Linear Algebra Notes Chapter 6 SOLUTIONS TO EXERCISES ...

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2 A picture to accompany this exercise will be drawn in class. Solution: a) Since (x − λ)(x − µ) = x 2 − x − 1, we have λ + µ = 1 and λµ = −1. Hence (y − λx)(y − µx) = y 2 − (λ + µ)xy + λµx = y 2 − xy − x 2 . b) f(x ′ , y ′ ) = (y ′ ) 2 −x ′ y ′ −(x ′ ) 2 = (x+y) 2 −(x+y)y −y 2 = −y 2 +xy +x 2 = −f(x, y). c) f(F n , F n+1 ) = −f(F n−1 , F n ) = (−1) 2 f(F n−2 , F n−1 ) = · · · = (−1) n f(F 0 , F 1 ) = (−1) n f(0, 1) = (−1) n . Exercise 6.4. The “Lucas numbers” are defined like the Fibonaccis, except the two starting values are L 0 = 2, L 1 = 1. Thus, L 2 = 3, L 3 = 4, L 4 = 7, etc. Using the method of this chapter, find a formula for L n that does not require computing any other Lucas numbers. (Note the same matrix Φ is used, but the initial vector is different). Solution: [ ] [ ] Ln = Φ n 2 L n+1 1 = √ 1 [ ] [ λ n−1 − µ n−1 λ n − µ n 2 5 λ n − µ n λ n+1 − µ n+1 1 ] , So L n = 1 √ 5 [2(λ n−1 − µ n−1 ) + (λ n − µ n )] = 1 √ 5 [λ n (1 + 2 λ ) − µn (1 + 2 µ ). Since we get 1 + 2 λ = −(1 + 2 µ ) = √ 5, L n = λ n + µ n .
3 Exercise 6.5. Consider the sequence of numbers a n defined by a 0 = 0, a 1 = 1, a n+1 = a n−1 + 2a n . Find a formula for a n that does not involve computing earlier terms in the sequence and compute a n+1 lim . n→∞ a n Solution: The matrix A = The eigenvalues of A are and the matrix B = [ ] 0 1 satisfies 1 2 [ ] A n 0 = 1 [ an a n+1 ] . λ = 1 + √ 2, µ = 1 − √ 2, [ ] 1 1 satisfies λ µ B −1 AB = Using the fact that λµ = −1, we get so Finally, A n = B as n → ∞, since |µ| < 1. [ ] λ 0 . 0 µ [ ] λ n 0 0 µ n B −1 = 1 [ ] λ 2 √ n−1 − µ n−1 λ n − µ n 2 λ n − µ n λ n+1 − µ n+1 , a n+1 = λn+1 − µ n+1 a n λ n − µ n a n = λn − µ n 2 √ 2 . [ ] 1 − (µ/λ) n+1 = λ 1 − (µ/λ) n → λ,
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Linear Algebra Notes Chapter 6 SOLUTIONS TO EXERCISES ...

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