# When doing a linear transformation for change of variables, why is a rectangle mapped to a parallelogram? [closed]

I recently started learning change of variables for (double/triple) integration. One concept that I've struggled to understand is why the linear transformation from one coordinate system to another changes the shape of our rectangle to a parallelogram. I've read through many articles and my textbook, but none give a direct explanation of why this phenomenon occurs.

The following webpage displays an example of this change in geometry:

Notice how the transformation causes the rectangle to transform into a parallelogram.

I've spent a lot of time reading various articles, but I still do not understand what causes this change.

I would greatly appreciate it if people could please take the time to explain why this phenomenon occurs and what the benefits of it are to us.

## closed as off-topic by TheGeekGreek, JMP, Juniven, hardmath, Daniel W. FarlowApr 4 '17 at 14:51

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, Juniven, Daniel W. Farlow
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• How is my question off-topic? I would appreciate advice on how I can improve my question. – The Pointer Apr 4 '17 at 7:14
• It is unclear what you mean by "why" a rectangle is transformed (by a linear transformation) into a parallelogram. You seem to say that you've seen examples where this happens, but this doesn't exclude that you might want a proof that the image of a rectangle is always a parallelogram (if not again a rectangle, or something degenerate like a line segment or a point). Asking "why this phenomenon occurs" could leave Readers wondering if you think a linear transformation might create a triangle (or a circle?) instead of a parallelogram, and "what the benefits" are is a bit subjective. – hardmath Apr 4 '17 at 14:14

Slightly more general than your statement: linear transformations also map parallelograms to parallelograms. Note that a parallelogram is completely determined by three (non-collinear) points (vertices), call them $\vec a$, $\vec b$ and $\vec c$. The parallelogram is then given by the set of all points $\vec x$: $$\vec x = \vec a + \lambda \bigl( \vec b - \vec a \bigr)+ \mu \bigl( \vec c - \vec a \bigr) \quad \quad \left( 0 \le \lambda, \mu \le 1 \right)$$ A linear transformation $L$ maps these points to: \begin{align} L(\vec x) & = L\left(\vec a + \lambda \bigl( \vec b - \vec a \bigr)+ \mu \bigl( \vec c - \vec a \bigr)\right) \quad \quad\quad \left( 0 \le \lambda, \mu \le 1 \right)\\ & = L(\vec a) + \lambda \bigl( L(\vec b) - L(\vec a) \bigr)+ \mu \bigl( L(\vec c) - L(\vec a) \bigr) \end{align} And the set of all points $L(\vec x)$ is now a parallelogram determined by the points $L(\vec a)$, $L(\vec b)$ and $L(\vec c)$. Note that these vertices are not collinear since $\vec a$, $\vec b$ and $\vec c$ weren't.