Introduction to eigenvalues and eigenvectors:
As we saw in the simple succession model, after a long enough time
period, the fraction of the landscape in each vegetation type
approaches some constant. In the simplest model, all the land moved to
the "climax" state - so the vector that described the landscaope was [0
0 1] for the case of 3 landscape types where the last type is the
climax state. However, when we added disturbance (fire) to the system,
the state of the system again approached some fixed fraction of each
vegetation type, but all types were present. What we want to do now is
define a mathematical method to find these "final" states - tghe jargon
for this is that we are finding an "eigenvector" for the system. This
means a vector that characterizes the system after a long time period
has elapsed.
One way to find an eigenvector is numerically. We can use Matlab to
find for any transition matrix P, P^n, where n is a large number (say
100), then multiply this times the initial vector for the landscape,
x0, to get a numerical answer for the long-term state of the landscape.
If the initial vector x0 contained the fractoion of the landscape in
each vegetation type, then P^100 x0 will be a vector giving the
long-term fraction of the landscape in each vegetation type. If the
initial vector x0 rather represents the number of hectares or acres of
each type, then divide each term in P^100 x0 by the sum of the
components of this vector to get the long-term fraction in each state
(the eigenvector is specified only up to a constant multiple).
A second way to get the eigenvector is to realize that it arises when
the long term structure of the system doesn't change. This is expressed
as P y = lambda y where lambda is a constant that represents how the
vector of vegetation types increases or decreases through one time
period. In our case of a fixed landscape, land area is neither created
nor destroyed, so there is no change from one time period to another
and so lambda = 1. This means to find the eigenvector all we need to do
is find a vector y that satisfies P y = y. this is easy to do using
simple algebra for small matrices, but for larger ones the mathemaics
becomes more difficult. In this case you either use the theory of
determinants, or else use Matlab to find the answer.
Consider the 2x2 matrix P = [ a b; c d ] and lets look at the
equations arising from P y = lambda y where y = [ y1; y2]. this holds
if a y1 + b y2 = lambda y1 and c y1 + d y2 = lambda y2. In order for
these to both hold, we need y1 = (b/(lambda-a)) y2 and
y1 = ((lambda-d)/c) y2. So the only way these can both hold is if y1=y2=0
(not interesting) or if b/(lambda-a) = (lambda-d)/c. This is a
quadratic equation in lambda: lambda^2 - (a+d) lambda + ad - bc = 0
and this equation is called the characteristic equation for this
matrix. We call a+d the Trace of the matrix P ( Tr(P) ) and ad-bc the
Determinant of the matrix P ( Det( P) ). By solving this quadratic for
the roots, we find the eigenvalues lambda (there will be two in general
for a 2x2 matrix). To find the eigenvector, we plug in one of these
lambda values to find the ratio of y1 to y2, and this gives us the
eigenvector up to a constant.
Note that in the case of our succession model where the landscape
doesn't change, we have lambda = 1, and so we must have a+d = ad-bc.
You should check for yourselves that this holds true because the
columns of the matrix P sum to one (so a+d=1 and b+c=1). So then
y1 = (b/(1-a)) y2 = ((1-d)/c) y2 gives the eigenvector. Normalize
this so they sum to 1.