Eigenvalues with a leslie population model

  • #1

Homework Statement


In the Leslie population model, suppose matrix A has a strictly dominant eigenvalue [tex]\lambda_1[/tex]. Age class evolution is given by: [tex]x^{(k)} = Ax^{(k-1)}[/tex]; initial population is x^{(0)}.

(i) Initially, let [tex]x^{(0)}[/tex] be the linear combination [tex]a_1x_1 + a_2x_2 + ... + a_nx_n[/tex], of A's linearly independent eigenvectors, with each constant [tex]a_i \neq 0[/tex]. Find two expressions for the population distribution [tex]x^{(1)}[/tex]after one generation; one using products involving matrix A, the other in terms of A's eigenvalues. Then, repeat this process to find the population distribution in the k-th generation.

(ii) Repeat (i) to find the population distribution [tex]x^{(k)}[/tex]

(iii) Use (ii) to find the consequence of [tex]\lambda_1[/tex] , being the strictly dominant
eigenvalue. For suitably large values of k , what can be said of the
relative sizes of the terms in the expression found in ii. Hint:Consider the limit [tex]lim_k\rightarrow\infty (\frac{x^{(k)}}{\lambda_1 ^{k}})[/tex].

(iv) Explain why the population approaches a stable population division into its classes as k gets large, with stable population growth rate of [tex]\lambda_1[/tex].

(v) If A has no strictly dominant eigenvalue, how will the conclusions differ?

Homework Equations



diagonalization of matrices

The Attempt at a Solution



(i) [tex]x^{(1)} = L^{-1} D L x^{(0)}[/tex]
[tex]x^{(1)} = \lambda_n a_nx_n[/tex]

(ii) [tex]x^{(k)} = \lambda_n ^{(k)} a_nx_n[/tex]

(iii) >>> I'm stuck on this question

(iv) as the [tex]x^k[/tex] the population age classes becomes stable, due to k getting bigger, [tex]\lambda_1[/tex] will become the more significant eigen value, therefore population growth rate approaches [tex] \lambda_1[/tex], which leads to a stable population division which is proportional to its eigenvector.

(v) >>> I'm stuck on this question

I'm not confident about any of my answers, I'd appreciate if you would also comment on my answers as well as giving tips on the questions i'm not sure of. Thank you ever so much :smile:
 

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