Exercise 6.11.7

Answers

1.
We prove that if T is an orthogonal operator with det ⁡ (T) = 1 on a three-dimensional space V , then T is a rotation. First, we know that the decomposition of T contains no reflections by Theorem 6.47. According to T, V could be decomposed into some subspaces. If V is decomposed into three one-dimensional subspace, then T is the identity mapping on V since it’s an identity mapping on each subspace. Otherwise, V should be decomposed into one one-dimensional and one two-dimensional subspace. Thus T is a rotation on the two-dimensional subspace and is an identity mapping on the one-dimensional space. Hence T must be a rotation.
Finally, we found that det ⁡ (A) = det ⁡ (B) = 1. Hence they are rotations.
2.
It comes from the fact that det ⁡ (AB) = det ⁡ (A)det ⁡ (B) = 1.
3.
It should be the null space of AB − I,
span ⁡ {((1 + cos ⁡ ϕ)(1 − cos ⁡ ψ),(1 + cos ⁡ ϕ)sin ⁡ ψ,sin ⁡ ϕsin ⁡ ψ)}.
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2011-06-27 00:00
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