OpenLLM-France/wikimedia huggingface dataset:
# Matrix/K/Transposed matrix/Definition

Let $K$ be a field, and let $M=(a_{ij})_{ij}$ be an $m\times n$-matrix over $K$. Then the $n\times m$-matrix
    ${M^{\text{tr}}}={\left(b_{ij}\right)}_{ij}{\text{ with }}b_{ij}:=a_{ji}$
is called the transposed matrix of $M$.
# Matrix/Definition/Explanations/Remark

For every $i\in I$, the family $a_{ij}$ , $j\in J$, is called the $i$-th row of the matrix, which is usually written as a row tuple
(or row vector)
    $(a_{i1},a_{i2},\ldots ,a_{in}).$
For every $j\in J$, the family $a_{ij}$ , $i\in I$, is called the $j$-th column of the matrix, usually written as a column tuple (or column vector)
    ${\begin{pmatrix}a_{1j}\\a_{2j}\\\vdots \\a_{mj}\end{pmatrix}}.$
The elements $a_{ij}$ are called the entries of the matrix. For $a_{ij}$, the number $i$ is called the row index, and $j$ is called the column index of the entry. The position of the entry $a_{ij}$ is where the $i$-th row meets the $j$-th column. A matrix with $m=n$ is called a square matrix. An $m\times 1$-matrix is simply a column tuple (or column vector) of length $m$, and an $1\times n$-matrix is simply a row tuple (or row vector) of length $n$. The set of all matrices with $m$ rows and $n$ columns (and with entries in $K$) is denoted by $\operatorname {Mat} _{m\times n}(K)$; in case $m=n$ we also write $\operatorname {Mat} _{n}(K)$.
# Real plane/Line equation/Sketch and intersection/1/Exercise

* Sketch the set $M={\left\{(x,y)\in \mathbb {R} ^{2}\mid 4x-7y=3\right\}}$ and the set $N={\left\{(x,y)\in \mathbb {R} ^{2}\mid 3x+2y=5\right\}}$.
* Determine the intersection $M\cap N$ geometrically and computationally.
# Linear equation/Integers/Possibilities/1/Exercise

* Show that the set is not empty.
    ${\left\{(x,y)\in \mathbb {Z} ^{2}\mid 3x+5y=1\right\}}$
* Show that the set is empty.
    ${\left\{(x,y)\in \mathbb {Z} ^{2}\mid 6x+9y=5\right\}}$
# Product set/Geometric examples/Exercise

Describe, for every combination (including the case where the product is taken with itself), the product set of the following geometric sets.
* A circle $K$.
* A line segment $I$.
* A line $G$.
* A parabola $P$.
Which product set can we realize as a subset in space?
# Product set/Intersection/Exercise

Let $M$ and $N$ denote sets, and let $A\subseteq M$ and $B\subseteq N$ be subsets. Show the identity
    ${\left(A\times N\right)}\cap {\left(M\times B\right)}=A\times B\,.$
# Real plane/Line equation/Sketch and intersection/2/Exercise

* Sketch the set $M={\left\{(x,y)\in \mathbb {R} ^{2}\mid -5x+2y=6\right\}}$ and the set $N={\left\{(x,y)\in \mathbb {R} ^{2}\mid 7x-5y=4\right\}}$.
* Determine the intersection $M\cap N$ by a drawing and computationally.
# Product set/Intersection/2/Exercise

Let $M$ and $N$ denote sets, and let $A_{1},A_{2}\subseteq M$ and $B_{1},B_{2}\subseteq N$ be subsets. Show the identity
    ${\left(A_{1}\times B_{1}\right)}\cap {\left(A_{2}\times B_{2}\right)}={\left(A_{1}\cap A_{2}\right)}\times {\left(B_{1}\cap B_{2}\right)}\,.$
# Set theory/4 sets/Sketch complete diagram/Exercise

Draw, for four sets, a diagram of sets that shows all possible intersecting sets.
# Finite sets/Explicit/Relationships/Exercise

Determine, for the sets
    $M=\{a,b,c,d,e\},\,N=\{a,c,e\},\,P=\{b\},\,R=\{b,d,e,f\},$
the following sets.
* $M\cap N$,
* $M\cap N\cap P\cap R$,
* $M\cup R$,
* ${\left(N\cup P\right)}\cap R$,
* $N\setminus R$,
* ${\left(M\cup P\right)}\setminus {\left(R\setminus N\right)}$,
* ${\left({\left(P\cup R\right)}\cap N\right)}\cap R$,
* ${\left(R\setminus P\right)}\cap {\left(M\setminus N\right)}$.
# Alphabets/Several sets/Exercise

Let $LA$ denote the set of capital letters in the Latin alphabet, $GA$ the set of capital letters in the Greek alphabet, and $RA$ the set of capital letters in the Russian alphabet. Determine the following sets.
* $GA\setminus RA$.
* ${\left(LA\cap GA\right)}\cup {\left(LA\cap RA\right)}$.
* $RA\setminus {\left(GA\cup RA\right)}$.
* $RA\setminus {\left(GA\cup LA\right)}$.
* ${\left(RA\setminus GA\right)}\cap {\left({\left(LA\cup GA\right)}\setminus {\left(GA\cap RA\right)}\right)}$.
# Finite set/Power set/Number/Fact/Proof/Exercise

Let $G$ denote a finite set with $n$ elements. Show that the power set ${\mathfrak {P}}\,(G)$ contains $2^{n}$ elements.
# Rules for functions/From school/Exercise

What functions (and their defining rules) do you know from school?
# Composition/Polynomial example/2/Exercise

Determine the compositions $\varphi \circ \psi$ and $\psi \circ \varphi$ for the mappings $\varphi ,\psi \colon \mathbb {R} \rightarrow \mathbb {R}$ that are defined by
    $\varphi (x)=x^{3}+3x^{2}-4\,\,{\text{ and }}\,\,\psi (x)=x^{2}+5x-3.$
# Mapping/Take preimage/Exercise

Let $F\colon L\rightarrow M$ denote a mapping. Show that taking preimages
    ${\mathfrak {P}}\,(M)\longrightarrow {\mathfrak {P}}\,(L),T\longmapsto F^{-1}(T),$
fulfills the following properties (for arbitrary subsets $T,T_{1},T_{2}\subseteq M$):
    $F^{-1}(T_{1}\cap T_{2})=F^{-1}(T_{1})\cap F^{-1}(T_{2})\,,$
    $F^{-1}(T_{1}\cup T_{2})=F^{-1}(T_{1})\cup F^{-1}(T_{2})\,,$
    $F^{-1}(M\setminus T)=L\setminus F^{-1}(T)\,.$
# Composition/Polynomial example/1/Exercise

Determine the
composite functions $\varphi \circ \psi$ and $\psi \circ \varphi$ for the functions $\varphi ,\psi \colon \mathbb {R} \rightarrow \mathbb {R}$, defined by
    $\varphi (x)=x^{4}+3x^{2}-2x+5\,\,{\text{ and }}\,\,\psi (x)=2x^{3}-x^{2}+6x-1.$
# Mapping/Bijection between N and Z/Exercise

Show that there exists a bijection between $\mathbb {N}$ and $\mathbb {Z}$.
# Mapping/Composition/Right surjective/Exercise

Let $L,M,N$ be sets, and let
    $f:L\longrightarrow M{\text{ and }}g:M\longrightarrow N$
be mappings with their composite mapping
    $g\circ f\colon L\longrightarrow N,x\longmapsto g(f(x)).$
Show that if $g\circ f$ is
surjective, then also $g$ is surjective.
# Mappings/Table of values/25614377/1003the composition/Exercise

Consider the set $M=\{1,2,3,4,5,6,7,8\}$, and the mapping
    $\varphi \colon M\longrightarrow M,x\longmapsto \varphi (x),$
defined by the following table
Compute $\varphi ^{1003}$, that is, the $1003$-rd composition (or iteration) of $\varphi$ with itself.
# Function/R/Strictly increasing/Definition/Injective/Exercise

A function
    $f\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto f(x),$
is called strictly increasing if for all $x_{1},x_{2}\in \mathbb {R}$ satisfying $x_{1}<x_{2}$, also $f(x_{1})<f(x_{2})$ holds. Show that a strictly increasing function $f$ is injective.
# Mapping/Composition/Leftinjective/Exercise

Let $L,M,N$ be sets and let
    $f:L\longrightarrow M{\text{ and }}g:M\longrightarrow N$
be mappings with their composition
    $g\circ f\colon L\longrightarrow N,x\longmapsto g(f(x)).$
Show that if $g\circ f$ is
injective, then also $f$ is injective.
# Four tuple/Modulus of differences/Introduction/Exercise

We consider the mapping
    $\Psi \colon \mathbb {N} ^{4}\longrightarrow \mathbb {N} ^{4}$
that assigns to a four tuple $(a,b,c,d)$ the four-tuple
    $(\vert {b-a}\vert ,\vert {c-b}\vert ,\vert {d-c}\vert ,\vert {a-d}\vert ).$
We denote by $\Psi ^{n}$ the $n$-th fold composition of $\Psi$ with itself.
* Compute until the result is the zero-tuple $(0,0,0,0)$.
    $\Psi (6,5,2,8),\,\Psi ^{2}(6,5,2,8),\,\Psi ^{3}(6,5,2,8),\,\Psi ^{4}(6,5,2,8)\,...$
* Compute until the result is the zero-tuple $(0,0,0,0)$.
    $\Psi (1,10,100,1000),\,\Psi ^{2}(1,10,100,1000),\,\Psi ^{3}(1,10,100,1000),\,\Psi ^{4}(1,10,100,1000)\,...$
* Show that $\Psi ^{4}(0,0,n,0)=(0,0,0,0)$ for every $n\in \mathbb {N}$.
# Four tuple/Modulus of differences/Mapping properties/Exercise

We consider the mapping
    $\Psi \colon \mathbb {N} ^{4}\longrightarrow \mathbb {N} ^{4}$
that assigns to a four-tuple $(a,b,c,d)$ the four-tuple
    $(\vert {b-a}\vert ,\vert {c-b}\vert ,\vert {d-c}\vert ,\vert {a-d}\vert ).$
Determine whether $\Psi$ is injective and whether $\Psi$ is surjective.
# Four tuple/Modulus of differences/Q/Descent/Exercise

We consider the mapping
    $\Psi \colon \mathbb {Q} _{\geq 0}^{4}\longrightarrow \mathbb {Q} _{\geq 0}^{4}$
that assigns to a four-tuple of nonnegative rational numbers $(a,b,c,d)$ the four-tuple
    $(\vert {b-a}\vert ,\vert {c-b}\vert ,\vert {d-c}\vert ,\vert {a-d}\vert )$
Show that after finitely many iterations, this mapping yields the zero-tuple.
# Four tuple/Modulus of differences/Descent/Exercise

We consider the mapping
    $\Psi \colon \mathbb {N} ^{4}\longrightarrow \mathbb {N} ^{4}$
that assigns to a four-tuple $(a,b,c,d)$ the four-tuple
    $(\vert {b-a}\vert ,\vert {c-b}\vert ,\vert {d-c}\vert ,\vert {a-d}\vert ).$
Show that for any initial value $(a,b,c,d)$, after finitely many iterations, this map reaches the zero-tuple.
# Four tuple/Modulus of differences/25/Exercise

We consider the mapping
    $\Psi \colon \mathbb {N} ^{4}\longrightarrow \mathbb {N} ^{4}$
that assigns to a four-tuple $(a,b,c,d)$ the four-tuple
    $(\vert {b-a}\vert ,\vert {c-b}\vert ,\vert {d-c}\vert ,\vert {a-d}\vert ).$
Find an example of a four-tuple $(a,b,c,d)$ with the property that all iterations $\Psi ^{n}(a,b,c,d)$ for $n\leq 25$ do not yield the zero-tuple. Check your result on http://www.vier-zahlen.bplaced.net/raetsel.php .
# Mapping/N/Injective not surjective/Exercise

Give examples of mappings
    $\varphi ,\psi \colon \mathbb {N} \longrightarrow \mathbb {N}$
such that $\varphi$ is injective but not surjective, and $\psi$ is surjective but not injective.
# Mapping/Mother of/Aspects/Exercise

* Let $H$ be the set of all (alive or dead) people. Study the mapping that assigns to every person the mother, with respect to injectivity and surjectivity.
    $\varphi \colon H\longrightarrow H$
* What is the meaning of the composition $\varphi ^{3}$?
* How does it look when we take the same mapping but, in the domain, restricted to the set $E$ of all only children, and, in the codomain, restricted to the set $M$ of all mothers?
* Be a bit hairsplitting and argue (referring to evolution or religion) that the mapping in (1) is not well-defined.
# Power set/Complement/Bijection/Exercise

Let $G$ be a set, and ${\mathfrak {P}}\,(G)$ its power set. Show that the mapping
    ${\mathfrak {P}}\,(G)\longrightarrow {\mathfrak {P}}\,(G),T\longmapsto \complement T,$
is bijective. How does the inverse mapping look?
# Arithmetic mean on lines/Not associative/Exercise

Show that the operation on a line that assigns to two points their midpoint, is commutative, but not associative. Does there exist a neutral element?
# AIEEE 2009 Q014 calculus

Reply to the thread, posted by 41.121.35.161 (discuss)
Thread post
Additional notes
These comments, along with the notes on the left, were contributed by 41.121.35.161 (discuss) 18:03, 27 October 2024 (UTC) :
# Death penalty

The death penalty, or capital punishment, is where a person is killed by the state as a punishment for a crime.

## In US law

* Legal Information Institute
* The Death Penalty - Exploring Constitutional Conflicts, University of Missouri-Kansas City
* Death penalty statistics from the US: which state executes the most people?, The Guardian

## Religious perspectives

* Theology Library page, Spring Hill College - primarily lists articles and resources related to the Catholic view of the death penalty, but also contains other faith groups, both Christian and non-Christian
* BBC Ethics: Capital punishment contains a section detailing the death penalty and views from Buddhism, Christianity, Hinduism, Islam and Judaism.
* religioustolerance.org
* Religion and capital punishment at Wikipedia

## Public opinion polls

* Pew Forum on Religion & Public Life: Death Penalty

## At Wikimedia sister projects

* Capital punishment on Wikipedia
* Wikinews
* Death penalty on Wiktionary
# Propositional logic/Zombie/Tautology/Exercise

We consider the following quotation of Sven Walter from the article "Zombies, Dualismus und Physikalismus“ (Zeitschrift for philosophische Forschung, Bd. 65, H. 2 (2011), pp. 241-254, https://www.jstor.org/stable/pdf/41346224.pdf).
" ($P_{1}$) Zombies are imaginable.
($P_{2}$) If Zombies are imaginable, then Zombies are possible.
($P_{3}$) If Zombies are possible, then the physicalism is wrong.
Hence: the physicalism is wrong.“
Formalize the propositional rules used here, and show, with the help of truth tables, that they are tautologies.
# Vector space/Complex numbers as real vector space/Example

The complex numbers $$ form a field, and therefore they form also a vector space over the field $\mathbb {C}$ itself. However, the set of complex numbers equals $\mathbb {R} ^{2}$ as an additive group. The multiplication of a complex number $a+b{\mathrm {i} }$ with a real number $s=(s,0)$ is componentwise, so this multiplication coincides with the scalar multiplication on $\mathbb {R} ^{2}$. Hence, the set of complex numbers is also a real vector space.
# Vector space/Real numbers over rational numbers/Example

We consider the inclusion $\mathbb {Q} \subseteq \mathbb {R}$ of the rational numbers inside the real numbers. Using the real addition and the multiplication of rational numbers with real numbers, we see that $\mathbb {R}$ is a $\mathbb {Q}$-vector space, as follows directly from the field axioms. This is a quite crazy vector space.
# Plane/Vector space/Intuitive/Example

Let $E$ be a "plane“ with a fixed "origin“ $Q\in E$. We identify a point $P\in E$ with the connecting vector ${\overrightarrow {QP}}$ (the arrow from $Q$ to $P$). In this situation, we can introduce an intuitive coordinate-free vector addition and a coordinate-free scalar multiplication. Two vectors ${\overrightarrow {QP}}$ and ${\overrightarrow {QR}}$ are added together by constructing the parallelogram of these vectors. The result of the addition is the corner of the parallelogram which lies in opposition to $Q$. In this construction, we have to draw a line parallel to ${\overrightarrow {QP}}$ through $R$ and a line parallel to ${\overrightarrow {QR}}$ through $P$. The intersection point is the point sought after. An accompanying idea is that we move the vector ${\overrightarrow {QP}}$ in a parallel way so that the new starting point of ${\overrightarrow {QP}}$ becomes the ending point of ${\overrightarrow {QR}}$.
In order to describe the multiplication of a vector ${\overrightarrow {QP}}$ with a scalar $s$, this number has to be given on a line $G$ that is also marked with a zero point $0\in G$ and a unit point $1\in G$. This line lies somewhere in the plane. We move this line
(by translating and rotating) in such a way that $0$ becomes $Q$, and we avoid that the line is identical with the line given by ${\overrightarrow {QP}}$ (which we call $H$).
Now we connect $1$ and $P$ with a line $L$, and we draw the line $L'$ parallel to $L$ through $s$. The intersecting point of $L'$ and $H$ is $s{\overrightarrow {QP}}$.
These considerations can also be done in higher dimensions, but everything takes place already in the plane spanned by these vectors.
# Vector space/mxn-matrices/Example

For a field $K$, and given natural numbers $m,n$, the set
    $\operatorname {Mat} _{m\times n}(K)$
of all $m\times n$-matrices, endowed with componentwise addition and componentwise scalar multiplication, is a $K$-vector space. The null element in this vector space is the null matrix
    $0={\begin{pmatrix}0&\ldots &0\\\vdots &\ddots &\vdots \\0&\ldots &0\end{pmatrix}}\,.$
# Vector space/Polynomials/Quick/Example

Let $R=K[X]$ be the polynomial ring in one variable over the field $K$, consisting of all polynomials, that is, expressions of the form
    $a_{n}X^{n}+a_{n-1}X^{n-1}+\cdots +a_{2}X^{2}+a_{1}X+a_{0},$
with $a_{i}\in K$. Using componentwise addition and componentwise multiplication with a scalar $s\in K$ (this is also multiplication with the constant polynomial $s$), the polynomial ring is a $K$-vector space.
# System of linear equations/Set of variables/Solution space is vector space/Fact

Let $K$ be a field, and let
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
be a homogeneous system of linear equations over $K$.
linear subspace
$K^{n}$
# Vector space/Linear subspace/Closed/Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. A subset $U\subseteq V$ is called a linear subspace if the following properties hold.
* $0\in U$.
* If $u,v\in U$, then also $u+v\in U$.
* If $u\in U$ and $s\in K$, then also $su\in U$ holds.
# Linear inhomogeneous system of equations/Elimination/2x+5y+2z-v, 3x-4y+u+2v, 4x -2z+2u/Example

We take a look at the homogeneous version of example, so we consider the homogeneous linear system
    ${\begin{matrix}2x&+5y&+2z&&-v&=&0\\\,3x&-4y&&+u&+2v&=&0\\\,4x&&-2z&+2u&&=&0\,\end{matrix}}$
over $\mathbb {R}$. Due to fact, the solution set $L$ is a linear subspace of $\mathbb {R} ^{5}$. We have described it explicitly in example as
    ${\left\{u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}}.$
This description also shows that the solution set is a vector space. Moreover, with this description, it is clear that $L$ is in bijection with $\mathbb {R} ^{2}$, and this bijection respects the addition and also the scalar multiplication (the solution set $L'$ of the inhomogeneous system is also in bijection with $\mathbb {R} ^{2}$, but there is no reasonable addition nor scalar multiplication on $L'$). However, this bijection depends heavily on the chosen "basic solutions“ ${\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}$ and ${\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}$, which depends on the order of elimination. There are several equally good basic solutions for $L$.
# Linear subspace/Solution space for linear system/Introduction/Section
## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. A subset $U\subseteq V$ is called a linear subspace if the following properties hold.
* $0\in U$.
* If $u,v\in U$, then also $u+v\in U$.
* If $u\in U$ and $s\in K$, then also $su\in U$ holds.
Addition and scalar multiplication can be restricted to such a linear subspace. Hence, the linear subspace is itself a vector space, see exercise. The simplest linear subspaces in a vector space $V$ are the null space $0$ and the whole vector space $V$.

## Lemma

Let $K$ be a field, and let
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
be a homogeneous system of linear equations over $K$. Then the set of all solutions to the system is a linear subspace of the standard space $K^{n}$.

### Proof

$\Box$
Therefore, we talk about the solution space of the linear system. In particular, the sum of two solutions of a system of linear equations is again a solution. The solution set of an inhomogeneous linear system is not a vector space. However, one can add, to a solution of an inhomogeneous system, a solution of the corresponding homogeneous system, and get a solution of the inhomogeneous system again.

## Example

We take a look at the homogeneous version of example, so we consider the homogeneous linear system
    ${\begin{matrix}2x&+5y&+2z&&-v&=&0\\\,3x&-4y&&+u&+2v&=&0\\\,4x&&-2z&+2u&&=&0\,\end{matrix}}$
over $\mathbb {R}$. Due to fact, the solution set $L$ is a linear subspace of $\mathbb {R} ^{5}$. We have described it explicitly in example as
    ${\left\{u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}}.$
This description also shows that the solution set is a vector space. Moreover, with this description, it is clear that $L$ is in bijection with $\mathbb {R} ^{2}$, and this bijection respects the addition and also the scalar multiplication (the solution set $L'$ of the inhomogeneous system is also in bijection with $\mathbb {R} ^{2}$, but there is no reasonable addition nor scalar multiplication on $L'$). However, this bijection depends heavily on the chosen "basic solutions“ ${\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}$ and ${\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}$, which depends on the order of elimination. There are several equally good basic solutions for $L$.
This example shows also the following: the solution space of a linear system over $K$ is "in a natural way“, that means, independent on any choice, a linear subspace of $K^{n}$ (where $n$ is the number of variables). For this solution space, there always exists a "linear bijection“ (an "isomorphism“) to some $K^{d}$ ($d\leq n$), but there is no natural choice for such a bijection. This is one of the main reasons to work with abstract vector spaces, instead of just $K^{n}$.
# Vector space/Linear combination/Generating system/Introduction/Section
## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. Let $v_{1},\ldots ,v_{n}$ denote a family of vectors in $V$. Then the vector
    $s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}{\text{ with }}s_{i}\in K$
is called a linear combination of this vectors
$(s_{1},\ldots ,s_{n})$
Two different coefficient tuples can define the same vector.

## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. A family $v_{i}\in V$, $i\in I$, is called a generating system (or spanning system) of $V$, if every vector $v\in V$ can be written as
    $v=\sum _{j\in J}s_{j}v_{j}\,,$
with a finite subfamily $J\subseteq I$, and with
$s_{j}\in K$
In $K^{n}$, the standard vectors $e_{i}$, $1\leq i\leq n$, form a generating system. In the polynomial ring $K[X]$, the powers $X^{n}$, $n\in \mathbb {N}$, form an (infinite) generating system.

## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. For a family $v_{i}$, $i\in I$, we set
    $\langle v_{i},\,i\in I\rangle ={\left\{\sum _{i\in J}s_{i}v_{i}\mid s_{i}\in K,\,J\subseteq I{\text{ finite subset}}\right\}}\,,$
The empty set generates the null space. The null space is also generated by the element $0$. A single vector $v$ spans the space $Kv={\left\{sv\mid s\in K\right\}}$. For $v\neq 0$, this is a line, a term we will make more precise in the framework of dimension theory. For two vectors $v$ and $w$, the "form“ of the spanned space depends on how the two vectors are related to each other. If they both lie on a line, say $w=sv$, then $w$ is superfluous, and the linear subspace generated by the two vectors equals the linear subspace generated by $v$. If this is not the case (and $v$ and $w$ are not $0$), then the two vectors span a "plane“.
We list some simple properties for generating systems and linear subspaces.

## Lemma

Let $K$ be a field, and let $V$ be a
$K$
* For a family $v_{i}$, $i\in I$, of elements in $V$, the linear span is a linear subspace of $V$.
* The family $v_{i}$, $i\in I$, is a spanning system of $V$, if and only if
    $\langle v_{i},\,i\in I\rangle =V\,.$

### Proof

$\Box$
* ↑ This follows from the definition, if we use the convention that the empty sum equals $0$.