Summary of week 5
This week we began class by reviewing the talking points from the week before. We revisited
determinants and saw that an even permutation occurs when it takes 2, 4, 6… ‘moves’ to get to the new
permutation and an odd permutation occurs when it takes 1, 3, 5… ‘moves’ to get to the new
permutation. Also, the reason why the determinant of a diagonal matrix consists of multiplying across
the diagonal, is because when you start choosing one element in each row and one element in each
column you inevitably pick up at least one zero except along the diagonal. The last important point from
the review of the talking points is that not every matrix has a dominant eigenvalue, and in order to have
a dominant eigenvalue you must have one eigenvalue greater in magnitude than all the others.
After we finished our discussion of the talking points we moved onto discrete population models
where we continued looking at the Fibonacci model created in the previous class where after three
months the rabbit no longer contributes to the population and is no longer counted as part of the
population either. The proposal was to calculate the population growth rate without using linear
algebra. To do this we considered the fact that the current age distribution at the beginning of the
observation time would be those listed in the table to the left below and the age distribution for the
next month would be represented by the equations listed in the table to the right below.
Current age distribution
newborns
a
1 month olds
b
2 month olds
c
3 month olds
d
Projected age distribution for next month
Newborns
𝑏 + 𝑐 + 𝑑 = 𝑎
1 month olds
𝑎 = 𝑏
2 month olds
𝑏 = 𝑐
3 month olds
𝑐 = 𝑑
Now the reason why lambda is in all of the equations in the projected age distribution is because every
age category grows by the same factor each month, which is represented by the lambda in each
equation. The next step we took was to put each of the variables, a, b, and c, and put them in terms of d
and lambda where we get 𝑎 = 3 , 𝑏 = 2 , 𝑎𝑛𝑑 𝑐 =  using substitution and when you substitute all of
that back into the equation for newborns for the next month you get that (4 − 2 −  − 1)𝑑 = 0.
Since allowing d to be equal to zero would imply that a, b, c are also equal to zero we say that 𝑑 ≠
0 implying that 4 − 2 −  − 1 = 0. Next, we defined 𝑝() = 4 − 2 −  − 1 graphed it and saw
that one of the roots of p() is 1.465571… and using this fact along with the fact that the percentages of
the age distribution for any given month must equal 1, we can set 𝑎 + 𝑏 + 𝑐 + 𝑑 = (3 + 2 +  +
1)𝑑 = 1 which tells us that 𝑑 ≈ 0.1288. Finally, using 𝑑 ≈ 0.1288 and  = 1.465571 … we calculated
𝑎 ≈ 0.4056, 𝑏 ≈ 0.2764, 𝑎𝑛𝑑 𝑐 ≈ 0.1888. It turned out that when we used the power method and
linear algebra to solve the same problem after defining  to be the dominant eigenvalue our
. 4056
. 2764
4
2
characteristic polynomial was 𝑝() =  −  −  − 1 and our eigenvector ended up being [
] or
. 1888
. 1288
in other words the exact same solution we found when we solved the equation not using linear algebra.
We ended the class with the introduction of Leslie models, named after Patrick Leslie, these
matrices are more complex population models that take into account birth rates, bk, survival rates, sk,
and fertility rates where only the females count. The reason for focusing only on the females is based
on the fact that one man with many wives can have several children born at the same time but one
woman with several husbands can only have one child at a time. Based on our previous population
model it is easy to see that
Newborns 𝑥0 𝑡 equal
The next cycle in the population 𝑥1 𝑡 equals
The next cycle after that 𝑥2 𝑡 equals
⋮
The final cycle in the population 𝑥𝑛 𝑡 equals
0
𝑠1
Which in matrix form is 0
0
0
[0
population distribution.
𝑏1
0
𝑠2
0
0
0
𝑏2
0
0
𝑠3
0
0
𝑏3
0
0
0
⋱
0
𝑏1 𝑥1 + 𝑏2 𝑥2 + ⋯ + 𝑏𝑛 𝑥𝑛
𝑠1 𝑥0
𝑠2 𝑥1
⋮
𝑠𝑛 𝑥𝑛−1
⋯
0
0
0
0
𝑠𝑛
𝑏𝑛
0
0 which is an example of a Leslie matrix for a
0
0
0 ]