# 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 $[\alpha]$ 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 $[\alpha]$ 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 $[\alpha]$ is.

###### Proposition 3.

A bijective affine transformation $\alpha:A_{1}\to A_{2}$ 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 $\alpha:A\to B$ is affine, then $\alpha:A_{P}\to B_{\alpha(P)}$ 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

 $\displaystyle f_{2}(\alpha(P),\alpha(S))$ $\displaystyle=$ $\displaystyle[\alpha](f_{1}(P,S))$ $\displaystyle=$ $\displaystyle[\alpha](f_{1}(P,Q)+f_{1}(P,R))$ $\displaystyle=$ $\displaystyle[\alpha](f_{1}(P,Q))+[\alpha](f_{1}(P,R))$ $\displaystyle=$ $\displaystyle f_{2}(\alpha(P),\alpha(Q))+f_{2}(\alpha(P),\alpha(R)),$

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

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

 $\displaystyle f_{2}(\alpha(P),\alpha(R))$ $\displaystyle=$ $\displaystyle[\alpha](f_{1}(P,R))$ $\displaystyle=$ $\displaystyle[\alpha](df_{1}(P,Q))$ $\displaystyle=$ $\displaystyle d[\alpha](f_{1}(P,Q))$ $\displaystyle=$ $\displaystyle df_{2}(\alpha(P),\alpha(Q)),$

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

###### Proposition 6.

If $(V,f)$ is an affine space associated with the vector space $V$, then the direction $f$ is given by $f(v,w)=T(w-v)$ 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 PropertiesOfAnAffineTransformation 2013-03-22 18:31:53 2013-03-22 18:31:53 CWoo (3771) CWoo (3771) 7 CWoo (3771) Definition msc 51A10 msc 15A04 msc 51A15