# properties of an affine transformation

In this entry, we prove some of the basic properties of affine transformations. Let $\alpha :{A}_{1}\to {A}_{2}$ be an affine transformation^{} and $[\alpha ]:{V}_{1}\to {V}_{2}$ its associated linear transformation.

###### Proposition 1.

$\alpha $ is one-to-one iff $\mathrm{[}\alpha \mathrm{]}$ is.

###### Proof.

Next, suppose $\alpha $ is one-to-one, and $T(v)=0$ for some $v\in {V}_{1}$. Let $P,Q\in {A}_{1}$ with ${f}_{1}(P,Q)=v$. Then $0=[\alpha ](v)=[\alpha ]({f}_{1}(P,Q))={f}_{1}(\alpha (P),\alpha (Q))$, which implies that $\alpha (P)=\alpha (Q)$, and therefore $P=Q$ by assumption^{}. Conversely, suppose $[\alpha ]$ is one-to-one, and $\alpha (P)=\alpha (Q)$. Then $[\alpha ]({f}_{1}(P,Q))={f}_{2}(\alpha (P),\alpha (Q))=0$, so that ${f}_{1}(P,Q)=0$, and consequently $P=Q$, showing that $\alpha $ is one-to-one.
∎

###### Proposition 2.

$\alpha $ is onto iff $\mathrm{[}\alpha \mathrm{]}$ is.

###### Proof.

Suppose $\alpha $ is onto. Let $w\in {V}_{2}$, so there are $X,Y\in {A}_{2}$ such that ${f}_{2}(X,Y)=w$. Since $\alpha $ is onto, there are $P,Q\in {A}_{1}$ with $\alpha (P)=X$ and $\alpha (Q)=Y$. So $w={f}_{2}(X,Y)={f}_{2}(\alpha (P),\alpha (Q))=[\alpha ]({f}_{1}(P,Q))$. Hence $[\alpha ]$ is onto. Conversely, assume $[\alpha ]$ be onto, and pick $Y\in {A}_{2}$. Take an arbitrary point $P\in {A}_{1}$ and set $X=\alpha (P)$. There is $v\in {V}_{1}$ such that $[\alpha ](v)={f}_{2}(X,Y)$, since $[\alpha ]$ is onto. Let $Q\in {A}_{1}$ such that ${f}_{1}(P,Q)=v$. Then ${f}_{2}(X,\alpha (Q))={f}_{2}(\alpha (P),\alpha (Q))=[\alpha ]({f}_{1}(P,Q))=[\alpha ](v)={f}_{2}(X,Y)$. But ${f}_{2}(X,-)$ is a bijection^{}, we must have $Y=\alpha (Q)$, showing that $\alpha $ is onto.
∎

###### Corollary 1.

$\alpha $ is a bijection iff $\mathrm{[}\alpha \mathrm{]}$ is.

###### Proposition 3.

A bijective^{} affine transformation $\alpha \mathrm{:}{A}_{\mathrm{1}}\mathrm{\to}{A}_{\mathrm{2}}$ is an affine isomorphism.

###### Proof.

Suppose an affine transformation $\alpha :{A}_{1}\to {A}_{2}$ is a bijection. We want to show that ${\alpha}^{-1}:{A}_{2}\to {A}_{1}$ is an affine transformation. Pick any $X,Y\in {A}_{2}$, then

$$[\alpha ]({f}_{1}({\alpha}^{-1}(X),{\alpha}^{-1}(Y)))={f}_{2}(X,Y).$$ |

By the corollary above, $[\alpha ]$ is bijective, and hence a linear isomorphism. So

$${f}_{1}({\alpha}^{-1}(X),{\alpha}^{-1}(Y))={[\alpha ]}^{-1}({f}_{2}(X,Y)).$$ |

This shows that ${\alpha}^{-1}$ is an affine transformation whose assoicated linear transformation is ${[\alpha ]}^{-1}$. ∎

###### Proposition 4.

Two affine spaces associated with the same vector space^{} $V$ are affinely isomorphic.

###### Proof.

In fact, all we need to do is to show that $(A,f)$ is isomorphic^{} to $(V,g)$, where $g$ is given by $g(v,w)=w-v$. Pick any $P\in A$, then $\alpha :=f(P,-):A\to V$ is a bijection. For any $v\in V$, there is a unique $Q\in A$ such that $v=f(P,Q)$. Then ${1}_{V}(f(X,Y))=f(X,Y)=f(P,Y)-f(P,X)=\alpha (Y)-\alpha (X)=g(\alpha (X),\alpha (Y))$, showing that ${1}_{V}$ is the associated linear transformation of $\alpha $.
∎

###### Proposition 5.

Any affine transformation is a linear transformation between the corresponding induced vector spaces. In other words, if $\alpha \mathrm{:}A\mathrm{\to}B$ is affine, then $\alpha \mathrm{:}{A}_{P}\mathrm{\to}{B}_{\alpha \mathit{}\mathrm{(}P\mathrm{)}}$ is linear.

###### Proof.

Suppose $Q,R,S\in A$ are such that $Q+R=S$, or ${f}_{1}(P,Q)+{f}_{1}(P,R)={f}_{1}(P,S)$. Then

${f}_{2}(\alpha (P),\alpha (S))$ | $=$ | $[\alpha ]({f}_{1}(P,S))$ | ||

$=$ | $[\alpha ]({f}_{1}(P,Q)+{f}_{1}(P,R))$ | |||

$=$ | $[\alpha ]({f}_{1}(P,Q))+[\alpha ]({f}_{1}(P,R))$ | |||

$=$ | ${f}_{2}(\alpha (P),\alpha (Q))+{f}_{2}(\alpha (P),\alpha (R)),$ |

which is equivalent^{} to $\alpha (Q)+\alpha (R)=\alpha (S)=\alpha (Q+R)$.

Next, suppose $dQ=R$, or $d{f}_{1}(P,Q)={f}_{1}(P,R)$, where $d\in D$. Then

${f}_{2}(\alpha (P),\alpha (R))$ | $=$ | $[\alpha ]({f}_{1}(P,R))$ | ||

$=$ | $[\alpha ](d{f}_{1}(P,Q))$ | |||

$=$ | $d[\alpha ]({f}_{1}(P,Q))$ | |||

$=$ | $d{f}_{2}(\alpha (P),\alpha (Q)),$ |

which is equivalent to $d\alpha (Q)=\alpha (R)=\alpha (dQ)$. ∎

###### Proposition 6.

If $\mathrm{(}V\mathrm{,}f\mathrm{)}$ is an affine space associated with the vector space $V$, then the direction $f$ is given by $f\mathit{}\mathrm{(}v\mathrm{,}w\mathrm{)}\mathrm{=}T\mathit{}\mathrm{(}w\mathrm{-}v\mathrm{)}$ for some linear isomorphism (invertible linear transformation) $T$.

###### Proof.

By proposition^{} 4, $(V,f)$ is affinely isomorphic to $(V,g)$ with $g(v,w)=w-v$. Suppose $\alpha :(V,f)\to (V,g)$ is the affine isomorphism. Then $[\alpha ](f(v,w))=g(\alpha (v),\alpha (w))=\alpha (w)-\alpha (v)$. Since $[\alpha ]$ is a linear isomorphism, $f(v,w)={[\alpha ]}^{-1}(\alpha (w))-{[\alpha ]}^{-1}(\alpha (v))$. By proposition 5, $\alpha $ itself is linear, so $f(v,w)=({[\alpha ]}^{-1}\circ \alpha )(w-v)$. Set $T={[\alpha ]}^{-1}\circ \alpha $. Then $T$ is linear and invertible^{} since $\alpha $ is, our assertion is proved.
∎

Title | properties of an affine transformation |
---|---|

Canonical name | PropertiesOfAnAffineTransformation |

Date of creation | 2013-03-22 18:31:53 |

Last modified on | 2013-03-22 18:31:53 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 7 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 51A10 |

Classification | msc 15A04 |

Classification | msc 51A15 |