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'properties of an affine transformation'
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| Title of object: |
properties of an affine transformation |
| Canonical Name: |
PropertiesOfAnAffineTransformation |
| Type: |
Definition |
| Created on: |
2008-11-05 17:23:37 |
| Modified on: |
2008-11-05 17:37:01 |
| Classification: |
msc:15A04, msc:51A15, msc:51A10 |
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Content:
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} |
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