If $A$ is an $n$ -by-$n$ matrix and $S\subset {\mathbf{R}}^{n}$ is the unit $n$ -dimensional cube then $|det(A)|=\mathrm{vol}(A(S))$ .

# 30. Geometry of determinants, 2

## 30. Geometry of determinants, 2

### Higher dimensions

Last time we saw a geometric interpretation of the determinant of a 2-by-2 matrix $A$ as the area of a parallelogram obtained by applying $A$ to a square. This result holds in all dimensions, though we will not prove it:

The absolute value signs around the determinant need to be there because areas and volumes are positive but determinants can be negative. In the proof of the 2-by-2 case, I hid the issue in the picture I drew: if I had switched the vectors $\left(\begin{array}{c}\hfill a\hfill \\ \hfill c\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill b\hfill \\ \hfill d\hfill \end{array}\right)$ in the picture, we would have ended up with $bc-ad$ , i.e. $-det(A)$ .

In higher dimensions, $A(S)$
is no longer a parallelogram, but a *parallelopiped* ("parallel-legged"): a shape whose faces are parallel pairs of congruent parallelograms.

### Tetrahedra/simplices

Another useful characterisation of the determinant (which we won't prove) is the following.

Let ${a}_{1},\mathrm{\dots},{a}_{n}$
be vectors in ${\mathbf{R}}^{n}$
. Consider the *simplex* with vertices at $0,{a}_{1},\mathrm{\dots},{a}_{n}$
. Then the volume of this simplex is $\frac{1}{n!}det(A)$
where $A$
is the matrix with columns ${a}_{1},\mathrm{\dots},{a}_{n}$
.

A simplex is the simplest convex shape you can draw with these vertices: you take the vertices, connect them by lines, and then fill in all the space in between. (More formally, this is called taking the convex hull of the vertices). In 2 dimensions this gives a triangle (3 vertices). In 3 dimensions you get a tetrahedron.

These two theorems are closely related: you can decompose a cube into $n!$ simplices. Rather than proving either of them, I want to use this second theorem to calculate the volumes of some tetrahedra in ${\mathbf{R}}^{3}$ .

Take ${a}_{1}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)$ , ${a}_{2}=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)$ and ${a}_{3}=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)$ . The corresponding tetrahedron has volume $\frac{1}{6}det\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\frac{1}{6}$ .

Take a regular tetrahedron (d4 for the roleplayers out there) with vertices at ${a}_{0}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)$ , ${a}_{1}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill -1\hfill \end{array}\right)$ , ${a}_{2}=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right)$ and ${a}_{3}=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill -1\hfill \\ \hfill 1\hfill \end{array}\right)$ . (These vertices are half of the vertices of a regular cube centred at the origin with sidelength 2.)

Translating by $-{a}_{0}$ , we get the vertices $0,{a}_{1}-{a}_{0},{a}_{2}-{a}_{0},{a}_{3}-{a}_{0}$ , so we can apply our theorem with $$A=\left(\begin{array}{ccc}\hfill {a}_{1}-{a}_{0}\hfill & \hfill {a}_{2}-{a}_{0}\hfill & \hfill {a}_{3}-{a}_{0}\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill -2\hfill & \hfill -2\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill -2\hfill \\ \hfill -2\hfill & \hfill -2\hfill & \hfill 0\hfill \end{array}\right).$$ We compute the determinant using the cofactor expansion along the top row: $$\mathrm{vol}=\left|\frac{1}{6}det\left(\begin{array}{ccc}\hfill 0\hfill & \hfill -2\hfill & \hfill -2\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill -2\hfill \\ \hfill -2\hfill & \hfill -2\hfill & \hfill 0\hfill \end{array}\right)\right|$$ $$=\frac{1}{6}|(-(-2)det\left(\begin{array}{cc}\hfill -2\hfill & \hfill -2\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right)-2det\left(\begin{array}{cc}\hfill -2\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill -2\hfill \end{array}\right)|$$ $$=\frac{2}{6}\left|(-4-4)\right|$$ $$=\frac{16}{6}$$

### Determinant as scale factor for volumes

$det(A)$ is the scale factor for volumes of cubes under the transformation $A$ . Since any volume is defined by subdividing/approximating by smaller and smaller cubes, $det(A)$ is the scale factor for any volume under the transformation $A$ . From this point of view, the formula $det(AB)=det(A)det(B)$ is just telling us that if we apply the transformation $B$ and then the transformation $A$ , the volumes rescale by $det(B)$ and then by $det(A)$ , giving a total scale factor of $det(A)det(B)$ for the composite transformation.