# Based on our understanding of what r ˆ w θ is we

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Based on our understanding of what R ( ˆ w, θ ) is, we are certain that this formula does not really depend on the choice of ˆ u and ˆ v . See Problem 7–16 for the explicit verification of this fact. You should notice that the inner matrix in the above formula is precisely R ( ˆ k, θ ). So the formula displays how we are able to move from a rotation about the x 3 -axis to a rotation about the axis determined by ˆ w . PROBLEM 7–12. Prove that R ( ˆ w, θ ) SO(3). PROBLEM 7–13. Using the above formula, show directly that R ( ˆ w, 0) = I. Then give a heuristic explanation of this equation. PROBLEM 7–14. Using the above formula, show directly that R ( - ˆ w, - θ ) = R ( ˆ w, θ ) . Then give a heuristic explanation of this equation. Also explain why we may thus assume without loss of generality that 0 θ π . PROBLEM 7–15. Using the above formula, calculate directly that R ( ˆ w, θ 1 ) R ( ˆ w, θ 2 ) = R ( ˆ w, θ 1 + θ 2 ) . Then give a heuristic explanation of this equation.

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16 Chapter 7 PROBLEM 7–16. Calculate the product and thus obtain from ( * ) R ( ˆ w, θ ) = (cos θ ) I + (1 - cos θ )( w i w j ) + sin θ 0 - w 3 w 2 w 3 0 - w 1 - w 2 w 1 0 . (This displays R ( ˆ w, θ ) in a form that shows its independence of the choice of ˆ u and ˆ v .) PROBLEM 7–17. Show that the action of R ( ˆ w, θ ) can also be expressed in the form R ( ˆ w, θ ) x = (cos θ ) x + (1 - cos θ )( ˆ w x ) ˆ w + (sin θ ) ˆ w × x, for all x R 3 . Do this two ways: a. by simply using Problem 7–16; b. by describing the action of R ( ˆ w, θ ) geometrically. PROBLEM 7–18. Prove that for all x ˆ w = 0, R ( ˆ w, π ) x = - x. Give a geometric interpretation of this result. Conversely, suppose that for a given rotation there exists a nonzero x R 3 such that R ( ˆ w, θ ) x = - x. Prove that θ = π and x ˆ w = 0. (Of course, θ could also be π +2 for any integer m .) PROBLEM 7–19. Prove that R ( ˆ w, θ ) is a symmetric matrix ⇐⇒ θ = 0 or π .
Cross product 17 PROBLEM 7–20. Backward Andy all his life has insisted upon using a left-handed coordinate system for R 3 . What formula must he use for R ( ˆ w, θ ) that is the analog of ours found in Problem 7–16? We now turn to the perhaps surprising converse of the above material, which states that the only matrices in SO(3) are rotations! The crucial background for this result is contained in the following THEOREM. Let A SO( n ), where n is odd. Then 1 is an eigenvalue of A . PROOF. Since A - 1 = A t , we have det( A - I ) = det( A - AA t ) = det( A ( I - A t )) = det A det( I - A t ) = det( I - A t ) = det( I - A ) = ( - 1) n det( A - I ) . Since n is odd, det( A - I ) = 0. QED PROBLEM 7–21. Give a different proof of this result for the case n = 3, based on the characteristic polynomial det( A - λI ) = - λ 3 + αλ 2 + βλ + 1 . PROBLEM 7–22. Let A O ( n ), but A / SO( n ). Prove that - 1 is an eigenvalue of A . THEOREM. Let A SO(3). Then there exists a unit vector ˆ w R 3 and 0 θ π such that A = R ( ˆ w, θ ). PROOF. Since A SO(3), we know from the preceding theorem that 1 is an eigenvalue of A . Therefore, there exists a unit vector ˆ w such that A ˆ w = ˆ w .

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18 Chapter 7 Choose an orthonormal basis ˆ u , ˆ v of the subspace of R 3 which is orthogonal to ˆ w ; arrange them so that the frame { ˆ u, ˆ v, ˆ w } has the standard orientation: ˆ w = ˆ u × ˆ v, ˆ v = ˆ w × ˆ u, ˆ u = ˆ v × ˆ w.
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