Thus 1 1 1 r ˆ w 2 π 3 problem 723 let a 1 3 1 2 1

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Thus 0 - 1 0 0 0 - 1 1 0 0 = R ( ˆ w, 2 π/ 3) . PROBLEM 7–23. Let A = 1 3 1 2 1 6 1 3 - 1 2 1 6 1 3 0 - 2 6 . Find ˆ w and θ such that A = R ( ˆ w, θ ). PROBLEM 7–24. Compute R ( ˆ w, θ ) - 1 = R (? , ?) .
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Cross product 21 PROBLEM 7–25. Show that the given matrix is in SO(3) and calculate the unique ˆ w R 3 and 0 θ π for it: 1 45 29 - 20 28 28 35 - 4 - 20 20 35 . G. Orientation reversal It is a relatively easy task to understand those matrices in O ( n ) which are not in SO( n ). Suppose A is such a matrix. Of course, det A = - 1. We conclude from Problem 7–22 that - 1 is an eigenvalue of A . Therefore, there exists a unit vector ˆ w : A ˆ w = - ˆ w. Any unit vector ˆ w can be used to define a “reflection” matrix S in that direction. Here’s a schematic description: w x Sx O The algebra is governed by ( Sx + x ) ˆ w = 0 , Sx = x + t ˆ w for some t R . Solving these equations gives t = - 2 x ˆ w so that Sx = x - 2( x ˆ w ) ˆ w.
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22 Chapter 7 This matrix S is called “reflection in the direction ˆ w ,” and we denote it S ( ˆ w ). PROBLEM 7–26. Prove that the matrix S = S ( ˆ w ) satisfies a. S ˆ w = - ˆ w , b. S 2 = I , c. S O ( n ), d. det S = - 1. PROBLEM 7–27. Prove that S ˆ w = I - 2( w i w j ) . Now we return to the situation of a matrix A O ( n ) with det A = - 1. We know from Problem 7–22 that there is a unit vector ˆ w satisfying A ˆ w = - ˆ w . Now just define B = S ( ˆ w ) A . Clearly, B SO( n ). Thus, A = S ( ˆ w ) B. Thus A is the product of a reflection S ( ˆ w ) in a direction ˆ w and a matrix B SO( n ) with B ˆ w = ˆ w . PROBLEM 7–28. It made no difference whether we chose to write B = SA or B = AS , for SA = AS . Prove this by noting that SA ˆ w = AS ˆ w and then considering those vectors x which are orthogonal to ˆ w . In particular, for n = 3 the matrix A has the representation A = S ( ˆ w ) R ( ˆ w, θ ) = R ( ˆ w, θ ) S ( ˆ w ) . PROBLEM 7–29. Show that in this n = 3 situation A = (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 .
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Cross product 23 PROBLEM 7–30. Show that in case n = 2 every matrix A O (2) with det A = - 1 is equal to a reflection matrix S ( ˆ w ). PROBLEM 7–31. Refer back to inversion in the unit sphere in R n , p. 6–34: f ( x ) = x k x k 2 . Prove that Df ( x ) = k x k - 2 S x k x k . Thus we say that inversion in the unit sphere is an orientation - reversing conformal mapping. PROBLEM 7–32. Let u , v be linearly independent vectors in R 3 and let u 0 , v 0 also be vectors in R 3 . a. Suppose that A SO(3) satisfies Au = u 0 , Av = v 0 . Prove that u 0 and v 0 are linearly independent and k u k = k u 0 k , k v k = k v 0 k , u v = u 0 v 0 . ( * ) b. Conversely, suppose that ( * ) is satisfied. Prove that u 0 and v 0 are linearly indepen- dent, and that there exists a unique A SO(3) such that Au = u 0 , Av = v 0 . (HINT: A ( u v u × v ) = ( u 0 v 0 u 0 × v 0 ).)
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24 Chapter 7 PROBLEM 7–33. Under the assumptions of the preceding problem prove that ( u v u × v ) - 1 = k u × v k - 2 ( v × ( u × v ) - u × ( u × v ) u × v ) t .
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