Prove $Area(T(\mathcal{R}))=|Det(A)|.Area(\mathcal{R})$

Question

Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be a linear transformation with matrix $A$. Let $\mathcal{R}$ be a region in $\mathbb{R}^2$. Then: $$ Area(T(\mathcal{R}))=|Det(A)|.Area(\mathcal{R}) $$ It is also statemed that this can be extended to higher dimensions too.

Is there a formal proof for this ? Or where does this actually come from ?

I just know that the above statements can be verified for specific examples and $|Det(A)|$ is the area of the parallelogram formed by the column vectors of $A$.

Answer 1

When you believe the formula for squares, and ${\cal R}$ is a nice region you can argue as follows (assume $\det T\ne 0$ for simplicity):

Given an $\epsilon>0$ draw a lattice in the plane which is so fine that there are two "buildings" $B_0$, $B_1$ of lattice squares such that $$B_0\subset {\cal R}\subset B_1,\qquad{\rm area}(B_1\setminus B_0)<\epsilon\ .\tag{1}$$ We then have $T(B_0)\subset T({\cal R})\subset T(B_1)$ and therefore $$|\det T|\ \>{\rm area}(B_0)\leq{\rm area}\bigl(T({\cal R})\bigr)\leq|\det T|\ {\rm area}(B_1)\ .$$ Because of $(1)$ this implies $${\rm area}({\cal R})-\epsilon\leq{{\rm area}\bigl(T({\cal R})\bigr)\over |\det T|}\leq{\rm area}({\cal R})+\epsilon\ ,$$ and as $\epsilon>0$ was arbitrary we obtain the claim.

Answer 2

As an illustration to my comment on your post I give you the calculation for any invertible linear transformation $T:\mathbb{R}^n\to\mathbb{R}^n$ with matrix $A$. In your case ($n = 2$) the n-dimensional "$Volume$" would be the "$Area$":

Startig with $y = Tx$ we have

\begin{eqnarray*} Volume(T(\mathcal{R})) & = & \int_{T(\mathcal{R})}\;dy \\ & = & \int_{\mathcal{R}}\;d(Tx) \\ & = & \int_{\mathcal{R}}|D(Tx)|\;dx\\ & = & \int_{\mathcal{R}}|det(A)|\;dx\\ & = & |\det(A)|\int_{\mathcal{R}}\;dx\\ & = & |\det(A)|Volume(\mathcal{R})\\ \end{eqnarray*}