# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 30/refcontrol

    Affine generating systems

## LemmaCreate referencenumber

Let $E$ be an affine spaceMDLD/affine space over the $K$-vector spaceMDLD/vector space $V$. In this situation, the intersectionMDLD/intersection (family) of a family of affine subspacesMDLD/affine subspaces $F_{i}\subseteq E$, $i\in I$,

### Proof

If the intersection is empty, then the statement holds by definition. So let $P\in \bigcap _{i\in I}F_{i}$. We may write the affine subspaces as
    $F_{i}=P+U_{i}\,$
with linear subspacesMDLD/linear subspaces $U_{i}\subseteq V$. Let
    $U=\bigcap _{i\in I}U_{i}\,,$
which is a linear subspace, due to fact. We claim that
    $\bigcap _{i\in I}F_{i}=P+U\,.$
From $Q\in \bigcap _{i\in I}F_{i}$, we can deduce
    $Q=P+u\,$
with $u\in \bigcap _{i\in I}U_{i}$, so that $Q\in P+U$ holds. If $Q\in P+U$ holds, then directly $Q\in P+U_{i}=F_{i}$ follows.
$\Box$
In particular, for every subset $T\subseteq E$ in an affine space $E$, there exist a smallest affine subspace containing $T$.

## LemmaCreate referencenumber

Let $E$ be an affine spaceMDLD/affine space over the $K$-vector spaceMDLD/vector space $V$, and let $T\subseteq E$ denote a subset. Then, the smallest affine subspaceMDLD/affine subspace $F\subseteq E$ of $E$, which contains $T$, consists of all barycentric combinationsMDLD/barycentric combinations
    $\sum _{i=1}^{n}a_{i}P_{i}{\text{ with }}P_{i}\in T{\text{ and }}\sum _{i=1}^{n}a_{i}=1.$

### Proof

The given set contains the points from $T$, as we can take a standard tuple as a barycentric coordinate tuple. Therefore, the claim follows from fact and exercise.
$\Box$

## Definition

Let $E$ be an affine spaceMDLD/affine space over the $K$-vector spaceMDLD/vector space $V$, and let $F\subseteq E$ be an affine subspace.MDLD/affine subspace A family of points $P_{i}\in F$, $i\in I$, is called an affine generating system
$F$
$F$
$E$
$P_{i}$
A point generates, as an affine space, the point itself, two points generate the connecting line.
    Affine independence

## Definition

Let $E$ be an affine spaceMDLD/affine space over a $K$-vector spaceMDLD/vector space $V$, and let
    $P_{1},\ldots ,P_{n}$
be a finite family of points in $E$. We say that this family of points is affinely independent,
if an equality
    $\sum _{i=1}^{n}a_{i}P_{i}=\sum _{i=1}^{n}b_{i}P_{i}\,$
wit
    $\sum _{i=1}^{n}a_{i}=\sum _{i=1}^{n}b_{i}=1\,$
is only possible if
    $a_{i}=b_{i}\,$
for all
$i=1,\ldots ,n$

## LemmaCreate referencenumber

Let $E$ be an affine spaceMDLD/affine space over a $K$-vector spaceMDLD/vector space $V$, and let
    $P_{1},\ldots ,P_{n}$
$E$
* The points $P_{1},\ldots ,P_{n}$ are affinely independent.MDLD/affinely independent
* For every $i\in \{1,\ldots ,n\}$, the family of vectors is linearly independent.MDLD/linearly independent
    ${\overrightarrow {P_{i}P_{1}}},\ldots ,{\overrightarrow {P_{i}P_{i-1}}},\,{\overrightarrow {P_{i}P_{i+1}}},\ldots ,{\overrightarrow {P_{i}P_{n}}}$
* There exists some $i\in \{1,\ldots ,n\}$ such that the family of vectors is linearly independent.
    ${\overrightarrow {P_{i}P_{1}}},\ldots ,{\overrightarrow {P_{i}P_{i-1}}},\,{\overrightarrow {P_{i}P_{i+1}}},\ldots ,{\overrightarrow {P_{i}P_{n}}}$
* The points $P_{1},\ldots ,P_{n}$ form an affine basisMDLD/affine basis in the affine subspaceMDLD/affine subspace generatedMDLD/generated (affine) by them.

### Proof

$\Box$
    Affine mappings

## Definition

Let $K$ be a fieldMDLD/field and let $E$ and $F$ denote affine spacesMDLD/affine spaces over the vector spacesMDLD/vector spaces $V$ and $W$, respectively. A mappingMDLD/mapping
    $\psi \colon E\longrightarrow F$
is called affine (or affine-linear), if there exists a linear mappingMDLD/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-linearMDLD/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
    $\psi _{0}({\overrightarrow {PQ}})={\overrightarrow {\psi (P)\psi (Q)}}\,.$

## LemmaCreate referencenumber

Let $K$ be a field,MDLD/field and let $E,F$ and $G$ denote affine spacesMDLD/affine spaces over the vector spacesMDLD/vector spaces
$U,V$
$W$
* The identity is affine-linear.
    $\operatorname {Id} _{E}\colon E\longrightarrow E$
* The compositionMDLD/composition (mapping) of affine-linear mappingsMDLD/affine-linear mappings and is again affine-linear.
    $E\longrightarrow F$
    $F\longrightarrow G$
* For a bijective affine-linear mapping also the inverse mappingMDLD/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 mappingMDLD/linear mapping is affine-linear.

### Proof

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

## LemmaCreate referencenumber

Let $E$ and $F$ denote affine spacesMDLD/affine spaces over a fieldMDLD/field $K$, and let
    $\psi \colon E\longrightarrow F$
denote a mapping. Then, $\psi$ is affine-linearMDLD/affine-linear if and only if for every barycentric combinationMDLD/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 $P$. 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,MDLD/field and let $E$ and $F$ denote affine spacesMDLD/affine spaces over the $K$-vector spacesMDLD/vector spaces $V$ and $W$. A bijectiveMDLD/bijective affine-linear mappingMDLD/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.

## LemmaCreate referencenumber

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

## TheoremCreate referencenumber

Let $K$ be a field,MDLD/field and let $E$ and $F$ denote affine spacesMDLD/affine spaces over the vector spacesMDLD/vector spaces $V$ and $W$. Let $P_{i}$, $i\in I$, denote an affine basisMDLD/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 mappingMDLD/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 mappingMDLD/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. 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.
$\Box$

## CorollaryCreate referencenumber

Let $K$ be a field,MDLD/field and let $E$ denote an affine spaceMDLD/affine space with an affine basisMDLD/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 coordinatesMDLD/barycentric coordinates of $P$, is an affine-linearMDLD/affine-linear mapping, which provides an affine isomorphismMDLD/affine isomorphism between $E$ and the affine subspaceMDLD/affine subspace (vs) $F\subset K^{n+1}$, 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 vectorMDLD/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.
$\Box$