A First Course in Linear Algebra, 2017a

378 Spectral Theory
Eigenvalues of Markov Matrices
The following is an important proposition.
Proposition 7.39: Eigenvalues of a Migration Matrix
Let A = [ ]
a ij be a migration matrix. Then 1 is always an eigenvalue for A.
Proof. Remember that the determinant of a matrix always equals that of its transpose. Therefore,
det(xI − A)=det
((xI ) − A) T = det ( xI − A T )
because I T = I. Thus the characteristic equation for A is the same as the characteristic equation for A T .
Consequently, A and A T have the same eigenvalues. We will show that 1 is an eigenvalue for A T and then
it will follow that 1 is an eigenvalue for A.
Remember that for a migration matrix, ∑ i a ij = 1. Therefore, if A T = [ ]
b ij with bij = a ji , it follows
that
Therefore, from matrix multiplication,
⎡
A T ⎢
⎣
1.
1
∑
j
⎤
⎥
⎦ =
b ij = ∑a ji = 1
j
⎡
⎢
⎣
⎤ ⎡
∑ j b ij
⎥ ⎢
. ⎦ = ⎣
∑ j b ij
⎡ ⎤
⎢ ⎥
Notice that this shows that ⎣
1. ⎦ is an eigenvector for A T corresponding to the eigenvalue, λ = 1. As
1
explained above, this shows that λ = 1isaneigenvalueforA because A and A T have the same eigenvalues.
♠
1.
1
⎤
⎥
⎦
7.3.4 Dynamical Systems
The migration matrices discussed above give an example of a discrete dynamical system. We call them
discrete because they involve discrete values taken at a sequence of points rather than on a continuous
interval of time.
An example of a situation which can be studied in this way is a predator prey model. Consider the
following model where x is the number of prey and y the number of predators in a certain area at a certain
time. These are functions of n ∈ N where n = 1,2,··· are the ends of intervals of time which may be of
interest in the problem. In other words, x(n) is the number of prey at the end of the n th interval of time.
An example of this situation may be modeled by the following equation
[ ] [ ][ ]
x(n + 1) 2 −3 x(n)
=
y(n + 1) 1 4 y(n)