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

${}U,V$ and ${}W$. Then the following statements hold.

1. The identity

   $\operatorname {Id} _{E}\colon E\longrightarrow E$

   is affine-linear.

2. The composition of affine-linear mappings

   $E\longrightarrow F$

   and

   $F\longrightarrow G$

   is again affine-linear.

3. For a bijective affine-linear mapping

   $\psi \colon E\longrightarrow F,$

   also the inverse mapping is affine-linear.

4. For ${}v\in V$, the translation

   $E\longrightarrow E,P\longmapsto P+v,$

   is affine-linear.

5. 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

${}\psi {\left(\sum _{i\in I}a_{i}P_{i}\right)}=\sum _{i\in I}a_{i}\psi (P_{i})\,$

holds.

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, 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

affine isomorphism.

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. 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.

$\Box$