Discrete Dynamical Systems
Suppose that A is an n  n matrix and suppose that x 0 is a vector in  n . Then
x 1  Ax 0
n
is a vector in  . Likewise,
x 2  Ax 1
is a vector in  n , and we can in fact generate an infinite sequence of vectors x k  k0 in
 n defined recursively by
x k1  Ax k .
When viewed in this context, we say that the matrix A defines a discrete
dynamical system (which we will also refer to more briefly as a dynamical system)
on  n . For a given initial vector x 0   n , the infinite sequence x k  k0 is called the
positive orbit (or, for our purposes, just the orbit) of the dynamical system A through
the initial vector x 0 . A fundamental problem is to determine the long term behavior of
the orbit through each vector x 0   n . Some particular questions of interest are
1. Does the orbit x k  
k0 approach some “limiting vector” as k  ?

2. If the orbit x k  k0 does not approach some limiting vector as k  , then is
the behavior of this orbit at least in some way predictable?
Example Suppose that A is the matrix
 12  14
A
1
0
and consider the dynamical system
x k1  Ax k .
Describe the long term behaviors of the orbits of this dynamical system through
the vectors
x0 
1
0
, x0 
1
, and x 0 
2
0
2
.
Solution First we study the orbit through
x0 
1
0
by computing the first few iterates:
1
x 1  Ax 0 
x 2  Ax 1 
x 3  Ax 2 
x 4  Ax 3 
 12  14
1
1
0
0

 12  14
 12
1
0
0
 12  14
1
4
1
0
0
 12  14
 18
1
0
0
 12
0

1
4
0
 18

0

1
16
.
0
There is a clear pattern here. It can be seen that
lim x k  0.
k
In fact, it can be see that for each k  0, we have
1 k 
xk 
1
2k
0
.
Next, we study the orbit through
x0 
1
2
by computing the first few iterates:
x 1  Ax 0 
x 2  Ax 1 
x 3  Ax 2 
x 4  Ax 3 
 12  14
1
1
2
0
 12  14
1
1
2
0
 12  14
1
1
2
0
 12  14
1
1
2
0




1
2
1
2
1
2
1
2
.
The pattern of this orbit is clear. It does not approach a limiting vector as
k   but, instead, it alternates back and forth between the two vectors
2
1
1
and
2
.
2
We call this a periodic orbit with period 2. An explicit formula for this orbit is
1 k  1
xk 
1 k  2
.
Finally, we study the orbit through
x0 
0
2
by computing the first few iterates:
x 1  Ax 0 
x 2  Ax 1 
x 3  Ax 2 
x 4  Ax 3 
 12  14
0
1
2
0
 12  14
 12
1
2
0
 12  14
3
4
1
2
0
 12  14
 78
1
2
0
 12

2



3
4
2
 78
2
15
16
2
.
For this orbit, it can be seen that the second component alternates between the
values 2 and 2. The first component alternates between positive and negative
values that are approaching the limiting values of 1 and 1, respectively. It is thus
fair to say that the limiting behavior of the orbit through the vector
x0 
0
2
is essentially the same as the limiting behavior of the orbit through the vector
x0 
1
2
.
With a little thought, we can write an explicit formula for the orbit through
x0 
0
2
.
It is
3
2 k 1
2k
1 k 
xk 
.
1 k  2
In the preceding example, we computed the orbits of a dynamical system, A,
through three specific initial vectors. This computations showed that there are at least
two basic “behaviors” that can be expected from orbits of this dynamical system. In
particular, we saw that some orbits (at least one!) satisfies
lim x k  0
k
and that some orbits approach the periodic orbit
1
2
1

2
as k  .
Are these the only two possible behaviors of orbits of this dynamical system, or
are there other possible behaviors. In order to answer this question, we need to use
an arbitrary initial vector
a
x0 
 n
b
and see if we can tell what happens to it. By performing a few computations, we see
that
x 1  Ax 0 
x 2  Ax 1 
x 3  Ax 2 
x 4  Ax 3 
 12
 14
a
0
1
b
 12
 14
 12 a 
0
1
b
 12
 14
0
1
 12
 14
 18 a 
0
1
b
1
4
 12 a 

a
3
8
1
4
b
b
1
4
b
b
7
16
b
a
3
8
b
b
 18 a 

b
1
4

7
16
b
b
1
16

a
15
32
b
.
b
These computations seem to show that
xk 
1 k 
1
2k
 a  1 k 
1 k  b
1
2

2 k 1
2k
b

1 k 
0
1
2k
1 k 
1
2

1 k
2 k 1
2k
a
b
for all k  0.
Since
4
lim 1 k  1k
k
2
0
and the quantity
k
1 k  1  2 k 1
2
2
alternates between positive and negative values that approach 12 and  12 ,
respectively, we conclude that if if k is very large and even, then
xk 
1
2
a
0 1
b
0
1
2

b
b
 1b
2
1
 1b
2
1
2
and if k is very large and odd, then
xk 
0  12
a
0 1
b
 12 b

b
2
.
Hence, if b  0, then x k alternates between a scalar multiple (whose magnitude
depends on b) of the vectors
1
1
and
2
2
as k  .
However, if b  0, then x k  0 for all large values of k and
lim x k  0.
k
For example, the analysis that we just performed shows that the orbit through the
initial vector
3
x0 
6
approaches the periodic orbit
3
3

6
6
as k  ; whereas the orbit through the initial vector
x0 
18
0
satisfies
lim x k  0.
k
The approach that we have taken in studying the dynamical system defined by
5
A
 12
 14
0
1
was, in some sense, completely successful, because we were able to determine the
long term behaviors of all orbits of this dynamical system. In particular, we determined
that there are essentially only two possible behaviors and we were able to identify
which orbits exhibit which of the behaviors. Clearly, however, the approach that we
used here is not efficient in general. This approach generally does not yield such a
pleasing and complete set of results if the matrix A is of size larger than 2  2 and/or
the matrix A is not triangular. Clearly, we need to make more efficient use of the
powerful theory of linear algebra that we have been developing in this course. The
main ideas that we need are those of eigenvalues, eigenvectors, similarity, and
diagonalization.
Dynamical Systems Defined by Diagonalizable
Matrices
If A is an n  n matrix, then the orbit of a vector x 0   n for the dynamical system
defined by A is
x 1  Ax 0
x 2  Ax 1  AAx 0   A 2 x 0
x 3  Ax 2  AA 2 x 0   A 3 x 0

and in general
xk  Akx0.
Thus, understanding the long term behavior of an orbit depends on understanding
the nature of A k for large values of k.
If A is diagonalizable, then A  PDP 1 where D is an n  n diagonal matrix whose
entries along the main diagonal are the eigenvalues of A and P is a matrix whose
columns are an eigenvector basis for  n (consisting of n linearly independent
eigenvectors of A that correspond to the eigenvalues of A). As we have seen in our
previous work,
A k  PD k P 1
for any positive integer k. Thus, for the orbit through the vector x 0 of the dynamical
system defined by A, we have
x k  PD k P 1 x 0
for all positive integers k.
Example Use the fact that the matrix
6
 12  14
A
1
0
is diagonalizable to compute A k (for any positive integer k). Then obtain an explicit
formula for the orbit of the dynamical system defined by A through an arbitrary
initial vector
a
x0 
.
b
Solution Since A is a triangular matrix, it can be seen that the eigenvalues of A are
   12 and   1. Since these eigenvalues are distinct (and there are two of
them), then A is diagonalizable. Furthermore, it can be seen that an eigenvector
corresponding to    12 is
1
v1 
0
and that an eigenvector corresponding to   1 is
1
v2 
.
2
This means that A  PDP 1 where
D
 12
0
0
1
and
P
P
1

1 1
0 2
1  12
0
1
2
.
Therefore
7
A k  PD k P 1

1 1
 12
0 2
0

 12

 12
k
0
k
k
1  12
0
1 k
1  12
21 k
0
1
2
1
2
1 k
k
 12  12

.
1 k
0
1
2
0
1 k
For the initial vector
a
x0 
,
b
we have
xk  Akx0


k
 12
k
 12  12

1 k
1
2
a
b
1 k
0
k
 12
a   12  12
k

1
2
1 k b
1 k b
a
 12
a
 12
k
b
0
k
0
 12  12
k

1
2
1 k
1 k
 12
 1b
2
k1
 1 k
k
1  2
.
Example In a previous example (in the Section 5.3 notes), we showed that the matrix
1
A
3
3
3 5 3
3
3
1
is diagonalizable. In particular, we showed that A  PDP 1 where
8
1 0
D
0
0 2 0
0 0 2
1 1 1
P
P
1

1 1
0
1
0
1
1
1 1
1
2 1
.
1 1 0
Use this information to determine the long term behavior of the orbit through
the initial vector
1
x0 
0
4
for the dynamical system
x k1  Ax k .
Solution We know that
xk  Akx0
for all positive integers k. Also,
A k  PD k P 1

1 1 1
1
1 1
0 2 k
1

0
0
1
0
0
0
2
k
1 2 k 2 k
1
1 1
1 2 k
1
2 1
1
0
1

0
0
0
2 k
1 1 0
1  2 k
1  2 k
1  2 k 1  22 k 1  2 k
1  2 k
1  2 k
1
1 1
1
2 1
1 1 0
.
1
This gives us
9
xk  Akx0
1  2 k
1

1  2
1  2
k
k
1  22
1  2
1  2 k
k
1  2
k
1
k
0
4
1
3  42 k

3  32 k
.
3  2 k
It can now be seen that the components of x k grow beyond all bounds and
alternate in sign as k  . For example
3  42 10
x 10 
3  32
3  2
x 11 
10
4, 093

10
1, 027
3  42 11
8, 195
3  32 11
6, 147
3  2
11
2, 045
3  42 12
x 12 
3, 069
3  32 12
3  2 12
16, 381

12, 285
.
4, 099
An Introductory Application of Discrete Dynamical
Systems in the Study of Age–Structured Population
Models
Suppose that we have a population of some type of animal that has an average
life span of two years. Suppose also that these animals are capable of reproduction,
on average, when they get to be one year old. Finally, suppose that each of these
animals produces two offspring during its lifetime but that, due to harsh conditions,
only about half of the offspring survive long enough to reach maturity (meaning
reproductive age). What do you think will happen to this animal population over the
long run?
We can write a simple mathematical model of this situation as follows:
Let x k be the juvenile population at time k, and let y k be the adult population at time
k. Then the population changes (approximately) according to the equations
10
x k1  2y k
y k1  1 x k .
2
This mathematical model can be written in matrix form:
x k1

y k1
0 2
xk
1
2
yk
0
or more simply as
x k1  Ax k
where
xk
xk 
yk
is called the age–structured population vector.
Now, to ask a specific question, suppose that we have an initial population
consisting of 200 juveniles and 1000 adults. What will happen to this population over
time?
To answer this question, we observe (via the usual diagonalization process) that
the matrix
0 2
A
1
2
0
is diagonalizable. In fact, after a little work, we can see that
A
0 2
1
2
0
 PDP 1 
2 2
1 0
1
0
1
1
 14
1
4
1
2
1
2
.
The population vector, starting with initial population vector
x0 
200
1000
thus satisfies
11
x k  PD k P 1 x 0




1
1
0
k
 14
1
2
1
2
1
4
1
k
1

2 1
 14 1 k 
1
4
1
21 k 2
1
 14
1 k 0
2 2
1
2
1
4
1
2
1
2
200
1000
200
1000
1 k  1
200
1 k 
1000
1
2
9001 k  1100
4501 k  550
1
2
.
Thus, the juvenile population in any year k  0 will be
x k  900  1 k  1100
and the adult population will be
y k  4501 k  550.
The total (juvenile and adult) population will be
x k  y k  450  1 k  1650.
The total population and the juvenile and adult populations fluctuate from year to
year. This is summarized in the table below
Year Juveniles Adults Total
0
200
1000
1200
1
2000
100
2100
2
200
1000
1200
3
2000
100
2100
4
200
1000
1200
.
Now let us change our model slightly by adding more age structure. In particular, let
us assume that juveniles take two years (instead of one year) to mature. This give us
two juvenile age classes, which we will call x and y, and one adult age class, which we
will call z. Let us further assume that only half of the juveniles in the younger juvenile
age class survive and reach the older juvenile age class, but that all juveniles who
reach the older juvenile age class mature into adults. Our model then becomes
12
x k1  2z k
y k1  1 x k
2
z k1  y k
which can be written in matrix form as
x k1

y k1
z k1
0 0 2
xk
1
2
0 0
yk
0 1 0
zk
or simply as
x k1  Ax k
where
xk
xk 
yk
zk
and
0 0 2
1
2
A
0 0
.
0 1 0
We would like to see whether or not the matrix A is diagonalizable. (If it is, then we
will be happy because we can proceed as before in trying to understand the long term
population dynamics.)
The characteristic polynomial of A is
0
0
2
1
2
0
0
0
1
0
 1  3,
which means that the characteristic equation of A is
1  3  0
and that the only (real) eigenvalue of A is   1. (The other two eigenvalues of A are
imaginary numbers.)
To find the eigenvectors of A that are associate with the eigenvalue   1, we
consider
1
0
2
x1
1
2
1
0
x2
0
1
1
x3
0

0
.
0
13
Since
1
0
2
1
2
1
0
0
1
1
1 0 2
~
,
0 1 1
0 0
0
we see that the general solution of the above matrix equation is
x1
2t

x2
2
t
t
t
x3
.
1
1
Therefore, eig1 is only one–dimensional and we do not have an eigenvector basis.
This means that the matrix A is not diagonalizable.
Fortunately, even though A is not diagonalizable, this model is sufficiently simple
to allow us to determine the long term behavior or populations by direct computation.
For example, suppose that we start with 100 juveniles in each of the younger and
older juvenile age classes and with 1000 adults: That is,
100
x0 
.
100
1000
Then
x 1  Ax 0 
x 2  Ax 1 
x 3  Ax 2 
x 4  Ax 3 
0 0 2
100
1
2
0 0
100
0 1 0
1000
100
0 0 2
2000
200
1
2
2000


50
0 0
50
0 1 0
100
50
0 0 2
200
100
1
2
0 0
1000

1000
100
0 1 0
50
1000
0 0 2
100
2000
1
2
0 0
100
0 1 0
1000

50
.
100
We see that the age distribution in this population repeats itself over and over
again in four year cycles.
14
The basic formulation for an age-structured population model is as follows:
Suppose that juveniles are divided into r age classes, x 1 , x 2 ,  , x r , with corresponding
survival rates s 1 , s 2 ,  , s r , and suppose that adults are divided into p age classes,
y 1 , y 2 ,  , y p , with corresponding fertility rates f 1 , f 2 ,  , f p , and survival rates t 1 , t 2 ,  t p
(where t p  0). Then the model for the population growth is
x 1  k1  f 1 y 1  k  f 2 y 2  k    f p y p  k
x 2  k1  s 1 x 1  k

x r  k1  s r1 x r1  k
y 1  k1  s r x r  k
y 2  k1  t 1 y 1  k

y p  k1  t p1 y p1  k .
In matrix form, this model becomes
x 1  k1
0
0

0
f1
f2

fp
x 1  k
x 2  k1
s1
0

0
0
0
0
0
x 2  k

x r  k1
y 1  k1
y 2  k1

y p  k1
       


0
0

0
0
0
0
0
x r  k
0
0
 sr
0
0
0
0
y 1  k
0
0
0
t1
0
0
0
y 2  k
0
       
0
0
0
0
0
0
t p1
0
.

y p  k
The r  p  r  p matrix used in this model is called a Leslie matrix.
15