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Geometria Lingotto.
LeLing15: Fibonacci, Eigenvalues and eigenvectors.
Contents:
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• Fibonacci’s sequence.
• nth power of a matrix.
• Eigenvalues and eigenvectors.
• Binet’s formula.
Recommended exercises: Geoling 16 .
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On the roof of the Mole Antonelliana.
On the roof of the Mole we can see the numbers:
1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987 , · · ·
Each one is the sum of two before it in the sequence, so we may write:
Fn = Fn−1 + Fn−2
If this is clear, you should have no problem in telling which number should follow
987. (Answer: 1597). Putting F1 = 1, F2 = 1, we have F16 = 987, so F17 = 1597.
Historical remark:
This sequence of numbers is known as Fibonacci sequence in honour of Leonardo
of Pisa, known by his father’s name, hence ”figlio di Bonaccio” (in Latin, filius Bonacci,
ie.e son of Bonaccio), contracted to Fibonacci (another name was Bigollo); Fibonacci
was studying this sequence around the year 1223 in relation to the following problem on
the growth of a rabbit population:
Determine the number of couples of adult rabbits if each couple generates a younger
couple each month, and newer generations can reproduce when they are two months old.
The solution to this problem is F12 = 144 after 12 months, so after one year there
are 144 couples of adult rabbits.
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Geometria
Geometria Lingotto.
Binet’s formula
A modern version of Fibonacci’s problem is the following:
Problem 0.1. How big is F100 ? For example, is F100 larger or smaller than 100000000 =
108 ??
To address the question, we would (perhaps) need a ”formula” to compute the nth
number Fn . The well-read student will know that such formula exists and is known as
Binet formula (ca. 1843), although it was certainly known to Euler and Daniel Bernoulli:
1
Fn = √
5
(
√ !n
1+ 5
−
2
√ !n )
1− 5
2
This√
identity is far from self-evident or trivial. How come, for instance, the appearance of 5, not an integer number? (Recall Fn is integer for it’s the number of couples
of rabbits...). If the formula is correct, mysterious simplifications must occur in it...
Exercise : Check that the formula works for n = 1 and n = 2.
The formula is usually proved using the Principle of Mathematical Induction:
Exercise : using the Induction Principle prove Binet’s formula for any n ≥ 1.
Still the proof doesn’t help much to understand the origin of this beautiful relation.
In the sequel you might find one answer (of the many available) that uses the notions of
eigenvalue and eigenvector.
Now let’s solve problem 0.1 using the fact that (courtesy of Binet...):
√ !n
n
1+ 5
1
3
1
≥√
.
Fn ≈ √
2
5
5 2
Then F100 ≥
bigger than 108 .
√1
5
3 100
2
>
√1 250
5
> 248 = 28 .240 > 28 .58 = 108 . Therefore F100 is
Fibonacci and matrices
The equation
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Geometria
Geometria Lingotto.
Fn = Fn−1 + Fn−2
is the same as the system:
so
Fn = Fn−1 + Gn−1
Gn = Fn−1
We already know that we can write the latter using matrices, so
1 1
Fn−1
Fn
=
1 0
Gn−1
Gn
1 1
Fn
comes from multiplying Cn−1 by the matrix A =
,
The column Cn =
Gn
1 0
A Cn−1 = Cn
Since Cn−1 = A Cn−2 ,
A Cn−1 = A A Cn−2 = A2 Cn−2 = Cn
Iterating the same idea we get:
An−2 C2 = Cn
where C2 =
1
1
.
Thus we must raise a matrix A to the power n − 2, i.e. compute
1 1
1 0
n−2
The nth power of a matrix
2 0
0 3
If a matrix D is diagonal, e.g. D =
it is easy to compute Dn . Namely,
n
2
0
n
D =
.
0 3n
n
λ 0
λ
0
n
More generally, if D =
, then D =
0 β
0 βn
Another unproblematic case is that in which A can be written as:
A = M DM −1
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Geometria
Geometria Lingotto.
where D is a diagonal matrix and M is arbitrary (but invertible, given the presence
of M −1 ).
Nota 0.2. When such D and M exist, one says that A is diagonalizable or that it
can be diagonalized 1 . This amounts to the same as :
M −1 AM = D
Now we should see why An is easy when A is diagonalizable:
−1
−1
−1
−1
−1
An = M
{z · · · M DM M DM }
| DM M DM M DM
n times
= M DDD
· · · DD} M −1 = M Dn M −1
|
{z
n times
because the intermediate factors M M −1 cancel out by definition. Therefore, the
knowledge of Dn and M yields An .
Diagonalizing Fibonacci
1 1
In order to compute
1 0
invertible one M such that:
n−2
the best would be to find a diagonal matrix D plus an
1 1
1 0
= M DM −1
How do we find them? The idea is to assume that M, D do exist, and try to find
them somehow... The previous equation implies
1 1
1 0
M = MD .
But as D is diagonal, the columns M1 and M2 of M satisfy
1
1 1
1 0
M1 = λM1
Note M and D don’t always exist: for instance, A =
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0
0
1
0
cannot be diagonalized.
Geometria
Geometria Lingotto.
and
1 1
1 0
M2 = βM2
λ 0
where D =
; this though we don’t know, otherwise we would solve the
0 β
systems (equivalent to the previous equations) in the unknowns M1 , M2 :
1−λ 1
1
−λ
M1 =
0
0
1−β 1
1
−β
M2 =
0
0
This system looks interesting, because it says that M1 ,M2 are solutions of a homogeneous system. Recall also that we want M1 (and M2 ) to be a column of an invertible
matrix, so in particular M1 6= 0.
This forces us to take λ (and β ) equal to a root of the determinant of the matrix
1−λ 1
, because the system has non-trivial solutions iff
1
−λ
1−λ 1
det(
) = λ2 − λ − 1 = 0
1
−λ
√
√
Thus λ √= 1+2 5 , 1−2 5 (see [Livio] for some interesting features of these numbers); the
number 1+2 5 is the so-called golden ratio.
√
Nota 0.3. In the jargon of Linear Algebra the numbers 1±2 5 are the eigenvalues of
1 1
the matrix
. More generally, given a matrix A one defines its characteristic
1 0
polynomial by P (λ) = det(A−λId), 2 and its roots are said eigenvalues of the matrix.
2
Historically the characteristic polynomial first appears in the work of Luigi Lagrange on the perturbations, over a period of 100 years, of planetary orbits close to Kepler’s orbits. The equation
det(A − λId) = 0 was therefore called secular equation [Arnold, p.45]. Nowadays, the equation is
known with that name in Astronomy, Physics and Chemistry [McSi].
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Geometria
Geometria Lingotto.
√
1+ 5
2
"
We can then take D to be
#
0√
.
1− 5
0
2
Now it is all but easy to determine the columns M1 and M2 of M , these being the
solutions of the homogeneous systems
#
"
#
"
√
√
1− 5
1+ 5
1− 2
1− 2
1√
1√
1+ 5
1− 5
1
− 2
1
− 2
1+√5 1−√5 2
2
Thus we take M1 =
and M2 =
.
1
1
1 1
Nota 0.4. Linear Algebra calls the columns M1 , M2 eigenvectors of the matrix
.
1 0
In general, a solution x 6= 0 of the homogeneous system with matrix A − λId, with
λ eigenvalue of A, is called eigenvector for A.
Eventually, we have what we were looking for,
#
1+√5 1−√5 " 1+√5
1 1
0√
2
2
2
=
1− 5
1 0
1
1
0
2
"
#
√
√
√
−1
The inverse
1+ 5
2
1− 5
2
1
1
1 1
1 0
√1
5
−1
√
5
is
√
1+ 5
2
√
1− 5
2
1
1
√
1+ 5
2
√
1− 5
2
1
1
=
"
5−1
√
2 √5
1+√ 5
2 5
√
1+ 5
2
0
√
1+ 5
2
√
1− 5
2
1
1
−1
. Thus
0√
#"
1− 5
2
√1
5
−1
√
5
√
5−1
√
2 √5
1+√ 5
2 5
#
and
1 1
1 0
n−2
1
=√
5
"
√
0
( 1+2 5 )n−2
√
1− 5 n−2
0
( 2 )
Recall that to find Fn we should calculate
Fn
Gn
1
=√
5
√
1+ 5
2
√
1− 5
2
1
1
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"
√
1 1
1 0
n−2 ( 1+2 5 )n−2
0
√
1− 5 n−2
0
( 2 )
6
1
1
#"
√
#"
1
−1
Fn
Gn
=
√
1
−1
5−1
2√
1+ 5
2
5−1
2√
1+ 5
2
#
i.e.
#
Geometria
1
1
=
REFERENCES
1
=√
5
Geometria Lingotto.
√
1+ 5
2
√
1− 5
2
1
1
"
√
( 1+2 5 )n−2
0
√
1− 5 n−2
0
( 2 )
#"
√
1+ 5
√2
5−1
2
#
1
=√
5
√
1+ 5
2
√
1− 5
2
1
1
"
Therefore
Fn
Gn
"
=
√
√
√1 {( 1+ 5 )n − ( 1− 5 )n }
2√
2 √
5
√1 {( 1+ 5 )n−1 − ( 1− 5 )n−1 }
2
2
5
#
and we obtain
1
Fn = √
5
(
√ !n
1+ 5
−
2
√ !n )
1− 5
2
References
[Arnold]
Arnold, V.I.: Lectures on Partial Differential Equations , Springer Universitext
2004.
[Livio]
Livio, M.: La sezione aurea - Storia di un numero e di un mistero che dura da
tremila anni , Traduzione di Stefano Galli, Rizzoli, Prima edizione: 2003.
[McSi]
McQuarrie, D.A. and Simon, J.D.: CHIMICA FISICA Un approcio molecolare
, Trad. di M. Roncaglia, revisione di C. Galli. Ed. Zanichelli, 2000.
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Geometria
√
( 1+2 √5 )n−1
−( 1−2 5 )n−1
#