PropertiesOfAnAffineTransformation 2008-11-05 17:23:37 2008-11-06 16:48:05 Definition affine transformation \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} % define commands here \newcommand*{\abs}[1]{\left\lvert #1\right\rvert} \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{cor}{Corollary} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} 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. \begin{prop} $\alpha$ is one-to-one iff $[\alpha]$ is.\end{prop} \begin{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. \end{proof} \begin{prop} $\alpha$ is onto iff $[\alpha]$ is.\end{prop} \begin{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. \end{proof} \begin{cor} $\alpha$ is a bijection iff $[\alpha]$ is. \end{cor} \begin{prop} A bijective affine transformation $\alpha:A_1\to A_2$ is an affine isomorphism. \end{prop} \begin{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 $[\alpha]^{-1}$. \end{proof} \begin{prop} Two affine spaces associated with the same vector space $V$ are affinely isomorphic. \end{prop} \begin{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 $\alpha$. \end{proof} \begin{prop} 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. \end{prop} \begin{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 \begin{eqnarray*} 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)), \end{eqnarray*} 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 \begin{eqnarray*} f_2(\alpha(P),\alpha(R)) &=& [\alpha](f_1(P,R)) \\ &=& [\alpha](df_1(P,Q)) \\ &=& d[\alpha](f_1(P,Q)) \\ &=& df_2(\alpha(P),\alpha(Q)), \end{eqnarray*} which is equivalent to $d\alpha(Q)=\alpha(R)=\alpha(dQ)$. \end{proof} \begin{prop} 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) $T$. \end{prop} \begin{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. \end{proof}