# Area of an arbitrary region under dilation

I have just begun learning about dilations. In my book (Lang's Geometry 2nd edition), dilation by $r$ relative to a point $0$ is defined to be the association which to each point $P$ of the plane associates the point $P'$ that lies on the ray $R_{op}$ at a distance from $0$ equal to $r$ times that of $P$ from $O$.

What I'm having trouble with is proving the followng thereom, which the book presents without a rigorous proof:

Let $S$ be an arbitrary region in the plane with area $A$. Let $rS$ be the image of $S$ under a dilation by a positive number $r$. Then the area of $rS$ is $r^2·A$.

I know the proof for arbitrary triangles and rectangles, but how to prove it for an arbitrary shape, be it a polygon or not? Thanks very much in advance.

• The proof for an arbitrary shape depends on the definition you have for the area of an arbitrary shape. If it comes from an approximation by unions of rectangles it's straightforward. For polygons you can just triangulate. Commented Jan 4, 2018 at 18:01
• @Ethan Bolker It's not necessay to triangulate (see my answer) Commented Jan 5, 2018 at 23:53
• @JeanMarie True - as long as you have a definition of area. The determinant does indeed scale the area of anything that has an area ... Commented Jan 6, 2018 at 1:36
• @Ethan Bolker I agree. I was considering here that the regions under consideration have a well-defined area. In fact, we are working with $\int 1_{f(A)} = \int 1_A det(f)$ where $1_X$ is the characteristic function of $X$ and $f$ is here the enlargment operation. Commented Jan 6, 2018 at 5:32
• @EthanBolker I'm not sure if that's what you asked, but I've always thought of the area of a 2d shape as being the number of square units that can ben fit into it. I know Jean Marie has already answer my question, but I couldn't stop thinking about your comment. Would you care to post your proof if the area comes from an approximation by unions of rectangles, if it's not too much trouble? I think that seeing it proven two different ways would be really profitable for my understanding. Thanks. Commented Jan 6, 2018 at 10:10

let us express the relationship between $(x,y)$, the generic point of $P$, and $(X,Y)$ its image in $P'$ with vector/matrix notations:

$$\tag{1}\begin{cases}X=rx\\Y=ry \end{cases} \ \ \iff \ \ \begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}r&0\\0&r\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$

Then it suffices, (in the spirit of Serge Lang's book), to use the geometrical interpretation of the determinant of the matrix of the linear transformation $f$, which is $r^2$, as:

$$\tag{2} \text{final area = initial area} \ \times \ r^2$$

$Edit:$ In fact formula (2) comes from what could consider as the "ultimate form" of the change of variables formula :

$$\int 1_{f(A)} = \int 1_A \det(f).$$

where $1_X$ is the characteristic function of (measurable!) set $A$ : $1$ on $A$, zero elsewhere).

Note that $\det(f)$ appears as the jacobian of transformation $f$.