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:

Theorem:

If A is an n -by- n matrix and S 𝐑 n is the unit n -dimensional cube then | det ( A ) | = vol ( A ( S ) ) .

Remark:

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 ( a c ) and ( b d ) in the picture, we would have ended up with b c - a d , i.e. - det ( A ) .

Remark:

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.

Theorem:

Let a 1 , , a n be vectors in 𝐑 n . Consider the simplex with vertices at 0 , a 1 , , a n . Then the volume of this simplex is 1 n ! det ( A ) where A is the matrix with columns a 1 , , 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)

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 𝐑 3 .

Example:

Take a 1 = ( 1 0 0 ) , a 2 = ( 0 1 0 ) and a 3 = ( 0 0 1 ) . The corresponding tetrahedron has volume 1 6 det ( 1 0 0 0 1 0 0 0 1 ) = 1 6 .

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

Take a regular tetrahedron (d4 for the roleplayers out there) with vertices at a 0 = ( 1 1 1 ) , a 1 = ( 1 - 1 - 1 ) , a 2 = ( - 1 1 - 1 ) and a 3 = ( - 1 - 1 1 ) . (These vertices are half of the vertices of a regular cube centred at the origin with sidelength 2.)

A regular tetrahedron (4-sided die!)

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 = ( a 1 - a 0 a 2 - a 0 a 3 - a 0 ) = ( 0 - 2 - 2 - 2 0 - 2 - 2 - 2 0 ) . We compute the determinant using the cofactor expansion along the top row: vol = | 1 6 det ( 0 - 2 - 2 - 2 0 - 2 - 2 - 2 0 ) | = 1 6 | ( - ( - 2 ) det ( - 2 - 2 - 2 0 ) - 2 det ( - 2 0 - 2 - 2 ) | = 2 6 | ( - 4 - 4 ) | = 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 ( A B ) = 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.