hcju/proofwikiqa huggingface dataset:
Let $\mathbf A = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{bmatrix}$. Then $\mathbf A^\intercal = \begin{bmatrix} a_{11} & a_{21} & \ldots & a_{n1} \\ a_{12} & a_{22} & \cdots & a_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \cdots & a_{nn} \\ \end{bmatrix}$. Let $b_{r s} = a_{s r}$ for $1 \le r, s \le n$. We need to show that $\map \det {\sqbrk a_n} = \map \det {\sqbrk b_n}$. By the definition of determinant and Permutation of Determinant Indices, we have: {{begin-eqn}} {{eqn | l = \map \det {\sqbrk b_n}       | r = \sum_\lambda \map {\sgn} \lambda b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n}       | c =  }} {{eqn | r = \sum_\lambda \map {\sgn} \lambda a_{\map \lambda 1 1} a_{\map \lambda 2 2} \cdots a_{\map \lambda n n}       | c =  }} {{eqn | r = \map \det {\sqbrk a_n}       | c =  }} {{end-eqn}} {{qed}}
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: {{begin-eqn}} {{eqn | l = \map \det {\mathbf B}       | r = \map \det {\mathbf E \mathbf A}       | c = Determinant of Matrix Product }} {{eqn | r = -\map \det {\mathbf A}       | c = as $\map \det {\mathbf E} = -1$ }} {{end-eqn}} Hence the result. {{qed}}
The proof proceeds by induction over $n$, the order of the square matrix. === Basis for the Induction === Let $n = 2$, which is the smallest natural number for which a square matrix of order $n$ can have two identical rows. Let $\mathbf A = \left[{a}\right]_2$ be a square matrix over $R$ with two identical rows. Then, by definition of determinant: :$\det \left({\mathbf A}\right) = a_{11}a_{22} - a_{12}a_{21} = a_{11}a_{22}-a_{22}a_{11} = 0$ {{qed|lemma}} This is the basis for the induction. === Induction Hypothesis === Assume for $n \in \N_{\ge 2}$ that any square matrices of order $n$ over $R$ with two identical rows has determinant equal to $0$. === Induction Step === Let $\mathbf A$ be a square matrix of order $n+1$ over $R$ with two identical rows. Let $i_1, i_2 \in \left\{ {1, \ldots, n+1}\right\}$ be the indices of the identical rows, and let $i_1 < i_2$. Let $i \in \left\{ {1, \ldots, n+1}\right\}$. Let $\mathbf A \left({i ; 1}\right)$ denote the submatrix obtained from $\mathbf A$ by removing row $i$ and column $1$. If $i \ne i_1$ and $i \ne i_2$, then $\mathbf A \left({i ; 1}\right)$ still contains two identical rows. By the induction hypothesis: : $\det \left({\mathbf A \left({i ; 1}\right) }\right) = 0$ Now consider the matrices $\mathbf A \left({i_1 ; 1}\right)$ and $\mathbf A \left({i_2 ; 1}\right)$. Let $r_j$ denote row $j$ of $\mathbf A \left({i_1 ; 1}\right)$. If we perform the following sequence of $i_2 - i_1 -1$ elementary row operations on $\mathbf A \left({i_1 ; 1}\right)$: :$r_{i_1} \leftrightarrow r_{i_1 +1} \ ; \ r_{i_1 + 1} \leftrightarrow r_{i_1 +2}  \ ;  \ \ldots  \  ;  \ r_{i_2 - 1} \leftrightarrow r_{i_2}$ we will transform $\mathbf A \left({i_1 ; 1}\right)$ into $\mathbf A \left({i_2 ; 1}\right)$. From Determinant with Rows Transposed, it follows that $\det \left({\mathbf A \left({i_1 ; 1}\right) }\right) = \left({-1}\right)^{i_2 - i_1 - 1} \det \left({\mathbf A \left({i_2 ; 1}\right) }\right)$ Then: {{begin-eqn}} {{eqn | l = \det \left({\mathbf A}\right)       | r = \sum_{k \mathop = 1 }^{n + 1} a_{k1} \left({-1}\right)^{k + 1} \det \left({\mathbf A \left({k ; 1}\right) }\right)       | c = expanding the determinant along column $1$ }} {{eqn | r = a_{i_1 1}\left({-1}\right)^{i_1 + 1} \det \left({\mathbf A \left({i_1 ; 1}\right) }\right) + a_{i_2 1} \left({-1}\right)^{i_2 + 1} \det \left({\mathbf A \left({i_2 ; 1}\right) }\right) }} {{eqn | r = a_{i_2 1} \left({-1}\right)^{i_1 + 1 + i_2 - i_1 - 1} \det \left({\mathbf A \left({i_2; 1}\right) }\right) + a_{i_2 1} \left({-1}\right)^{i_2 + 1} \det \left({\mathbf A \left({i_2 ; 1}\right) }\right) }} {{eqn | r = 0 }} {{end-eqn}} {{qed}}
Suppose that $\forall x \in R: x + x = 0 \implies x = 0$. From Determinant with Rows Transposed, if you exchange two rows of a square matrix, the sign of its determinant changes. If you exchange two identical rows of a square matrix, then the sign of its determinant changes from $D$, say, to $-D$. But the matrix stays the same. So $D = -D$ and so $D = 0$. {{qed}}
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: {{begin-eqn}} {{eqn | l = \map \det {\mathbf B}       | r = \map \det {\mathbf E \mathbf A}       | c = Determinant of Matrix Product }} {{eqn | r = c \map \det {\mathbf A}       | c = as $\map \det {\mathbf E} = c$ }} {{end-eqn}} Hence the result. {{qed}}
Let: : $B = \begin{vmatrix}   a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{r1} + a'_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} + a'_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix} = \begin{vmatrix}   b_{11} & \cdots & b_{1s} & \cdots & b_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   b_{r1} & \cdots & b_{rs} & \cdots & b_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   b_{n1} & \cdots & b_{ns} & \cdots & b_{nn} \end{vmatrix}$ : $A_1 = \begin{vmatrix}   a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix}$ : $A_2 = \begin{vmatrix}   a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a'_{r1} & \cdots & a'_{rs} & \cdots & a'_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\   a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix}$ Then:  {{begin-eqn}} {{eqn | l = B       | r = \sum_\lambda \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k=1}^n b_{k \lambda \left({k}\right)} }\right)       | c =  }} {{eqn | r = \sum_\lambda \operatorname{sgn} \left({\lambda}\right) a_{1 \lambda \left({1}\right)} \cdots \left({a_{r \lambda \left({r}\right)} + a'_{r \lambda \left({r}\right)} }\right) \cdots a_{n \lambda \left({n}\right)}       | c =  }} {{eqn | r = \sum_\lambda \operatorname{sgn} \left({\lambda}\right) a_{1 \lambda \left({1}\right)} \cdots a_{r \lambda \left({r}\right)} \cdots a_{n \lambda \left({n}\right)} + \sum_\lambda \operatorname{sgn} \left({\lambda}\right) a_{1 \lambda \left({1}\right)} \cdots a'_{r \lambda \left({r}\right)} \cdots a_{n \lambda \left({n}\right)}       | c =  }} {{eqn | r = A_1 + A_2       | c =  }} {{end-eqn}} {{qed}} The result for columns follows directly from Determinant of Transpose. {{qed}} Category:Determinants 444jdmhgv95gdwkmqke2otcqw0v698t
Consider two cases: :$(1): \quad \mathbf A$ is not invertible. :$(2): \quad \mathbf A$ is invertible. === Proof of case $1$ === Assume $\mathbf A$ is not invertible.  Then: :$\map \det {\mathbf A} = 0$ Also if $\mathbf A$ is not invertible then neither is $\mathbf A \mathbf B$. Indeed, if $\mathbf A \mathbf B$ has an inverse $\mathbf C$, then $\mathbf A \mathbf B \mathbf C = \mathbf I$, whereby $\mathbf B \mathbf C$ is a right inverse of $\mathbf A$. It follows by Left or Right Inverse of Matrix is Inverse that in that case $\mathbf B \mathbf C$ is the inverse of $A$. It follows that: :$\map \det {\mathbf A \mathbf B} = 0$ Thus: :$0 = 0 \cdot \map \det {\mathbf B}$ :$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \cdot \map \det {\mathbf B}$ {{qed|lemma}} === Proof of case $2$ === Assume $\mathbf A$ is invertible. Then $\mathbf A$ is a product of elementary row matrices, $\mathbf E$. Let $\mathbf A = \mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1$. So: :$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1 \mathbf B}$ It remains to be shown that for any square matrix $\mathbf D$ of order $n$: :$\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \cdot \map \det {\mathbf D}$ Let $e_i \paren {\mathbf I} = \mathbf E_i$ for all $i \in \closedint 1 k$, then using Elementary Row Operations as Matrix Multiplications and Effect of Sequence of Elementary Row Operations on Determinant yields: :$\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E_k \mathbf E_{k - 1} \dotsm \mathbf {E_1} \mathbf D} = \map \det {e_k e_{k - 1} \cdots e_1 \paren {\mathbf D} } = \alpha \map \det {\mathbf D}$ Using Elementary Row Operations as Matrix Multiplications and Effect of Sequence of Elementary Row Operations on Determinant, '''and''' Unit Matrix is Unity of Ring of Square Matrices:   :$\map \det {\mathbf E} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf {E_1} \mathbf I} = \map \det {e_k e_{k - 1} \cdots e_1 \paren {\mathbf I} } = \alpha \map \det {\mathbf I}$ From Determinant of Unit Matrix: :$\map \det {\mathbf E} = \alpha$ And so $\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \cdot \map \det {\mathbf D}$ {{qed|lemma}} Therefore: :$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$ as required. {{qed}}
The Cauchy-Binet Formula gives: :$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m}}\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$ where: :$\mathbf A$ is an $m \times n$ matrix :$\mathbf B$ is an $n \times m$ matrix. :For $1 \le j_1, j_2, \ldots, j_m \le n$: ::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$. ::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$. When $m = n$, the only set $j_1, j_2, \ldots, j_m$ that fulfils $1 \le j_1 < j_2 < \cdots < j_m \le n$ is $\left\{ {1, 2, \ldots, n}\right\}$. Hence the result. {{qed}}
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: {{begin-eqn}} {{eqn | l = D       | r = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n b_{k \map \lambda k} }       | c =  }} {{eqn | r = \sum_{\lambda} \map \sgn \lambda b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n}       | c =  }} {{end-eqn}} Now we note: {{begin-eqn}} {{eqn | l = \map \lambda 1 = 1       | o = \implies       | r = b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n} = 1       | c =  }} {{eqn | l = \map \lambda 1 \ne 1       | o = \implies       | r = b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n} = 0       | c =  }} {{end-eqn}} So only those permutations on $\N^*_n$ such that $\map \lambda 1 = 1$ contribute towards the final summation. Thus we have: :$\displaystyle D = \sum_\mu \map \sgn \mu b_{2 \map \mu 2} \cdots b_{n \map \mu n}$ where $\mu$ is the collection of all permutations on $\N^*_n$ which fix $1$. Hence the result. {{Qed}}
Let: :$V_n = \begin{vmatrix} 1      & x_1    & x_1^2  & \cdots & x_1^{n - 2} & x_1^{n - 1} \\ 1      & x_2    & x_2^2  & \cdots & x_2^{n - 2} & x_2^{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots      & \vdots \\ 1      & x_n    & x_n^2  & \cdots & x_n^{n - 2} & x_n^{n - 1} \end{vmatrix}$ Let $\map f x$ be '''any''' monic polynomial of degree $n - 1$: :$\ds \map f x = x^{n - 1} + \sum_{i \mathop = 0}^{n - 2} a_i x^i$ Apply elementary column operations to $V_n$ repeatedly to show: :$V_n = W$ where: :$ W = \begin{vmatrix} 1      & x_1    & x_1^2  & \cdots & x_1^{n - 2} & \map f {x_1} \\ 1      & x_2    & x_2^2  & \cdots & x_2^{n - 2} & \map f {x_2} \\ \vdots & \vdots & \vdots & \ddots & \vdots      & \vdots \\ 1      & x_n    & x_n^2  & \cdots & x_n^{n - 2} & \map f {x_n} \end{vmatrix}$ Select a specific degree $n - 1$ monic polynomial: :$\ds \map f x = \prod_{k \mathop = 1}^{n - 1} \paren {x - x_k}$ The selected polynomial is zero at all values $x_1, \ldots, x_{n - 1}$. Then the last column of $W$ is all zeros except the entry $\map f {x_n}$. Expand $\map \det W$ by cofactors along the last column to prove: {{begin-eqn}} {{eqn | n = 1       | l = V_n       | r = \map f {x_n} V_{n - 1}       | c = Expansion Theorem for Determinants for columns }} {{eqn | r = V_{n - 1} \prod_{k \mathop = 1}^{n - 1} \paren {x_n - x_k} }} {{end-eqn}} For $n \ge 2$, let $\map P n$ be the statement: :$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$ Mathematical induction will be applied. === Basis for the Induction === By definition, determinant $V_1 = 1$. To prove $\map P 2$ is true, use equation $(1)$ with $n = 2$: :$V_2 = \paren {x_2 - x_1} V_1$ This is the basis for the induction. === Induction Step === This is the induction step: Let $\map P n$ is be assumed true. We are to prove that $\map P {n + 1}$ is true. As follows: {{begin-eqn}} {{eqn | l = V_{n + 1}       | r = V_n \prod_{k \mathop = 1}^n \paren {x_{n + 1} - x_k}       | c = from $(1)$, setting $n \to n + 1$ }} {{eqn | r = \prod_{1 \mathop \le m \mathop < k \mathop \le n} \paren {x_k - x_m} \prod_{k \mathop = 1}^n \paren {x_{n + 1} - x_k}       | c = Induction hypothesis with new indexing symbols: $i \to m, j \to k$ }} {{eqn | r = \prod_{1 \mathop \le i \mathop < j \mathop \le n + 1} \paren {x_j - x_i}       | c = simplifying }} {{end-eqn}} Thus $\map P {n + 1}$ has been shown to be true. The induction is complete. {{qed}}
Let $\mathbf A = \begin{bmatrix}  a_{11} & a_{12} & \cdots & a_{1n} \\   a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\   a_{n1} & a_{n2} & \cdots & a_{nn}\end{bmatrix}$ and $\mathbf A^* = \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\   A_{21} & A_{22} & \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\   A_{n1} & A_{n2} & \cdots & A_{nn}\end{bmatrix}$. Thus $\left({\mathbf A^*}\right)^\intercal = \begin{bmatrix} A_{11} & A_{21} & \cdots & A_{n1} \\   A_{12} & A_{22} & \cdots & A_{n2} \\ \vdots & \vdots & \ddots & \vdots \\   A_{1n} & A_{2n} & \cdots & A_{nn}\end{bmatrix}$ is the transpose of $\mathbf A^*$. Let $c_{ij}$ be the typical element of $\mathbf A \left({\mathbf A^*}\right)^\intercal$. Then $\displaystyle c_{ij} = \sum_{k \mathop = 1}^n a_{ik} A_{jk}$ by definition of matrix product. Thus by the corollary of the Expansion Theorem for Determinants, $c_{ij} = \delta_{ij} D$. So $\det \left({\mathbf A \left({\mathbf A^*}\right)^\intercal}\right) = \begin{vmatrix} D & 0 & \cdots & 0 \\   0 & D & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\   0 & 0 & \cdots & D\end{vmatrix} = D^n$ by Determinant of Diagonal Matrix. From Determinant of Matrix Product, $\det \left({\mathbf A}\right) \det \left({\left({\mathbf A^*}\right)^\intercal}\right) = \det \left({\mathbf A \left({\mathbf A^*}\right)^\intercal}\right)$ From Determinant of Transpose: : $\det \left({\left({\mathbf A^*}\right)^\intercal}\right) = \det \left({\mathbf A^*}\right)$ Thus as $D = \det \left({\mathbf A}\right)$ and $D^* = \det \left({\mathbf A^*}\right)$ it follows that $DD^* = D^n$. Now if $D \ne 0$, the result follows. However, if $D = 0$ we need to show that $D^* = 0$. Let $D^* = \begin{vmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\   A_{21} & A_{22} & \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\   A_{n1} & A_{n2} & \cdots & A_{nn}\end{vmatrix}$. Suppose that at least one element of $\mathbf A$, say $a_{rs}$, is non-zero (otherwise the result follows immediately). By the Expansion Theorem for Determinants and its corollary, we can expand $D$ by row $r$, and get: :$\displaystyle D = 0 = \sum_{j \mathop = 1}^n A_{ij} t_j, \forall i = 1, 2, \ldots, n$ for all $t_1 = a_{r1}, t_2 = a_{r2}, \ldots, t_n = a_{rn}$. But $t_s = a_{rs} \ne 0$. So, by '''(work in progress)''': :$D^* = \begin{vmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\   A_{21} & A_{22} & \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\   A_{n1} & A_{n2} & \cdots & A_{nn}\end{vmatrix} = 0$ {{WIP|One result to document, I've got to work out how best to formulate it.}} Category:Determinants bkm55tefjheohcpkb9z8u2t6jx7dz5p
As a diagonal matrix is also a triangular matrix (both upper and lower), the result follows directly from Determinant of Triangular Matrix. {{qed}} Category:Determinants Category:Diagonal Matrices 46j7gt7zea6di1bocalbw6n08ipn7hx
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 $r_i$ denote an arbitrary row of $\mathbf A$. Let $e$ denote the elementary row operation $r_i \to 1 r_i$ applied to $\mathbf A$. Then trivially: :$\map e {\mathbf A} = \mathbf A$ and so $\mathbf A$ is trivially row equivalent to itself. So row equivalence has been shown to be reflexive. {{qed|lemma}} === Symmetric === Let $\mathbf A$ be row equivalent to $\mathbf B$. Let $\Gamma$ be the row operation that transforms $\mathbf A$ into $\mathbf B$. From Row Operation has Inverse there exists a row operation $\Gamma'$ which transforms $\mathbf B$ into $\mathbf A$. Thus $\mathbf B$ is row equivalent to $\mathbf A$. So row equivalence has been shown to be symmetric. {{qed|lemma}} === Transitive === Let $\mathbf A$ be row equivalent to $\mathbf B$, and let $\mathbf B$ be row equivalent to $\mathbf C$. Let $\Gamma_1$ be the row operation that transforms $\mathbf A$ into $\mathbf B$. Let $\Gamma_2$ be the row operation that transforms $\mathbf B$ into $\mathbf C$. From Sequence of Row Operations is Row Operation, $\mathbf C$ is row equivalent to $\mathbf A$. So row equivalence has been shown to be transitive. {{qed|lemma}} Row equivalence has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
Let the first column of $\mathbf A$ containing a non-zero element be column $j$. Let such a non-zero element be in row $i$. Take element $a_{i j} \ne 0$ and perform the elementary row operations: :$(1): \quad r_i \to \dfrac {r_i} {a_{i j}}$ :$(2): \quad r_1 \leftrightarrow r_i$ This gives a matrix with $1$ in the $\tuple {1, j}$ position: :$\begin {bmatrix}      0 & \cdots &      0 &       1 & b_{1, j + 1} & \cdots & b_{1 n} \\      0 & \cdots &      0 & b_{2 j} & b_{2, j + 1} & \cdots & b_{2 n} \\ \vdots & \ddots & \vdots &  \vdots &       \vdots & \ddots &  \vdots \\      0 & \cdots &      0 & b_{m j} & b_{m, j + 1} & \cdots & b_{m n} \\ \end {bmatrix}$ Now the elementary row operations $r_k \to r_k - b_{k j} r_1, k \in \set {2, 3, \ldots, m}$ gives the matrix: :$\begin{bmatrix} 0      & \cdots &      0 &      1 & c_{1, j + 1} & \cdots & c_{1 n} \\ 0      & \cdots &      0 &      0 & c_{2, j + 1} & \cdots & c_{2 n} \\ \vdots & \ddots & \vdots & \vdots &       \vdots & \ddots &  \vdots \\ 0      & \cdots &      0 &      0 & c_{m, j + 1} & \cdots & c_{m n} \\ \end{bmatrix}$ If some zero rows have appeared, do some further elementary row operations, that is row interchanges, to put them at the bottom. We now repeat the process with the remaining however-many-there-are rows: :$\begin{bmatrix} \cdots &      0 &      1 & d_{1, j + 1} & \cdots & d_{1, k - 1} & d_{1 k} & d_{1, k + 1} & \cdots & d_{1 n} \\ \cdots &      0 &      0 &            0 & \cdots &            0 &       1 & d_{2, k + 1} & \cdots & d_{2 n} \\ \cdots &      0 &      0 &            0 & \cdots &            0 & d_{3 k} & d_{3, k + 1} & \cdots & d_{3 n} \\ \ddots & \vdots & \vdots &       \vdots & \ddots &       \vdots &  \vdots &       \vdots & \ddots & \vdots  \\ \cdots &      0 &      0 &            0 & \cdots &            0 & d_{n k} & d_{m, k + 1} & \cdots & d_{m n} \\  \end{bmatrix}$ Then we can get the reduced echelon form by: :$r_i \to r_i - d_{i k} r_2, i \in \set {1, 3, 4, \ldots, m}$ as follows: :$\begin{bmatrix} \cdots &      0 &      1 & {e_{1, j + 1 } } & \cdots & {e_{1, k - 1} } &      0 & {e_{1, k + 1} } & \cdots & {e_{1 n} } \\ \cdots &      0 &      0 &                0 & \cdots &               0 &      1 & {e_{2, k + 1} } & \cdots & {e_{2 n} } \\ \cdots &      0 &      0 &                0 & \cdots &               0 &      0 & {e_{3, k + 1} } & \cdots & {e_{3 n} } \\ \ddots & \vdots & \vdots &           \vdots & \ddots &          \vdots & \vdots &          \vdots & \ddots & \vdots     \\ \cdots &      0 &      0 &                0 & \cdots &               0 &      0 & {e_{m, k + 1} } & \cdots & {e_{m n} } \\ \end{bmatrix}$ Thus we progress, until the entire matrix is in reduced echelon form. {{Qed}}
Let $\mathbf A$ be a square matrix of order $n$. We proceed by induction on $n$, the number of rows of $\mathbf A$. === Basis for the Induction === For $n = 1$, we have a matrix of just one element, which is trivially diagonal, hence both upper and lower triangular. This is the basis for the induction. === Induction Hypothesis === Fix $n \in \N$, and assume all $n \times n$-matrices can be upper triangularised by elementary row operations.  If $R$ is a field, assume all $n \times n$-matrices can be upper triangularised by elementary row operations of type 2. This forms our induction hypothesis. === Induction Step === Let $\mathbf A = \left[{a}\right]_{n+1}$ be a square matrix of order $n + 1$. When the first column of $\mathbf A$ contains only zeroes, it is upper triangularisable iff the submatrix $\mathbf A \left({1; 1}\right)$ is. Each elementary row operation used in triangularisation process of the submatrix $\mathbf A \left({1; 1}\right)$ will not change the zeros of the first column of $\mathbf A$. So when $\mathbf A \left({1; 1}\right)$ is upper triangularised, then $A$ will also be upper triangularised. From the induction hypothesis, we conclude that $\mathbf A$ can be upper triangularised by elementary row operations. Now suppose that its first column contains a non-zero value. Suppose that $a_{11} = 0$. Let $j$ be the smallest row index such that $a_{j1} \ne 0$, and note that $j$ exists by assumption. Now apply the following operation of type 2: :$r_1 \to r_1 + r_j$ As $a_{j1} \ne 0$, this enforces $a_{11} \ne 0$, and we continue as in the case below. Suppose $a_{11} \ne 0$.  We use the following operations for all $j \in \left\{{2, \ldots, n + 1}\right\}$: :$(1): \quad$ Put $c = a_{j1}$. :$(2): \quad r_j \to a_{11} r_j$ :$(3): \quad r_j \to r_j - c r_1$ This will put the first column to zero (except for the first element, $a_{11}$). It follows that $\mathbf A$ can be upper triangularised precisely when the submatrix $\mathbf A \left({1; 1}\right)$ can. Again, the induction hypothesis renders $\mathbf A$ upper triangularisable by elementary row operations. This completes the case distinction, and hence the result follows by induction. To put the matrix $\mathbf A$ into lower triangular form, just do the same thing, but start with the last column and the last diagonal element $a_{n + 1 \; n + 1}$. {{qed}}