# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 29/refcontrol

    Affine spaces
Linear subspaces of a vector space always contain $0$. Hence, a line $G\subset \mathbb {R} ^{2}$ not running through the zero point, is not a linear subspace. However, it is still a "linear object“, which should be studied in the framework of linear algebra.

## Definition

Let $V$ be a vector space.MDLD/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 subspaceMDLD/linear subspace and $P\in V$
The point $P$ is called a starting point, and the linear subspace $U$ is called the translation space, or simply the underlying linear subspace. The points in an affine space should be thought of as locations, the points of $U$ should be thought of as translation vectors. One might discuss whether the empty set should be allowed as an affine (sub)space, the following remark, the definitionMDLD/definition and fact speak in favor of allowing this.

## Remark

The solution set of a system of inhomogeneous linear equations in $n$ variables is an affine subspaceMDLD/affine subspace of $K^{n}$. The underlying vector space in the solution space of the corresponding homogeneous system.

## Example Create referencenumber

For a linear mappingMDLD/linear mapping
    $\varphi \colon V\longrightarrow W$
between $K$-vector spacesMDLD/vector spaces $V$ and $W$ and an element $Q\in W$, the preimageMDLD/preimage for $Q$ (the fiberMDLD/fiber over $Q$)
    $\varphi ^{-1}(Q)={\left\{P\in V\mid \varphi (P)=Q\right\}}\,$
is an affine subspaceMDLD/affine subspace (vs) of $V$. If this is non-empty, then we can take any point $P_{0}\in V$ with
    $\varphi (P_{0})=Q\,$
as starting pointMDLD/starting point The translation space is the kernelMDLD/kernel (vs) of $\varphi$. When a linear mapping is given, then $V$ is partitioned in a layered family of parallel affine subspaces.
A further reasoning yields a somewhat more abstract concept. The natural space can be given coordinates and then it is identified with $\mathbb {R} ^{3}$. For this, we have to choose arbitrarily a point in the space as $0$. The natural space does not contain a natural zero, and also no natural addition of points. Despite of this, the natural space is tightly related to a vector space, namely the vector space of all translations of the space. Such a translation is an elementary geometric construction, every point of the space is translated by a certain translating vector. A translation is determined by every point together with its image point. The set of all these translations form a vector space, where the addition is given by composing translations. The zero translation is the identity. If a point $P$ of the space is fixed, then we get a bijection between the space and the vector space of translations, by attaching a translation vector at $P$ and determining the resulting point. Such a fixation is also called a choice of an origin.

## Definition

An affine space over a $K$-vector spaceMDLD/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 $Q=P+v$.
This addition is called affine addition or Translation. For two given points $P,Q\in E$, the uniquely determined translation vector is denoted by ${\overrightarrow {PQ}}$. Beside $P+{\overrightarrow {PQ}}=Q$, the following rules holds
* ${\overrightarrow {PP}}=0$ for $P\in E$.
* ${\overrightarrow {PQ}}=-{\overrightarrow {QP}}$ for $P,Q\in E$.
* ${\overrightarrow {PQ}}+{\overrightarrow {QR}}={\overrightarrow {PR}}$ for $P,Q,R\in E$,
where these identities live in the vector space $V$, see exercise.
The mapping
    $+\colon V\times E\longrightarrow E$
exhibits several aspects. For every point $P\in E$, the mapping
    $V\longrightarrow E,v\longmapsto P+v,$
is a bijection between the underlying vector space and the affine space. This bijection is not canonical, as it depends on the chosen point. Every vector $v\in V$ defines the mapping
    $E\longrightarrow E,P\longmapsto P+v,$
which is called the translation on $E$ for the vector $v$. The mapping
    $E\times E\longrightarrow V,(P,Q)\longmapsto {\overrightarrow {PQ}},$
assigns to a pair of points their (uniquely determined) translating vector. Instead of ${\overrightarrow {PQ}}$, we sometimes write $Q-P$.
Every vector space $V$ is also an affine space over itself, with the vector space addition as addition. An affine subspace $P+U\subseteq V$ in the sense of DefinitionMDLD/Definition is an affine space over $U$.

## Example Create referencenumber

The solution space of the homogeneous linear equationMDLD/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 additionMDLD/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 bases
For the following concepts, we do not loose much if we always assume that the index set $I$ is finite. In the non-finite case, the coefficient tuples are to be interpreted that, up to finitely many exceptions, all entries are $0$.

## Definition

A family of points $P_{i}\in E$, $i\in I$, in an affine spaceMDLD/affine space $E$ over a $K$-vector spaceMDLD/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 basisMDLD/basis (vs)
$V$
Because of
    ${\overrightarrow {P_{i_{0}}P_{i}}}={\overrightarrow {P_{i_{0}}P_{i_{1}}}}+{\overrightarrow {P_{i_{1}}P_{i}}}=-{\overrightarrow {P_{i_{1}}P_{i_{0}}}}+{\overrightarrow {P_{i_{1}}P_{i}}}\,,$
the basis vectors ${\overrightarrow {P_{i_{0}}P_{i}}}$ with respect to the origin $P_{i_{0}}$ can be expressed as Linear combinationsMDLD/Linear combinations of the corresponding vectors with respect to any other origin point $P_{i_{1}}$ of the family. Therefore, the property of being an affine basis is independent from the chosen $P_{i_{0}}$.

## Definition

For a family $P_{i}$, $i\in I$, of points in an affine spaceMDLD/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
    $\sum _{i\in I}a_{i}P_{i}=Q+\sum _{i\in I}a_{i}{\overrightarrow {QP_{i}}}\,,$
$Q$
$E$

## LemmaCreate referencenumber

For a family $P_{i}$, $i\in I$, of points in an affine spaceMDLD/affine space $E$, a
barycentric combinationMDLD/barycentric combination
    $\sum _{i\in I}a_{i}P_{i}$
$E$

### Proof

$\Box$

## TheoremCreate referencenumber

Let $P_{i}$, $i\in I$, denote an affine basisMDLD/affine basis in an affine spaceMDLD/affine space $E$ over the $K$-vector spaceMDLD/vector space $V$. Then, for every point $P\in E$, there exists a unique barycentric representationMDLD/barycentric representation
    $P=\sum _{i\in I}a_{i}P_{i}\,.$

### 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.
$\Box$

## Definition

Let $P_{i}$, $i\in I$, denote an affine basisMDLD/affine basis in an affine spaceMDLD/affine space $E$ over the $K$-vector spaceMDLD/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
$P$

## Example Create referencenumber

Let $P_{i}$, $i\in I$, be an affine basisMDLD/affine basis in an affine spaceMDLD/affine space $E$ over the $K$-vector spaceMDLD/vector space $V$. Then the point $P_{j}$ ($j=I$) has the barycentric coordinatecMDLD/barycentric coordinatec $(0,\ldots ,0,1,0,\ldots ,0)$, where the $1$ is at the $j$-th place ($I$ being finite and ordered).

## Definition

Let $E$ be an affine spaceMDLD/affine space with an affine basisMDLD/affine basis
    $P_{1},P_{2},\ldots ,P_{n},P_{n+1}.$
Then, $n$ is called the dimension
$E$
Hence, the dimension of a non-empty affine space equals the dimension of the corresponding translation space. This observation shows also that the dimension is well-defined. The empty affine space has dimension $-1$.
    Affine subspaces

## Definition

Let $E$ be an affine spaceMDLD/affine space over the $K$-vector spaceMDLD/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 subspaceMDLD/linear subspace
$U\subseteq V$
This definition is compatible with the definition of affine subspaces in a vector space mentioned at the beginning.

## LemmaCreate referencenumber

Let $E$ be an affinen spaceMDLD/affinen space over the $K$-vector spaceMDLD/vector space $V$. For a subset $F\subseteq E$,
* $F$ is an affine subspaceMDLD/affine subspace of $E$.
* For $P_{1},\ldots ,P_{n}\in F$ and numbers $a_{1},\ldots ,a_{n}$ satisfying $\sum _{i=1}^{n}a_{i}=1$, we have $\sum _{i=1}^{n}a_{i}P_{i}\in F$.
* For two points $P,Q\in F$ and numbers $r,s\in K$ with $r+s=1$, we have $rP+sQ\in F$.

### Proof

If $F$ is empty, then all three conditions are true. So we assume that $F$ is not empty. $(1)\Rightarrow (2)$. Let
$F=P+U$ with $P\in F$ and a linear subspaceMDLD/linear subspace $U\subseteq V$. Then, $P_{i}=P+u_{i}$ with some $u_{i}\in U$. Due to the definition of a barycentric combination, it follows that
    $\sum _{i=1}^{n}a_{i}P_{i}=P+\sum _{i=1}^{n}a_{i}{\overrightarrow {PP_{i}}}=P+\sum _{i=1}^{n}a_{i}u_{i}\,$
is an element of $F$.
$(2)\Rightarrow (3)$. This is a weakening of the condition.
$(3)\Rightarrow (1)$. We choose a point $P\in F$, and consider
    $U:={\left\{{\overrightarrow {PQ}}\mid Q\in F\right\}}\subseteq V\,.$
We have $0\in U$. For $Q,Q'\in F$, due to the condition, also $-P+2Q$ and $-P+2Q'$ belong to $F$. Therefore, also
    ${\frac {1}{2}}{\left(-P+2Q\right)}+{\frac {1}{2}}{\left(-P+2Q'\right)}=-P+Q+Q'\,$
belongs to $F$, where this equality rest on exercise. This point equals
    $P-{\overrightarrow {PP}}+{\overrightarrow {PQ}}+{\overrightarrow {PQ'}}=P+{\overrightarrow {PQ}}+{\overrightarrow {PQ'}}\,,$
so that ${\overrightarrow {PQ}}+{\overrightarrow {PQ'}}$ belongs to $U$. Hence, $U$ is closed under the vector addition. Let $Q\in F$
and $s\in K$. Then, due to the condition, also
    $(1-s)P+sQ=P+s{\overrightarrow {PQ}}\,$
belongs to $F$, and, therefore, $s{\overrightarrow {PQ}}$ belongs to $U$. Thus, $F=P+U$ with a linear subspace $U$.
$\Box$
    Footnotes
* ↑ Affine subspaces are called parallel, if there is an inclusion between the corresponding linear subspaces.