30. Geometry of determinants, 2

Higher dimensions

0.00 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:

Theorem:

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

Remark:

1.15 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 $\begin{pmatrix}a\\ c\end{pmatrix}$ and $\begin{pmatrix}b\\ d\end{pmatrix}$ in the picture, we would have ended up with $bc-ad$ , i.e. $-\det(A)$ .

Remark:

2.06 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

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

Theorem:

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

Remark:

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.

A triangle and a tetrahedron (2- and 3-dimensional simplices)

5.17 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 $\mathbf{R}^{3}$ .

Example:

5.40 Take $a_{1}=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$ , $a_{2}=\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}$ and $a_{3}=\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}$ . The corresponding tetrahedron has volume $\frac{1}{6}\det\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\frac{1}{6}$ .

The tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0) and (0,0,1)

Example:

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

8.18 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

$A=\begin{pmatrix}a_{1}-a_{0}&a_{2}-a_{0}&a_{3}-a_{0}\end{pmatrix}=\begin{% pmatrix}0&-2&-2\\ -2&0&-2\\ -2&-2&0\end{pmatrix}.$ 9.22 We compute the determinant using the cofactor expansion along the top row:

$\mathrm{vol}=\left|\frac{1}{6}\det\begin{pmatrix}0&-2&-2\\ -2&0&-2\\ -2&-2&0\end{pmatrix}\right|$

$=\frac{1}{6}\left|(-(-2)\det\begin{pmatrix}-2&-2\\ -2&0\end{pmatrix}-2\det\begin{pmatrix}-2&0\\ -2&-2\end{pmatrix}\right|$

$=\frac{2}{6}\left|(-4-4)\right|$

$=\frac{16}{6}$

Determinant as scale factor for volumes

11.13 $\det(A)$ 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.