Linear Algebra Notes Chapter 6 SOLUTIONS TO EXERCISES ...

Linear Algebra Notes
Chapter 6
SOLUTIONS TO EXERCISES
Exercise 6.1.
(a) Use this method to compute F 20 , F 21 and F 30 without computing any other
Fibonnaci numbers.
(b) Compute F 30 again by starting with your values for F 20 and F 21 from
part (a) and then computing F 22 , F 23 , . . . , F 30 using the recursive formula
F n+1 = F n−1 + F n .
Solution:
a) λ 20 / √ 5 = 6765.000029 . . . , so F 20 = 6765.
λ 21 / √ 5 = 10945.999981 . . . , so F 21 = 10946.
λ 30 / √ 5 = 832040.00000024 . . . , so F 30 = 832040.
b) F 22 = 17711, F 23 = 28657, F 24 = 46368, F 25 = 75025, F 26 = 121393, F 27 =
196418, F 28 = 317811, F 29 = 514229, F 30 = 832040.
Exercise 6.2. Show that
Solution:
since |µ/λ| < 1.
F n+1
= λn+1 − µ n+1
F n λ n − µ n
F n+1
lim = λ.
n→∞ F n
= λ 1 − (µ/λ)n+1
1 − (µ/λ) n → λ,
Exercise 6.3. The eigenvectors u, v of Φ live on the lines with equations y = λx,
y = µx, respectively. Consider the function
which is zero on these lines.
f(x, y) = (y − λx)(y − µx)
(a) Show that f(x, y) = y 2 − xy − x 2 .
(b) Given a point (x, y), define (x ′ , y ′ ) by the equation
Show that
[ ] [ ]
x x
′
Φ =
y y ′ .
f(x ′ , y ′ ) = −f(x, y).
(c) Let C + be the graph of the equation y 2 − xy − x 2 = 1, and let C − be the
graph of the equation y 2 − xy − x 2 = −1. These graphs are hyperbolas,
having the lines y = λx, y = µx as asymptotes. Use part (b) to show that
the point (F n , F n+1 ) lives on C + if n is even and on C − if n is odd.
1

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,