avewright/proofwiki-math huggingface dataset:
Theorem: Empty Group Word is Reduced

Let $S$ be a set
Let $\epsilon$ be the empty group word on $S$.
Then $\epsilon$ is reduced.

Proof:
By definition, a group word $w = w_1 \cdots w_i \cdots w_n$ is reduced {{iff}}:
:$w_i \ne {w_{i + 1} }^{-1}$ for all $i \in \set {1, \ldots, n - 1}$
where $w_1, w_2, \ldots$ are elements of $S$.
We have {{hypothesis}} that $\epsilon$ is the empty group word on $S$.
Hence by definition it has no elements of $S$ in it.
Hence the condition for $\epsilon$ to be reduced is vacuously true.
{{qed}}
Category:Group Words
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Theorem: Scalar Multiplication Corresponds to Multiplication by 1x1 Matrix

Let $\map \MM 1$ denote the matrix space of square matrices of order $1$.
Let $\map \MM {1, n}$ denote the matrix space of order $1 \times n$.
Let $\mathbf A = \begin {pmatrix} a \end {pmatrix} \in \map \MM 1$ and $\mathbf B = \begin {pmatrix} b_1 & b_2 & \cdots & b_n \end{pmatrix} \in \map \MM {1, n}$.
Let $\mathbf C = \mathbf A \mathbf B$ denote the (conventional) matrix product of $\mathbf A$ with $\mathbf B$.
Let $\mathbf D = a \mathbf B$ denote the matrix scalar product of $a$ with $\mathbf B$.
Then $\mathbf C = \mathbf D$.

Proof:
By definition of (conventional) matrix product, $\mathbf C$ is of order $1 \times n$.
By definition of matrix scalar product, $\mathbf D$ is also of order $1 \times n$.
Consider arbitrary elements $c_i \in \mathbf C$ and $d_i \in \mathbf D$ for some index $i$ where $1 \le i \le n$.
We have:
{{begin-eqn}}
{{eqn | l = c_i
      | r = \sum_{j \mathop = 1}^i a_{j j} b_j
      | c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = a b_j
      | c = Definition of $\mathbf A$
}}
{{end-eqn}}
and: 
{{begin-eqn}}
{{eqn | l = d_i
      | r = a b_j
      | c = {{Defof|Matrix Scalar Product}}
}}
{{eqn | r = c_i
      | c = from above
}}
{{end-eqn}}
{{qed}}
Theorem: Closed Unit Ball is Convex Set

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\map {B_1^-} 0$ be the closed unit ball in $X$.
Then $\map {B_1^-} 0$ is convex.

Proof:
Let $x, y \in \map {B_1^-} 0$.
Let $\alpha \in \closedint 0 1$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \norm {\paren {1 - \alpha} x + \alpha y}
      | o = \le
      | r = \norm {\paren {1 - \alpha} x} + \norm {\alpha y}
      | c = {{NormAxiomVector|3}}
}}
{{eqn | r = \size {1 - \alpha} \norm x + \size \alpha \norm y
      | c = {{NormAxiomVector|2}}
}}
{{eqn | r = \paren {1 - \alpha} \norm x + \alpha \norm y
      | c = {{Defof|Convex Set (Vector Space)|Convex Set}}: $0 \le \alpha \le 1$
}}
{{eqn | o = \le
      | r = \paren {1 - \alpha} + \alpha
      | c = $x, y \in \map {B_1^-} 0$
}}
{{eqn | r = 1
}}
{{end-eqn}}
Therefore, $\paren {1 - \alpha}x + \alpha y \in \map {B_1^-} 0$.
By definition, $\map {B_1^-} 0$ is convex.
{{qed}}
Theorem: Uncountable Sum as Series/Corollary

Let $f: X \to \closedint 0 {+\infty}$ have uncountably infinite support.
Then:
:$\ds \sum_{x \mathop \in X} \map f x = +\infty$

Proof:
This is the first case of Uncountable Sum as Series.
{{qed}}
Theorem: Identity Matrix from Upper Triangular Matrix

Let $\mathbf A = \sqbrk a_{m n}$ be an upper triangular matrix of order $m \times n$ with no zero diagonal elements.
Let $k = \min \set {m, n}$.
Then $\mathbf A$ can be transformed into a matrix such that the first $k$ rows and columns form the unit matrix of order $k$.

Proof:
By definition of $k$:
:if $\mathbf A$ has more rows than columns, $k$ is the number of columns of $\mathbf A$.
:if $\mathbf A$ has more columns than rows, $k$ is the number of rows of $\mathbf A$.
Thus let $\mathbf A'$ be the square matrix of order $k$ consisting of the first $k$ rows and columns of $\mathbf A$:
:<nowiki>$\mathbf A' = \begin {bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} & \cdots &     a_{1, k - 1} &      a_{1 k} \\
      0 & a_{2 2} & a_{2 3} & \cdots &     a_{2, k - 1} &      a_{2 k} \\
      0 &       0 & a_{3 3} & \cdots &     a_{3, k - 1} &      a_{3 k} \\
 \vdots &  \vdots &  \vdots & \ddots &           \vdots &      \vdots  \\
      0 &       0 &       0 & \cdots & a_{k - 1, k - 1} & a_{k - 1, k} \\
      0 &       0 &       0 & \cdots &                0 &      a_{k k} \\
\end {bmatrix}$</nowiki>
$\mathbf A$ can be transformed into echelon form $\mathbf B$ by using the elementary row operations:
:$\forall j \in \set {1, 2, \ldots, k}: e_j := r_j \to \dfrac 1 {a_{j j} } r_j$
Again, let $\mathbf B'$ be the square matrix of order $k$ consisting of the first $k$ rows and columns of $\mathbf B$:
:<nowiki>$\mathbf B' = \begin {bmatrix}
     1 & b_{1 2} & b_{1 3} & \cdots &     b_{1, k - 1} &      b_{1 k} \\
     0 &       1 & b_{2 3} & \cdots &     b_{2, k - 1} &      b_{2 k} \\
     0 &       0 &       1 & \cdots &     b_{3, k - 1} &      b_{3 k} \\
\vdots &  \vdots &  \vdots & \ddots &           \vdots &      \vdots  \\
     0 &       0 &       0 & \cdots &                1 & b_{k - 1, k} \\
     0 &       0 &       0 & \cdots &                0 &            1 \\
\end {bmatrix}$</nowiki>
$\mathbf B$ is then transformed into reduced echelon form $\mathbf C$ by means of the elementary row operations:
:$\forall j \in \set {1, 2, \ldots, k - 1}: e_{j k} := r_j \to r_j - b_{j k} r_k$
:$\forall j \in \set {1, 2, \ldots, k - 2}: e_{j, k - 1} := r_j \to r_j - b_{j, k - 1} r_{k - 1}$
and so on, until:
:$e_{1 2} := r_1 \to r_1 - b_{1 2} r_2$
Again, let $\mathbf C'$ be the square matrix of order $k$ consisting of the first $k$ rows and columns of $\mathbf C$:
:<nowiki>$\mathbf C' = \begin {bmatrix}
     1 &      0 &      0 & \cdots &      0 &      0 \\
     0 &      1 &      0 & \cdots &      0 &      0 \\
     0 &      0 &      1 & \cdots &      0 &      0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
     0 &      0 &      0 & \cdots &      1 &      0 \\
     0 &      0 &      0 & \cdots &      0 &      1 \\
\end {bmatrix}$</nowiki>
By inspection, $\mathbf C$ is seen to be the unit matrix of order $k$.
{{qed}}
Theorem: Simultaneous Linear Equations have Solution iff Ranks of Matrix of Coefficients and Augmented Matrix are Equal

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 $S$ be expressed in matrix form as:
:$\mathbf {A x} = \mathbf b$
where:
:$\mathbf A = \begin {pmatrix}
\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n} \\
\end {pmatrix}$,  $\mathbf x = \begin {pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$, $\mathbf b = \begin {pmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_m \end {pmatrix}$
Then $S$ has at least one solution {{iff}}:
:$\map \rho {\mathbf A} = \map \rho {\begin {array} {c|c} \mathbf A & \mathbf b \end {array} }$
where:
:$\map \rho {\mathbf A}$ denotes the rank of $\mathbf A$
:$\paren {\begin {array} {c|c} \mathbf A & \mathbf b \end {array} }$ denotes the augmented matrix of $S$.

Proof:
{{ProofWanted|tedious}}
Theorem: Simultaneous Linear Equations has Unique Solution iff Rank of Matrix of Coefficients equals Number of Columns

Let $S$ be a system of $m$ simultaneous linear equations in $n$ variables:
:$\ds \forall i \in \set {1, 2, \ldots, m} : \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$
Let $S$ be expressed in matrix form as:
:$\mathbf A \mathbf x = \mathbf b$
where:
:$\mathbf A = \begin {pmatrix}
\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n} \\
\end {pmatrix}$,  $\mathbf x = \begin {pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$, $\mathbf b = \begin {pmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_m \end {pmatrix}$
Then $S$ has exactly one solution {{iff}}:
:$\map \rho {\mathbf A} = n$
where $\map \rho {\mathbf A}$ denotes the rank of $\mathbf A$.

Proof:
{{ProofWanted|tedious}}
Theorem: Max Operation Equals an Operand

Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
Then:
:$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$

Proof:
We will prove the result by induction on the number of operands $n$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$
Theorem: Trace of Sum of Matrices is Sum of Traces

Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be square matrices of order $n$.
let $\mathbf A + \mathbf B$ denote the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
Then:
:$\map \tr {\mathbf A + \mathbf B} = \map \tr {\mathbf A} + \map \tr {\mathbf B}$
where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$.

Proof:
{{begin-eqn}}
{{eqn | l = \map \tr {\mathbf A} + \map \tr {\mathbf B}
      | r = \sum_{k \mathop = 1}^n a_{kk} + \sum_{k \mathop = 1}^n b_{kk}
      | c = {{Defof|Trace of Matrix}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {a_{kk} + b_{kk} }
      | c = Sum of Summations equals Summation of Sum
}}
{{eqn | r = \map \tr {\mathbf A + \mathbf B}
      | c = {{Defof|Matrix Entrywise Addition}}, {{Defof|Trace of Matrix}}
}}
{{end-eqn}}
{{qed}}
Theorem: Similar Matrices have same Traces

Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be square matrices of order $n$.
Let $\mathbf A$ and $\mathbf B$ be similar.
Then:
:$\map \tr {\mathbf A} = \map \tr {\mathbf B}$
where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$.

Proof:
By definition of similar matrices:
:$\exists \mathbf P: \mathbf P^{-1} \mathbf A \mathbf P = \mathbf B$
where $\mathbf P$ is an nonsingular matrix of order $n$.
Therefore:
{{begin-eqn}}
{{eqn | l = \map \tr {\mathbf B}
      | r = \map \tr {\mathbf P^{-1} \mathbf A \mathbf P}
}}
{{eqn | r = \map \tr {\mathbf P \paren {\mathbf P^{-1} \mathbf A} }
      | c = Trace of Product of Matrices
}}
{{eqn | r = \map \tr {\paren {\mathbf P \mathbf P^{-1} } \mathbf A}
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \map \tr {\mathbf I_n \mathbf A}
      | c = {{Defof|Inverse Matrix}}
}}
{{eqn | r = \map \tr {\mathbf A}
      | c = Unit Matrix is Identity for Matrix Multiplication
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Integer Multiple of Argument/Formulation 1

{{begin-eqn}}
{{eqn | l = \cos n \theta
      | r = \dfrac 1 2 \paren {\paren {2 \cos \theta}^n - \dfrac n 1 \paren {2 \cos \theta}^{n - 2} + \dfrac n 2 \dbinom {n - 3} 1 \paren {2 \cos \theta}^{n - 4} - \dfrac n 3 \dbinom {n - 4} 2 \paren {2 \cos \theta}^{n - 6} + \cdots}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac n k \dbinom {n - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n - 2 k} }
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \cos n \theta = \dfrac 1 2 \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac n k \dbinom {n - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n - 2 k} }$
Theorem: Cosine of Integer Multiple of Argument/Formulation 2

{{begin-eqn}}
{{eqn | l = \cos n \theta
      | r = \cos^n \theta \paren {1 - \dbinom n 2 \paren {\tan \theta}^2 + \dbinom n 4 \paren {\tan \theta}^4 - \cdots}
      | c = 
}}
{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\tan^{2 k } \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 even, the expression being summed is real.
Equating the real parts of both sides of the equation, replacing $k$ with $2 k$ to make $k$ even, gives:
{{begin-eqn}}
{{eqn | l = \cos n \theta
      | r = \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}
      | c = 
}}
{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\tan^{2 k } \theta}
      | c = factor out $\cos^n \theta$
}}
{{end-eqn}}
{{qed}}
Theorem: Polygamma Reflection Formula

Let $z \in \C \setminus \Z$.
Let $\psi_n$ denote the $n$th polygamma function.
Then:
:$\map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z} = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z$

Proof:
By definition:
:$\map {\psi_n} z = \dfrac {\d^n} {\d z^n} \map \psi z$
where:
:$\psi$ denotes the digamma function
:$z \in \C \setminus \Z_{\le 0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \psi z - \map \psi {1 - z}
      | r = -\pi \cot \pi z
      | c = Digamma Reflection Formula
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d^n} {\d z^n} \map \psi z - \dfrac {\d^n} {\d z^n} \map \psi {1 - z}
      | r = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z
      | c = taking $n$th derivative
}}
{{eqn | ll= \leadsto
      | l = \map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z}
      | r = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z
      | c = {{Defof|Polygamma Function}}
}}
{{end-eqn}}
Finally, from the {{Lemma|Polygamma Reflection Formula}}, we note that:
:$\map \psi z - \map \psi {1 - z}$
is defined on the domain $\C \setminus \Z$.
{{qed}}
Theorem: Polygamma Reflection Formula

Let $z \in \C \setminus \Z$.
Let $\psi_n$ denote the $n$th polygamma function.
Then:
:$\map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z} = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma z \map \Gamma {1 - z}
      | r = \dfrac \pi {\sin \pi z}
      | c = Euler's Reflection Formula
}}
{{eqn | ll= \leadsto
      | l = \map \ln {\map \Gamma z \map \Gamma {1 - z} }
      | r = \map \ln {\dfrac \pi {\sin \pi z} }
      | c = applying $\ln$ on both sides
}}
{{eqn | ll= \leadsto
      | l = \map \ln {\map \Gamma z} + \map \ln {\map \Gamma {1 - z} }
      | r = \map \ln \pi - \map \ln {\sin \pi z}
      | c = Sum of Logarithms and Difference of Logarithms
}}
{{eqn | ll= \leadsto
      | l = \dfrac \d {\d z} \map \ln {\map \Gamma z} + \dfrac \d {\d z} \map \ln {\map \Gamma {1 - z} }
      | r = \dfrac \d {\d z} \map \ln \pi - \dfrac \d {\d z} \map \ln {\sin \pi z}
      | c = differentiation {{WRT|Differentiation}} $z$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\map {\Gamma'} z} {\map \Gamma z} - \dfrac {\map {\Gamma'} {1 - z} } {\map \Gamma {1 - z} }
      | r = 0 - \pi \cot \pi z
      | c = Derivative of Natural Logarithm Function, Derivative of Sine Function, Chain Rule for Derivatives, Derivative of Constant
}}
{{eqn | ll= \leadsto
      | l = \map \psi z - \map \psi {1 - z}
      | r = -\pi \cot \pi z
      | c = {{Defof|Digamma Function}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d^n} {\d z^n} \map \psi z - \dfrac {\d^n} {\d z^n} \map \psi {1 - z}
      | r = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z
      | c = taking $n$th derivative
}}
{{eqn | ll= \leadsto
      | l = \map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z}
      | r = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z
      | c = {{Defof|Polygamma Function}}
}}
{{end-eqn}}
Finally, from the {{Lemma|Polygamma Reflection Formula}}, we note that:
:$\map \psi z - \map \psi {1 - z}$
is defined on the domain $\C \setminus \Z$.
{{qed}}
Theorem: Polygamma Function in terms of Hurwitz Zeta Function

:$\map {\psi_n} z = \paren {-1}^{n + 1} \map \Gamma {n + 1} \map \zeta {n + 1, z}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \psi z
      | r = \dfrac {\map {\Gamma'} z} {\map \Gamma z}
      | c = {{Defof|Digamma Function}}
}}
{{eqn | r = -\gamma + \sum_{k \mathop = 1}^\infty \paren {\dfrac 1 k - \dfrac 1 {z + k - 1} }
      | c = Reciprocal times Derivative of Gamma Function
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d^n} {\d z^n} \map \psi z
      | r = -\dfrac {\d^n} {\d z^n} \gamma + \dfrac {\d^n} {\d z^n} \sum_{k \mathop = 1}^\infty \paren {\dfrac 1 k - \dfrac 1 {z + k - 1} }
      | c = taking $n$th derivative
}}
{{eqn | ll= \leadsto
      | l = \map {\psi_n} z
      | r = -\dfrac {\d^n} {\d z^n} \sum_{k \mathop = 1}^\infty \dfrac 1 {z + k - 1}
      | c = {{Defof|Polygamma Function}}, Derivative of Constant
}}
{{eqn | r = -\dfrac {\d^n} {\d z^n} \sum_{k \mathop = 0}^\infty \dfrac 1 {z + k}
      | c = reindexing $k$ from $1$ to $0$
}}
{{eqn | r = -\sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^n n!} {\paren {z + k}^{n + 1} }
      | c = $n$th Derivative of Reciprocal
}}
{{eqn | r = \paren {-1}^{n + 1} \map \Gamma {n + 1} \sum_{k \mathop = 0}^\infty \dfrac 1 {\paren {z + k}^{n + 1} }
      | c = Gamma Function Extends Factorial and simplification
}}
{{eqn | r = \paren {-1}^{n + 1} \map \Gamma {n + 1} \map \zeta {n + 1, z}
      | c = {{Defof|Hurwitz Zeta Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Parallelogram from Determinant

Let $OABC$ be a parallelogram in the Cartesian plane whose vertices are located at:
{{begin-eqn}}
{{eqn | l = O
      | r = \tuple {0, 0}
}}
{{eqn | l = A
      | r = \tuple {a, c}
}}
{{eqn | l = B
      | r = \tuple {a + b, c + d}
}}
{{eqn | l = C
      | r = \tuple {b, d}
}}
{{end-eqn}}
The area of $OABC$ is given by:
:$\map \Area {OABC} = \begin {vmatrix} a & b \\ c & d \end {vmatrix}$
where $\begin {vmatrix} a & b \\ c & d \end {vmatrix}$ denotes the determinant of order $2$.

Proof:
Arrange for the parallelogram to be situated entirely in the first quadrant.
:500px
First need we establish that $OABC$ is actually a parallelogram in the first place.
Indeed:
{{begin-eqn}}
{{eqn | l = \vec {AB}
      | r = \tuple {a + b - a, c + d - c}
      | c = 
}}
{{eqn | r = \tuple {b, d}
      | c = 
}}
{{eqn | r = \vec {OC}
      | c = 
}}
{{eqn | l = \vec {CB}
      | r = \tuple {a + b - b, c + d - d}
      | c = 
}}
{{eqn | r = \tuple {a, c}
      | c = 
}}
{{eqn | r = \vec {OA}
      | c = 
}}
{{end-eqn}}
Thus:
:$OA = CB$
:$OC = AB$
and it follows from Opposite Sides Equal implies Parallelogram that $OABC$ is indeed a parallelogram.
Now we calculate the area of $OABC$ as equal to:
:the area occupied by the large rectangle in the diagram above
less:
:the $4$ triangles
:the $2$ small rectangles.
Thus:
{{begin-eqn}}
{{eqn | l = \map \Area {OABC}
      | r = \paren {a + b} \paren {c + d}
      | c = the large rectangle
}}
{{eqn | o =
      | ro= -
      | r = \paren {\dfrac {a c} 2} - \paren {\dfrac {\paren {a + b - b} \paren {c + d - d} } 2}
      | c = the $2$ triangles at top and bottom
}}
{{eqn | o =
      | ro= -
      | r = \paren {\dfrac {b d} 2} - \paren {\dfrac {\paren {a + b - a} \paren {c + d - c} } 2}
      | c = the $2$ triangles at left and right
}}
{{eqn | o =
      | ro= -
      | r = \paren {a + b - a} c - b \paren {c + d - d}
      | c = the $2$ small rectangles
}}
{{eqn | r = a c + a d + b c + b d - \dfrac {a c} 2 - \dfrac {a c} 2 - \dfrac {b d} 2 - \dfrac {b d} 2 - 2 b c
      | c = multiplying out and simplifying
}}
{{eqn | r = a d - b c
      | c = simplifying
}}
{{eqn | r = \begin {vmatrix} a & b \\ c & d \end {vmatrix}
      | c = Determinant of Order $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Matrix is Nonsingular iff Rank equals Order

Let $R$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.
Then $\mathbf A$ is nonsingular {{iff}} its rank also equals $n$.

Proof:
This is an immediate consequence of Square Matrix has Full Rank iff Nonsingular and the definition of full rank.
{{qed}}
Theorem: Determinant of Upper Triangular Matrix

Let $\mathbf T_n$ be an upper triangular matrix of order $n$.
Let $\map \det {\mathbf T_n}$ be the determinant of $\mathbf T_n$.
Then $\map \det {\mathbf T_n}$ is equal to the product of all the diagonal elements of $\mathbf T_n$.
That is:
:$\ds \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$

Proof:
Let $\mathbf T_n$ be an upper triangular matrix of order $n$.
We proceed by induction on $n$, the number of rows of $\mathbf T_n$.
Theorem: Cosine of Integer Multiple of Argument/Formulation 3

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

Proof:
{{begin-eqn}}
{{eqn | l = \cos n \theta 
      | r = \map \cos {\paren {n - 1} \theta + \theta}
      | c =
}}
{{eqn | r = \cos \paren {n - 1} \theta \cos \theta - \sin \paren {n - 1} \theta \sin \theta
      | c = Cosine of Sum
}}
{{eqn | r = \cos \paren {n - 1} \theta \cos \theta - \paren {\sin \theta \cos^{n - 2} \theta \sum_{k \mathop = 0}^{n - 2} \frac {\cos k \theta} {\cos^k \theta} } \sin \theta
      | c = {{Formulation|Sine of Integer Multiple of Argument|3}}
}}
{{eqn | r = \cos \paren {n - 1} \theta \cos \theta - \paren {\sin^2 \theta \cos^{n - 2} \theta \sum_{k \mathop = 0}^{n - 2} \frac {\cos k \theta} {\cos^k \theta} }
      | c = 
}}
{{eqn | r = \cos \paren {n - 1} \theta \cos \theta - \paren {\paren {1 - \cos^2 \theta} \cos^{n - 2} \theta \sum_{k \mathop = 0}^{n - 2} \frac {\cos k \theta} {\cos^k \theta} }
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \cos \paren {n - 1} \theta \cos \theta + \paren {\paren {\cos^2 \theta - 1} \cos^{n - 2} \theta \sum_{k \mathop = 0}^{n - 2} \frac {\cos k \theta} {\cos^k \theta} }
      | c = 
}}
{{eqn | r = \cos \paren {n - 1} \theta \cos \theta + \paren {1 - \sec^2 \theta} \cos^n \theta \sum_{k \mathop = 0}^{n - 2} \frac {\cos k \theta} {\cos^k \theta}
      | c = Secant is Reciprocal of Cosine
}}
{{end-eqn}}
{{qed}}
Category:Cosine of Integer Multiple of Argument
346t4ldp7cc3cli3119fjq69457q78s
Theorem: Determinant of Elementary Row Matrix/Scale Row

Let $e_1$ be the elementary row operation $\text {ERO} 1$:
{{begin-axiom}}
{{axiom | n = \text {ERO} 1
        | t = For some $\lambda \ne 0$, multiply row $k$ by $\lambda$
        | m = r_k \to \lambda r_k
}}
{{end-axiom}}
which is to operate on some arbitrary matrix space.
Let $\mathbf E_1$ be the elementary row matrix corresponding to $e_1$.
The determinant of $\mathbf E_1$ is:
:$\map \det {\mathbf E_1} = \lambda$

Proof:
By Elementary Matrix corresponding to Elementary Row Operation: Scale Row, the elementary row matrix corresponding to $e_1$ is of the form:
:$E_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \lambda \cdot \delta_{a b} & : a = k \end{cases}$
where:
:$E_{a b}$ denotes the element of $\mathbf E_1$ whose indices are $\tuple {a, b}$
:$\delta_{a b}$ is the Kronecker delta:
::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$
Thus when $a \ne b$, $E_{a b} = 0$.
This means that $\mathbf E_1$ is a diagonal matrix.
{{begin-eqn}}
{{eqn | l = \map \det {\mathbf E_1}
      | r = \prod_i E_{i i}
      | c = Determinant of Diagonal Matrix
      | cc= where the index variable $i$ ranges over the order of $\mathbf E_1$
}}
{{eqn | r = \prod_i \paren {\begin {cases} 1 & : i \ne k \\ \lambda & : a = k \end{cases} }
      | c = 
}}
{{eqn | r = \prod_{i \mathop \ne k} 1 \times \prod_{i \mathop = k} \lambda
      | c = 
}}
{{eqn | r = 1 \times \lambda
      | c = 
}}
{{eqn | r = \lambda
      | c = 
}}
{{end-eqn}}
{{qed}}