Nth power of a square matrix and the Binet Formula for Fibonacci sequence
Yue Kwok Choy
Given
A=(
4
−12
−12
).
11
We begin to investigate how to find
An .
(1) The story begins in finding the eigenvalue(s) and eigenvector(s) of A .
A real number  is said to be an eigenvalue of a matrix A if there exists a non-zero
column vector v such that A𝐯 = λ𝐯 or (A − λI)𝐯 = 0
(a) Eigenvalues
A=(
4
−12
−12
),
11
x
4−λ
𝐯 = (y), (A − λI)𝐯 = (
−12
−12
)=0
11 − λ
Now, (A − λI)𝐯 = 0 has non-zero solution, |A − λI| = 0
4 − λ −12
|
|=0
−12 11 − λ
(4 − λ)(11 − λ) − 144 = 0
λ2 − 15λ − 100 = 0
(λ − 20) (λ + 5) = 0
 λ = 20 or λ2 = −5 , and these are the eigenvalues.
1
(b) Eigenvectors
We usually would like to find the unit eigenvector corresponding to each eigenvalue.
The process is called normalization.
For λ = 20,
1
x1
4
−12 x1
) (y ) = 20 (y )
1
1
{ −12 11
2
2
x1 + y1 = 1
(
{

4x1 − 12y1 = 20x1
−12x
{
1 + 11y1 = 20y1
x1 2 + y1 2 = 1
4x1 + 3y1 = 0
x1 2 + y1 2 = 1
x1
3/5
 (y ) = (
) , which is a unit eigenvector.
−4/5
1
For λ = −5,
2
x2
4
−12 x2
) (y ) = −5 (y )
2
2
{ −12 11
x2 2 + y2 2 = 1
(
{

4x2 − 12y2 = −5x2
{−12x2 + 11y2 = −5y2
x2 2 + y2 2 = 1
3x1 − 4y1 = 0
x2 2 + y2 2 = 1
x2
4/5
 (y ) = (
) , which is another unit eigenvector.
3/5
2
1
(c) Diagonalization of matrix
x1
P = (y
Let
x2
3/5
y2 ) = (−4/5
1
4/5
)
3/5
Consider another matrix:
B=P
−1
3/5 4/5 −1 4
3/5 4/5
−12
AP = (
) (
)(
)
−4/5 3/5
−4/5 3/5
−12 11
=(
3/5
4/5
=(
20 0
)
0 −5
Note that
−4/5
4
)(
3/5
−12
3/5
−12
)(
−4/5
11
4/5
)
3/5
B is a diagonal matrix with eigenvalues as entries in the main diagonal.
The nth power of a diagonal matrix is much easier to find than the original matrix.
20n
0 n
) =(
0
−5
20
Bn = (
0
0
)
(−5)n
Note:
The diagonalization of a matrix may not be a simple subject since |A − λI| = 0
may have equal roots or even complex roots. Although most matrices are not
diagonal matrix, most matrix can be diagonalized. A thorough study of
diagonalization of a matrix is not discussed here.
(d) Finding the nth power of the original matrix A
B = P −1 AP, we have A = PBP −1
An = (PBP −1 )n = (PBP −1 )(PBP −1 ) … (PBP −1 ) = PBn P −1
Since
n
A
3/5
=(
−4/5
=(
4/5 20n
)(
3/5
0
3/5 4/5 20n
)(
−4/5 3/5
0
1
25
1
3/5 −4/5
0
)(
)
n
(−5)
4/5 3/5
3 4 20n
)(
0
−4 3
0
3 −4
)(
)
(−5)n 4 3
3 4 20n
)(
0
−4 3
3(20n ) 4(−5)n 3
0
1
3 −4
)
(
)
=
(
)(
25 −4(20n ) 3(−5)n
(−5)n 4 3
4
1
= 25 (
=
3/5 4/5 −1
0
)(
)
(−5)n −4/5 3/5
(
= 25 (
9(20n ) + 12(−5)n
−12(20n ) + 12(−5)n
−4
)
3
−12(20n ) + 12(−5)n
)
−16(20n ) + 9(−5)n
(2) (a) Fibonacci sequence
The sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...,
defined by F(1) = 1, F(2) = 1, and
F(n) = F(n-1) + F(n-2) for n = 3, 4, 5, ...
is named the Fibonacci sequence.
2
Fibonacci, also known as Leonardo of Pisa, was born in Pisa, home of
the
famous leaning tower (inclined at an angle of 16.5 degrees to the
vertical).
Little is known of his life, and no portrait exists. However, a
statue of Fibonacci was erected by the citizens of Pisa. A picture, seen here,
of the head of the Fibonacci statue, is some fifteen feet above ground. Since
then, the statue has been moved to its present (and original) location at
Camposanto Monumentale (at Piazza dei Miracoli, where the Leaning Tower
stands.)
(b) Binet’s formula for finding the nth term of Fibonacci sequence
The general term of the Fibonacci sequence was derived by
Jacques Binet in 1843, although the result was known to Euler, Daniel
Bernoulli, and de Moivre more than a century earlier.
F(n) =
1
√5
[(
1+√5
2
n
) −(
n
1−√5
2
) ]
There are many different proof for this formula. One way
is to use mathematical induction. Here we are going to use
the concept of diagonalization of matrix to prove this
formula.
(c) Redefine the problem
(F(n+2)
) = (F(n+1)+F(n)
)
F(n+1)
F(n+1)
, where (F(2)
) = (11)
F(1)
=(
1 1 F(n + 1)
)(
)
F(n)
1 0
=(
F(n)
1 1 2
) (
)
F(n − 1)
1 0
= ….
=(
1 1 n F(2)
) (
)
F(1)
1 0
=(
1 1 n 1
) ( )
1 0
1
The problem is reduced to finding
…
(1)
1 1
)
1 0
An , where A = (
3
(d) Diagonalize A
x
1−λ
𝐯 = (y), (A − λI)𝐯 = (
1
1 1
),
1 0
A=(
1
)=0
−λ
Now, (A − λI)𝐯 = 0 has non-zero solution, |A − λI| = 0
1−λ 1
|
|=0
1
−λ
λ2 − λ − 1 = 0
λ1 =
∴
1+√5
2
or λ =
…
(2)
…
(3)
1−√5
2
2
x
1 1 x
) (y) = λ1 (y),
1 0
(
, which are the eigenvalues .
1
1
(
x
1 x
) (y) = λ2 (y)
0
Since the sum of roots of eq (2), λ1 + λ = 1 ,
2
(2) is reduced to
x = λ1 y, x = λ2 y
x1
x2
λ
λ
(y ) = ( 1 ) , (y ) = ( 2 )
1
2
1
1
And
are the corresponding eigenvectors.
The eigenvectors are not normalized, that is, are not unit vectors, but is alright in
working for diagonalization.
x1
P = (y
Take
x2
λ1
y2 ) = ( 1
1
λ
B = P −1 AP = ( 1
1
=λ
1
(
1 −λ2
λ2
).
1
λ2 −1 1
) (
1
1
1 λ1
)(
0 1
1 −λ2 1 1 λ1
)(
)(
1 0 1
−1 λ1
λ2
)
1
λ2
)
1
=
1 −λ2 λ1 + 1 λ2 + 1
(
)(
)
√5 −1
λ1
λ1
λ2
=
λ1 + 1 − λ1 λ2
(
√5 −λ1 − 1 + λ1 2
=
0
λ +2
( 1
)
√5
−λ2 − 2
0
1
1
λ2 + 1 − λ2 2
)
−λ2 − 1 + λ1 λ2
1
, since the product of roots of eq (2) = λ1 λ2 = −1 and λ1 , λ2
Substitute the values of
B=
1
√5
1+√5
(
1+√5
=(
2
0
Note that
λ1 , λ2 ,
0
0
5+√5
1
+2
1−√5
)= ( 2
√5
− 2 −2
0
0
0
0
λ
1−√5) = ( 1
)
λ
0
2
2
2
−
5−√5)
2
are roots of (2).
0
5+√5
= ( 2√5
0
−
5−√5)
2√5
B is a diagonal matrix with eigenvalues as entries in the main diagonal.
4
n
0
).
λ2 n
Bn = (λ1
0
We easily get
Now,
An = (PBP −1 )n = (PBP −1 )(PBP −1 ) … (PBP −1 ) = PBn P −1
=(
λ2 −1
)
1
λ1
1
λ2 λ1 n
)(
1
0
0
λ
)( 1
λ2 n 1
λ1
1
λ2 λ1 n
)(
1
0
0 1 1 −λ2
) (
)
λ2 n √5 −1 λ1
An = (
=
λ n+1
( 1 n
√5
λ1
=
λ n+1 − λ2 n+1
( 1 n
√5
λ1 − λ2 n
−λ1 n+1 λ2 + λ1 λ2 n+1
)
−λ1 n λ2 + λ1 λ2 n
=
λ n+1 − λ2 n+1
( 1 n
√5
λ1 − λ2 n
λ1 n − λ2 n
)
λ1 n−1 − λ2 n−1
1
λ2 n+1
1
n )(
−1
λ2
1
1
−λ2
)
λ1
, since λ1 λ2 = −1 .
Lastly, from (1),
(F(n+2)
)
F(n+1)
=(
=
1 1 n 1
) ( )
1 0
1
λ n+1 − λ2 n+1
( 1 n
√5
λ1 − λ2 n
1
λ1 n − λ2 n
1
n−1
n−1 ) ( )
1
λ1
− λ2
λ n+1 − λ2 n+1 + λ1 n − λ2 n
( 1n
)
n
n−1
√5 λ
− λ2 n−1
1 − λ2 + λ1
λ1 n (λ1 + 1) − λ2 n (λ2 + 1)
1
= ( n−1
)
√5 λ
(λ1 + 1) − λ2 n−1 (λ2 + 1)
1
=
1
=
λ1 n (λ1 2 ) − λ2 n (λ2 2 )
( n−1
)
√5 λ
(λ1 2 ) − λ2 n−1 (λ2 2 )
1
=
λ n+2 − λ2 n+2
( 1 n+1
)
√5 λ
− λ2 n+1
1
1
, since λi 2 − λi − 1 = 0 , i = 1,2 .
1
In the end of our journey, we get our beautiful Binet’s formula for Finonacci sequence:
F(n) =
1
√5
(λ1 n − λ2 n ) =
1
√5
1+√5
[(
2
n
1−√5
) −(
2
n
)
5