We want to calculate an expression for:
(m)n
where:
- n is an integer.
- m is a multivector.
When we calculated the powers of vectors we saw that, if the vectors anti-commute (as they do in euclidean space) then the result was not necessarily closed within the vector space, if 'n' is even then the result will be a pure scalar value and if 'n' is odd then the result will be a vector. So:
| grade | n | result |
|---|---|---|
| vector | even | scalar |
| vector | odd | vector |
If we now go on to try this for pure bivectors we can see that:
| grade | n | result |
|---|---|---|
| bivector | even | scalar |
| bivector | odd | bivector |
So we can see that when m is any pure grade then:
- n=even then result is scalar
- n=odd then result is same grade as input
So what if m is any mixed grade multivector?
Square
Lets start by assuming that the multivector squares to a scalar value 's' that is: (m)²=s this assumption only applies when the real part of the multivector is zero. This value may be positive or negative since if vectors square to +ve then bivectors square to -ve so depending on which is greater will determine the sign of the square.
Other powers
So even powers will have scalar values, here are the first few powers:
| power | value | type |
|---|---|---|
| m² | s | scalar (+ve or -ve) |
| m³ | m*s | multivector |
| m4 | s² | scalar (+ve) |
| m5 | m*s² | multivector |
I cant think of a way to analyse the more general case of a non-zero real part, I've tried this with the specific case of 2D vectors here but things get too messy.
