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I thought that an affine transformation that maps an affine subspace on a parallel subspace is a dilation and in the other direction.

I tried to prove this.

So first I assumed that $F$ is an affine transformation that $S=p+V$ maps on $T=q+V$ (because $S$ and $T$ are parallel $(V=V)$). So $F(S)=T$.

Now I tried to prove that $F$ must be a dilation, so have an linear part $A=\lambda I$

I don't really know how to move on because I tried : $F(S)=A(p+V)+b=q+V$. So $Ap+AV+b=q+V$ and so $A=\frac{q+V-b}{p+V}$ and I'm not sure I can do this.

For the other way I assumed that $F(p)=\lambda Ip+b$ and I took an $S=p+V$. Now is $F(S)=\lambda (p+V)+b= \lambda p+b+(\lambda V)$. I'm not sure if this proves that $\lambda p+b+( \lambda V)$ is an affine subspace with the same direction as $S$.

Can someone correct and help me?

Alessio K
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questmath
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2 Answers2

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Notice that an affine transformation $F$ (such as a dilation) maps an affine supspace to an affine supspace. To be more concrete: if $S = p + V$ is an affine supspace of $\mathbb{A}^n$, then $F(S) = F(p) + F_*(V)$. Here $F_*$ denotes the linear part of the affine transformation.

First I will prove the other way around because this is rather trivial. Let $F$ be a dilation, i.e. $F_* = \lambda I$ for a $\lambda \in \mathbb{R}_0$. We have to prove that for every linear supspace $V$ of $\mathbb{R}^n$ we have $F_*(V) = V$. Let $x \in F_*(V)$. Then there exists a $v \in V$ such that $x = \lambda v$, and thus $x \in V$. For the other way around, let $x \in V$. Because $\frac{1}{\lambda}x \in V$, we have $ \lambda * \frac{1}{\lambda}x = x \in F_*(V)$. So we can conclude that $F_*(V) = V$ and thus every dilation maps an affine supspace to a parallel affine supspace.

Now for the "hard" part: Let $F$ be an affine transformation that maps every affine supspace to a parallel affine supspace. Let $v \in \mathbb{R}^n_0$. Define $V := \operatorname{span}(\{v\})$, which is a vector supspace of $\mathbb{R}^n$. Because $F$ maps every affine supspace to a parallel affine supspace, we have $F_*(V) = V$. Then $F_*(v) = \lambda v$ for some $\lambda \in \mathbb{R}$. Since $v$ was arbitrarily chosen, this means that every non-zero vector is an eigenvector of $F_*$. Now this precisely means that the linear part of the affine transformation is of the form $F_* = \lambda 'I$ with $\lambda' \in \mathbb{R}_0$. For a proof: Demonstration: If all vectors of $V$ are eigenvectors of $T$, then there is one $\lambda$ such that $T(v) = \lambda v$ for all $v \in V$.

Einsteinwasmyfather
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Write $G(x)=F(x)-F(0)$ it is a linear map which maps a line to a line, it is of the form $G(x)=cx$.

Classical proof: Let $u,v$ independent, $G(u)=au, G(v)=bv, G(u+v)=c(u+v)=au+bv$ implies $(c-a)u+(c-b)v=0$, since $u$ and $v$ are linearly independent, $c=a=b$.

Tsemo Aristide
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