Maths - Inverse

Can we use the same trick that we used with complex numbers, that is multiply top and bottom of equation by the conjugate, which will convert the denominator into a scalar making the division possible.

Terminology:

	input	result (1/input)
scalar	a.e	e
vector	a.e1 , a.e2 , a.e3	e1 , e2 , e3
bivector	a.e12 , a.e31 , a.e23	e12 , e31 , e23
trivector	a.e123	e123

Since I'm not yet sure how to calculate the inverse of a multivector, I'll first try an easier problem, how to calculate the inverse of parts of a multivector (scalar, vector, bivector and trivector).

inverse of scalar

e^-1 = 1/a.e

inverse of vector

(e1 + e2 + e3 )^-1

use conjugate (-e1 - e2 - e3 )

so multiplying a vector by its conjugate gives:

(e1 + e2 + e3 ) (-e1 -e2 -e3 ) = -e1² -e2² -e3²

so (e1 + e2 + e3 )^-1 = (-e1 - e2 - e3 )/(-e1² -e2² -e3²)

e1 = a.e1/(a.e1² +a.e2² +a.e3²)
e2 = a.e2/(a.e1² +a.e2² +a.e3²)
e3 = a.e3/(a.e1² +a.e2² +a.e3²)

inverse of bivector

(e12 + e31 + e23)^-1

use conjugate (-e12 - e31 - e23)

so multiplying a bivector by its conjugate gives:

(e12 + e31 + e23) (-e12 - e31 - e23) = e12² + e31² + e23²

so (e12 + e31 + e23)^-1 = (-e12 - e31 - e23)/(e12² + e31² + e23²)

e12 = - a.e12/(a.e1² +a.e2² +a.e3²)
e31 = - a.e31/(a.e1² +a.e2² +a.e3²)
e23 = - a.e23/(a.e1² +a.e2² +a.e3²)

inverse of trivector

e123^-1 = a.e123 / (a.e123 * a.e123) = -1/a.e123

inverse of multivector

If it follows the pattern then it would be:

(e1 + e2 + e3 + e12 + e31 + e23 + e123)^-1 = (e -e1 -e2 -e3 - e12 - e31 - e23 + a.e123)/(e² - ex² - ey² - ez² + e12² + e31² + e23² - e123²)

but is it? this would make the conjugate e -e1 -e2 -e3 - e12 - e31 - e23 + a.e123

so if we multiply a multivector by this possible conjugate we get:

(a.e1 + a.e2 + a.e3 + a.e12 + a.e31 + a.e23 + a.e123)(a.e -a.e1 -a.e2 -a.e3 - a.e12 - a.e31 - a.e23 + a.e123)

which gives:

e = a.e² - a.e1² - a.e2² - a.e3² + a.e12² + a.e31² + a.e23² - a.e123²
e123 = 2*a.e * a.e123 - 2*a.e1 *a.e23 -2*a.e2*a.e31 - 2*a.e3*a.e12

so due to this extra e123 term this will only work if:

a.e * a.e123 = a.e1 *a.e23 + a.e2*a.e31 + a.e3*a.e12

which would make the extra term zero.

Although not necessary to calculate the inverse, the calculation would be a lot simpler if:

1 = a.e² - a.e1² - a.e2² - a.e3² + a.e12² + a.e31² + a.e23² - a.e123²

so if,

a.e * a.e123 = a.e1 *a.e23 + a.e2*a.e31 + a.e3*a.e12

and

1 = a.e² - a.e1² - a.e2² - a.e3² + a.e12² + a.e31² + a.e23² - a.e123²

then,

(e1 + e2 + e3 + e12 + e31 + e23 + e123)^-1 = e -e1 -e2 -e3 - e12 - e31 - e23 + a.e123

Prerequisites

To check the conventions and notation look at the following page first:

• Introduction to Clifford / Geometric Algebra

Generating Geometric Algebras