Linear Algebra Notes Chapter 6 SOLUTIONS TO EXERCISES ...

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 .