# Affine mapping/Introduction/Section
## Definition

Let $K$ be a field and let $E$ and $F$ denote affine spaces over the vector spaces $V$ and $W$, respectively. A mapping
    $\psi \colon E\longrightarrow F$
is called affine (or affine-linear), if there exists a linear mapping
    $\varphi \colon V\longrightarrow W$
such that
    $\psi (P+v)=\psi (P)+\varphi (v)\,$
holds for all $P\in E$ and
$v\in V$
It suffices to check this condition for just one point and all vectors, see exercise.

## Remark

A mapping
    $\psi \colon E\longrightarrow F$
is affine-linear with linear part $\varphi$ if and only if the diagram
    ${\begin{matrix}V\times E&{\stackrel {+}{\longrightarrow }}&E&\\\!\!\!\!\!\varphi \times \psi \downarrow &&\downarrow \psi \!\!\!\!\!&\\W\times F&{\stackrel {+}{\longrightarrow }}&F&\!\!\!\!\!\\\end{matrix}}$
commutes. For an affine-linear mapping
    $\psi \colon E\longrightarrow F,$
the linear part (assume $E\neq \emptyset$)
    $\varphi \colon V\longrightarrow W$
is uniquely determined. This is because we must have
    $\varphi (v)={\overrightarrow {\psi (P)\psi (P+v)}}\,$
for an arbitrary point $P\in E$. Therefore, we denote the linear part with $\psi _{0}$. in particular, for two points $P,Q\in E$, we have
    $\psi _{0}({\overrightarrow {PQ}})={\overrightarrow {\psi (P)\psi (Q)}}\,.$

## Lemma

Let $K$ be a field, and let $E,F$ and $G$ denote affine spaces over the vector spaces
$U,V$
$W$
* The identity is affine-linear.
    $\operatorname {Id} _{E}\colon E\longrightarrow E$
* The composition of affine-linear mappings and is again affine-linear.
    $E\longrightarrow F$
    $F\longrightarrow G$
* For a bijective affine-linear mapping also the inverse mapping is affine-linear.
    $\psi \colon E\longrightarrow F,$
* For $v\in V$, the translation is affine-linear.
    $E\longrightarrow E,P\longmapsto P+v,$
* A linear mapping is affine-linear.

### Proof

These properties follow immediately from the definition.
$\Box$

## Lemma

Let $E$ and $F$ denote affine spaces over a field $K$, and let
    $\psi \colon E\longrightarrow F$
denote a mapping. Then, $\psi$ is affine-linear if and only if for every barycentric combination $\sum _{i\in I}a_{i}P_{i}$ with $P_{i}\in E$, the equality
    $\psi {\left(\sum _{i\in I}a_{i}P_{i}\right)}=\sum _{i\in I}a_{i}\psi (P_{i})\,$

### Proof

Let $V$ and $W$ denote the vector spaces for $E$ and for $F$, respectively. Suppose first that $\psi$ is affine-linear with linear part
    $\psi _{0}\colon V\longrightarrow W.$
Let a barycentric combination $\sum _{i\in I}a_{i}P_{i}$ with $P_{i}\in E$ and $\sum _{i\in I}a_{i}=1$ be given. Then we have (with an arbitrary point $Q\in E$)
    ${\begin{aligned}\psi {\left(\sum _{i\in I}a_{i}P_{i}\right)}&=\psi {\left(Q+\sum _{i\in I}a_{i}{\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\psi _{0}{\left(\sum _{i\in I}a_{i}{\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\sum _{i\in I}a_{i}\psi _{0}{\left({\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\sum _{i\in I}a_{i}{\overrightarrow {\psi (Q)\psi (P_{i})}}\\&=\sum _{i\in I}a_{i}\psi (P_{i}).\end{aligned}}$
Now, suppose that the mapping $\psi$ is compatible with barycentric combinations. We set
    $\varphi (v):={\overrightarrow {\psi (P)\psi (P+v)}}\,$
for $v\in V$, where $P\in E$ is any point. We first show that this is independent of the chosen point $P$. For another point $Q\in E$, the sum
    $(P+v)-(Q+v)+Q$
is a barycentric combination of the point $P$, see exercise. Therefore, we have in $F$ the equality
    $\psi (P+v)-\psi (Q+v)+\psi (Q)=\psi (P)\,.$
Hence, we have in $V$ the equality
    ${\overrightarrow {\psi (P)\psi (P+v)}}-{\overrightarrow {\psi (P)\psi (Q+v)}}+{\overrightarrow {\psi (P)\psi (Q)}}=0\,,$
and, therefore,
    ${\begin{aligned}{\overrightarrow {\psi (P)\psi (P+v)}}&={\overrightarrow {\psi (P)\psi (Q+v)}}-{\overrightarrow {\psi (P)\psi (Q)}}\\&={\overrightarrow {\psi (P)\psi (Q+v)}}+{\overrightarrow {\psi (Q)\psi (P)}}\\&={\overrightarrow {\psi (Q)\psi (Q+v)}}.\end{aligned}}$
We have to show that $\varphi$ is linear. For $u={\overrightarrow {PQ}}$ and $v={\overrightarrow {PR}}$, we have
    ${\begin{aligned}\psi (P)+\varphi (au+bv)&=\psi (P+au+bv)\\&=\psi {\left(P+a{\overrightarrow {PQ}}+b{\overrightarrow {PR}}\right)}\\&=\psi ((1-a-b)P+aQ+bR)\\&=(1-a-b)\psi (P)+a\psi (Q)+b\psi (R)\\&=\psi (P)+a{\overrightarrow {\psi (P)\psi (Q)}}+b{\overrightarrow {\psi (P)\psi (R)}}\\&=\psi (P)+a\varphi {\left({\overrightarrow {PQ}}\right)}+b\varphi {\left({\overrightarrow {PR}}\right)}\\&=\psi (P)+a\varphi (u)+b\varphi (v).\end{aligned}}$
Thus, we have
    $\varphi (au+bv)=a\varphi (u)+b\varphi (v)\,.$
$\Box$

## Definition

Let $K$ be a field, and let $E$ and $F$ denote affine spaces over the $K$-vector spaces $V$ and $W$. A bijective affine-linear mapping
    $\psi \colon E\longrightarrow F$
is called an
In a certain sense, affine-linear mappings are built from translations and linear mappings.

## Lemma

Let $K$ be a field and let $E$ be an affine space over the vector space $V$. Let $P\in E$. Then the affine-linear mappings
    $\psi \colon E\longrightarrow E$
having $P$ as a fixed point correspond to the linear mappings
    $\varphi \colon V\longrightarrow V.$

### Proof

The assignment is given by $\psi \mapsto \psi _{0}$. We have to show that for every linear mapping $\varphi \colon V\rightarrow V$, there is a unique affine-linear mapping
    $\psi \colon E\longrightarrow E$
with this linear part. Because of
    $\psi (Q)=\psi (P)+\varphi ({\overrightarrow {PQ}})=P+\varphi ({\overrightarrow {PQ}})\,,$
there can exist at most one such an affine-linear mapping, and, by this rule, we can define such a mapping.
$\Box$
The following theorem is called Determination theorem for affine mappings, and is analogous to fact.

## Theorem

Let $K$ be a field, and let $E$ and $F$ denote affine spaces over the vector spaces $V$ and $W$. Let $P_{i}$, $i\in I$, denote an affine basis of $E$, and let $Q_{i}$, $i\in I$, denote a family of points in $F$. Then, there exists a uniquely determined affine-linear mapping
    $\psi \colon E\longrightarrow F$
such that
    $\psi (P_{i})=Q_{i}\,$
for all
$i\in I$

### Proof

Let $i_{0}\in I$. Due to fact, there exists a uniquely determined linear mapping
    $\varphi \colon V\longrightarrow W$
such that
    $\varphi ({\overrightarrow {P_{i_{0}}P_{i}}})={\overrightarrow {Q_{i_{0}}Q_{i}}}\,$
for all $i\in I\setminus \{i_{0}\}$. Therefore,
    $\psi (R)=Q_{i_{0}}+\varphi ({\overrightarrow {P_{i_{0}}R}})\,$
is an affine-linear mapping with the properties looked for. Note that
    $\psi (P_{i_{0}})=Q_{i_{0}}+\varphi ({\overrightarrow {P_{i_{0}}P_{i_{0}}}})=Q_{i_{0}}\,$
and
    $\psi (P_{i})=Q_{i_{0}}+\varphi ({\overrightarrow {P_{i_{0}}P_{i}}})=Q_{i_{0}}+{\overrightarrow {Q_{i_{0}}Q_{i}}}=Q_{i}\,$
holds. Such an affine mapping $\psi$ is uniquely determined by its linear part and the image of just one point, so that
    $\psi _{0}=\varphi \,$
must hold.
$\Box$

## Corollary

Let $K$ be a field, and let $E$ denote an affine space with an affine basis $P_{1},\ldots ,P_{n},P_{n+1}$. Then the mapping
    $E\longrightarrow K^{n+1},P\longmapsto (a_{1},\ldots ,a_{n},a_{n+1}),$
where $a_{i}$ denotes the barycentric coordinates of $P$, is an affine-linear mapping, which provides an affine isomorphism between $E$ and the affine subspace $F\subset K^{n+1}$, guven by
    $F={\left\{x\in K^{n+1}\mid \sum _{i=1}^{n+1}x_{i}=1\right\}}\,.$
The translating vector space of $F$ is
    $W={\left\{x\in K^{n+1}\mid \sum _{i=1}^{n+1}x_{i}=0\right\}}\,.$

### Proof

Due to fact, there exists a uniquely determined affine-linear mapping
    $E\longrightarrow K^{n+1}$
sending $P_{i}$ to the $i$-th standard vector $e_{i}$. Because of fact, the point
    $P=\sum _{i=1}^{n+1}a_{i}P_{i}\,$
is sent to
    $\sum _{1=1}^{n+1}a_{i}e_{i}.$
Because of
    $\sum _{i=1}^{n+1}a_{i}=1\,,$
this image point belongs to $F$. Bijectivity is clear.
$\Box$