hcju/proofwikiqa huggingface dataset:
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: :$\displaystyle \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$ 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}} Category:Cosine of Integer Multiple of Argument bcldz9phe21keab8wzscyuh1xqpihc1
{{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 = taking first derivative }} {{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, 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}} {{qed}}
{{begin-eqn}} {{eqn   | ll= \leadsto 	| l = \map \psi z 	| r = -\gamma + \sum_{k \mathop = 1}^\infty \paren {\dfrac 1 n - \dfrac 1 {z + k - 1} }         | c = Series for digamma 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   | ll= \leadsto 	| l =  	| r = - \dfrac {\d^n} {\d z^n} \sum_{k \mathop = 0}^\infty \dfrac 1 {z + k}         | c = reindexing $k$ from $1$ to $0$ }} {{eqn   | ll= \leadsto 	| l =  	| r = - \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^{n} \map \Gamma {n + 1} } {\paren {z + k}^{n + 1} }         | c = $n$th derivative of reciprocal function }} {{eqn   | ll= \leadsto 	| l =  	| r = \paren {-1}^{n + 1} \map \Gamma {n + 1} \map \zeta {n + 1, z}         | c = {{Defof|Hurwitz Zeta Function}} }} {{end-eqn}} {{qed}}
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 {CB}       | c =  }} {{eqn | l = \vec {OA}       | 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 - c}       | 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 c - b d       | c = simplifying }} {{eqn | r = \begin {vmatrix} a & b \\ c & d \end {vmatrix}       | c = {{Defof|Determinant of Order 2}} }} {{end-eqn}} {{qed}}
Let $\mathbf T_n$ be an upper triangular matrix of order $n$. We proceed by induction on $n$, the number of rows of $\mathbf T_n$. === Basis for the Induction === For $n = 1$, the determinant is $a_{11}$, which is clearly also the diagonal element. This forms the basis for the induction. === Induction Hypothesis === Fix $n \in \N$. Then, let: :$\mathbf T_n = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\       0 & a_{2 2} & \cdots & a_{2 n} \\  \vdots &  \vdots & \ddots &  \vdots \\       0 &       0 & \cdots & a_{n n} \\ \end {bmatrix}$ be an upper triangular matrix. Assume that: :$\displaystyle \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$ This forms our induction hypothesis. === Induction Step === Let $\mathbf T_{n + 1}$ be an upper triangular matrix of order $n + 1$. Then, by the Expansion Theorem for Determinants (expanding across the $n + 1$th row): :$\displaystyle D = \map \det {\mathbf T_{n + 1} } = \sum_{k \mathop = 1}^{n + 1} a_{n + 1, k} T_{n + 1, k}$ Because $\mathbf T_{n + 1}$ is upper triangular, $a_{n + 1, k} = 0$ when $k < n + 1$. Therefore: :$\map \det {\mathbf T_{n + 1} } = a_{n + 1 \, n + 1} T_{n + 1, n + 1}$ By the definition of the cofactor: :$T_{n + 1, n + 1} = \paren {-1}^{n + 1 + n + 1} D_{n + 1, n + 1} = D_{n n}$ where $D_{n n}$ is the order $n$ determinant obtained from $D$ by deleting row $n + 1$ and column $n + 1$.  But $D_{n n}$ is just the determinant of an upper triangular matrix $\mathbf T_n$. Therefore: :$\map \det {\mathbf T_{n + 1} } = a_{n + 1, n + 1} \map \det {\mathbf T_n}$ and the result follows by induction. {{qed}}
{{begin-eqn}} {{eqn | l = \map \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 = Sine of Integer Multiple of Argument/Formulation 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 = Definition:Secant Function/Complex }} {{end-eqn}} {{qed}} Category:Cosine of Integer Multiple of Argument nk013c8qxqdfsoyzcmu83w1f7ktbpof
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 = \displaystyle \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}}
By Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, $\mathbf E_2$ is of the form: :$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$ where: :$E_{a b}$ denotes the element of $\mathbf E$ 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}$ Because $i \ne j$ it follows that: :if $a = i$ and $b = j$ then $a \ne b$ Hence when $a = b$ we have that: :$\delta_{a i} \cdot \delta_{j b} = 0$ Hence the diagonal elements of $\mathbf E_2$ are all equal to $1$. We also have that $\delta_{a i} \cdot \delta_{j b} = 1$ {{iff}} $a = i$ and $b = j$. Hence, all elements of $\mathbf E_2$ apart from the diagonal elements and $a_{i j}$ are equal to $0$. Thus $\mathbf E_2$ is a triangular matrix (either upper or lower). From Determinant of Triangular Matrix, $\map \det {\mathbf E_2}$ is equal to the product of all the diagonal elements of $\mathbf E_2$. But as we have seen, these are all equal to $1$. Hence the result. {{qed}}
Let $\mathbf I$ denote the unit matrix of arbitrary order $n$. By Determinant of Unit Matrix: :$\map \det {\mathbf I} = 1$ Let $\rho$ be the permutation on $\tuple {1, 2, \ldots, n}$ which transposes $i$ and $j$. From Parity of K-Cycle, $\map \sgn \rho = -1$. By definition we have that $\mathbf E_3$ is $\mathbf I$ with rows $i$ and $j$ transposed. By the definition of a determinant: :$\displaystyle \map \det {\mathbf I} = \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$ By Permutation of Determinant Indices: :$\displaystyle \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }$ We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get: {{begin-eqn}} {{eqn | l = \map \det {\mathbf E_3}       | r = \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }       | c =  }} {{eqn | r = -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }       | c =  }} {{eqn | r = -\map \det {\mathbf I}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
By definition of the unit matrix: :$I_{a b} = \delta_{a b}$ where: :$I_{a b}$ denotes the element of $\mathbf I$ whose indices are $\tuple {a, b}$. By definition, $\mathbf E$ is the square matrix of order $m$ formed by applying $e$ to the unit matrix $\mathbf I$. That is, all elements of column $k$ of $\mathbf I$ are to be multiplied by $\lambda$. By definition of unit matrix, all elements of column $k$ are $0$ except for element $I_{k k}$, which is $1$. Thus in $\mathbf E$: :$E_{k k} = \lambda \cdot 1 = \lambda$ The elements in all the other columns of $\mathbf E$ are the same as the corresponding elements of $\mathbf I$. Hence the result. {{qed}} Category:Elementary Matrix corresponding to Elementary Column Operation jb6yvslnfj0bla5ztd3536s3d31e4o2
By definition of the unit matrix: :$I_{a b} = \delta_{a b}$ where: :$I_{a b}$ denotes the element of $\mathbf I$ whose indices are $\tuple {a, b}$. By definition, $\mathbf E$ is the square matrix of order $m$ formed by applying $e$ to the unit matrix $\mathbf I$. That is, all elements of column $i$ of $\mathbf I$ are to have the corresponding elements of column $j$ added to them after the latter have been multiplied by $\lambda$. By definition of unit matrix: :all elements of column $i$ are $0$ except for element $I_{i i}$, which is $1$. :all elements of column $j$ are $0$ except for element $I_{j j}$, which is $1$. Thus in $\mathbf E$: :where $b \ne i$, $E_{a b} = \delta_{a b}$ :where $b = i$: ::$E_{a b} = \delta_{a b}$ where $a \ne j$ ::$E_{a b} = \delta_{a b} + \lambda \cdot 1$ where $a = j$ That is: :$E_{a b} = \delta_{a b}$ for all elements of $\mathbf E$ except where $b = i$ and $a = j$, at which element: :$E_{a b} = \delta_{a b} + \lambda$ That is: :$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{b i} \cdot \delta_{j a}$ Hence the result. {{qed}}
By definition of the unit matrix: :$I_{a b} = \delta_{a b}$ where: :$I_{a b}$ denotes the element of $\mathbf I$ whose indices are $\tuple {a, b}$. By definition, $\mathbf E$ is the square matrix of order $n$ formed by applying $e$ to the unit matrix $\mathbf I$. That is, all elements of column $i$ of $\mathbf I$ are to be exchanged with the corresponding elements of column $j$. By definition of unit matrix: :all elements of column $i$ are $0$ except for element $I_{i i}$, which is $1$. :all elements of column $j$ are $0$ except for element $I_{j j}$, which is $1$. Thus in $\mathbf E$: :where $a \ne i$ and $a \ne j$, $E_{a b} = \delta_{a b}$ (all columns except $i$ and $j$ are unchanged) :where $a = i$, $E_{a b} = \delta_{a j}$ (the contents of column $j$) :where $a = j$, $E_{a b} = \delta_{a i}$ (the contents of column $i$) That is: :$E_{a b} = \begin {cases} \delta_{a b} & : \text {if $b \ne i$ and $b \ne j$} \\ \delta_{a j} & : \text {if $b = i$} \\ \delta_{a i} & : \text {if $b = j$} \end {cases}$ Hence the result. {{qed}} Category:Elementary Matrix corresponding to Elementary Column Operation 3d6p4m50lbveddqbkwxoje3483kcrsf
In the following, $\mathbf A$, $\mathbf B$ and $\mathbf C$ denote arbitrary matrices in a given matrix space $\map \MM {m, n}$ for $m, n \in \Z{>0}$. We check in turn each of the conditions for equivalence: === Reflexive === Let $\kappa_i$ denote an arbitrary column of $\mathbf A$. Let $e$ denote the elementary column operation $\kappa_i \to 1 \kappa_i$ applied to $\mathbf A$. Then trivially: :$\map e {\mathbf A} = \mathbf A$ and so $\mathbf A$ is trivially column equivalent to itself. So column equivalence has been shown to be reflexive. {{qed|lemma}} === Symmetric === Let $\mathbf A$ be column equivalent to $\mathbf B$. Let $\Gamma$ be the column operation that transforms $\mathbf A$ into $\mathbf B$. From Column Operation has Inverse there exists a column operation $\Gamma'$ which transforms $\mathbf B$ into $\mathbf A$. Thus $\mathbf B$ is column equivalent to $\mathbf A$. So column equivalence has been shown to be symmetric. {{qed|lemma}} === Transitive === Let $\mathbf A$ be column equivalent to $\mathbf B$, and let $\mathbf B$ be column equivalent to $\mathbf C$. Let $\Gamma_1$ be the column operation that transforms $\mathbf A$ into $\mathbf B$. Let $\Gamma_2$ be the column operation that transforms $\mathbf B$ into $\mathbf C$. From Sequence of Column Operations is Column Operation, the application of $\mathbf C$ is column equivalent to $\mathbf A$. So column equivalence has been shown to be transitive. {{qed|lemma}} Column equivalence has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary row operations that compose $\Gamma$. Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary row matrices. From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, we have: :$\mathbf R \mathbf A = \mathbf B$ where $\mathbf R$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$: :$\mathbf R = \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$ By Elementary Row Matrix is Invertible, each of $\mathbf E_i$ is invertible. By Product of Matrices is Invertible iff Matrices are Invertible, it follows that $\mathbf R$ is likewise invertible. Thus $\mathbf R$ has an inverse $\mathbf R^{-1}$. Hence: {{begin-eqn}} {{eqn | l = \mathbf R \mathbf A       | r = \mathbf B       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf R^{-1} \mathbf R \mathbf A       | r = \mathbf R^{-1} \mathbf B       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A       | r = \mathbf R^{-1} \mathbf B       | c =  }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \mathbf R^{-1}       | r = \paren {\mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1}^{-1}       | c =  }} {{eqn | r = {\mathbf E_1}^{-1} {\mathbf E_2}^{-1} \dotsb {\mathbf E_{k - 1} }^{-1} {\mathbf E_k}^{-1}       | c = Inverse of Matrix Product }} {{end-eqn}} From Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse, each of ${\mathbf E_i}^{-1}$ is the elementary row matrix corresponding to the inverse $e'_i$ of the corresponding elementary row operation $e_i$. Let $\Gamma'$ be the row operation composed of the finite sequence of elementary row operations $\tuple {e'_k, e'_{k - 1}, \ldots, e'_2, e'_1}$. Thus $\Gamma'$ is a row operation which transforms $\mathbf B$ into $\mathbf A$. Hence the result. {{qed}}
We have that: :Real Numbers form Vector Space :Absolute Value is Norm By definition, $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space. {{qed}}
We have that: :Real Vector Space is Vector Space :By Euclidean Space is Normed Space, $\norm {\, \cdot \,}_2$ is a norm on $\R^n$ By definition, $\struct {\R^n, \norm {\, \cdot \,}_2}$ is a normed vector space. {{qed}}
=== Necessary Condition === Let both $\mathbf A$ and $\mathbf B$ be invertible. By Matrix is Invertible iff Determinant has Multiplicative Inverse: :$\map \det {\mathbf A} \ne 0$ and $\map \det {\mathbf B} \ne 0$ where $\map \det {\mathbf A}$ denotes the determinant of $\mathbf A$. By Determinant of Matrix Product: :$\map \det {\mathbf A} \map \det {\mathbf B} = \map \det {\mathbf A \mathbf B}$ Thus as both $\map \det {\mathbf A} \ne 0$ and $\map \det {\mathbf B} \ne 0$, it follows that: :$\map \det {\mathbf A \mathbf B} = \ne 0$ Hence by Matrix is Invertible iff Determinant has Multiplicative Inverse, $\map \det {\mathbf A \mathbf B}$ is invertible. === Sufficient Condition === Let $\mathbf A \mathbf B$ be invertible. {{AimForCont}} it is not the case that both $\mathbf A$ and $\mathbf B$ are invertible. Then by Matrix is Invertible iff Determinant has Multiplicative Inverse, either: :$\map \det {\mathbf A} = 0$ or: :$\map \det {\mathbf B} = 0$ By Determinant of Matrix Product: :$\map \det {\mathbf A} \map \det {\mathbf B} = \map \det {\mathbf A \mathbf B}$ and so: :$\map \det {\mathbf A \mathbf B} = 0$ Hence by Matrix is Invertible iff Determinant has Multiplicative Inverse, $\map \det {\mathbf A \mathbf B}$ is not invertible. This contradicts the assumption that $\mathbf A \mathbf B$ is invertible. Hence by Proof by Contradiction it follows that both $\mathbf A$ and $\mathbf B$ are invertible. {{qed}}
From Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse it is demonstrated that: :if $\mathbf E$ is the elementary row matrix corresponding to an elementary row operation $e$ then: :the inverse of $e$ corresponds to an elementary row matrix which is the inverse of $\mathbf E$. So as $\mathbf E$ has an inverse, a priori it is invertible. {{qed}}
Let us take each type of elementary column 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: :$\kappa_k$ denote column $k$ of $\mathbf A$ :$\kappa'_k$ denote column $k$ of $\mathbf A'$ :$\kappa''_k$ denote column $k$ of $\mathbf A''$ for arbitrary $k$ such that $1 \le k \le m$. By definition of elementary column operation: :only the column or columns directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$ and similarly: :only the column or columns 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 columns directly affected will be under consideration when showing that $\mathbf A = \mathbf A''$. === $\text {ECO} 1$: Scalar Product of Column === {{:Existence of Inverse Elementary Column Operation/Scalar Product of Column}}{{qed|lemma}} === $\text {ECO} 2$: Add Scalar Product of Column to Another === {{:Existence of Inverse Elementary Column Operation/Add Scalar Product of Column to Another}}{{qed|lemma}} === $\text {ECO} 3$: Exchange Columns === {{:Existence of Inverse Elementary Column Operation/Exchange Columns}}{{qed|lemma}} Thus in all cases, for each elementary column operation which transforms $\mathbf A$ to $\mathbf A'$, we have constructed the only possible elementary column operation which transforms $\mathbf A'$ to $\mathbf A$. Hence the result. {{qed}}
Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary column operations that compose $\Gamma$. Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary column matrices. From Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices, we have: :$\mathbf A \mathbf K = \mathbf B$ where $\mathbf K$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$: :$\mathbf K = \mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k$ By Elementary Column Matrix is Invertible, each of $\mathbf E_i$ is invertible. By Product of Matrices is Invertible iff Matrices are Invertible, it follows that $\mathbf K$ is likewise invertible. Thus $\mathbf K$ has an inverse $\mathbf K^{-1}$. Hence: {{begin-eqn}} {{eqn | l = \mathbf A \mathbf K       | r = \mathbf B       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A \mathbf K \mathbf K^{-1}       | r = \mathbf B \mathbf K^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A       | r = \mathbf B \mathbf K^{-1}       | c =  }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \mathbf K^{-1}       | r = \paren {\mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k}^{-1}       | c =  }} {{eqn | r = {\mathbf E_k}^{-1} {\mathbf E_{k - 1} }^{-1} \dotsb {\mathbf E_2}^{-1} {\mathbf E_1}^{-1}       | c = Inverse of Matrix Product }} {{end-eqn}} From Elementary Column Matrix for Inverse of Elementary Column Operation is Inverse, each of ${\mathbf E_i}^{-1}$ is the elementary column matrix corresponding to the inverse $e'_i$ of the corresponding elementary column operation $e_i$. Let $\Gamma'$ be the column operation composed of the finite sequence of elementary column operations $\tuple {e'_k, e'_{k - 1}, \ldots, e'_2, e'_1}$. Thus $\Gamma'$ is a column operation which transforms $\mathbf B$ into $\mathbf A$. Hence the result. {{qed}}
From Elementary Column Matrix for Inverse of Elementary Column Operation is Inverse it is demonstrated that: :if $\mathbf E$ is the elementary column matrix corresponding to an elementary column operation $e$ then: :the inverse of $e$ corresponds to an elementary column matrix which is the inverse of $\mathbf E$. So as $\mathbf E$ has an inverse, a priori it is invertible. {{qed}}