Multivector differentiation and Linear Algebra

17th Santalo´ Summer School 2016, Santander

Joan Lasenby

Signal Processing Group, Engineering Department, Cambridge, UK and Trinity College Cambridge [email protected], www-sigproc.eng.cam.ac.uk/ ∼ jl

23 August 2016

1 / 78 Examples of differentiation wrt multivectors.

Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra.

Functional Differentiation: very briefly...

Summary

Overview

The Multivector Derivative.

2 / 78 Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra.

Functional Differentiation: very briefly...

Summary

Overview

The Multivector Derivative.

Examples of differentiation wrt multivectors.

3 / 78 Functional Differentiation: very briefly...

Summary

Overview

The Multivector Derivative.

Examples of differentiation wrt multivectors.

Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra.

4 / 78 Summary

Overview

The Multivector Derivative.

Examples of differentiation wrt multivectors.

Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra.

Functional Differentiation: very briefly...

5 / 78 Overview

The Multivector Derivative.

Examples of differentiation wrt multivectors.

Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra.

Functional Differentiation: very briefly...

Summary

6 / 78 We now want to generalise this idea to enable us to find the derivative of F(X), in the A ‘direction’ – where X is a general mixed grade multivector (so F(X) is a general multivector valued function of X). Let us use ∗ to denote taking the scalar part, ie P ∗ Q ≡ hPQi. Then, provided A has same grades as X, it makes sense to define:

F(X + τA) − F(X) A ∗ ∂XF(X) = lim τ→0 τ

The Multivector Derivative

Recall our definition of the directional derivative in the a direction F(x + ea) − F(x) a·∇ F(x) = lim e→0 e

7 / 78 Let us use ∗ to denote taking the scalar part, ie P ∗ Q ≡ hPQi. Then, provided A has same grades as X, it makes sense to define:

F(X + τA) − F(X) A ∗ ∂XF(X) = lim τ→0 τ

The Multivector Derivative

Recall our definition of the directional derivative in the a direction F(x + ea) − F(x) a·∇ F(x) = lim e→0 e

We now want to generalise this idea to enable us to find the derivative of F(X), in the A ‘direction’ – where X is a general mixed grade multivector (so F(X) is a general multivector valued function of X).

8 / 78 The Multivector Derivative

Recall our definition of the directional derivative in the a direction F(x + ea) − F(x) a·∇ F(x) = lim e→0 e

We now want to generalise this idea to enable us to find the derivative of F(X), in the A ‘direction’ – where X is a general mixed grade multivector (so F(X) is a general multivector valued function of X). Let us use ∗ to denote taking the scalar part, ie P ∗ Q ≡ hPQi. Then, provided A has same grades as X, it makes sense to define:

F(X + τA) − F(X) A ∗ ∂XF(X) = lim τ→0 τ

9 / 78 With the definition on the previous slide, eJ ∗ ∂X is therefore the partial derivative in the eJ direction. Giving

J ∂X ≡ ∑ e eJ ∗ ∂X J I [since eJ ∗ ∂X ≡ eJ ∗ {e (eI ∗ ∂X}]. Key to using these definitions of multivector differentiation are several important results:

The Multivector Derivative cont...

Let {eJ} be a basis for X – ie if X is a bivector, then {eJ} will be the basis bivectors.

10 / 78 Key to using these definitions of multivector differentiation are several important results:

The Multivector Derivative cont...

Let {eJ} be a basis for X – ie if X is a bivector, then {eJ} will be the basis bivectors.

With the definition on the previous slide, eJ ∗ ∂X is therefore the partial derivative in the eJ direction. Giving

J ∂X ≡ ∑ e eJ ∗ ∂X J I [since eJ ∗ ∂X ≡ eJ ∗ {e (eI ∗ ∂X}].

11 / 78 The Multivector Derivative cont...

Let {eJ} be a basis for X – ie if X is a bivector, then {eJ} will be the basis bivectors.

With the definition on the previous slide, eJ ∗ ∂X is therefore the partial derivative in the eJ direction. Giving

J ∂X ≡ ∑ e eJ ∗ ∂X J I [since eJ ∗ ∂X ≡ eJ ∗ {e (eI ∗ ∂X}]. Key to using these definitions of multivector differentiation are several important results:

12 / 78 We can see this by going back to our definitions:

h(X + τeJ)Bi − hXBi hXBi + τheJBi − hXBi eJ ∗ ∂XhXBi = lim = lim τ→0 τ τ→0 τ

τheJBi lim = heJBi τ→0 τ

Therefore giving us J J ∂XhXBi = e (eJ ∗ ∂X)hXBi = e heJBi ≡ PX(B)

The Multivector Derivative cont...

If PX(B) is the projection of B onto the grades of X (ie J PX(B) ≡ e heJBi), then our first important result is

∂XhXBi = PX(B)

13 / 78 τheJBi lim = heJBi τ→0 τ

Therefore giving us J J ∂XhXBi = e (eJ ∗ ∂X)hXBi = e heJBi ≡ PX(B)

The Multivector Derivative cont...

If PX(B) is the projection of B onto the grades of X (ie J PX(B) ≡ e heJBi), then our first important result is

∂XhXBi = PX(B)

We can see this by going back to our definitions:

h(X + τeJ)Bi − hXBi hXBi + τheJBi − hXBi eJ ∗ ∂XhXBi = lim = lim τ→0 τ τ→0 τ

14 / 78 Therefore giving us J J ∂XhXBi = e (eJ ∗ ∂X)hXBi = e heJBi ≡ PX(B)

The Multivector Derivative cont...

If PX(B) is the projection of B onto the grades of X (ie J PX(B) ≡ e heJBi), then our first important result is

∂XhXBi = PX(B)

We can see this by going back to our definitions:

h(X + τeJ)Bi − hXBi hXBi + τheJBi − hXBi eJ ∗ ∂XhXBi = lim = lim τ→0 τ τ→0 τ

τheJBi lim = heJBi τ→0 τ

15 / 78 The Multivector Derivative cont...

If PX(B) is the projection of B onto the grades of X (ie J PX(B) ≡ e heJBi), then our first important result is

∂XhXBi = PX(B)

We can see this by going back to our definitions:

h(X + τeJ)Bi − hXBi hXBi + τheJBi − hXBi eJ ∗ ∂XhXBi = lim = lim τ→0 τ τ→0 τ

τheJBi lim = heJBi τ→0 τ

Therefore giving us J J ∂XhXBi = e (eJ ∗ ∂X)hXBi = e heJBi ≡ PX(B)

16 / 78 ∂XhXBi = PX(B)

∂XhXB˜ i = PX(B˜ )

˜ ∂X˜ hXBi = PX˜ (B) = PX(B)

−1 −1 −1 ∂ψhMψ i = −ψ Pψ(M)ψ

X, B, M, ψ all general multivectors.

Other Key Results

Some other useful results are listed here (proofs are similar to that on previous slide and are left as exercises):

17 / 78 Other Key Results

Some other useful results are listed here (proofs are similar to that on previous slide and are left as exercises):

∂XhXBi = PX(B)

∂XhXB˜ i = PX(B˜ )

˜ ∂X˜ hXBi = PX˜ (B) = PX(B)

−1 −1 −1 ∂ψhMψ i = −ψ Pψ(M)ψ

X, B, M, ψ all general multivectors.

18 / 78 Exercises 1

1 By noting that hXBi = h(XB)˜i, show the second key result ∂XhXB˜ i = PX(B˜ )

2 ˜ Key result 1 tells us that ∂X˜ hXBi = PX˜ (B). Verify that PX˜ (B) = PX(B), to give the 3rd key result.

3 to show the 4th key result

−1 −1 −1 ∂ψhMψ i = −ψ Pψ(M)ψ

−1 use the fact that ∂ψhMψψ i = ∂ψhMi = 0. Hint: recall that XAX has the same grades as A.

19 / 78 One possible way forward is to find the plane that minimises the sum of the squared perpendicular distances of the points from the plane.

X2

X3 X1

X0 1 X0 0 2 X3

A Simple Example

Suppose we wish to fit a set of points {Xi} to a plane Φ – where the Xi and Φ are conformal representations (vector and 4 vector respectively).

20 / 78 A Simple Example

Suppose we wish to fit a set of points {Xi} to a plane Φ – where the Xi and Φ are conformal representations (vector and 4 vector respectively).

One possible way forward is to find the plane that minimises the sum of the squared perpendicular distances of the points from the plane.

X2

X3 X1

0 X1 0 X2 X0 3 21 / 78 Now use the result ∂XhXBi = PX(B) to differentiate this expression wrt Φ

˙ ˙ ˙ ˙ ∂ΦS = − ∑ ∂ΦhXiΦXiΦi = − ∑ ∂ΦhXiΦXiΦi + ∂ΦhXiΦXiΦi i i

= −2 ∑ PΦ(XiΦXi) = −2 ∑ XiΦXi i i

=⇒ solve (via linear algebra techniques) ∑i XiΦXi = 0.

Plane fitting example, cont....

Recall that ΦXΦ is the reflection of X in Φ, so that −X·(ΦXΦ) is the distance between the point and the plane. Thus we could take as our cost function:

S = − ∑ Xi·(ΦXiΦ) i

22 / 78 = −2 ∑ PΦ(XiΦXi) = −2 ∑ XiΦXi i i

=⇒ solve (via linear algebra techniques) ∑i XiΦXi = 0.

Plane fitting example, cont....

Recall that ΦXΦ is the reflection of X in Φ, so that −X·(ΦXΦ) is the distance between the point and the plane. Thus we could take as our cost function:

S = − ∑ Xi·(ΦXiΦ) i

Now use the result ∂XhXBi = PX(B) to differentiate this expression wrt Φ

˙ ˙ ˙ ˙ ∂ΦS = − ∑ ∂ΦhXiΦXiΦi = − ∑ ∂ΦhXiΦXiΦi + ∂ΦhXiΦXiΦi i i

23 / 78 =⇒ solve (via linear algebra techniques) ∑i XiΦXi = 0.

Plane fitting example, cont....

Recall that ΦXΦ is the reflection of X in Φ, so that −X·(ΦXΦ) is the distance between the point and the plane. Thus we could take as our cost function:

S = − ∑ Xi·(ΦXiΦ) i

Now use the result ∂XhXBi = PX(B) to differentiate this expression wrt Φ

˙ ˙ ˙ ˙ ∂ΦS = − ∑ ∂ΦhXiΦXiΦi = − ∑ ∂ΦhXiΦXiΦi + ∂ΦhXiΦXiΦi i i

= −2 ∑ PΦ(XiΦXi) = −2 ∑ XiΦXi i i

24 / 78 Plane fitting example, cont....

Recall that ΦXΦ is the reflection of X in Φ, so that −X·(ΦXΦ) is the distance between the point and the plane. Thus we could take as our cost function:

S = − ∑ Xi·(ΦXiΦ) i

Now use the result ∂XhXBi = PX(B) to differentiate this expression wrt Φ

˙ ˙ ˙ ˙ ∂ΦS = − ∑ ∂ΦhXiΦXiΦi = − ∑ ∂ΦhXiΦXiΦi + ∂ΦhXiΦXiΦi i i

= −2 ∑ PΦ(XiΦXi) = −2 ∑ XiΦXi i i

=⇒ solve (via linear algebra techniques) ∑i XiΦXi = 0.

25 / 78 Another example is differentiating wrt rotors or bivectors.

Suppose we wished to create a Kalman filter-like system which tracked bivectors (not simply their components in some basis) – this might involve evaluating expressions such as

L h i i ˜ i ∂Bn ∑ vnRnun−1Rn i=1 −B where Rn = e n , u, v s are vectors.

Differentiation cont....

Of course we can extend these ideas to other geometric fitting problems and also to those without closed form solutions, using gradient information to find solutions.

26 / 78 Suppose we wished to create a Kalman filter-like system which tracked bivectors (not simply their components in some basis) – this might involve evaluating expressions such as

L h i i ˜ i ∂Bn ∑ vnRnun−1Rn i=1 −B where Rn = e n , u, v s are vectors.

Differentiation cont....

Of course we can extend these ideas to other geometric fitting problems and also to those without closed form solutions, using gradient information to find solutions.

Another example is differentiating wrt rotors or bivectors.

27 / 78 Differentiation cont....

Of course we can extend these ideas to other geometric fitting problems and also to those without closed form solutions, using gradient information to find solutions.

Another example is differentiating wrt rotors or bivectors.

Suppose we wished to create a Kalman filter-like system which tracked bivectors (not simply their components in some basis) – this might involve evaluating expressions such as

L h i i ˜ i ∂Bn ∑ vnRnun−1Rn i=1 −B where Rn = e n , u, v s are vectors.

28 / 78 1 ˜ = − ( ) + h ( ) ˜ i ∂Bn vnRnun−1Rn Γ Bn 2 BnΓ Bn RnBnRn 2 |Bn|

sin(|b |) B Γ(B )R˜ B  + n n n n n + ( ) ˜ 2 Γ Bn Rn |Bn| |Bn| 2

1 ˜ where Γ(Bn) = 2 [un−1∧RnvnRn]Rn.

Differentiation cont....

Using just the standard results given, and a page of algebra later (but one only needs to do it once!) we find that

29 / 78 Differentiation cont....

Using just the standard results given, and a page of algebra later (but one only needs to do it once!) we find that

1 ˜ = − ( ) + h ( ) ˜ i ∂Bn vnRnun−1Rn Γ Bn 2 BnΓ Bn RnBnRn 2 |Bn|

sin(|b |) B Γ(B )R˜ B  + n n n n n + ( ) ˜ 2 Γ Bn Rn |Bn| |Bn| 2

1 ˜ where Γ(Bn) = 2 [un−1∧RnvnRn]Rn.

30 / 78 We can now extendf to act on any order blade by (outermorphism)

f(a1∧a2∧...∧an) = f(a1)∧f(a2)∧....∧f(an)

Note that the resulting blade has the same grade as the original blade. Thus, an important property is that these extended linear functions are grade preserving, ie

f(Ar) = hf(Ar)ir

Linear Algebra

A linear function,f, mapping vectors to vectors asatisfies f(λa + µb) = λf(a) + µf(b)

31 / 78 Note that the resulting blade has the same grade as the original blade. Thus, an important property is that these extended linear functions are grade preserving, ie

f(Ar) = hf(Ar)ir

Linear Algebra

A linear function,f, mapping vectors to vectors asatisfies f(λa + µb) = λf(a) + µf(b)

We can now extendf to act on any order blade by (outermorphism)

f(a1∧a2∧...∧an) = f(a1)∧f(a2)∧....∧f(an)

32 / 78 Linear Algebra

A linear function,f, mapping vectors to vectors asatisfies f(λa + µb) = λf(a) + µf(b)

We can now extendf to act on any order blade by (outermorphism)

f(a1∧a2∧...∧an) = f(a1)∧f(a2)∧....∧f(an)

Note that the resulting blade has the same grade as the original blade. Thus, an important property is that these extended linear functions are grade preserving, ie

f(Ar) = hf(Ar)ir

33 / 78 Fij = ei·f(ej)

Where {ei} is the basis in which the vectors the matrix acts on are written. As with matrix multiplication, where we obtain a 3rd matrix (linear function) from combining two other matrices (linear functions), ieH = FG, we can also write

h(a) = f[g(a)] = fg(a)

The product of linear functions is associative.

Linear Algebra cont....

Matrices are also linear functions which map vectors to vectors. IfF is the matrix corresponding to the linear functionf, we obtain the elements ofF via

34 / 78 As with matrix multiplication, where we obtain a 3rd matrix (linear function) from combining two other matrices (linear functions), ieH = FG, we can also write

h(a) = f[g(a)] = fg(a)

The product of linear functions is associative.

Linear Algebra cont....

Matrices are also linear functions which map vectors to vectors. IfF is the matrix corresponding to the linear functionf, we obtain the elements ofF via

Fij = ei·f(ej)

Where {ei} is the basis in which the vectors the matrix acts on are written.

35 / 78 The product of linear functions is associative.

Linear Algebra cont....

Matrices are also linear functions which map vectors to vectors. IfF is the matrix corresponding to the linear functionf, we obtain the elements ofF via

Fij = ei·f(ej)

Where {ei} is the basis in which the vectors the matrix acts on are written. As with matrix multiplication, where we obtain a 3rd matrix (linear function) from combining two other matrices (linear functions), ieH = FG, we can also write

h(a) = f[g(a)] = fg(a)

36 / 78 Linear Algebra cont....

Matrices are also linear functions which map vectors to vectors. IfF is the matrix corresponding to the linear functionf, we obtain the elements ofF via

Fij = ei·f(ej)

Where {ei} is the basis in which the vectors the matrix acts on are written. As with matrix multiplication, where we obtain a 3rd matrix (linear function) from combining two other matrices (linear functions), ieH = FG, we can also write

h(a) = f[g(a)] = fg(a)

The product of linear functions is associative.

37 / 78 First take a blade a1∧a2∧...∧ar and note that

h(a1∧a2∧...∧ar) = fg(a1)∧fg(a2)∧...∧fg(ar)

= f[g(a1)∧g(a2)∧.....∧g(ar)] = f[g(a1∧a2∧...∧ar)]

from which we get the first result.

Linear Algebra

We now need to verify that

h(A) = f[g(A)] = fg(A)

for any multivector A.

38 / 78 = f[g(a1)∧g(a2)∧.....∧g(ar)] = f[g(a1∧a2∧...∧ar)]

from which we get the first result.

Linear Algebra

We now need to verify that

h(A) = f[g(A)] = fg(A)

for any multivector A.

First take a blade a1∧a2∧...∧ar and note that

h(a1∧a2∧...∧ar) = fg(a1)∧fg(a2)∧...∧fg(ar)

39 / 78 from which we get the first result.

Linear Algebra

We now need to verify that

h(A) = f[g(A)] = fg(A)

for any multivector A.

First take a blade a1∧a2∧...∧ar and note that

h(a1∧a2∧...∧ar) = fg(a1)∧fg(a2)∧...∧fg(ar)

= f[g(a1)∧g(a2)∧.....∧g(ar)] = f[g(a1∧a2∧...∧ar)]

40 / 78 Linear Algebra

We now need to verify that

h(A) = f[g(A)] = fg(A)

for any multivector A.

First take a blade a1∧a2∧...∧ar and note that

h(a1∧a2∧...∧ar) = fg(a1)∧fg(a2)∧...∧fg(ar)

= f[g(a1)∧g(a2)∧.....∧g(ar)] = f[g(a1∧a2∧...∧ar)]

from which we get the first result.

41 / 78 Exercises 2

1 For a matrixF F F  F = 11 12 F21 F22 T T Verify thatF ij = ei·f(ej), where e1 = [1, 0] and e2 = [0, 1] , for i, j = 1, 2.

2 Rotations are linear functions, so we can writeR (a) = RaR˜ , where R is the rotor. If Ar is an r-blade, show that

RArR˜ = (Ra1R˜ )∧(Ra2R˜ )∧...∧(RarR˜ )

Thus we can rotate any element of our algebra with the same rotor expression.

42 / 78 The unit cube I = e1∧e2∧e3 is transformed to a parallelepiped, V V = f(e1)∧f(e2)∧f(e3) = f(I)

The Determinant

Consider the action of a linear functionf on an orthogonal basis in 3d:

43 / 78 The Determinant

Consider the action of a linear functionf on an orthogonal basis in 3d:

The unit cube I = e1∧e2∧e3 is transformed to a parallelepiped, V V = f(e1)∧f(e2)∧f(e3) = f(I)

44 / 78 Let us define the determinant of the linear functionf as the volume scale factor V. So that

f(I) = det(f) I

This enables us to find the form of the determinant explicitly (in terms of partial derivatives between coordinate frames) very easily in any dimension.

The Determinant cont....

So, sincef (I) is also a pseudoscalar, we see that if V is the magnitude ofV, then f(I) = VI

45 / 78 This enables us to find the form of the determinant explicitly (in terms of partial derivatives between coordinate frames) very easily in any dimension.

The Determinant cont....

So, sincef (I) is also a pseudoscalar, we see that if V is the magnitude ofV, then f(I) = VI

Let us define the determinant of the linear functionf as the volume scale factor V. So that

f(I) = det(f) I

46 / 78 The Determinant cont....

So, sincef (I) is also a pseudoscalar, we see that if V is the magnitude ofV, then f(I) = VI

Let us define the determinant of the linear functionf as the volume scale factor V. So that

f(I) = det(f) I

This enables us to find the form of the determinant explicitly (in terms of partial derivatives between coordinate frames) very easily in any dimension.

47 / 78 So we have proved that

det(fg) = det(f) det(g)

A very easy proof!

A Key Result

As before, leth = fg, then

h(I) = det(h) I = f(g(I)) = f(det(g) I)

= det(g) f(I) = det(g) det(f)(I)

48 / 78 A very easy proof!

A Key Result

As before, leth = fg, then

h(I) = det(h) I = f(g(I)) = f(det(g) I)

= det(g) f(I) = det(g) det(f)(I)

So we have proved that

det(fg) = det(f) det(g)

49 / 78 A Key Result

As before, leth = fg, then

h(I) = det(h) I = f(g(I)) = f(det(g) I)

= det(g) f(I) = det(g) det(f)(I)

So we have proved that

det(fg) = det(f) det(g)

A very easy proof!

50 / 78 In GA we can write this in terms of linear functions as

a·f(b) = f¯(a)·b

This reverse linear function, f¯, is called the adjoint.

The Adjoint/Transpose of a Linear Function

For a matrixF and its transpose,F T we have (for any vectors a, b) aTFb = bTFTa = φ (scalar)

51 / 78 This reverse linear function, f¯, is called the adjoint.

The Adjoint/Transpose of a Linear Function

For a matrixF and its transpose,F T we have (for any vectors a, b) aTFb = bTFTa = φ (scalar)

In GA we can write this in terms of linear functions as

a·f(b) = f¯(a)·b

52 / 78 The Adjoint/Transpose of a Linear Function

For a matrixF and its transpose,F T we have (for any vectors a, b) aTFb = bTFTa = φ (scalar)

In GA we can write this in terms of linear functions as

a·f(b) = f¯(a)·b

This reverse linear function, f¯, is called the adjoint.

53 / 78 The Adjoint/Transpose of a Linear Function

For a matrixF and its transpose,F T we have (for any vectors a, b) aTFb = bTFTa = φ (scalar)

In GA we can write this in terms of linear functions as

a·f(b) = f¯(a)·b

This reverse linear function, f¯, is called the adjoint.

54 / 78 See following exercises to show that a·f(b∧c) = f[f¯(a)·(b∧c)]

This can now be generalised to

Ar·f¯(Bs) = f¯[f(Ar)·Bs] r ≤ s

f(Ar)·Bs = f[Ar·f¯(Bs)] r ≥ s

The Adjoint cont....

It is not hard to show that the adjoint extends to blades in the expected way

f¯(a1∧a2∧...∧an) = f¯(a1)∧f¯(a2)∧....∧f¯(an)

55 / 78 This can now be generalised to

Ar·f¯(Bs) = f¯[f(Ar)·Bs] r ≤ s

f(Ar)·Bs = f[Ar·f¯(Bs)] r ≥ s

The Adjoint cont....

It is not hard to show that the adjoint extends to blades in the expected way

f¯(a1∧a2∧...∧an) = f¯(a1)∧f¯(a2)∧....∧f¯(an)

See following exercises to show that a·f(b∧c) = f[f¯(a)·(b∧c)]

56 / 78 The Adjoint cont....

It is not hard to show that the adjoint extends to blades in the expected way

f¯(a1∧a2∧...∧an) = f¯(a1)∧f¯(a2)∧....∧f¯(an)

See following exercises to show that a·f(b∧c) = f[f¯(a)·(b∧c)]

This can now be generalised to

Ar·f¯(Bs) = f¯[f(Ar)·Bs] r ≤ s

f(Ar)·Bs = f[Ar·f¯(Bs)] r ≥ s

57 / 78 Exercises 3

1 For any vectors p, q, r, show that

p·(q∧r) = (p·q)r − (p·r)q

2 By using the fact that a·f(b∧c) = a·[f(b)∧f(c)], use the above result to show that

a·f(b∧c) = (f¯(a)·b)f(c) − (f¯(a)·c)f(b)

and simplify to get the final result

a·f(b∧c) = f[f¯(a)·(b∧c)]

58 / 78 Now put Bs = I in this formula:

Ar·f¯(I) = Ar·det(f)(I) = det(f)(ArI)

= f¯[f(Ar)·I] = f¯[f(Ar)I]

We can now write this as

−1 −1 Ar = f¯[f(Ar)I]I [det(f)]

The Inverse

Ar·f¯(Bs) = f¯[f(Ar)·Bs] r ≤ s

59 / 78 We can now write this as

−1 −1 Ar = f¯[f(Ar)I]I [det(f)]

The Inverse

Ar·f¯(Bs) = f¯[f(Ar)·Bs] r ≤ s

Now put Bs = I in this formula:

Ar·f¯(I) = Ar·det(f)(I) = det(f)(ArI)

= f¯[f(Ar)·I] = f¯[f(Ar)I]

60 / 78 The Inverse

Ar·f¯(Bs) = f¯[f(Ar)·Bs] r ≤ s

Now put Bs = I in this formula:

Ar·f¯(I) = Ar·det(f)(I) = det(f)(ArI)

= f¯[f(Ar)·I] = f¯[f(Ar)I]

We can now write this as

−1 −1 Ar = f¯[f(Ar)I]I [det(f)]

61 / 78 −1 The next stage is to put Ar = f (Br) in this equation: −1 −1 −1 f (Br) = f¯[BrI]I [det(f)]

This leads us to the important and simple formulae for the inverse of a function and its adjoint

f−1(A) = [det(f)]−1f¯[AI]I−1

−1 f¯ (A) = [det(f)]−1f[AI]I−1

The Inverse cont...

Repeat this here: −1 −1 Ar = f¯[f(Ar)I]I [det(f)]

62 / 78 This leads us to the important and simple formulae for the inverse of a function and its adjoint

f−1(A) = [det(f)]−1f¯[AI]I−1

−1 f¯ (A) = [det(f)]−1f[AI]I−1

The Inverse cont...

Repeat this here: −1 −1 Ar = f¯[f(Ar)I]I [det(f)]

−1 The next stage is to put Ar = f (Br) in this equation: −1 −1 −1 f (Br) = f¯[BrI]I [det(f)]

63 / 78 −1 f¯ (A) = [det(f)]−1f[AI]I−1

The Inverse cont...

Repeat this here: −1 −1 Ar = f¯[f(Ar)I]I [det(f)]

−1 The next stage is to put Ar = f (Br) in this equation: −1 −1 −1 f (Br) = f¯[BrI]I [det(f)]

This leads us to the important and simple formulae for the inverse of a function and its adjoint

f−1(A) = [det(f)]−1f¯[AI]I−1

64 / 78 The Inverse cont...

Repeat this here: −1 −1 Ar = f¯[f(Ar)I]I [det(f)]

−1 The next stage is to put Ar = f (Br) in this equation: −1 −1 −1 f (Br) = f¯[BrI]I [det(f)]

This leads us to the important and simple formulae for the inverse of a function and its adjoint

f−1(A) = [det(f)]−1f¯[AI]I−1

−1 f¯ (A) = [det(f)]−1f[AI]I−1

65 / 78 So, putting this in our inverse formula:

R−1(A) = [det(f)]−1R¯ (AI)I−1

= [det(f)]−1R˜ (AI)RI−1 = RAR˜

since det(R) = 1. Thus the inverse is the adjoint ... as we know from RR˜ = 1.

An Example

Let us see if this works for rotations

R(a) = RaR˜ and R¯ (a) = RaR˜

66 / 78 = [det(f)]−1R˜ (AI)RI−1 = RAR˜

since det(R) = 1. Thus the inverse is the adjoint ... as we know from RR˜ = 1.

An Example

Let us see if this works for rotations

R(a) = RaR˜ and R¯ (a) = RaR˜

So, putting this in our inverse formula:

R−1(A) = [det(f)]−1R¯ (AI)I−1

67 / 78 An Example

Let us see if this works for rotations

R(a) = RaR˜ and R¯ (a) = RaR˜

So, putting this in our inverse formula:

R−1(A) = [det(f)]−1R¯ (AI)I−1

= [det(f)]−1R˜ (AI)RI−1 = RAR˜

since det(R) = 1. Thus the inverse is the adjoint ... as we know from RR˜ = 1.

68 / 78 Symmetric(f (a) = f¯(a)) and antisymmetric(f (a) = −f¯(a)) functions. In particular, antisymmetric functions are best studied using bivectors.

Decompositions such as Singular Value Decomposition

Tensors - we can think of tensors as linear functions mapping r-blades to s-blades. Thus we retain some physical intuition that is generally lost in index notation.

More Linear Algebra...

That we won’t look at.....

The idea of eigenblades – this becomes possible with our extension of linear functions to act on blades.

69 / 78 Decompositions such as Singular Value Decomposition

Tensors - we can think of tensors as linear functions mapping r-blades to s-blades. Thus we retain some physical intuition that is generally lost in index notation.

More Linear Algebra...

That we won’t look at.....

The idea of eigenblades – this becomes possible with our extension of linear functions to act on blades.

Symmetric(f (a) = f¯(a)) and antisymmetric(f (a) = −f¯(a)) functions. In particular, antisymmetric functions are best studied using bivectors.

70 / 78 Tensors - we can think of tensors as linear functions mapping r-blades to s-blades. Thus we retain some physical intuition that is generally lost in index notation.

More Linear Algebra...

That we won’t look at.....

The idea of eigenblades – this becomes possible with our extension of linear functions to act on blades.

Symmetric(f (a) = f¯(a)) and antisymmetric(f (a) = −f¯(a)) functions. In particular, antisymmetric functions are best studied using bivectors.

Decompositions such as Singular Value Decomposition

71 / 78 More Linear Algebra...

That we won’t look at.....

The idea of eigenblades – this becomes possible with our extension of linear functions to act on blades.

Symmetric(f (a) = f¯(a)) and antisymmetric(f (a) = −f¯(a)) functions. In particular, antisymmetric functions are best studied using bivectors.

Decompositions such as Singular Value Decomposition

Tensors - we can think of tensors as linear functions mapping r-blades to s-blades. Thus we retain some physical intuition that is generally lost in index notation.

72 / 78 ∂f(a)(f(b)·c) = (a·b)c

In engineering, this, in particular, enables us to differentiate wrt to structured matrices in a way which is very hard to do otherwise.

Functional Differentiation

We will only touch on this briefly, but it is crucial to work in physics and has hardly been used at all in other fields.

73 / 78 In engineering, this, in particular, enables us to differentiate wrt to structured matrices in a way which is very hard to do otherwise.

Functional Differentiation

We will only touch on this briefly, but it is crucial to work in physics and has hardly been used at all in other fields.

∂f(a)(f(b)·c) = (a·b)c

74 / 78 Functional Differentiation

We will only touch on this briefly, but it is crucial to work in physics and has hardly been used at all in other fields.

∂f(a)(f(b)·c) = (a·b)c

In engineering, this, in particular, enables us to differentiate wrt to structured matrices in a way which is very hard to do otherwise.

75 / 78 Linear Algebra : we will see applications of the GA approach to linear algebra – using, in particular, the beautiful expressions for the inverse of a function.

Functional Differentiation : used widely in physics, scope for much more use in engineering.

Summary

Multivector differentiation : can usually differentiate most expressions using a set of standard rules.

76 / 78 Functional Differentiation : used widely in physics, scope for much more use in engineering.

Summary

Multivector differentiation : can usually differentiate most expressions using a set of standard rules.

Linear Algebra : we will see applications of the GA approach to linear algebra – using, in particular, the beautiful expressions for the inverse of a function.

77 / 78 Summary

Multivector differentiation : can usually differentiate most expressions using a set of standard rules.

Linear Algebra : we will see applications of the GA approach to linear algebra – using, in particular, the beautiful expressions for the inverse of a function.

Functional Differentiation : used widely in physics, scope for much more use in engineering.

78 / 78