- #1

missbooty87

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## 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