HW7_sol - 80 Chapter 5 17(a A is symmetric since AT =(xyT yxT)T =(xyT)T(yxT)T =(yT)T xT(xT)T yT = yxT xyT = A(b For any vector z in Rn Az = xyT z

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80 Chapter 5 17. (a) A is symmetric since A T =( xy T + yx T ) T xy T ) T +( T ) T y T ) T x T x T ) T y T = T + xy T = A (b) For any vector z in R n A z = xy T z + yx T z = c 1 x + c 2 y where c 1 = y T z and c 2 = x T z .If z is in N ( A ) then 0 = A z = c 1 x + c 2 y and since x and y are linearly independent we have y T z = c 1 = 0 and x T z = c 2 = 0. So z is orthogonal to both x and y . Since x and y span S it follows that z S . Conversely, if z is in S then z is orthogonal to both x and y .It follows that A z = c 1 x + c 2 y = 0 since c 1 = y T z = 0 and c 2 = x T z = 0. Therefore z is in N ( A ) and hence N ( A )= S . (c) Clearly dim S = 2 and by Theorem 5.2.2, dim S + dim S = n . Using our result from part (a) we have dim N ( A ) = dim S = n - 2 So A has nullity n - 2. It follows from the Rank-Nullity Theorem that the rank of A must be 2. SECTION 3 1. (b) A T A = 6 - 1 - 16 and A T b = 20 - 25 The solution to the normal equations A T A x = A T b is x = 19 / 7 - 26 / 7 2. (Exercise 1b.) (a) p = 1 7 ( - 45 , 12 , 71) T (b) r = 1 7 (115 , 23 , 69) T (c) A T r = - 12 1 11 - 2 115 7 23 7 69 7 = 0 0 0 Therefore r is in N ( A T ).
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Section 3 81 6. A = 1 - 11 10 0 1 12 4 , b = 0 1 3 9 A T A = 42 6 26 8 681 8 ,A T b = 13 21 39 The solution to A T A x = A T b is (0 . 6 , 1 . 7 , 1 . 2) T . Therefore the best least squares Ft by a quadratic polynomial is given by p ( x )=0 . 6+1 . 7 x +1 . 2 x 2 7. To Fnd the best Ft by a linear function we must Fnd the least squares solution to the linear system 1 x 1 1 x 2 . . . . . . 1 x n c 0 c 1 = y 1 y 2 . . . y n If we form the normal equations the augmented matrix for the system will be n n ± i =1 x i n ± i =1 y i n ± i =1 x i n ± i =1 x 2 i n ± i =1 x i y i If x = 0 then n ± i =1 x i = n x =0 and hence the coefficient matrix for the system is diagonal. The solution is easily obtained. c 0 = n ± i =1 y i n = y and c 1 = n ± i =1 x i y i n ± i =1 x 2 i = x T y x T x
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82 Chapter 5 8. To show that the least squares line passes through the center of mass, we introduce a new variable z = x - x . If we set z i = x i - x for i =1 ,...,n , then z = 0. Using the result from Exercise 7 the equation of the best least squares Ft by a linear function in the new zy -coordinate system is y = y + z T y z T z z If we translate this back to xy -coordinates we end up with the equation y - y = c
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This note was uploaded on 04/18/2010 for the course FINANCE 1231854365 taught by Professor Wuyiling during the Spring '10 term at Nashville State Community College.

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HW7_sol - 80 Chapter 5 17(a A is symmetric since AT =(xyT yxT)T =(xyT)T(yxT)T =(yT)T xT(xT)T yT = yxT xyT = A(b For any vector z in Rn Az = xyT z

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