64 a a b t b a t c d t d c t a c b t d t b d a t c t

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Nature of Mathematics
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Chapter 4 / Exercise 51
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64. a. A - B T B A T + C - D T D C T = A + C - B T - D T B + D A T + C T = A + C - ( B + D ) T B + D ( A + C ) T is of the required form. b. k A - B T B A T = kA - kB T kB kA T = kA - ( kB ) T kB ( kA ) T is of the required form. c. The general element of H is M = p - q - r - s q p s - r r - s p q s r - q p , with four arbitrary constants, r,s,p , and q . Thus dim( H ) = 4; use the strategy outlined in Summary 4.1.6 to construct a basis. d. A - B T B A T C - D T D C T = AC - B T D - AD T - B T C T BC + A T D - BD T + A T C T = AC - B T D - ( BC + A T D ) T BC + A T D ( AC - B T D ) T is of the required form. 263
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Nature of Mathematics
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Chapter 4 / Exercise 51
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Chapter 5 ISM: Linear Algebra Note that A, B, C, D , and their transposes are rotation-dilation matrices, so that they all commute. e. A - B T B A T T = A T B T - B ( A T ) T is of the required form. f. Note that the columns v 1 , v 2 , v 3 , v 4 or M are orthogonal, and they all have length p 2 + q 2 + r 2 + s 2 . Now M T M is the 4 × 4 matrix whose ij th entry is v i · v j , so that M T M = ( p 2 + q 2 + r 2 + s 2 ) I 4 . g. If M = 0, then k = p 2 + q 2 + r 2 + s 2 > 0, and ( 1 k M T ) M = I 4 , so that M is invertible, with M - 1 = 1 p 2 + q 2 + r 2 + s 2 M T . By parts b and e, M - 1 is in H as well. h. No! A = 0 - 1 0 0 1 0 0 0 0 0 0 1 0 0 - 1 0 and B = 0 0 - 1 0 0 0 0 - 1 1 0 0 0 0 1 0 0 do not commute ( AB = - BA ). 65. Write 10 A = a b c d ; it is required that a, b, c and d be integers. Now A = a 10 b 10 c 10 d 10 must be an orthogonal matrix, implying that ( a 10 ) 2 +( c 10 ) 2 = 1, or a 2 + c 2 = 100. Checking the squares of all integers from 1 to 9, we see that there are only two ways to write 100 as a sum of two positive perfect squares: 100 = 36+64 = 64+36. Since a and c are required to be positive, we have either a = 6 and c = 8 or a = 8 and c = 6. In each case we have two options for the second column of A , namely, the two unit vectors perpendicular to the first column vector. Thus we end up with four solutions: A = . 6 - . 8 . 8 . 6 , . 6 . 8 . 8 - . 6 , . 8 - . 6 . 6 . 8 or . 8 . 6 . 6 - . 8 . 66. One approach is to take one of the solutions from Exercise 65, say, the rotation matrix B = 0 . 8 - 0 . 6 0 . 6 0 . 8 , and then let A = B 2 = 0 . 28 - 0 . 96 0 . 96 0 . 28 . Matrix A is orthogonal by Fact 5.3.4a. 67. a. We need to show that A T Ac = A T x , or, equivalently, that A T ( x - Ac ) = 0. But A T ( x - Ac ) = A T ( x - c 1 v 1 - · · · - c m v m ) is the vector whose i th component is 264
ISM: Linear Algebra Section 5.4 ( v i ) T ( x - c 1 v 1 - · · · - c m v m ) = v i · ( x - c 1 v 1 - · · · - c m v m ), which we know to be zero. b. The system A T Ac = A T x has a unique solution c for a given x , since c is the coordinate vector of proj V x with respect to the basis v 1 , . . . , v m . Thus the coe ffi cient matrix A T A must be invertible, so that we can solve for c and write c = ( A T A ) - 1 A T x . Then proj V x = c 1 v 1 + · · · + c m v m = Ac = A ( A T A ) - 1 A T x . 68. If A = QR, then A ( A T A ) - 1 A T = QR ( R T Q T QR ) - 1 R T Q T = QR ( R T R ) - 1 R T Q T = QRR - 1 ( R T ) - 1 R T Q T = QQ T , as in Fact 5.3.10. The equation Q T Q = I m holds since the columns of Q are orthonormal. 5.4 1. A basis of ker( A T ) is - 3 2 . (See Figure 5.15.) Figure 5.15: for Problem 5.4.1. 2. A basis of ker( A T ) is 1 - 2 1 . im( A ) is the plane perpendicular to this line. 265
Chapter 5 ISM: Linear Algebra 3. We will first show that the vectors v 1 , . . . , v p , w 1 , . . . , w q span R n . Any vector v in R n can be written as v = v + v , where v is in V and v is in V (by definition of a projection, Fact 5.1.4).

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