Linear Algebra, 2020a

Topic: Linear Recurrences 475
Write T for the matrix and ⃗v n for the vector with components F(n) and F(n − 1)
so that ⃗v n = T n−1 ⃗v 1 for n 1. If we diagonalize T then we have a fast way
to compute its powers: where T = PDP −1 then T n = PD n P −1 and the n-th
power of the diagonal matrix D is the diagonal matrix whose entries are the
n-th powers of the entries of D.
The characteristic equation of T is λ 2 − λ − 1 = 0. The quadratic formula
gives its roots as (1+ √ 5)/2 and (1− √ 5)/2. (These are sometimes called “golden
ratios;” see [Falbo].) Diagonalizing gives this.
( )
1 1
=
1 0
(
1+ √ 5
2
1− √ 5
2
1 1
)( √ )(
1+ 5
1
2
0
1−
0 √ 5
2
Introducing the vectors and taking the n-th power, we have
( ) ( ) n−1 ( )
F(n) 1 1 f(1)
=
F(n − 1) 1 0 f(0)
) ⎛ (
⎜ ⎝
=
(
1+ √ 5
2
1− √ 5
2
1 1
1+ √ ) n−1
5
2
0
(
The calculation is ugly but not hard.
( )
F(n)
=
F(n − 1)
(
1+ √ 5
2
0
1− √ 5
2
1 1
(
= √ 1 1+ √ 5
2
5
) ⎛ ⎜ ⎝
(
1− √ 5
2
1− √ 5
2
√5 −( 1−√ 5
2 √ ) 5
√−1
1+ √ 5
5 2 √ 5
√−1
1+ √ 5
5 2 √ 5
)
⎞
( 1√5
⎟ −( 1−√ 5
)
2 √ 5 n−1 ⎠
)
)( )
1+ √ ) n−1
5
2
0
(
0
) ⎛ ⎜ ⎝
(
(
1 1 −
⎛ (
= √ 1 1+ √ ) n (
5
2 −
1− √ 5
2
⎝(
) n−1 (
5 −
1+ √ 5
2
We want the first component.
F(n) = 1 √
5
[(
1− √ 5
2
1+ √ ) n−1
5
2
1− √ 5
2
) n
1 + √ ) n (
5
−
2
) n−1
⎞
⎠
1− √ 5
2
) n−1
⎞
⎟
⎠
1 − √ ) n ]
5
2
⎞
⎟
) n−1 ⎠
1
0
( 1 √5
− 1 √
5
)
This formula gives the value of any member of the sequence without having to
first find the intermediate values.
Because (1 − √ 5)/2 ≈ −0.618 has absolute value less than one, its powers
go to zero and so the F(n) formula is dominated by its first term. Although we