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