# Transformation Magic

How do people come up with transformations??

It can't be a coincidence when the textbook has you turn a weird shape into a perfect rectangle!!

My question: What is the procedure for transforming a parallelogram into a rectangle?

I believe that it should be very easy, and involves some type of $x=\frac{1}{2}(au+bv)$ and $y=\frac{1}{2}(au-bv)$.

Hint:

Let the parallelogram be defined by the vectors $(p,q)$ and $(r,s)$. You want to turn it to a "perfect rectangle", i.e. presumably map to the vectors $(w,0)$ and $(0,h)$.

Write

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}p&r\\q&s\end{bmatrix}=\begin{bmatrix}w&0\\0&h\end{bmatrix}$$

expand and solve for the unknown coefficients $a,b,c,d$. This is a system of four equations in four unknowns, but that splits in two systems of two equations.

Let the parallelogram have the vertices $ABCD$ and the rectangle the vertices $A'B'C'D'$.

Assuming the transformation is the net result of linear and affine transformations, you could stick to homogeneous coordinates and try to solve $$T(OA, OB, OC, OD) = (OA', OB', OC', OD')$$ where $OX$ is the vector from the origin to point $X$.

You should provide the dimensions (2D, 3D, nD?) and additional constraints (e.g. particular sides oriented parallel to some coordinate axis) that might simplify the problem.

An example would help.

I'm not very sure if this answers your question, but if you define a linear isomorphism $T: \mathbb{R^2} \to \mathbb{R^2}$ such that $T(0,1) = v$ and $T(1,0) = w$, the image of the square $[0,1] \times [0,1]$ gets mapped to a parallelogram, with a vertex in $(0,0)$ and such that the segments $\vec{0v}$ and $\vec{0w}$ are two of its sides. Now you can take the inverse function to map the parallelogram to $[0,1] \times [0,1]$. (You can calculate $T^{-1}$ since $T$ is an isomorphism,it all just comes down to inverting a 2 by 2 matrix).

You can further generalize this by considering linear isomorpshisms in $\mathbb{R}^k$, or more in general a special family of functions which are injective "almost everywhere".