The components of v 1 give the distribution of the

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Applied Calculus for the Managerial, Life, and Social Sciences
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Chapter 9 / Exercise 26
Applied Calculus for the Managerial, Life, and Social Sciences
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The components of v 1 give the distribution of the population among the age groups in the long run, assuming that current trends continue. λ 1 gives the factor by which the population will grow in the long run in a period of 15 years; this translates to an annual growth factor of 15 1 . 908 1 . 044, or an annual growth of about 4.4%. 37. a. Use that w + z = w + z and w z = wz . w 1 - z 1 z 1 w 1 + w 2 - z 2 z 2 w 2 = w 1 + w 2 - ( z 1 + z 2 ) z 1 + z 2 w 1 + w 2 is in H . w 1 - z 1 z 1 w 1 w 2 - z 2 z 2 w 2 = w 1 w 2 - z 1 z 2 - ( z 1 w 2 + w 1 z 2 ) z 1 w 2 + w 1 z 2 w 1 w 2 - z 1 z 2 is in H . b. If A in H is nonzero, then det( A ) = w w + z z = | w | 2 + | z | 2 > 0, so that A is invertible. c. Yes; if A = w - z z w , then A - 1 = 1 | w | 2 + | z | 2 w z - z w is in H . 381
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Applied Calculus for the Managerial, Life, and Social Sciences
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Chapter 9 / Exercise 26
Applied Calculus for the Managerial, Life, and Social Sciences
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Chapter 7 ISM: Linear Algebra d. For example, if A = i 0 0 - i and B = 0 - 1 1 0 , then AB = 0 - i - i 0 and BA = 0 i i 0 . 38. a. C 2 4 = 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 , C 3 4 = 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 , C 4 4 = I 4 , then C 4+ k 4 = C k 4 . Figure 7.31 illustrates how C 4 acts on the basis vectors e i . Figure 7.31: for Problem 7.5.38a. b. The eigenvalues are λ 1 = 1 , λ 2 = - 1 , λ 3 = i , and λ 4 = - i , and for each eigenvalue λ k , v k = λ 3 k λ 2 k λ k 1 is an associated eigenvector. c. M = aI 4 + bC 4 + cC 2 4 + dC 3 4 If v is an eigenvector of C 4 with eigenvalue λ , then Mv = av + b λ v + c λ 2 v + d λ 3 v = ( a + b λ + c λ 2 + d λ 3 ) v , so that v is an eigenvector of M as well, with eigenvalue a + b λ + c λ 2 + d λ 3 . The eigenbasis for C 4 we found in part b is an eigenbasis for all circulant 4 × 4 matrices. 39. Figure 7.32 illustrates how C n acts on the standard basis vectors e 1 , e 2 , . . . , e n of R n . a. Based on Figure 7.9, we see that C k n takes e i to e i + k “modulo n ,” that is, if i + k exceeds n then C k n takes e i to e i + k - n (for k = 1 , . . . , n - 1). To put it di ff erently: C k n is the matrix whose i th column is e i + k if i + k n and e i + k - n if i + k > n (for k = 1 , . . . , n - 1). 382
ISM: Linear Algebra Section 7.5 Figure 7.32: for Problem 7.5.39. b. The characteristic polynomial is 1 - λ n , so that the eigenvalues are the n distinct solutions of the equation λ n = 1 (the so-called n th roots of unity), equally spaced points along the unit circle, λ k = cos ( 2 π k n ) + i sin ( 2 π k n ) , for k = 0 , 1 , . . ., n - 1 (compare with Exercise 5 and Figure 7.7.). For each eigenvalue λ k , v k = λ n - 1 k . . . λ 2 k λ k 1 is an associated eigenvector. c. The eigenbasis v 0 , v 1 , . . . , v n - 1 for C n we found in part b is in fact an eigenbasis for all circulant n × n matrices. 40. In Exercise 7.2.50 we derived the formula x = 3 q 2 + ( q 2 ) 2 + ( p 3 ) 3 + 3 q 2 - ( q 2 ) 2 + ( p 3 ) 3 for the solution of the equation x 3 + px = q . Here ( q 2 ) 2 + ( p 3 ) 3 is negative, and we can write x = 3 q 2 + i - ( q 2 ) 2 + ( - p 3 ) 3 + 3 q 2 - i - ( q 2 ) 2 + ( - p 3 ) 3 . Let us write this solution in polar coordinates: x = 3 ( - p 3 ) 3 / 2 (cos α + i sin α ) + 3 ( - p 3 ) 3 / 2 (cos α - i sin α ) = - p 3 ( cos α +2 π k 3 + i sin α +2 π k 3 ) + - p 3 ( cos α +2 π k 3 - i sin α +2 π k 3 ) = 2 - p 3 cos ( α +2 π k 3 ) , k = 0 , 1 , 2. See Figure 7.33.

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