Prove that the area of an image in $2d$ cartesian coordinates is equal to the determinant of the linear transformation times the area of the initial shape.

I've tried to formulate general expression for area given lots of points, but it feels like that's barking up the wrong tree.

I've also proved it for transformations which are combinations of rotations and enlargements. If that is the case, the distance between each point in the shape will increase by a constant, which has to be the same no matter what the initial shape is. So we can take a unit square which is the easiest case, and it is trivial to show that the resulting area is $\det(T)$. But because these transformations stretch lengths by the same constant, it must stretch areas by the same constant - $\det(T)$. Similarly, is it possible to show that all linear transformations (i.e. shears and compressions) have some property that allows us to deduce that the area of the image must transform by some constant for any image?

  • $\begingroup$ Related: math.stackexchange.com/questions/427528 $\endgroup$ – Watson Dec 28 '16 at 12:24
  • $\begingroup$ More generally: math.stackexchange.com/questions/1923303 $\endgroup$ – Watson Dec 28 '16 at 12:30
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    $\begingroup$ How do you define the area of the shape in the first place? $\endgroup$ – amd Dec 28 '16 at 20:03
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    $\begingroup$ area enclosed by perimeter that connects every point and doesn't intersect itself $\endgroup$ – ggmate Dec 29 '16 at 11:36
  • $\begingroup$ all we have to do is show that for any given linear transformation, the area of the image is directly proportional to the area of the object $\endgroup$ – ggmate Dec 29 '16 at 11:49

Assuming the linear transformation $T$ is bijective (otherwise the question loses its meaning) this proposition can be proven using calculus as follows:

Let $ U \subseteq \mathbb{R}^2 $ be the set representing the initial shape to be transformed by T. The usual area of U can be defined using the Riemann Integral in $ \mathbb{R}^2 $ as

$$ area(U) := \int_{ U }{ f(u) \, du } $$

where $ f(u) = 1 $ for every $ u \in \mathbb{R}^2 $. So the area of U is $ \int_{ U }{du } $ for short.

Let $ dT(u) $ denote the Jacobian matrix of T at the point $u$. Because T is a linear transformation it follows that $ dT(u) = T $. Skipping the finer details, the change of variables theorem for multiple integrals implies that

$$ \int_{ T(U) }{ f(v) \, dv } = \int_{ U }{ f(T(u)) \, |det(\,dT(u)\,)| \, du } $$

Hence the area of the transformed image is

$$ area(\,T(U)\,) = \int_{ T(U) }{ dv } = \int_{ U }{ |det(\,dT(u)\,)| \, du } = \int_{ U }{ |det(T)| \, du } $$

Since $ |det(T)| $ is constant, it follows that

$$ \int_{ U }{ |det(T)| \, du } = |det(T)| \int_{ U }{ du } = |det(T)| \, area(U) $$


$$ area(\,T(U)\,) = area(U) \, |det(T)| $$

as we wanted to show.

Notice that this is a more general result that holds for $ \mathbb{R}^n $.


If - as stated in a comment by the OP - the area is enclosed by a perimeter that connects every point and doesn't intersect itself, then we might have a picture like this:

enter image description here

What we can do is to draw ($\color{red}{red}$) lines from the origin $(0)$ to the vertices of the polygon (six in our example) and sum up the areas of the triangles $(0,1,2),(0,2,3),(0,3,4),(0,4,5),(0,5,6),(0,6,1)$ : see figure on the left.
It is shown in the figure on the right how the area of just one of these triangles is calculated, using a determinant: $$ \mbox{area}\,\Delta = \frac{1}{2} \det\begin{bmatrix}(x_1-x_0) & (x_2-x_0)\\(y_1-y_0) & (y_2-y_0)\end{bmatrix} $$ Note that, in general, the triangle areas thus calculated can be positive as well as negative; and the latter is essential. Now continue for the vertex coordinates of the polygon, in anti-clockwise order, to calculate the area of the whole 2-D object.
Closed perimeters in an image are likely to be the product of a contouring procedure. Such contours or isolines can be clockwise or anti-clockwise oriented, corresponding respectively with a negative or a positive enclosed area. It's wise not to destroy that information prematurely. As an example: negative areas enable the existence of objects with holes in it.
In principle, the origin can be chosen at will. But for best results (numerically) the vertex centroid $\,\sum_k(x_k,y_k)/N\,$ of the polygon may be a good choice.

It's easy to prove the OP's conjecture for one of these triangles. One of the vertices is located at the origin, take $\,(x_0,y_0) = (0,0)$ . Therefore the triangle is defined by two column vectors, that can be summarized into a matrix. And the area of that triangle is half the determinant of that matrix, as we have seen. Symbolically: $$ \Delta = \begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix} $$ Now let the transformation be given by: $$ T = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ Then the deformed triangle is represented by: $$ \Delta' = \begin{bmatrix} x'_1 & x'_2 \\ y'_1 & y'_2 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix} = \begin{bmatrix} ax_1+by_1 & ax_2+by_2\\ cx_1+dy_1 & cx_1+dy_2\end{bmatrix} $$ The deformed area, for one triangle, is: $$ \frac{1}{2} \det\begin{bmatrix} x'_1 & x'_2 \\ y'_1 & y'_2 \end{bmatrix} = \frac{1}{2} \det\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix}\right) = \det(T)\left(\mbox{area}\,\Delta\right) $$ And for all triangle areas summed: $$ \sum_\Delta \left(\mbox{area}\,\Delta\right) \det(T) = \det(T) \sum_\Delta \left(\mbox{area}\,\Delta\right) = \mbox{Area of object} \cdot \det(T) $$ Almost as conjectured. It is not guaranteed, though, that $\det(T)$ is positive and that the absolute value of the determinant can be taken, unless such is specified explicitly for the transformation $T$. For example, if we have mirroring in the y-axis: $$ T = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \quad \Longrightarrow \quad \det(T) = -1 $$ If our perimeter has been counter-clockwise then it becomes clockwise, and the positive area of the original is transformed into a negative area. We should put $\;\left|\det(T)\right|\;$ instead of $\;\det(T)\;$ only if it is decided that the sign of the area of an object is not relevant.

  • $\begingroup$ One thing I don't understand is why you're allowed to move the origin without it messing everything up? $\endgroup$ – ggmate Jan 5 '17 at 18:31
  • $\begingroup$ @gebra: You can see it graphically by considering the area of the polygon enclosed by (1)-(2)-(3)-(4)-(5)-(6), coloring it gray for example. That doesn't change with another choice of the origin: make another picture with the same polygon, different triangles and you shall see. $\endgroup$ – Han de Bruijn Jan 5 '17 at 18:54
  • $\begingroup$ ah ok, I think I see $\endgroup$ – ggmate Jan 5 '17 at 19:00
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    $\begingroup$ @gebra: Good point ! It makes indeed a difference where the origin is chosen, for e.g. performing a rotation around it. I think that the problem is solved if the origin is simply determined according to the transformation at hand, because areas will remain independent of that choice. Thus: (1) determine origin according to the transformation, (2) determine area, (3) do the transformation, (4) determine transformed area. $\endgroup$ – Han de Bruijn Jan 6 '17 at 15:55

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