Idempotents are important in CliffordAlgebra. They are quantities which square to themselves i.e. for some **p**
**e1**, though I would prefer the 1 to be a suffix.
**p1** = (1 + **e1**)/2, which is idempotent.
**p2** = (1 - **e1**)/2, which is also idempotent.
**p1** and **p2** is zero. This makes the *BinomialTheorem?*, as usually understood, very simple. If *a* and *b* are scalar constants then
**p1** and **p2**.
What is the use of all this? Well **e1** can be represented as the difference of **p1** and **p2**.
**e1**.
*BinomialTheorem?*
**p1** nor **p2** can be inverted. However, **e1**^-1 can be evaluated, see below, but not following directly from (1), giving the result
**e1** is its own inverse.
I have an extended version of this called **Vector Exponential** on my web site: http://www.ceac.aston.ac.uk/research/staff/jpf/clifford/vecexp/index.php

**Proof of why it works for negative power -1**
As **e1**^2 = 1 the characteristic polynomial of **e1** is
*m1* = +1 and *m2* = -1
In fact (see Snygg) **p1** and **p2** are the eigenfunctions of **e1** given by
**p1**, right multiplying by **e1**^-1 and dividing by *m1* (scalar and nonzero), and using **p1 p2** = 0, then
**p2**, right multiplying by **e1**^-1 and dividing by *m2* (scalar and nonzero)
**Check**

Reference: John Snygg "Functions of Clifford Numbers or Square Matrices", Chapter 8 in Dorst Doran and Lasenby (eds) "Applications of Geometric Algebra in Computer Science and Engineering", Birkhauser, 2002, ISBN 0817642676 He shows how to find the expression of a Clifford Variable as a sum of multiples of idempotents.

This all depends on something which Hestenes mentions, in HestenesOerstedMedalLecture at the bottom of page 16, as a misleading piece of teaching: "it is meaningless to add scalars to vectors". This was one of the comments I had when I circulated the**Vector Exponential** to my colleagues as a Christmas greeting.
-- JohnFletcher

See also CliffordAlgebra

**Note: IdempotentDesign** is something completely different.

CategoryMath

The obvious ones are 0 and 1, but there are others. Consider a unit vector, which I will callp^2 =p

so that is not idempotent. Instead look ate1^2 = 1

Now tryp1^2 = ((1 +e1)/2)^2 = (1 +e1)(1 +e1)/4 = (1 + 2e1+1)/4 = (1 +e1)/2 =p1

Now tryp2^2 = ((1 -e1)/2)^2 = (1 -e1)(1 -e1)/4 = (1 - 2e1+1)/4 = (1 -e1)/2 =p2

so that the product ofp1 p2= (1 +e1)(1 -e1)/4 = (1 +e1-e1-1)/4 = 0

(as all the other terms are zero, containing at least one of each ofap1+bp2)^n =a^np1+b^np2

so that the previous result can be used to show the structure of the powers ofe1=p1-p2

This can be extended in various ways. It works for negative powers, where the expression in terms of thee1^2 = (p1-p2)^2 = 1^2p1+ (-1)^2p2=p1+p2= 1

cannot be manipulated directly because neithere1^-1 = (p1-p2)^-1 -- (1)

showing thate1^-1 = 1^-1p1+ (-1)^-1p2=p1-p2=e1

which has rootsm^2 = 1

and alsop1= (e1-m2) / (m1-m2) = (1 +e1)/2p2= (e1-m1) / (m2-m1) = (1 -e1)/2

Multiplying (2) on the left byp1+p2= 1 --- (1)m1p1+m2p2=e1--- (2)

Similarly, multiplying (2) on the left byp1 e1^-1 =p1 e1 e1^-1/m1=p1/m1--- (3)

Adding (3) and (4)p2 e1^-1 =p2 e1 e1^-1/m2=p2/m2--- (4)

(and because of (1), the result isp1+p2)e1^-1 =p1/m1+p2/m2

e1^-1 =p1/m1+p2/m2

e1 e1^-1 = (m1p1+m2p2) (p1/m1+p2/m2) = (m1/m1)p1+ (m2/m2)p2=p1+p2= 1

Reference: John Snygg "Functions of Clifford Numbers or Square Matrices", Chapter 8 in Dorst Doran and Lasenby (eds) "Applications of Geometric Algebra in Computer Science and Engineering", Birkhauser, 2002, ISBN 0817642676 He shows how to find the expression of a Clifford Variable as a sum of multiples of idempotents.

This all depends on something which Hestenes mentions, in HestenesOerstedMedalLecture at the bottom of page 16, as a misleading piece of teaching: "it is meaningless to add scalars to vectors". This was one of the comments I had when I circulated the

See also CliffordAlgebra

CategoryMath

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