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The coordinate rotation matrices Rz(a) and Ry(a): values in SO(3), composition law, and derivatives

Statement

For a∈R define, in the standard basis of R3, Rz(a):=(cos⁡a−sin⁡a0sin⁡acos⁡a0001),Ry(a):=(cos⁡a0sin⁡a010−sin⁡a0cos⁡a)∈M3(R). Then:

  1. Rz(a)∈SO(3) and Ry(a)∈SO(3) for every a (The orthogonal group O(3) and the rotation group SO(3)), Rz(a)e3=e3 and Ry(a)e2=e2, and Rz(0)=Ry(0)=I3.
  2. Rz(a)Rz(b)=Rz(a+b) and Ry(a)Ry(b)=Ry(a+b) for all a,b∈R.
  3. For every differentiable a:I→R on an interval I⊆R (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, Intervals of R: the nine order-convex forms, nondegeneracy, and length), the matrix-valued curves t↦Rz(a(t)) and t↦Ry(a(t)) are differentiable with (Rz∘a)′=a′ Rz(a) [e3]×=a′ [e3]×Rz(a),(Ry∘a)′=a′ Ry(a) [e2]×=a′ [e2]×Ry(a), where [e3]×,[e2]× are the cross-product matrices of The skew-matrix/cross-product correspondence [ω]×z=ω×z and its conjugation law R[ω]×RT=[Rω]×.

Facts & Assumptions

Given: Reals a,b, matrices Rz(a),Ry(a) as displayed, and a differentiable a:I→R.

[L1]

sin⁡ and cos⁡ are differentiable with (sin⁡)′=cos⁡, (cos⁡)′=−sin⁡, and satisfy the addition formulas sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b, cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b and the Pythagorean identity sin⁡2a+cos⁡2a=1; also sin⁡(−a)=−sin⁡a and cos⁡(−a)=cos⁡a The derivatives of sine and cosine are cosine and minus sine, The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Sine and cosine defined by their real power series.

[L3]

The determinant is multiplicative and det⁡I3=1; for a block-triangular matrix the determinant is the product of the determinant of the leading block and the remaining diagonal entry For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B), For n≥1, the determinant over a commutative ring by the Leibniz formula, and ∣det⁡A∣ for a real matrix.

[L5]

[e3]× and [e2]× are the skew matrices of ω=e3 and ω=e2; in particular [e3]×=(0−10100000) and [e2]×=(001000−100) The skew-matrix/cross-product correspondence [ω]×z=ω×z and its conjugation law R[ω]×RT=[Rω]×, The vector space Mm×n(F):=F m×n of m by n matrices over a field, with entrywise operations.

Proof

technique · direct
1.1givenL1L2L3algebra

Entrywise, Rz(a)TRz(a) has diagonal entries cos⁡2a+sin⁡2a=1, 1, 1 and off-diagonal entries ∓cos⁡asin⁡a±sin⁡acos⁡a=0; hence Rz(a)∈O(3), and expanding the determinant along the third row gives det⁡Rz(a)=det⁡(cos⁡a−sin⁡asin⁡acos⁡a)=(cos⁡a)(cos⁡a)−(−sin⁡a)(sin⁡a)=1, so Rz(a)∈SO(3), with Rz(a)e3=e3 read off the third column.

1.2givenL1L2L3algebra

Similarly Ry(a)TRy(a)=I3 and det⁡Ry(a)=det⁡(cos⁡asin⁡a−sin⁡acos⁡a)=1, so Ry(a)∈SO(3), and Ry(a)e2=e2 is its second column. Rz(0)=Ry(0)=I3 because cos⁡0=1 and sin⁡0=0.

1.3givenL1L2algebra

The (1,1) entry of Rz(a)Rz(b) is cos⁡acos⁡b−sin⁡asin⁡b=cos⁡(a+b), the (1,2) entry is −cos⁡asin⁡b−sin⁡acos⁡b=−sin⁡(a+b), the (2,1) entry is sin⁡acos⁡b+cos⁡asin⁡b=sin⁡(a+b), the (2,2) entry is −sin⁡asin⁡b+cos⁡acos⁡b=cos⁡(a+b), and the remaining entries are 0,0,1; hence Rz(a)Rz(b)=Rz(a+b) by [L1].

1.4givenL1L2algebra

The analogous four planar entries of Ry(a)Ry(b) equal cos⁡(a+b),0,sin⁡(a+b) and 1 in the corresponding positions, so Ry(a)Ry(b)=Ry(a+b) by [L1].

1.5givenL1L4L5algebra

If a:I→R is differentiable, entrywise differentiation using [L1] and [L4] gives (Rz∘a)′=a′(−sin⁡a−cos⁡a0cos⁡a−sin⁡a0000). Direct multiplication of a′Rz(a) with [e3]× from [L5] gives the same matrix: its (1,1),(1,2),(2,1),(2,2) entries are respectively −a′sin⁡a,−a′cos⁡a,a′cos⁡a,−a′sin⁡a, and all remaining entries are zero. Thus (Rz∘a)′=a′Rz(a)[e3]×.

1.6givenL1L4L5algebra

The same comparison for Ry(a(t)): the derivative matrix has entries −a′sin⁡a at (1,1), a′cos⁡a at (1,3), −a′cos⁡a at (3,1), −a′sin⁡a at (3,3) and 0 in the second row and column, which is exactly a′Ry(a)[e2]× by [L5]; hence (Ry∘a)′=a′Ry(a)[e2]×.

2.1step 1.51.6givenL1L2L5algebra∎

Since Rz(a) commutes with [e3]× and Ry(a) commutes with [e2]×, as direct entrywise multiplication using [L1] shows, the two alternative forms a′[e3]×Rz(a) and a′[e2]×Ry(a) of clause 3 hold as well.

Depends on

Dependency tree · 2 levels

Sources