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

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. 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.

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 definition and Lemma 30.1 speak in favor of allowing this.

Remark

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

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.

The action of a translation in the plane on a subset.

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 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:

1. ${}P+0=P$ for all ${}P\in E$,
2. ${}(P+v)+w=P+(v+w)$ for all ${}v,w\in V$ and ${}P\in E$,
3. 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

1. ${}{\overrightarrow {PP}}=0$ for ${}P\in E$.
2. ${}{\overrightarrow {PQ}}=-{\overrightarrow {QP}}$ for ${}P,Q\in E$.
3. ${}{\overrightarrow {PQ}}+{\overrightarrow {QR}}={\overrightarrow {PR}}$ for ${}P,Q,R\in E$,

where these identities live in the vector space ${}V$, see Exercise 29.10 .

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 Definition is an affine space over ${}U$.

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 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 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 ${}V$.

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 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 ${}P_{i_{0}}$.

The barycentric coordinates in the plane, where the affine basis points are the corners of a triangle.

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 ${}E$.

Lemma

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 ${}E$.

Proof

See Exercise 29.14 .

$\Box$

Theorem

Let ${}P_{i}$, ${}i\in I$, denote an affine basis in an affine space ${}E$ over the ${}K$-vector space ${}V$. Then, for every point ${}P\in E$, there exists a unique 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.

$\Box$

The colors for additive mixing with primary colors red, blue, green (this corresponds to the three cone cells of the human eye). The eye is only interested in the mixing of the three colors, therefore, only linear combinations ${}(r,g,b)$ with ${}r+g+b=1$ (and non-negative coefficients) are relevant. Thus, colors are described by barycentric coordinates, saving one dimension.

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 ${}P$.

Example

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

Definition

Let ${}E$ be an affine space with an affine basis

$P_{1},P_{2},\ldots ,P_{n},P_{n+1}.$

Then, ${}n$ is called the dimension

of ${}E$.

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 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

${}U\subseteq V$.

This definition is compatible with the definition of affine subspaces in a vector space mentioned at the beginning.

Lemma

Let ${}E$ be an affine space over the ${}K$-vector space ${}V$. For a subset ${}F\subseteq E$,

the following conditions are equivalent.

1. ${}F$ is an affine subspace of ${}E$.
2. 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 ${}\sum _{i=1}^{n}a_{i}P_{i}\in F$.
3. 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 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 rests on Exercise 29.20 . 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

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