Biology Reference
In-Depth Information
across stage classes. Suppose that
A
is a projection matrix that meets the assumptions
of the Perron-Frobenius theorem and that
v
a
is any vector. We can describe the vector
v
a
using the eigenvectors that form a
basis
for this vector space. A set of vectors
is a basis for a vector space V provided any vector in V can be written as a linear
combination of
{
v
1
,
v
2
,...,
v
n
}
in exactly one way. Suppose that
{
v
1
,
v
2
,...,
v
n
}
is
a basis of eigenvectors and let
λ
i
denote the eigenvalue of
v
i
. Thus,
v
a
=
c
1
v
1
+
c
2
v
2
+···+
c
n
v
n
,
where
c
1
,
c
n
are real-valued constants.
Multiplying by the matrix
A
from the left, we get:
c
2
,...,
Av
a
=
(
c
1
v
1
+
c
2
v
2
+···+
c
n
v
n
)
=
c
1
Av
1
+
c
2
Av
2
+···+
c
n
Av
n
.
A
Because
v
i
is an eigenvector of
A
and
Av
i
=
λ
i
v
i
,wehave:
c
n
λ
n
v
n
.
Recall fromEq. (
7.7
) that we can raise
A
to the
k
th
power to determine the distribution
of individuals across stages after units of time. Similarly,
c
1
Av
1
+
c
2
Av
2
+···+
c
n
Av
n
=
c
1
λ
1
v
1
+
c
2
λ
2
v
2
+···+
A
k
v
a
=
k
k
k
c
1
λ
1
v
1
+
c
2
λ
2
v
2
+···+
c
n
λ
n
v
n
.
(7.9)
Now we need to determine what happens when
k
goes to infinity. Since the
dominant
eigenvalue
is greater than the absolute value of each of the other eigenvalues, dividing
another eigenvalue by the dominant eigenvalue will always generate a value less
than one. If we raise that value to the
k
th
power, we expect it to approach zero as
k
approaches infinity. Expressed mathematically, we mean:
Since
|
(
λ
i
(
λ
i
k
λ
1
>
|
λ
i
|
for
i
>
1, we have
λ
1
)
|
<
1for
i
>
1 and so
λ
1
)
→
0as
k
.
Therefore, if we divide both sides of Eq. (
7.9
)by
→∞
λ
1
k
, we get
c
2
λ
2
λ
1
k
c
n
λ
n
λ
1
k
1
λ
1
k
A
k
v
a
=
c
1
v
1
+
v
2
+···+
v
n
.
When we take the limit as
k
, all terms except the one containing the dominant
eigenvector will give a limit of zero. Thus,
→∞
1
λ
A
k
v
a
=
lim
k
→∞
c
1
v
1
.
(7.10)
k
1
of the number of individuals at each of the stages at
time
t
0
,Eq.(
7.10
) shows that the eigenvector associated with the dominant eigenvalue
describes a long-range, and stable, distribution of individuals across stages. More
specifically, if
Applied to the vector
n
(
t
0
)
λ
1
is the dominant eigenvalue for
n
(
t
0
)
, combining Eqs. (
7.7
) and
(
7.10
) yields
n
(
t
0
+
k
)
1
λ
A
k
n
lim
=
lim
(
t
0
)
=
c
1
v
1
.
(7.11)
k
1
k
1
λ
k
→∞
k
→∞
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