# Uniqueness of affine transformations

I'm trying to find an affine transformation from the unit square $S = [0,1]\times[0,1]$ onto the parallelogram $P$ with vertices $(0,3)$, $(2,1)$, $(1,6)$, and $(3,4)$. I've done this two different ways, and gotten two different answers.

Way #1 is to first map $S$ onto the parallelogram $P_1$ with vertices $(0,0)$, $(1,3)$, $(-2,2)$, and $(-1,5)$. This can be done via the linear map $L_1\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}1&-2\\3&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$. Then, since $P$ is $P_1$ shifted by the vector $\begin{bmatrix}2\\1\end{bmatrix}$, the desired affine map would be $A_1\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}1&-2\\3&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} + \begin{bmatrix}2\\1\end{bmatrix}$.

Way #2 is to first map $S$ onto the parallelogram $P_2$ with vertices $(0,0)$, $(3,1)$, $(2,-2)$, and $(1,3)$. This can be done via the linear map $L_2\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}2&1\\-2&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$. Then, since $P$ is $P_2$ shifted up by the vector $\begin{bmatrix}0\\3\end{bmatrix}$, the desired affine map would be $A_2\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}2&1\\-2&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} + \begin{bmatrix}0\\3\end{bmatrix}$.

These both seem to be valid answers, so how do I reconcile them? They seem like different functions, but both are affine and both successfully (I think) map $S$ onto $P$.

• Yes, it definitely is. – Ron Dec 12 '16 at 22:34
• The line (segment) between $(0,3)$ and $(2,1)$ is parallel to the line between $(1,6)$ and $(3,4)$. Both have slope $-1$. At the same time, the line between $(0,3)$ and $(1,6)$ is parallel to the line between $(2,1)$ and $(3,4)$. Both have slope $3$. – Ron Dec 12 '16 at 22:37
• $y = 3x+3$, $y = 3x-5$ is one set of parallel lines; the other is $y = -x+3$, $y = -x+7$ – Ron Dec 12 '16 at 22:40

As you’ve discovered, there is not a unique affine map that takes $S$ to $P$. In fact, there are eight different ones: You have four choices for the vertex that is the image of the origin, and for each of these choices, there are two choices (orientations) for the edge mappings. You’ve found two of them, one that maps the origin to $(2,1)$ and one that maps it to $(0,3)$.