### Application

##### 1. Discrete dynamical system

We have already seen the use of eigenvalue and eigenvector in the computation of powers of matrix such as

 [ 13 -4 ]k. -4 7

In a general linear discrete dynamical system, we are given an initial vector x0V and a linear transformation T: VV. Then we have a sequence

x0,
x1 = T(x0),
x2 = T(x1) = T2(x0),
x3 = T(x2) = T3(x0),
:
xk = T(xk-1) = Tk(x0),
:

Suppose T is diagonalizable, with a basis v1, v2, ..., vn of eigenvectors and the corresponding eigenvalues λ1, λ2, ..., λn. Then we may express x0 as a linear combination of the basis vectors

x0 = c1v1 + c2v2 + ... + cnvn,

and obtain the formula for xk:

xk = c1Tk(v1) + c2Tk(v2) + ... + cnTk(vn) = c1λ1kv1 + c2λ1kv2 + ... + cnλnkvn.

Example Suppose the three sequences

x1, x2, ..., xk, ...
y1, y2, ..., yk, ...
z1, z2, ..., zk, ...

start with x1 = y1 = z1 = 1 and are related by the recursive relation

 xk+1 = xk + 3yk - 3zk yk+1 = - 3xk + 7yk - 3zk zk+1 = - 6xk + 6yk - 2zk

Then we denote

 xk = [ xk ], A = [ 1 3 -3 ] yk -3 7 -3 zk -6 6 -2

and get xk = Axk-1 = Ak-1x1. In an earlier example, we found a basis of eigenvectors (1, 1, 0), (1, 0, -1), (1, 1, 2) with eigenvalues 4, 4, -2. Since

 x1 = [ 1 ] = (1/2)[ 1 ] + 0[ 1 ] + (1/2)[ 1 ], 1 1 0 1 1 0 -1 2

we conclude that

 xk = Ak-1[ 1 ] = (1/2)4k-1[ 1 ] + (1/2)(-2)k-1[ 1 ] = [ 2×4k-2 + (-1)k-12k-2 ]. 1 1 1 2×4k-2 + (-1)k-12k-2 1 0 2 2×4k-2 + (-1)k-12k-1

In other words, the three sequences are given by

xk = 2×4k-2 + (-1)k-12k-2, yk = 2×4k-2 + (-1)k-12k-2, zk = 2×4k-2 + (-1)k-12k-1.

Some problems may be converted to discrete dynamical systems.

Example The Fibonacci series

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...

is determined by the initial data a0 = a1 = 1 and the recursive relation ak+2 = ak+1 + ak. We would like to find the general formula for ak.

Denote

 xk = [ ak+1 ]. ak

Then we have

 x0 = [ 1 ], xk+1 = [ ak+2 ] = [ ak+1 + ak ] = [ 1 1 ][ ak+1 ] = Axk. 1 ak+1 ak+1 1 0 ak

The characteristic polynomial of the matrix is det(A - λI) = (1 - λ)(- λ) - 1 = -1 - λ + λ2, and A has two eigenvalues

 λ1 = 1 + √5 , λ2 = 1 - √5 . 2 2

We also find the corresponding eigenvectors

 v1 = ( 1 + √5 , 1), v2 = ( 1 - √5 , 1). 2 2

From

 x0 = (1, 1) = (1 + √5)v1 - (1 - √5)v2 , 2√5

we then get

 xk = Akx0 = (1 + √5)λ1kv1 - (1 - √5)λ2kv2 . 2√5

Since ak is the second coordinate of xk, we only need to calculate the second coordinate of the combination on the right:

 ak = (1 + √5)λ1k - (1 - √5)λ2k = 1 [ ( 1 + √5 )k+1 - ( 1 - √5 )k+1 ]. 2√5 √5 2 2