OpenLLM-France/wikimedia huggingface dataset:
# Linear mapping/Finite dimensional/Change of basis/Fact/Proof

The linear standard mappings $K^{n}\rightarrow V$ and $K^{m}\rightarrow W$ for the various bases are denoted by $\Psi _{\mathfrak {v}},\,\Psi _{\mathfrak {u}},\,\Psi _{\mathfrak {w}},\,\Psi _{\mathfrak {z}}$. We consider the commutative diagram
    ${\begin{matrix}K^{n}&&&{\stackrel {M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )}{\longrightarrow }}&&&K^{m}\\&\searrow \Psi _{\mathfrak {v}}\!\!\!\!\!&&&&\Psi _{\mathfrak {w}}\swarrow \!\!\!\!\!&\\\!\!\!\!\!M_{\mathfrak {u}}^{\mathfrak {v}}\downarrow &&V&{\stackrel {\varphi }{\longrightarrow }}&W&&\,\,\,\,\downarrow M_{\mathfrak {z}}^{\mathfrak {w}}\\&\nearrow \Psi _{\mathfrak {u}}\!\!\!\!\!&&&&\Psi _{\mathfrak {z}}\nwarrow \!\!\!\!\!&\\K^{n}&&&{\stackrel {M_{\mathfrak {z}}^{\mathfrak {u}}(\varphi )}{\longrightarrow }}&&&K^{m},\!\!\!\!\!\end{matrix}}$
where the commutativity rests on fact and fact. In this situation, we have altogether
    ${\begin{aligned}M_{\mathfrak {z}}^{\mathfrak {u}}(\varphi )&=\Psi _{\mathfrak {z}}^{-1}\circ \varphi \circ \Psi _{\mathfrak {u}}\\&=\Psi _{\mathfrak {z}}^{-1}\circ (\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ \Psi _{\mathfrak {v}}^{-1})\circ \Psi _{\mathfrak {u}}\\&=(\Psi _{\mathfrak {z}}^{-1}\circ \Psi _{\mathfrak {w}})\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (\Psi _{\mathfrak {v}}^{-1}\circ \Psi _{\mathfrak {u}})\\&=(\Psi _{\mathfrak {z}}^{-1}\circ \Psi _{\mathfrak {w}})\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (\Psi _{\mathfrak {u}}^{-1}\circ \Psi _{\mathfrak {v}})^{-1}\\&=M_{\mathfrak {z}}^{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1}.\end{aligned}}$
# Vector space/Isomorphic iff equal dimension/Fact/Proof/Exercise

Let $K$ denote a field, and let $V$ and $W$ denote finite-dimensional $K$-vector spaces. Show that $V$ and $W$ are isomorphic to each other if and only if their dimension coincides.
# MDLD/transformation matrix (basis)

Start=transformation matrix (basis) Target=Base change/Transformation matrix/Definition
# Homogeneous linear system/Dimension estimate/Fact

Let a homogeneous system of linear equations with $k$ equations in $n$ variables be given.
dimension
$n-k$
# MDLD/direct sum decomposition

Start=direct sum decomposition Target=Vector space/Linear subspaces/Finite/Direct sum/Definition
# MDLD/polynomial ring (1)

Start=polynomial ring (1) Target=Polynomial ring/Field/One variable/Definition
# MDLD/finite-dimensional

Start=finite-dimensional Target=Vector space/Finitely generated/Dimension/Definition
# MDLD/family of subsets

Start=family of subsets Target=Set theory/Indexed family of sets/Definition
# MDLD/neutral element

Start=neutral element Target=Binary operation/Neutral element/Definition
# MDLD/inverse element

Start=inverse element Target=Binary operation/Inverse element/Definition
# MDLD/associative

Start=associative Target=Binary operation/Associative/Definition
# MDLD/commutative

Start=commutative Target=Binary operation/Commutative/Definition
# MDLD/composition of mappings

Start=composition of mappings Target=Mapping/Composition/Definition
# MDLD/identity

Start=identity Target=Mapping/Introduction/Section
# MDLD/group

Start=group Target=Group/Direct/Definition
# MDLD/ring

Start=ring Target=Algebra/Ring/Direct/Definition
# MDLD/commutative ring

Start=commutative ring Target=Ring/Commutative/Definition
# Vector space/K^n component wise/Example

Let $K$ denote a field, and let $n\in \mathbb {N} _{+}$. Then the product set
    $K^{n}=\underbrace {K\times \cdots \times K} _{n{\text{-times}}}={\left\{(x_{1},\ldots ,x_{n})\mid x_{i}\in K\right\}}\,,$
with componentwise addition and with scalar multiplication given by
    $s(x_{1},\ldots ,x_{n})=(sx_{1},\ldots ,sx_{n})\,,$
is a vector space. This space is called the $n$-dimensional standard space. In particular, $K^{1}=K$ is a vector space.
# K^n/Linear subspace/Sum is 0/Basis/Example

We consider the $K$-linear subspace $U\subset K^{n}$ given by
    $U={\left\{v\in K^{n}\mid \sum _{i=1}^{n}v_{i}=0\right\}}\,.$
A basis for $U$ is given by the $n-1$ vectors
    $u_{1}=(1,-1,0,0,\ldots ,0),\,u_{2}=(0,1,-1,0,\ldots ,0),\,,\ldots ,u_{n-1}=(0,0,\ldots ,0,1,-1),\,.$
These vectors belong evidently to $U$. The linear independence can be checked in $K^{n}$. From an equation
    $\sum _{i=1}^{n-1}a_{i}u_{i}=0\,$
we can deduce step by step $a_{1}=0$, $a_{2}=0$, etc. That the system is a generating system follows from
    ${\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\\\vdots \\v_{n-1}\\v_{n}\end{pmatrix}}=v_{1}{\begin{pmatrix}1\\-1\\0\\\vdots \\0\\0\end{pmatrix}}+(v_{1}+v_{2}){\begin{pmatrix}0\\1\\-1\\\vdots \\0\\0\end{pmatrix}}+(v_{1}+v_{2}+v_{3}){\begin{pmatrix}0\\0\\1\\-1\\\vdots \\0\end{pmatrix}}+\cdots +(v_{1}+v_{2}+v_{3}+\cdots +v_{n-1}){\begin{pmatrix}0\\0\\0\\\vdots \\1\\-1\end{pmatrix}}\,,$
where the equality in the last row rests on the condition
    $\sum _{i=1}^{n}v_{i}=0\,.$
# MDLD/describing matrix (linear)

Start=describing matrix (linear) Target=Linear mapping/Matrix to basis and reverse/Definition
# MDLD/isomorphism (vs)

Start=isomorphism (vs) Target=Linear mapping/Field/Isomorphism/Definition
# MDLD/isomorphic (vs)

Start=isomorphic (vs) Target=Vector spaces/Isomorphic/Definition
# Vector space/Basis and isomorphic to K^n/Remark

An isomorphism between an $n$-dimensional vector space $V$ and the standard space $K^{n}$ is essentially equivalent with the choice of a basis of $V$. For a basis
    ${\mathfrak {v}}=v_{1},\ldots ,v_{n}\,,$
we associate the linear mapping
    $\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,e_{i}\longmapsto v_{i},$
which maps from the standard space to the vector space by sending the $i$-th standard vector to the $i$-th basis vector of the given basis. This defines a unique linear mapping due to fact. Due to exercise, this mapping is bijective. It is just the mapping
    $(a_{1},\ldots ,a_{n})\longmapsto \sum _{i=1}^{n}a_{i}v_{i}.$
The inverse mapping
    $x=\Psi _{\mathfrak {v}}^{-1}\colon V\longrightarrow K^{n}$
is also linear, and it is called the coordinate mapping for this basis. The $i$-th component of this map, that is, the composed mapping
    $x_{i}=p_{i}\circ x\colon V\longrightarrow K,v\longmapsto (\Psi _{\mathfrak {v}}^{-1}(v))_{i},$
is called the $i$-th coordinate function. It is denoted by $v_{i}^{*}$. It assigns to vector $v\in V$ with the unique representation
    $v=\sum _{i=1}^{n}\lambda _{i}v_{i}\,$
the coordinate $\lambda _{i}$. Note that the linear mapping $v_{i}^{*}$ depends on the basis, not only on the vector $v_{i}$.
If an isomorphism
    $\Psi \colon K^{n}\longrightarrow V$
is given, then the images
    $\Psi (e_{i}),i=1,\ldots ,n,$
form a basis of $V$.
# MDLD/inverse

Start=inverse Target=Mapping/Bijective/Inverse mapping/Definition
# MDLD/preimage

Start=preimage Target=Mapping/Preimage under a mapping/Definition
# MDLD/kernel

Start=kernel Target=Linear mapping/Kernel/Definition
# MDLD/kernel (vs)

Start=kernel (vs) Target=Linear mapping/Kernel/Definition
# Linear mapping/Dimension formula/Fact/Proof

Set $n=\dim _{\left(V\right)}$. Let $U=\operatorname {kern} \varphi \subseteq V$ denote the kernel of the mapping and let $k=\dim _{\left(U\right)}$ denote its dimension ($k\leq n$). Let
    $u_{1},\ldots ,u_{k}$
be a basis of $U$. Due to fact, there exist vectors
    $v_{1},\ldots ,v_{n-k}$
such that
    $u_{1},\ldots ,u_{k},\,v_{1},\ldots ,v_{n-k}$
is a basis of $V$. We claim that
    $w_{j}=\varphi (v_{j}),\,j=1,\ldots ,n-k,$
is a basis of the image. Let $w\in W$ be an element of the image $\varphi (V)$. Then there exists a vector $v\in V$ such that $\varphi (v)=w$. We can write $v$ with the basis as
    $v=\sum _{i=1}^{k}s_{i}u_{i}+\sum _{j=1}^{n-k}t_{j}v_{j}\,.$
Then we have
    ${\begin{aligned}w&=\varphi (v)\\&=\varphi {\left(\sum _{i=1}^{k}s_{i}u_{i}+\sum _{j=1}^{n-k}t_{j}v_{j}\right)}\\&=\sum _{i=1}^{k}s_{i}\varphi (u_{i})+\sum _{j=1}^{n-k}t_{j}\varphi (v_{j})\\&=\sum _{j=1}^{n-k}t_{j}w_{j},\end{aligned}}$
which means that $w$ is a linear combination in terms of the $w_{j}$. In order to prove that the family $w_{j}$, $j=1,\ldots ,n-k$, is linearly independent, let a representation of zero be given,
    $0=\sum _{j=1}^{n-k}t_{j}w_{j}\,.$
Then
    $\varphi {\left(\sum _{j=1}^{n-k}t_{j}v_{j}\right)}=\sum _{j=1}^{n-k}t_{j}\varphi {\left(v_{j}\right)}=0\,.$
Therefore, $\sum _{j=1}^{n-k}t_{j}v_{j}$ belongs to the kernel of the mapping. Hence, we can write
    $\sum _{j=1}^{n-k}t_{j}v_{j}=\sum _{i=1}^{k}s_{i}u_{i}\,.$
Since this is altogether a basis of $V$, we can infer that all coefficients are $0$, in particular, $t_{j}=0$.
# Linear system/Kernel/Relation/Remark

For an $m\times n$-matrix $M$, the kernel of the linear mapping
    $K^{n}\longrightarrow K^{m},x\longmapsto Mx,$
given by $M$ is just the solution space of the homogeneous linear system
    $Mx=0\,.$
# Linear mapping/Dimension formula/Simple cases/Remark

Let $\varphi \colon V\rightarrow W$ be a linear mapping, where $V$ has finite dimension. The dimension formula can be illustrated with the following special cases. If $\varphi$ is the zero mapping, then $\operatorname {kern} \varphi =V$ and
    $\operatorname {rk} \,\varphi =0\,.$
If $\varphi$ is injective, then $\operatorname {kern} \varphi =0$ and
    $\operatorname {rk} \,\varphi =\dim _{\left(V\right)}\,.$
The rank is always between $0$ and the dimension of the source space $V$. If $\varphi$ is surjective, then
    $\operatorname {rk} \,\varphi =\dim _{\left(W\right)}\,$
and
    $\dim _{\left(V\right)}=\dim _{\left(\operatorname {kern} \varphi \right)}+\dim _{\left(W\right)}\,.$
# MDLD/rank (linear mapping)

Start=rank (linear mapping) Target=Linear mapping/On finite dimensional space/Rank/Definition
# MDLD/finite-dimensional (vs)

Start=finite-dimensional (vs) Target=Vector space/Finitely generated/Dimension/Definition
# MDLD/direct sum decompositions

Start=direct sum decompositions See=direct sum decomposition MDLD/direct sum decomposition Target=Vector space/Linear subspaces/Finite/Direct sum/Definition
# MDLD/dimensions (vs)

Start=dimensions (vs) See=dimension (vs) MDLD/dimension (vs) Target=Vector space/Finitely generated/Dimension/Definition
# MDLD/isomorphic (linear)

Start=isomorphic (linear) Target=Vector spaces/Isomorphic/Definition
# MDLD/direct complement

Start=direct complement Target=Vector space/Direct sum/Introduction/Section
# MDLD/linear form

Start=linear form Target=Linear mapping/Linear form/Definition
# MDLD/space of homomorphisms

Start=space of homomorphisms Target=Linear Algebra/K/Homomorphism space/Definition
# MDLD/linear forms

Start=linear forms See=linear form MDLD/linear form Target=Linear mapping/Linear form/Definition
# Finite-dimensional vector space/Dual basis/Definition

Let $V$ denote a finite-dimensional $K$-vector space, endowed with a basis $v_{1},\ldots ,v_{n}$. Then the linear forms
    $v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,,$
defined by
    $v_{i}^{*}(v_{j})=\delta _{ij}={\begin{cases}1,{\text{ falls }}i=j\,,\\0,{\text{ falls }}i\neq j\,,\end{cases}}\,$
are called the dual basis of the given basis.
# MDLD/dual basis

Start=dual basis Target=Finite-dimensional vector space/Dual basis/Definition
# MDLD/dual space

Start=dual space Target=Vector space/Dual space/Definition
# Vector space/Dual basis/Base change/Fact

Let $V$ be a finite-dimensional $K$-vector space. Let $v_{1},\ldots ,v_{n}$ be a basis of $V$ with the dual basis $v_{1}^{*},\ldots ,v_{n}^{*}$, and let $w_{1},\ldots ,w_{n}$ be another basis with the dual basis $w_{1}^{*},\ldots ,w_{n}^{*}$, and with
    $w_{r}=\sum _{k=1}^{n}a_{kr}v_{k}\,.$
    $w_{j}^{*}=\sum _{i=1}^{n}b_{ij}v_{i}^{*}\,,$
where ${\left(b_{ij}\right)}_{ij}={{\left(A^{-1}\right)}^{\text{tr}}}$ is the transposed matrix of the
inverse matrix of
$A={\left(a_{kr}\right)}_{kr}$
# R^2/(2,1),(-1,3)/Dual basis/Standard dual basis/Example

We consider $\mathbb {R} ^{2}$ with the standard basis $e_{1},e_{2}$, its dual basis $e_{1}^{*},e_{2}^{*}$, and the basis consisting in $u_{1}={\begin{pmatrix}2\\1\end{pmatrix}}$ and $u_{2}={\begin{pmatrix}-1\\3\end{pmatrix}}$. We want to express the dual basis $u_{1}^{*}$ and $u_{2}^{*}$ as a linear combination of the standard dual basis, that is, we want to determine the coefficients $a$ and $b$ (and $c$ and $d$) in
    $u_{1}^{*}=ae_{1}^{*}+be_{2}^{*}\,$
(and in $u_{2}^{*}=ce_{1}^{*}+de_{2}^{*}$). Here, $a=u_{1}^{*}(e_{1})$ and $b=u_{1}^{*}(e_{2})$. In order to compute this, we have to express $e_{1}$ and $e_{2}$ as a linear combination of $u_{1}$ and $u_{2}$. This is
    $e_{1}={\frac {3}{7}}{\begin{pmatrix}2\\1\end{pmatrix}}-{\frac {1}{7}}{\begin{pmatrix}-1\\3\end{pmatrix}}\,$
and
    $e_{2}={\frac {1}{7}}{\begin{pmatrix}2\\1\end{pmatrix}}+{\frac {2}{7}}{\begin{pmatrix}-1\\3\end{pmatrix}}\,.$
Therefore, we have
    $a=u_{1}^{*}(e_{1})=u_{1}^{*}{\left({\frac {3}{7}}u_{1}-{\frac {1}{7}}u_{2}\right)}={\frac {3}{7}}\,$
and
    $b=u_{1}^{*}(e_{2})=u_{1}^{*}{\left({\frac {1}{7}}u_{1}+{\frac {2}{7}}u_{2}\right)}={\frac {1}{7}}\,.$
Hence,
    $u_{1}^{*}={\frac {3}{7}}e_{1}^{*}+{\frac {1}{7}}e_{2}^{*}\,.$
With similar computations we get
    $u_{2}^{*}=-{\frac {1}{7}}e_{1}^{*}+{\frac {2}{7}}e_{2}^{*}\,.$
The transformation matrix from $u^{*}$ to $e^{*}$ is thus
    $M_{\mathfrak {e^{*}}}^{\mathfrak {u^{*}}}={\begin{pmatrix}{\frac {3}{7}}&-{\frac {1}{7}}\\{\frac {1}{7}}&{\frac {2}{7}}\end{pmatrix}}\,.$
The transposed matrix of this is
    ${{\left(M_{\mathfrak {e^{*}}}^{\mathfrak {u^{*}}}\right)}^{\text{tr}}}={\begin{pmatrix}{\frac {3}{7}}&{\frac {1}{7}}\\-{\frac {1}{7}}&{\frac {2}{7}}\end{pmatrix}}={\begin{pmatrix}2&-1\\1&3\end{pmatrix}}^{-1}={\left(M_{\mathfrak {e}}^{\mathfrak {u}}\right)}^{-1}\,.$
The inverse task to express the standard dual basis with $u_{1}^{*}$ and $u_{2}^{*}$, is easier to solve, because we can read of directly the representations of the $u_{i}$ with respect to the standard basis. We have
    $e_{1}^{*}=2u_{1}^{*}+u_{2}^{*}\,$
and
    $e_{2}^{*}=-u_{1}^{*}+3u_{2}^{*}\,,$
as becomes clear by evaluation $u_{1},u_{2}$ on both sides.
# Linear subspace/Dual space/1/Example

We consider the linear subspace
    $U=\langle {\begin{pmatrix}8\\6\\5\end{pmatrix}},{\begin{pmatrix}4\\7\\-3\end{pmatrix}}\rangle \subseteq \mathbb {R} ^{3}\,.$
The orthogonal space of $U$ consists of all linear forms
    $f\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R}$
with $f{\begin{pmatrix}8\\6\\5\end{pmatrix}}=0$ and $f{\begin{pmatrix}4\\7\\-3\end{pmatrix}}=0$. Because a linear form is described by a row matrix $\left(a,\,b,\,c\right)$ with respect to the standard basis, we are dealing with the solution set of the linear system
    $8a+6b+5c=0\,,$
    $4a+7b-3c=0\,.$
The solution space is
    $U^{\perp }={\left\{s\left({\frac {17}{4}},\,1,\,-8\right)\mid s\in K\right\}}\,.$
# MDLD/orthogonal space (dual space)

Start=orthogonal space (dual space) Target=Linear subspace/Orthogonal space in dual space/Definition
# MDLD/dual mapping

Start=dual mapping Target=Linear mapping/Dual mapping/Definition
# MDLD/dual bases

Start=dual bases See=dual basis MDLD/dual basis Target=Finite-dimensional vector space/Dual basis/Definition
# MDLD/transposed matrix

Start=transposed matrix Target=Matrix/K/Transposed matrix/Definition
# Should Wikipedia and related wikis allow anonymous IP editing?

Wikipedia and related wikis allow anonymous IP editing. Is it a good thing or rather something to be abandoned?

## Wikipedia and related wikis should allow anonymous IP editing
### Pro

* Pro Allowing anonymous IP addresses to edit creates something like a beneficial soft drug (of editing), which can lure editors to become serious registered editors later.
** Objection It is unclear whether luring people to spend their time and attention online on wiki without significant act of volition is actually ethical.
* Pro Disallowing IP editing is likely to somewhat reduce the volume of useful edits.
* Pro As per W:Wikipedia:IP editors are human too, "You are an IP too", which can be interpreted as saying that registered accounts are also IPs.
** Objection Untrue. A registered user account is not an IP account, and a person who has a registered user account is a person, not an IP address. (Ontology matters.)
** Objection Being a human is not an automatic indication of goodness (skill, intent, etc.), and therefore, there does not seem to be any deep significance of the predicate in "IP editors are human too". (Stupid people are human too; undisciplined people are human too; unskilled people are human too; malevolent/ill-wishing people are human too; etc.)

### Con

* Con Anyone serious about substantially improving the wiki should have enough time and attention on their hands to create an account. While there would be some loss of, say, minor corrections, arguably, the loss is tolerable.
* Con Presenting oneself only via an IP creates a burden on the memory and cognition of those who view edits by the person or interact with the person. Put differently, numerical identifiers are not as cognitively friendly as alphanumeric names.
* Con Requiring the hurdle of an account registration provides a disincentive for vandals to vandalize, thereby preventing "dirtying" revision histories.
* Con Anonymity contributes to poor behavior online. People all too often say things that they would not say in a face-to-face interaction.
** Objection That may be true, but registered accounts are also de facto anonymous (they are pseudonymous, but that come out as nearly the same).
*** Objection Fair point. However, psychologically, an IP may feel more anonymous than a registered user account.
**** Objection An interesting point. However, the stated hypothesis is an empirical one and requires proper verification. Absent verification, the above is a speculation.
* Comment Medical MDWiki run by Wiki Project Med Foundation does not allow anonymous IPs: "Editing is restricted to registered and approved individuals". It remains to be seen how well it works for the project.
# Affine subspace/Translated linear subspace/Definition

Let $V$ be a vector space. An affine subspace of $V$ is (the empty set or) a subset of the form
    $P+U={\left\{P+u\mid u\in U\right\}}\,,$
where $U\subseteq V$ is a linear subspace and $P\in V$ is a vector.
# Linear mapping/Fiber/Affine subspace/Kernel/Example

For a linear mapping
    $\varphi \colon V\longrightarrow W$
between $K$-vector spaces $V$ and $W$ and an element $Q\in W$, the preimage for $Q$ (the fiber over $Q$)
    $\varphi ^{-1}(Q)={\left\{P\in V\mid \varphi (P)=Q\right\}}\,$
is an affine subspace of $V$. If this is non-empty, then we can take any point $P_{0}\in V$ with
    $\varphi (P_{0})=Q\,$
as starting point The translation space is the kernel of $\varphi$. When a linear mapping is given, then $V$ is partitioned in a layered family of parallel affine subspaces.
# Affine space/Over vector space/Definition

An affine space over a $K$-vector space $V$ is a set $E$, together with a mapping
    $V\times E\longrightarrow E,(v,P)\longmapsto P+v,$
which satisfies the following three conditions:
* $P+0=P$ for all $P\in E$,
* $(P+v)+w=P+(v+w)$ for all $v,w\in V$ and $P\in E$,
* For two points $P,Q\in E$, there exists exactly one vector $v\in V$ such that $Q=P+v$.
# Linear system/Homogeneous and inhomogeneous/Affine space/1/Example

The solution space of the homogeneous linear equation
    $7x-3y+4z=0\,$
is
    $U=\langle {\begin{pmatrix}3\\7\\0\end{pmatrix}},\,{\begin{pmatrix}0\\4\\3\end{pmatrix}}\rangle \,,$
the solution set of the inhomogeneous linear equation
    $7x-3y+4z=2\,$
is
    $E={\begin{pmatrix}0\\0\\{\frac {1}{2}}\end{pmatrix}}+U={\begin{pmatrix}0\\0\\{\frac {1}{2}}\end{pmatrix}}+\langle {\begin{pmatrix}3\\7\\0\end{pmatrix}},\,{\begin{pmatrix}0\\4\\3\end{pmatrix}}\rangle \,.$
The affine addition is the mapping
    $U\times E\longrightarrow E,(u,P)\longmapsto (u+P),$
which assigns to a pair consisting in a solution of the homogeneous equation and a solution of the inhomogeneous equation their sum, which is a solution of the inhomogeneous equation. For two solutions of the inhomogeneous equation, their difference is a solution of the homogeneous equation. For example, for
    $u=-2{\begin{pmatrix}3\\7\\0\end{pmatrix}}+3{\begin{pmatrix}0\\4\\3\end{pmatrix}}={\begin{pmatrix}-6\\-2\\9\end{pmatrix}}\in U\,,$
the point
    ${\begin{pmatrix}0\\0\\{\frac {1}{2}}\end{pmatrix}}+{\begin{pmatrix}-6\\-2\\9\end{pmatrix}}={\begin{pmatrix}-6\\2\\{\frac {19}{2}}\end{pmatrix}}\,$
is another solution in $E$. The two solutions ${\begin{pmatrix}1\\3\\1\end{pmatrix}}$ and ${\begin{pmatrix}2\\4\\0\end{pmatrix}}$ from $E$ are related by the translating vector
    ${\begin{pmatrix}2\\4\\0\end{pmatrix}}-{\begin{pmatrix}1\\3\\1\end{pmatrix}}={\begin{pmatrix}1\\1\\-1\end{pmatrix}}\,.$
# Affine space/Affine basis/Definition

A family of points $P_{i}\in E$, $i\in I$, in an affine space $E$ over a $K$-vector space $V$ is called an affine basis of $E$, if there exists an $i_{0}\in I$ such that the family of vectors
    ${\overrightarrow {P_{i_{0}}P_{i}}},i\in I\setminus \{i_{0}\},$
is a basis of $V$.
# Affine space/Point set/Barycentric combination/Definition

For a family $P_{i}$, $i\in I$, of points in an affine space $E$ and a tuple $a_{i}$, $i\in I$, in $K$ satisfying
    $\sum _{i\in I}a_{i}=1\,$
(for $I$ infinite, only finitely many of the $a_{i}$ are allowed to be different from $0$), the sum $\sum _{i\in I}a_{i}P_{i}$ is called a barycentric combination of the $P_{i}$. The corresponding point in $E$ is given by
    $\sum _{i\in I}a_{i}P_{i}=Q+\sum _{i\in I}a_{i}{\overrightarrow {QP_{i}}}\,,$
where $Q$ is an arbitrary point in $E$.
# Affine space/Point set/Barycentric combination/Point/Fact

For a family $P_{i}$, $i\in I$, of points in an affine space $E$, a
barycentric combination
    $\sum _{i\in I}a_{i}P_{i}$
$E$
# Affine space/Point set/Barycentric combination/Point/Fact/Proof/Exercise

Show that, for a family $P_{i}$, $i\in I$, of points in an affine space $E$, a
barycentric combination
    $\sum _{i\in I}a_{i}P_{i}$
defines a unique point in $E$.
# Affine space/Affine Basis/Barycentric coordinates/Fact

Let $P_{i}$, $i\in I$, denote an affine basis in an affine space $E$ over the $K$-vector space $V$.
$P\in E$, there exists a unique barycentric representation
    $P=\sum _{i\in I}a_{i}P_{i}\,.$
# Affine space/Affine Basis/Barycentric coordinates/Fact/Proof

Let $i_{0}\in I$ be fixed. In $V$, we have a unique representation
    ${\overrightarrow {P_{i_{0}}P}}=\sum _{i\in I\setminus \{i_{0}\}}a_{i}{\overrightarrow {P_{i_{0}}P_{i}}}\,.$
We set
    $a_{i_{0}}:=1-\sum _{i\in I\setminus \{i_{0}\}}a_{i}\,.$
Then $\sum _{i\in I}a_{i}=1$, and
    ${\begin{aligned}P&=P_{i_{0}}+{\overrightarrow {P_{i_{0}}P}}\\&=P_{i_{0}}+\sum _{i\in I\setminus \{i_{0}\}}a_{i}{\overrightarrow {P_{i_{0}}P_{i}}}\\&=P_{i_{0}}+a_{i_{0}}{\overrightarrow {P_{i_{0}}P_{i_{0}}}}+\sum _{i\in I\setminus \{i_{0}\}}a_{i}{\overrightarrow {P_{i_{0}}P_{i}}}\\&=P_{i_{0}}+\sum _{i\in I}a_{i}{\overrightarrow {P_{i_{0}}P_{i}}}.\end{aligned}}$
Therefore, there exists such a representation with $P_{i_{0}}$ as origin. Uniqueness follows from the facts that the $a_{i}$, $i\neq i_{0}$, are uniquely determined as the coefficients of the vector space basis, and that $a_{i_{0}}$ is determined by the baryzentric condition.
# Affine space/Barycentric coordinates/Definition

Let $P_{i}$, $i\in I$, denote an affine basis in an affine space $E$ over the $K$-vector space $V$. For a point $P\in E$, the uniquely determined numbers
    $(a_{i},i\in I){\text{ with }}\sum _{i\in I}a_{i}=1$
such that
    $P=\sum _{i\in I}a_{i}P_{i}\,$
holds, are called the barycentric coordinates of $P$.
# Affine space/Dimension/Basis number/Definition

Let $E$ be an affine space with an affine basis
    $P_{1},P_{2},\ldots ,P_{n},P_{n+1}.$
Then, $n$ is called the dimension of $E$.
# Affine space/Subspace/Definition

Let $E$ be an affine space over the $K$-vector space $V$. A subset $F\subseteq E$ is called an affine subspace, if ($F$ is empty or)
    $F=P+U\,$
with a point $P\in E$ and a $K$-linear subspace $U\subseteq V$.
# MDLD/affine subspace (vs)

Start=affine subspace (vs) Target=Affine subspace/Translated linear subspace/Definition
# MDLD/affine space

Start=affine space Target=Affine space/Over vector space/Definition