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How can I prove that a $2\times2$ matrix $A$ is area-preserving iff $\det(A)=1$ or $\det(A)=-1$?

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You need to check that the area of a parallelogram $ABCD$ is the absolute value of the determinant of the matrix whose columms (or rows) are $\vec{AB}$ and $\vec{AD}$. Then use multiplicativity of det. – Joel Cohen Oct 12 '11 at 11:59

A cute argument (in any dimension $n$) is to recall that the signed volume of a parallelepiped spanned by $v_1, \ldots, v_n$ is given by the determinant of the matrix $$V = \left[ v_1, \ldots, v_n\right],$$ and that a matrix $A$ transforms the parallelepiped simply by $V \to AV$, i.e. by acting on the columns of $V$. Thus $A$ preserves area when

$$|\det V | = |\det(AV)| = |\det A||\det V|.$$

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But how can I be sure that the area of any (not just a parallelepiped) region on the plane will be preserved? – bass Oct 12 '11 at 12:07
Area is defined to be the integral over the region of the constant function 1 times the area form: $\int_{\Omega} 1 dA$. You can think of this integral as tiling the region with parallelepipeds, and approximating the area of the region by the sum of the areas of the parallelepipeds; as you shrink the size of tile you use you shrink the error also, until in the limit of infinitesimally small parallelepipeds $dA$ you get exactly the area of the region. So if a map preserves areas of parallelepipeds it preserves areas of other regions (defied as sums of smaller and smaller parallelepipeds) too. – user7530 Oct 12 '11 at 12:21

[This was going to be just a comment, but it got too large]

The answers of @Phira and @user7530 are both correct.

In fact, it suffices to see that $A$ preserves the areas of parallelograms. For an arbitrary region $E$, I may approximate $E$ by small rectangles (parallelograms), each centered at a specific vector $\mathbf{v}_i$ (think of integration to find area of a region in $\mathbb{R}^2$). Now say each small rectangle has area $a_i$. Then if $A$ is parallelogram-preserving, applying $A$ to the small rectangle centered at $\mathbb{v}_i$ will result in a parallelogram of size $a_i$ centered at $A\mathbb{v}_i$.

There is the small matter of showing that rectangles do not get mapped so that their images overlap, but that is easy to rule out because if true, then the area of a larger parallelogram containing the two would necessarily map to a parallelogram of less area.

Hope this helps!

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The answer to this question has two parts:

  1. The difficult part: A linear map $A$ multiplies the area of all shapes with the same factor $\chi_A$. To prove this one has to set up a clear notion of "area". Then one proves that this "area" has the expected properties (finite additivity, various invariances, etc.) by means of an approximation argument, as hinted at in Shaun Alt's answer.

  2. The easy part: The factor $\chi_A$ described in 1. is given by $\chi_A=|\det(A)|$. This can be proven, e.g., in the following way: It is a theorem of linear algebra that any $2\times2$-matrix $A$ is the product of matrices of the following types: $$\left[\matrix{\lambda & 0\cr 0& 1\cr}\right],\ \left[\matrix{1 & 0\cr 0& \mu\cr}\right],\ \left[\matrix{1 & 1\cr 0& 1\cr}\right],\ \left[\matrix{1 & 0\cr 1& 1\cr}\right].$$ For each of these types $T$ one can easily verify that $\chi_T=|\det(T)|$, and as $\chi_{\cdot}$ as well as $|\det(\cdot)|$ is multiplicative the general statement follows.

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