Affine Maps, proving that any two lines on $\mathbb{R}^2$ are affine related.

Question

Consider affine maps, this time on $\mathbb{R}^2$.

1. Show that any two lines in $\mathbb{R}^2$ are affinely related.
2. Show that no triangle is affine related to a rectangle.

I will define the affine relation as follows: For two subsets, $S$, $T$ of $\mathbb{R}^n$, write $S \sim T$ if there exists an affine map $f$ such that $f(S) =T$. We then say that $S, T$ are affine-related.

1. I am having trouble proving that Vertical lines are affinely related to one another, and that Vertical lines are affinely related to Non-vertical lines. There must be a small trick that I am just not seeing..
2. I get caught up in the equations of a triangle and a rectangle and hit a rut early in the attempted proof... I believe I am going about the proof in the wrong way.

Answer 1

1) Suppose lines $S$ and $T$ are parallel in $\mathbb{R}^2$. So, for lines $S$ and $T$ in $\mathbb{R}^2$ defined by $S=Mx+s_1$ and $T=M'x+s_2$, $M=M'$.

The following affine transformation $f:S \rightarrow T$ can be defined $f(S)=T$ outlines by the following. $f(S)=f(Mx+s_1)=Mx+s_2=T$. So, for two parallel lines $S$ and $T$, $S \sim T$. This shows that any two parallel lines have an affine transformation and thus are affine related.

Now, it needs to be shown that any two lines in $\mathbb{R}^2$ are affine related. So, suppose now that $S$ and $T$ are not parallel. I claim that there is a rotational transformation of a line such that changes the slope of the line. Specifically, the rotation of any line $R$, for line $Z$, $R(Z)$ is also a line. So, for any two lines $S$ and $T$ in $\mathbb{R}^2$ defined by $S=Mx+s_1$ and $T=M'x+s_2$, $M \neq M'$, there is a rotation $R$ of line $S$, defined by $R(S)=R(Mx+s_1)=M'x+s_1$, which makes lines $S$ and $T$ parallel. And two parallel lines are affine related, thus, any two lines in $\mathbb{R}^2$ are affine related.

2) Let $T$ be a triangle, let $R$ be a rectangle, and let $f$ be an affine transformation. $f$ is invertible, as shown from problem 1 above. (Note: for any $f(T)=R$, there exists another affine transformation, namely $f^{-1}$ such that $f^{-1}(R)=T$.)

But, affine transformations take a line $\in \mathbb{R}^2$ to another distinct line, as seen from #1.

So, for any $f(T)$, the three lines (edges) of the triangle $T$ will be sent to another three lines- but no rectangle is made up of only three lines- 4 lines are required. So there can be no affine transformation where the edges of $T$ span the edges of $R$. So, $T$ is not affine related to $R$.

Answer 2

Try doing it in steps!

---

For $(1)$:

• First, show that if two lines $S$ and $T$ are parallel, then there is a translation $f$ with $f(S)=T$.
• Then, show that for any line $S$ and rotation $r$, $r(S)$ is a line.
• Finally, show that for any two lines $S$ and $T$, there is a rotation $g$ about the origin such that $g(S)$ and $T$ are parallel.

What can you conclude?

---

For $(2)$, let $T$ be a triangle, $R$ a rectangle, and $f$ an affine transformation.

• Show that if an affine transformation is not invertible, then its image lies in a line. Therefore, if $f$ is not invertible, is it possible that $f(T)=R$?
• Now, show that if an affine transformation $f$ is invertible, then $f^{-1}$ is also affine. Hence, if $f(T)=R$ there must be an affine $g=f^{-1}$ with $g(R)=T$.
• Finally, show that any affine transformation $h$ takes distinct lines to distinct lines. How many distinct lines are there in $R$? How many are there in $T$? Conclude.