + 2} & \cdots & c_{m n} \\
\end{bmatrix}$</nowiki>
Thus we progress, until the entire matrix is in echelon form.
{{Qed}}
Theorem: Singleton is Independent implies Rank is One/Corollary

:$\set x$ is an independent subset {{iff}} $\map \rho {\set x} = 1$

Proof:
By definition of an independent subset:
:$x$ is an independent subset {{iff}} $\set x \notin \mathscr I$
From Singleton is Independent implies Rank is One:
:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
From Singleton is Dependent implies Rank is Zero:
:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
It follows that:
:$\set x \in \mathscr I$ {{iff}} $\map \rho {\set x} = 1$
{{qed}}
Category:Matroid Independent Subsets
Category:Matroid Rank Functions
3d96so20xxbjlwkhn0pd5heh9xp5mkn
Theorem: Singleton is Independent implies Rank is One

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x \in S$.
Let $\set x$ be independent.
Then:
:$\map \rho {\set x} = 1$
where $\rho$ denotes the rank function of $M$.

Proof:
From Rank of Independent Subset Equals Cardinality:
:$\map \rho {\set x} = \size {\set x}$
From Cardinality of Singleton:
:$\size {\set x} = 1$
The result follows.
{{qed}}
Category:Matroid Independent Subsets
Category:Matroid Rank Functions
mo4v8tljdjkn871iojpx49jz0d441ih
Theorem: Singleton is Dependent implies Rank is Zero

:$\map \rho {\set x} = 0$

Proof:
By definition of a dependent subset:
:$\set x \notin \mathscr I$
Then:
{{begin-eqn}}
{{eqn | l = \map \rho {\set x}
      | r = \max \set{\size A : A \in \powerset {\set x} \land A \in \mathscr I} 
      | c = {{Defof|Rank Function (Matroid)|Rank Function}}
}}
{{eqn | r = \max \set {\size A : A \in \set {\O, \set x} \land A \in \mathscr I} 
      | c = Power Set of Singleton
}}
{{eqn | r = \max \set {\size \O}
      | c = As $\set x \notin \mathscr I$ and Matroid axiom $(\text I 1)$ : $\O \in \mathscr I$
}}
{{eqn | r = \max \set 0
      | c = Cardinality of Empty Set
}}
{{eqn | r = 0
      | c = {{Defof|Max Operation}}
}}
{{end-eqn}}
{{qed}}
Category:Matroid Dependent Subsets
Category:Matroid Rank Functions
a5zbinu9qv767k5icxzie4w3xp442xd
Theorem: System of Simultaneous Equations may have No Solution

Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is empty.

Proof:
Consider this system of simultaneous linear equations:
{{begin-eqn}}
{{eqn | n = 1
      | l = x_1 + x_2
      | r = 2
}}
{{eqn | n = 2
      | l = 2 x_1 + 2 x_2
      | r = 3
}}
{{end-eqn}}
From its evaluation it is seen to have no solutions.
Hence the result.
{{qed}}
Theorem: System of Simultaneous Equations may have Unique Solution

Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is a singleton.

Proof:
Consider this system of simultaneous linear equations:
{{begin-eqn}}
{{eqn | n = 1
      | l = x_1 - 2 x_2 + x_3
      | r = 1
}}
{{eqn | n = 2
      | l = 2 x_1 - x_2 + x_3
      | r = 2
}}
{{eqn | n = 3
      | l = 4 x_1 + x_2 - x_3
      | r = 1
}}
{{end-eqn}}
From its evaluation it has the following unique solution:
{{begin-eqn}}
{{eqn | l = x_1
      | r = -\dfrac 1 2
}}
{{eqn | l = x_2
      | r = \dfrac 1 2
}}
{{eqn | l = x_3
      | r = \dfrac 3 2
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: System of Simultaneous Equations may have Multiple Solutions

Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is a singleton.

Proof:
Consider this system of simultaneous linear equations:
{{begin-eqn}}
{{eqn | n = 1
      | l = x_1 - 2 x_2 + x_3
      | r = 1
}}
{{eqn | n = 2
      | l = 2 x_1 - x_2 + x_3
      | r = 2
}}
{{end-eqn}}
From its evaluation it has the following solutions:
{{begin-eqn}}
{{eqn | l = x_1
      | r = 1 - \dfrac t 3
}}
{{eqn | l = x_2
      | r = \dfrac t 3
}}
{{eqn | l = x_3
      | r = t
}}
{{end-eqn}}
where $t$ is any number.
Hence the are as many solutions as the cardinality of the domain of $t$.
{{qed}}
Theorem: Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then $X$ is a Banach space {{iff}}:
:every absolutely convergent series $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.

Proof:
=== Necessary Condition ===
{{:Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach/Necessary Condition}}{{qed|lemma}}
Theorem: Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach/Necessary Condition

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $X$.
Suppose $X$ is a Banach space.
Then $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.

Proof:
That $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\ds \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges in $\R$.
Hence by Convergent Sequence in Normed Vector Space is Cauchy Sequence:
:the sequence of partial sums is a Cauchy sequence.
Now let $\epsilon > 0$.
Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:
:$\ds \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$
This $N$ exists because the sequence is Cauchy.
Now observe that, for $m \ge n \ge N$, one also has:
{{begin-eqn}}
{{eqn | l = \norm {\sum_{k \mathop = 1}^m a_k - \sum_{k \mathop = 1}^n a_k}
      | r = \norm {\sum_{k \mathop = n + 1}^m a_k}
}}
{{eqn | o = \le
      | r = \sum_{k \mathop = n + 1}^m \norm {a_k}
      | c = Triangle inequality for $\norm {\, \cdot \,}$
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
It follows that the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$ is Cauchy.
As $X$ is a Banach space, this implies that $\ds \sum_{n \mathop = 1}^\infty a_n$ converges.
Theorem: Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach/Sufficient Condition

Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $X$.
Suppose $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
Then $X$ is a Banach space.

Proof:
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $X$.
We have that:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m} < \epsilon$
We will prove the existence of a subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ such that:
:$n > n_k \implies \norm {x_n - x_{n_k} } < \dfrac 1 {2^k}$
=== Basis for the Induction ===
Let $\epsilon = \dfrac 1 2$.
Choose $n_1$ such that:
:$n, m \ge n_1 \implies \norm {x_n - x_m} < \dfrac 1 2$
In particular, when $m = n_1$:
:$n \ge n_1 \implies \norm {x_n - x_{n_1} } < \dfrac 1 2$
=== Induction Step ===
Suppose $x_{n_1}, \dots x_{n_k}$ have been constructed.
Let $\epsilon = \dfrac 1 {2^{k + 1} }$.
Choose $n_{k + 1}$ such that $n_{k + 1} > n_k$ and:
:$n, m \ge n_{k + 1} \implies \norm {x_n - x_m} < \dfrac 1 {2^{k + 1} }$
In particular, when $m = n_{k + 1}$:
:$n \ge n_{k + 1} \implies \norm {x_n - x_{n_{k + 1} } } < \dfrac 1 {2^{k + 1}}$
{{qed|lemma}}
Define:
:$u_1 := x_{n_1}$
:$u_{k + 1} := x_{n_{k + 1} } - x_{n_k}$
Now we have a sequence $\sequence {u_k}_{k \mathop \in \N}$.
Consider the series $\ds \sum_{k \mathop = 1}^\infty \norm {u_k}$:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^\infty \norm {u_k}
      | r = \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {u_k}
}}
{{eqn | r = \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {x_{n_k} - x_{n_{k - 1} } }
}}
{{eqn | r = \norm {u_1} + \sum_{k \mathop = 1}^\infty \norm {x_{n_{k + 1} } - x_{n_k} }
      | c = Relabeling: $k \to k + 1$
}}
{{eqn | o = \le
      | r = \norm {x_{n_1} } + \sum_{k \mathop = 1}^\infty \frac 1 {2^k}
}}
{{eqn | o = \le
      | r = \norm {x_{n_1} } + 1
}}
{{eqn | o = <
      | r = \infty
}}
{{end-eqn}}
Thus, $\ds \sum_{k \mathop = 1}^\infty u_k$ is absolutely convergent.
By assumption in the theorem, $\ds \sum_{k \mathop = 1}^\infty u_k$ is convergent.
In other words:
:$\ds \lim_{k \mathop \to \infty} \sum_{j \mathop = 1}^k u_j = u$.
On the other hand:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^k u_j
      | r = x_{n_1} + \sum_{j \mathop = 2}^k \paren {x_{n_j} - x_{n_{j \mathop - 1} } }
}}
{{eqn | r = x_{n_k}
      | c = Telescoping Series
}}
{{end-eqn}}
Therefore:
:$\ds \lim_{k \mathop \to \infty} x_{n_k} = u =: x$
So $\sequence {x_{n_k} }_{k \mathop \in \N}$ converges in $X$.
We have that $\sequence {x_{n_k} }_{k \mathop \in \N}$ is a convergent subsequence of a Cauchy sequence$\sequence {x_n}_{n \mathop \in \N}$.
By Convergent Subsequence of Cauchy Sequence in Normed Vector Space, $\sequence {x_n}_{n \mathop \in \N}$ is convergent with the same limit $x$.
By definition, the underlying space is Banach.
Theorem: Sine of Integer Multiple of Argument/Formulation 2

{{begin-eqn}}
{{eqn | l = \sin n \theta
      | r = \cos^n \theta \paren {\dbinom n 1 \paren {\tan \theta} - \dbinom n 3 \paren {\tan \theta}^3 + \dbinom n 5 \paren {\tan \theta}^5 - \cdots}
      | c = 
}}
{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\tan^{2 k + 1} \theta}
      | c = 
}}
{{end-eqn}}

Proof:
By De Moivre's Formula:
:$\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$
As $n \in \Z_{>0}$, we use the Binomial Theorem on the {{RHS}}, resulting in:
:$\ds \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$
When $k$ is odd, the expression being summed is imaginary.
Equating the imaginary parts of both sides of the equation, replacing $k$ with $2 k + 1$ to make $k$ odd, gives:
{{begin-eqn}}
{{eqn | l = \sin n \theta
      | r = \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\cos^{n - \paren {2 k + 1} } \theta} \paren {\sin^{2 k + 1} \theta}
      | c = 
}}
{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\tan^{2 k + 1} \theta}
      | c = factor out $\cos^n \theta$
}}
{{end-eqn}}
{{qed}}
Theorem: Trivial Solution to System of Homogeneous Simultaneous Linear Equations is Solution

Let $S$ be a '''system of homogeneous simultaneous linear equations''':
:$\ds \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = 0$
Consider the trivial solution to $A$:
:$\tuple {x_1, x_2, \ldots, x_n}$
such that:
:$\forall j \in \set {1, 2, \ldots, n}: x_j = 0$
Then the trivial solution is indeed a solution to $S$.

Proof:
Let $i \in \set {1, 2, \ldots, m}$.
We have:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \alpha_{i j} x_j
      | r = \sum_{j \mathop = 1}^n \alpha_{i j} \times 0
      | c = 
}}
{{eqn | r = \sum_{j \mathop = 1}^n 0
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
This holds for all $i \in \set {1, 2, \ldots, m}$.
Hence:
:$\ds \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = 0$
and the result follows.
{{qed}}
Category:Simultaneous Linear Equations
trp5jkqh1tfl9xj9dp818w8llihagbq
Theorem: Sine of Integer Multiple of Argument/Formulation 3

{{begin-eqn}}
{{eqn | l = \sin n \theta
      | r = \sin \theta \cos^{n - 1} \theta \paren {1 + 1 + \frac {\cos 2 \theta} {\cos^2 \theta} + \frac {\cos 3 \theta} {\cos^3 \theta} + \cdots + \frac {\cos \paren {n - 1} \theta} {\cos^{n - 1} \theta} }
      | c = 
}}
{{eqn | r = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta}
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop \ge 0} \frac {\cos k \theta} {\cos^k \theta}$
Theorem: Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations

Let $S$ be a system of simultaneous linear equations:
:$\ds \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$
Let $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ denote the augmented matrix of $S$.
Let $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$ be obtained from $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ by means of an elementary row operation.
Let $S'$ be the system of simultaneous linear equations of which $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$ is the augmented matrix.
Then $S$ and $S'$ are equivalent.

Proof:
We have that an elementary row operation $e$ is used to transform $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ to $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$.
Now, whatever $e$ is, $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$ is the augmented matrix of a system of simultaneous linear equations $S'$.
We investigate each type of elementary row operation in turn.
In the below, let:
:$r_k$ denote row $k$ of $\mathbf A$
:$r'_k$ denote row $k$ of $\mathbf A'$
for arbitrary $k$ such that $1 \le k \le m$.
By definition of elementary row operation, only the row or rows directly operated on by $e$ is or are different between $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ and $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$.
Hence it is understood that in the following, only those equations corresponding to those rows directly affected will be under consideration.
Theorem: Existence of Inverse Elementary Row Operation

Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be an elementary row operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is an elementary row operation which always exists and is unique.

Proof:
Let us take each type of elementary row operation in turn.
For each $\map e {\mathbf A}$, we will construct $\map {e'} {\mathbf A'}$ which will transform $\mathbf A'$ into a new matrix $\mathbf A' ' \in \map \MM {m, n}$, which will then be demonstrated to equal $\mathbf A$.
In the below, let:
:$r_k$ denote row $k$ of $\mathbf A$
:$r'_k$ denote row $k$ of $\mathbf A'$
:$r' '_k$ denote row $k$ of $\mathbf A' '$
for arbitrary $k$ such that $1 \le k \le m$.
By definition of elementary row operation:
:only the row or rows directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the row or rows directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A' '$.
Hence it is understood that in the following, only those rows directly affected will be under consideration when showing that $\mathbf A = \mathbf A' '$.
Theorem: Superset of Dependent Set is Dependent

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $A, B \subseteq S$ such that $A \subseteq B$
If $A$ is a dependent subset then $B$ is a dependent subset.

Proof:
From the contrapositive statement of matroid axiom $(\text I 2)$:
:$A \notin \mathscr I \implies B \notin \mathscr I$
By the definition of a dependent subset:
:If $A$ is not an dependent subset then $B$ is not an dependent subset.
{{qed}}
Category:Matroid Dependent Subsets
ed2sk63don3viuma9rtyxgx1bx7b39f
Theorem: Powers of 16 Modulo 20

Let $n \in \Z_{> 0}$ be a strictly positive integer.
Then:
:$16^n \equiv 16 \pmod {20}$

Proof:
Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$16^n \equiv 16 \pmod {20}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$16^1 \equiv 16 \pmod {20}$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$16^k \equiv 16 \pmod {20}$
from which it is to be shown that:
:$16^{k + 1} \equiv 16 \pmod {20}$
=== Induction Step ===
This is the induction step:
We have:
{{begin-eqn}}
{{eqn | l = 16^{k + 1}
      | r = 16^k \times 16
}}
{{eqn | o = \equiv
      | r = 16 \times 16
      | rr= \pmod {20}
      | c = Induction Hypothesis
}}
{{eqn | o = \equiv
      | r = 256
      | rr= \pmod {20}
}}
{{eqn | o = \equiv
      | r = 16
      | rr= \pmod {20}
      | c = 
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that:
:$\forall n \in \Z_{> 0}: 16^n \equiv 16 \pmod {20}$
{{qed}}
Theorem: Powers of 16 Modulo 20

Let $n \in \Z_{> 0}$ be a strictly positive integer.
Then:
:$16^n \equiv 16 \pmod {20}$

Proof:
{{begin-eqn}}
{{eqn | l = 16
      | o = \equiv
      | r = 16
      | rr= \pmod {20}
}}
{{eqn | ll= \leadsto
      | l = 16
      | o = \equiv
      | r = 0
      | rr= \pmod 4
}}
{{eqn | lo= \text {and}
      | l = 16
      | o = \equiv
      | r = 1
      | rr= \pmod 5
}}
{{eqn | ll= \leadsto
      | l = 16^n
      | o = \equiv
      | r = 0
      | rr= \pmod 4
}}
{{eqn | lo= \text {and}
      | l = 16^n
      | o = \equiv
      | r = 1
      | rr= \pmod 5
}}
{{eqn | ll= \leadsto
      | l = 16^n
      | o = \equiv
      | r = 16
      | rr= \pmod {20}
      | c = Chinese Remainder Theorem
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of Integer Multiple of Argument/Formulation 1/Lemma

:For $n \in \Z$:
{{begin-eqn}}
{{eqn | l = \map \cos {n \theta} \map \sin {\theta} 
      | r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta} 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \map \cos {n \theta} \map \sin {\theta} 
      | r = \map \cos {n \theta} \map \sin {\theta} 
}}
{{eqn | r = \paren {\map \sin {n \theta} \map \cos {\theta} - \map \sin {n \theta} \map \cos {\theta} } + \map \cos {n \theta} \map \sin {\theta} 
      | c = add zero
}}
{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \paren {\map \sin {n \theta} \map \cos {\theta} - \map \cos {n \theta} \map \sin {\theta} }
      | c = regroup
}}
{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {n \theta - \theta}
      | c = Sine of Difference
}}
{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1} \theta}
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Category:Sine of Integer Multiple of Argument
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Theorem: Conjugacy Class of Identity is only Conjugacy Class which is Subgroup

Let $G$ be a group.
Let $e$ denote the identity of $G$.
Let $\conjclass g$ denote the conjugacy class of the element $g$.
Then conjugacy class of identity is the only conjugacy class which is a subgroup of $G$:
:$\conjclass g < G \iff g = e$

Proof:
=== Necessary Condition ===
Assume $g = e$.
Then by Identity of Group is in Singleton Conjugacy Class, $\conjclass e = \set e$, which is the trivial subgroup.
{{qed|lemma}}