hcju/proofwikiqa huggingface dataset:
Using the operation Row Operation to Clear First Column of Matrix, $\mathbf A$ is converted to $\mathbf B$, which will be in the form: :$\begin{bmatrix} 0      & \cdots &      0 &      1 & b_{1, j + 1} & \cdots & b_{1 n} \\ 0      & \cdots &      0 &      0 & b_{2, j + 1} & \cdots & b_{2 n} \\ \vdots & \ddots & \vdots & \vdots &       \vdots & \ddots &  \vdots \\ 0      & \cdots &      0 &      0 & b_{m, j + 1} & \cdots & b_{m n} \\ \end{bmatrix}$ 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: :$\begin{bmatrix} b_{2, j + 1} & b_{2, j + 2} & \cdots & b_{2 n} \\ b_{3, j + 1} & b_{3, j + 2} & \cdots & b_{3 n} \\       \vdots &       \vdots & \ddots &  \vdots \\ b_{m, j + 1} & b_{m, j + 2} & \cdots & b_{m n} \\ \end{bmatrix}$ and perform the same operation on that. This results in the submatrix being transformed into the form: :$\begin{bmatrix}      1 & c_{2, j + 2} & \cdots & c_{2 n} \\      0 & c_{3, j + 2} & \cdots & c_{3 n} \\ \vdots &       \vdots & \ddots &  \vdots \\      0 & c_{m, j + 2} & \cdots & c_{m n} \\ \end{bmatrix}$ Again, we process the submatrix: :$\begin{bmatrix} c_{3, j + 2} & \cdots & c_{3 n} \\       \vdots & \ddots &  \vdots \\ c_{m, j + 2} & \cdots & c_{m n} \\ \end{bmatrix}$ Thus we progress, until the entire matrix is in echelon form. {{Qed}}
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 Theory rkwlr71vzp2a4bgrb7osuudni8p1ilv
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 Theory nmy3shzkoghepoempvcg6zhnep4tw1q
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 Theory 51xabv2osf8qf1m5mgl08foem8ue5b2
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}}
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}}
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}}
=== Necessary Condition === {{:Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Sufficient Condition}}{{qed}}
That $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\displaystyle \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges in $\R$. Hence the sequence of partial sums is a Cauchy sequence by Convergent Sequence in Normed Vector Space is 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: :$\displaystyle \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$ 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 $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is Cauchy. As $X$ is a Banach space, this implies that $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges.
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.
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 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}} Category:Sine of Integer Multiple of Argument ixx0qmzthb6kv3ovxu3cq4c13kqywd1
The proof proceeds by induction. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop \ge 0} \frac {\cos k \theta} {\cos^k \theta}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sin \theta       | r = \sin \theta       | c =  }} {{eqn | r = \sin \theta \cos^{1 - 1} \theta \paren 1       | c =  }} {{end-eqn}} So $\map P 1$ is seen to hold. $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = \sin 2 \theta       | r = 2 \sin \theta \cos \theta       | c = Double Angle Formula for Sine }} {{eqn | r = \sin \theta \cos^{2 - 1} \theta \paren {1 + 1}       | c =  }} {{end-eqn}} So $\map P 2$ is also seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P n$ is true, where $n > 2$, then it logically follows that $\map P {n + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \map \sin {n \theta} = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta } {\cos^k \theta}$ from which we are to show: :$\displaystyle \map \sin {\paren {n + 1} \theta} = \sin \theta \cos^n \theta \sum_{k \mathop = 0}^n \frac {\cos k \theta} {\cos^k \theta}$ === Induction Step === This is our induction step: For the first part: {{begin-eqn}} {{eqn | l = \map \sin {\paren {n + 1} \theta}        | r = \map \sin {n \theta + \theta}       | c = }} {{eqn | r = \sin n \theta \cos \theta + \cos n \theta \sin \theta       | c = Sine of Sum }} {{eqn | r = \paren {\sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta} } \cos \theta + \cos n \theta \sin \theta       | c =  }} {{eqn | r = \sin \theta \paren {\cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta} \cos \theta + \cos n \theta}       | c = Factor out $\sin \theta$ }} {{eqn | r = \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta}  + \cos n \theta}       | c = {{Defof|Integer Power}} }} {{eqn | r = \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta}  + \cos n \theta \frac {\cos^n \theta} {\cos^n \theta} }       | c = multiply by $1$ }} {{eqn | r = \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^n \frac {\cos k \theta} {\cos^k \theta} }       | c =  }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{>0}: \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1 } \frac {\cos k \theta} {\cos^k \theta}$ {{qed}} Category:Sine of Integer Multiple of Argument p9h2uid0r5dmsd684420g9d55w5otf2
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 $\mathbf A = \mathbf A''$. === $\text {ERO} 1$: Scalar Product of Row === {{:Existence of Inverse Elementary Row Operation/Scalar Product of Row}}{{qed|lemma}} === $\text {ERO} 2$: Add Scalar Product of Row to Another === {{:Existence of Inverse Elementary Row Operation/Add Scalar Product of Row to Another}}{{qed|lemma}} === $\text {ERO} 3$: Exchange Rows === {{:Existence of Inverse Elementary Row Operation/Exchange Rows}}{{qed|lemma}} Thus in all cases, for each elementary row operation which transforms $\mathbf A$ to $\mathbf A'$, we have constructed the only possible elementary row operation which transforms $\mathbf A'$ to $\mathbf A$. Hence the result. {{qed}}
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 Theory 37pbg3iuv6mpq72x8ovqrpjny6b0uit
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}}
{{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}}
{{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 9qv5qxvp6c5iv9vkuw2i3uqmbjdxrh4
=== 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}} === Sufficient Condition === Assume $g \neq e$. Then by Conjugacy Classes are Disjoint, $e \notin \conjclass g$. Since $\conjclass g$ does not contain the identity element, it could not be a subgroup. {{qed}}
{{improve|Rather than mentioning them in passing and expecting the reader to understand what is meant, we need $w_i$ etc. to be defined. I know from experience that you can't rely on people involved in this website to have adequate knowledge, understanding or deductive abilities to be able to work it out for themselves.}} 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 $i \in \set {1, \ldots, n - 1}$, which is vacuously true for $\epsilon$. {{qed}} Category:Abstract Algebra i739mtq52b4yqdd9s8x4u2bfbp5t96q
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}}
Let $x, y \in \map {B_1} 0$. Then: {{begin-eqn}} {{eqn | l = \norm {\paren {1 - \alpha} x + \alpha y}       | o = \le       | r = \norm {\paren {1 - \alpha} x} + \norm {\alpha y}       | c = Norm Axiom $(\text N 3)$: Triangle Inequality }} {{eqn | r = \size {1 - \alpha} \norm x + \size \alpha \norm y       | c = Norm Axiom $(\text N 2)$: Positive Homogeneity }} {{eqn | r = \paren {1 - \alpha} \norm x + \alpha \norm y       | c = {{Defof|Convex Set (Vector Space)}}: $0 \le \alpha \le 1$ }} {{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}}
Let $n$ be the cardinality of $T$. The proof goes by induction on $n$. === Basis for the Induction === Let $n = 0$. {{finish}} === Induction Step === Let $n > 0$. Let $t \in T$. Use Cardinality of Set minus Singleton {{ProofWanted}}
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 $\mathbf A$: :$\mathbf A' = \begin {bmatrix} a_{1 1} & a_{1 2} & a_{1 3} & \cdots &     a_{1, k - 1} &      a_{1 k} \\       0 & a_{2 2} & a_{2 3} & \cdots &     a_{2, k - 1} &      a_{2 k} \\       0 &       0 & a_{3 3} & \cdots &     a_{3, k - 1} &      a_{3 k} \\  \vdots &  \vdots &  \vdots & \ddots &           \vdots &      \vdots  \\       0 &       0 &       0 & \cdots & a_{k - 1, k - 1} & a_{k - 1, k} \\       0 &       0 &       0 & \cdots &                0 &      a_{k k} \\ \end {bmatrix}$ $\mathbf A$ 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 $\mathbf B$: :$\mathbf B' = \begin {bmatrix}      1 & b_{1 2} & b_{1 3} & \cdots &     b_{1, k - 1} &      b_{1 k} \\      0 &       1 & b_{2 3} & \cdots &     b_{2, k - 1} &      b_{2 k} \\      0 &       0 &       1 & \cdots &     b_{3, k - 1} &      b_{3 k} \\ \vdots &  \vdots &  \vdots & \ddots &           \vdots &      \vdots  \\      0 &       0 &       0 & \cdots &                1 & b_{k - 1, k} \\      0 &       0 &       0 & \cdots &                0 &            1 \\ \end {bmatrix}$ $\mathbf B$ 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 $\mathbf C$: :$\mathbf C' = \begin {bmatrix}      1 &      0 &      0 & \cdots &      0 &      0 \\      0 &      1 &      0 & \cdots &      0 &      0 \\      0 &      0 &      1 & \cdots &      0 &      0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\      0 &      0 &      0 & \cdots &      1 &      0 \\      0 &      0 &      0 & \cdots &      0 &      1 \\ \end {bmatrix}$ By inspection, $\mathbf C$ is seen to be the unit matrix of order $k$. {{qed}}
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: :$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$ === Basis for the Induction === $\map P 1$ is the case: :$\exists i \in \closedint 1 1 : x_i = \max \set {x_1}$ By definition of the max operation: :$\max \set {x_1} = x_1$ 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: :$\exists i \in \closedint 1 k : x_i = \max \set {x_1, x_2, \dotsc, x_k}$ from which it is to be shown that: :$\exists i \in \closedint 1 {k + 1} : x_i = \max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$ === Induction Step === This is the induction step. By the definition of Max Operation: :$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \max \set {\max \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }$ By the Induction hypothesis: :$\exists i \in \closedint 1 k : x_i = \max \set {x_1, x_2, \dotsc, x_k}$ So: :$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \max \set {x_i, x_{k + 1} }$ As $\struct {S, \preceq}$ is a totally ordered set, all elements of $S$ are comparable by $\preceq$. Therefore there are two cases to consider: ==== Case 1: $x_{k+1} \preceq x_i$ ==== By definition of the max operation: :$\max \set {x_i, x_{k + 1} } = x_i$ {{qed|lemma}} ==== Case 2: $x_i \preceq x_{k + 1}$ ==== By definition of the max operation: :$\max \set {x_i, x_{k + 1} } = x_{k + 1}$ {{qed|lemma}} In either case, the result holds. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$ {{qed}} Category:Max and Min Operations jsficvqcwn813hmtxanb6d7e6u4q5n0
{{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}}
By definition of similar matrices :$\exists \mathbf P: \mathbf P^{-1} \mathbf A \mathbf P = \mathbf B$ where $\mathbf P$ is an invertible matrix of order $n$. Thus it remains to show that: :$\map \tr {\mathbf P^{-1} \mathbf A \mathbf P} = \map \tr {\mathbf A}$ {{ProofWanted}}