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Derivative rules for differentiable matrix-valued curves, and skewness of RTR′ for a curve in SO(3)

Statement

Let I⊆R be an interval with at least two points, let t0∈I and let A,B:I→M3(R) be differentiable at t0, meaning (The vector space Mm×n(F):=F m×n of m by n matrices over a field, with entrywise operations) that every entry function t↦A(t)ij and t↦B(t)ij is differentiable at t0 in the sense of The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set; write A′(t0) for the matrix with entries A′(t0)ij:=(t↦A(t)ij)′(t0), and similarly for B.

  1. A+B and AT are differentiable at t0, with (A+B)′(t0)=A′(t0)+B′(t0) and (AT)′(t0)=(A′(t0))T.
  2. AB is differentiable at t0, with (AB)′(t0)=A′(t0)B(t0)+A(t0)B′(t0).
  3. If A(t)∈SO(3) for every t∈I (The orthogonal group O(3) and the rotation group SO(3)), then the matrix A(t0)TA′(t0) is skew-symmetric, i.e. (ATA′)T=−ATA′ at t0, and likewise A′(t0)A(t0)T is skew-symmetric. In particular the trace of each of these two matrices vanishes.

Facts & Assumptions

Given: An interval I⊆R with at least two points, a point t0∈I, and maps A,B:I→M3(R) differentiable at t0 in the entrywise sense of the statement.

[L1]

If f,g:I→R are differentiable at t0 and α∈R, then f+g, αf and fg are differentiable at t0 with (f+g)′(t0)=f′(t0)+g′(t0), (αf)′(t0)=αf′(t0) and (fg)′(t0)=f′(t0)g(t0)+f(t0)g′(t0) Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0.

[L2]

M3(R) is the function space of entries indexed by {0,1,2}×{0,1,2} with pointwise sum, so (A+B)ij=Aij+Bij; the transpose is (AT)ij=Aji; and the product is (AB)ij=∑k=02AikBkj The vector space Mm×n(F):=F m×n of m by n matrices over a field, with entrywise operations, The transpose AT of a matrix, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes.

[L3]

Transposition is involutive and reverses products: (AT)T=A and (AB)T=BTAT Transpose is linear and involutive, and (AB)T=BTAT.

[L4]

For R∈SO(3) one has RTR=I3, det⁡R=1, R invertible with R−1=RT, and also RRT=I3 The orthogonal group O(3) and the rotation group SO(3), SO(3) is a subgroup of the general linear group, with R−1=RT.

[L5]

A constant function I→R is differentiable at every point of I with derivative 0: its difference quotient at any limit point of I is the constant function 0, whose limit is 0 The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set.

[L6]

The identity matrix I3 has entries 1 for equal indices and 0 otherwise, and the matrix all of whose entries are 0 is the zero matrix Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes, The vector space Mm×n(F):=F m×n of m by n matrices over a field, with entrywise operations.

Proof

technique · direct
1.1givenL1L2

For all i,j the entry (A+B)ij=Aij+Bij is differentiable at t0 with derivative Aij′(t0)+Bij′(t0) by [L1] and [L2]; these are the entries of A′(t0)+B′(t0), so A+B is differentiable with the asserted derivative.

1.2givenL1L2

For all i,j the entry (AT)ij=Aji is differentiable at t0 with derivative Aji′(t0), which is the (i,j) entry of (A′(t0))T; so AT is differentiable with the asserted derivative.

1.3givenL1L2

For all i,j the entry (AB)ij=∑k=02AikBkj is a finite sum of products of functions differentiable at t0, hence differentiable at t0 with derivative ∑k=02(Aik′Bkj+AikBkj′) by two applications of [L1]; this sum is the (i,j) entry of A′(t0)B(t0)+A(t0)B′(t0) by [L2]. Hence AB is differentiable at t0 with (AB)′=A′B+AB′.

2.1step 1.21.3givenL4L5L6

Assume A(t)∈SO(3) for all t∈I. By steps 1.2 and 1.3 the map t↦A(t)TA(t) is differentiable at t0, with derivative (A′)TA+ATA′ at t0. By [L4] this map is the constant map t↦I3, and every entry of this constant entry function is constant, so by [L5] and [L2] its derivative at t0 is the zero matrix [L6]. Hence (A′)TA+ATA′=0 at t0.

2.2step 1.32.1L3L4algebra

By [L4] and [L3] the map t↦A(t)A(t)T is also the constant map t↦I3, so step 1.3 gives A′AT+A(A′)T=0 at t0, and transposing with [L3] gives (A′AT)T=−A′AT.

3.1step 2.1L3algebra

Transposing the identity of step 2.1 and using [L3] gives ATA′+(ATA′)T=0, hence (ATA′)T=−ATA′, so ATA′ is skew-symmetric.

4.1step 3.12.2L2algebra∎

A skew-symmetric matrix has zero diagonal, hence trace 0: the (i,i) entry satisfies Mii=−Mii by step 3.1 applied entrywise, so 2Mii=0 in R; this applies to the two matrices of steps 3.1 and 2.2.

Depends on

Dependency tree · 2 levels

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