# Proof that the area of a region in the plane is unchanged under shearing transformations

My Geometry book (Lang's Geometry 2nd edition) presents the following two theorems concerning shearing transformations without proof:

1. The area of a region in the plane is unchanged under shearing transformations.

2. The volume of a region in 3-space is unchanged under shearing transformations.

I would like to know how these results can be proved. The book defines area and volume to be the amount of space enclosed by a 2/3-dimensional figure, expressed in unit squares and cubes respectively. As to what a "shearing transformation" is, it simply states that it's a "stretching in some direction", so you are free to pick the definition you find best suited to the proof. Thanks.

• Could you define the shearing transformations as matrices, and use the determinant $=\pm1$? Commented Mar 9, 2018 at 20:41
• en.wikipedia.org/wiki/Shear_mapping Commented Mar 9, 2018 at 20:49
• – dxiv
Commented Mar 10, 2018 at 0:04

The easiest way to prove it would be geometrically.

All a shear mapping is is a way to transform a rectangle into a parallelogram.

As you learned in elementary school, a parallelogram with a given vertical and horizontal extent (base and height) has the same area as a rectangle with the same base and height, because you can just chop them up and rearrange them into one another.

This simple argument can be extended to any number of dimensions, and any combination of shear transforms in any direction.

The second, somewhat more rigorous way to prove it would be to use determinants, as @David-Quinn suggested. (If you don't know much linalg yet, a matrix is a way to transform space, and the determinant of a matrix represents how much the space is scaled during the transformation -- that is how much area/volume is lost or gained)

A shear matrix in 2D looks like this, which stretches space across the x-axis by a value $\lambda$:

$$\begin{bmatrix} 1 & \lambda \\ 0 & 1 \end{bmatrix}$$

Or this, which stretches spaces across the y-axis by a value $\lambda$:

$$\begin{bmatrix} 1 & 0 \\ \lambda & 1 \end{bmatrix}$$

The determinant of a 2D matrix is:

$$\begin{vmatrix} a & b\\c & d \end{vmatrix}=ad - bc$$

Substituting in our values gets us:

\begin{align}\begin{vmatrix} 1 & \lambda \text{ or } 0\\0 \text{ or } \lambda & 1 \end{vmatrix} & =1\cdot1 - \lambda\cdot0\\ \\& = 1\end{align}

Since our determinant is one, space has been scaled by exactly one, and therefore the area/volume of the space has not changed. This works pretty much the same in 3D, but taking the determinant of a 3D matrix is a long and painful procedure, so I'm not writing that one out.