[Math] How to find the affine transformation(s) ( if any ) that maps one quadrilateral into another.

euclidean-geometrygeometrylinear algebra

Given the

Fundamental theorem of affine geometry. Let P,Q,R be any three non-collinear points in R2, and let U, V,W be any three other such points. Then there is exactly one affine transformation that maps P to U, Q to V and R to W.

and that I know how to calculate the aforementioned affine transformation, I have the following question.

Suppose I have two quadrilaterals q1 and q2 in R2. Quadrilateral q1 is allegedly transformed into quadrilateral q2. Is there in this case also one affine transformation that maps q1 to q2? Are there perhaps more? What is the strategy to find the translation and linear transformation of the affine transformation?

Best Answer

By insisting that a quadrilateral be sent to another quadrilateral, you are insisting that four non-colinear points be sent to four non-colinear points.

By a translation, we can take any vertex of $Q_1$ on to any vertex of $Q_2$. That leaves three free vertices for both $Q_1$ and $Q_2$ to be paired up. Given one free vertex of $Q_1$, there are three free vertices of $Q_2$ for it to be sent to. Given another, remaining, free vertex of $Q_1$, there are two free vertices of $Q_2$ remaining. Hence, there are exactly $4 \times 3 \times 2 = 24$ distinct affine transformations which match three of the vertices of $Q_1$ with three of the vertices of $Q_2$.

At this point, all of the degrees of freedom have been used up. You can only specify the image of three points, and we have done that. The fourth vertex of $Q_1$ will, or will not be paired with the fourth vertex of $Q_2$. In very exceptional circumstances it will. Remember that parallel lines must remain parallel under affine transformations, and conversely, non-parallel lines cannot be made parallel under affine transformation.

(Think of an ordinary linear space with a fixed origin. Parallelagrams are sent to parallelograms. If you have a rhombus as $Q_1$ and a rectangle as $Q_2$ then you can pair three of the vertices (including the origin), but the fourth is only possible if the rhombus is actually a rectangle of an area prescribed by the pairing of the first three vertices.)

In short, for almost all quadrilaterals, there is no affine transformation from one to the other. In very exceptional circumstances, the fourth vertices will line up (by chance) and you will have 24 different affine transformations from $Q_1$ to $Q_2$, depending on how you wish to order the pairing of vertices.