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Chapter 8: Problem 31

The Leslie matrix for an age-structured population is given by $$\left[ \begin{array}{lll}{1} & {2} & {4} \\ {\frac{1}{2}} & {0} & {0} \\\ {0} & {\frac{1}{3}} {0}\end{array}\right]$$ Find its characteristic polynomial.

Short Answer

Expert verified
The characteristic polynomial is \(-\lambda^3 + \lambda^2 + \lambda + \frac{2}{3} = 0\).

Step by step solution

01

Understand the Leslie Matrix

The Leslie matrix \( L \) for an age-structured model is a matrix used in population dynamics. It consists of parameters that describe birth rates and survival rates across different age classes. In this exercise, the given Leslie matrix is: \[\left[\begin{array}{ccc} 1 & 2 & 4 \ \frac{1}{2} & 0 & 0 \ 0 & \frac{1}{3} & 0 \end{array}\right]\]
02

Write the General Formula for Characteristic Polynomial

The characteristic polynomial of a matrix \( L \) is given by the determinant of \( L - \lambda I \), where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix of the same dimension as \( L \).
03

Set Up the Matrix \(L - \lambda I\)

Subtract \( \lambda \) times the identity matrix from the Leslie matrix: \[L - \lambda I = \left[ \begin{array}{ccc} 1-\lambda & 2 & 4 \ \frac{1}{2} & -\lambda & 0 \ 0 & \frac{1}{3} & -\lambda \end{array} \right]\]
04

Calculate the Determinant

To find the characteristic polynomial, calculate the determinant of the matrix \( L - \lambda I \). Use the formula for a 3x3 matrix determinant:\[\det(A) = a(ei-fh) - b(di-fg) + c(dh-eg) \] Applying it here gives:\[(1-\lambda)\left((-\lambda)(-\lambda) - (0)\left(\frac{1}{3}\right)\right) - 2\left(\frac{1}{2}(-\lambda) - (0)(0)\right) + 4\left(\frac{1}{2}\frac{1}{3} - (0)\right)\]Simplifying, we find:\[(1-\lambda)(\lambda^2) + \lambda + \frac{2}{3}\]
05

Simplify the Determinant Expression

Continue to simplify the expression:\[(1-\lambda)\lambda^2 + \lambda + \frac{2}{3} = \lambda^2 - \lambda^3 + \lambda + \frac{2}{3}\]Combine terms:\[= -\lambda^3 + \lambda^2 + \lambda + \frac{2}{3}\]
06

Write the Characteristic Polynomial

The characteristic polynomial of the matrix is obtained from the simplified determinant expression:\[-\lambda^3 + \lambda^2 + \lambda + \frac{2}{3} = 0\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Polynomial
The characteristic polynomial is a foundational concept in linear algebra, especially when working with matrices in models, like the Leslie matrix. It helps us understand how the matrix behaves in terms of eigenvalues and plays a critical role in identifying stability and long-term behavior of systems.
The characteristic polynomial comes from the equation \(L - \lambda I\), where \(L\) is our matrix, \(\lambda\) represents an eigenvalue, and \(I\) is the identity matrix.
By subtracting \(\lambda I\) from \(L\), we create a new matrix whose determinant we need to set to zero to find the polynomial.
  • The determinant of the resulting matrix gives us a polynomial function of \(\lambda\).
  • This is the characteristic polynomial, giving us important information like potential eigenvalues when factored.
Eigenvalues
Eigenvalues are deeply intertwined with the characteristic polynomial and matrix theory. In an age-structured model like the Leslie matrix, eigenvalues give insights into the dynamics of the population over time.
After finding the characteristic polynomial, we solve for \(\lambda\) when the polynomial is set to zero.
These solutions are the eigenvalues of the matrix.
  • Each eigenvalue provides insights into the scaling factors of the matrix vectors.
  • Eigenvalues can indicate growth or decline trends in population structures over time.
Understanding eigenvalues helps predict the changes in population size and structure.
Age-Structured Population
An age-structured population deals with understanding how different age cohorts within a population contribute to overall growth and sustainability. The Leslie matrix is a popular tool for such analysis.
It reflects real-world conditions by incorporating birth rates and survival rates into different segments of a population.
  • Top row of the Leslie matrix: birth rates for each age group.
  • Subsequent rows: survival rates from one age group to the next.
This structured approach helps to formulate strategies for managing natural resources, conserving wildlife, or even making economic predictions, by understanding age-specific contributions.
Population Dynamics
Population dynamics is the study of how and why populations change over time. It involves exploring factors that contribute to growth, decline, or stability within a population. The Leslie matrix, combined with characteristic polynomials and eigenvalues, aids in analyzing these dynamics effectively.
Using the Leslie matrix, researchers can gain insights into:
  • The growth rate of the population and its subgroups.
  • Potential long-term behavior or trends, such as equilibrium or extinction of certain age groups.
  • Effects of different birth and survival rates on overall population health.
By studying population dynamics, one can make predictions and form decisions related to conservation biology, resource management, or public policy.

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Most popular questions from this chapter

Darwin's finches have been used to study how differences in bird morphology are related to differences in diet. Morphological measurements (in mm) of three species are given in the table for three traits. $$\begin{array}{|c|c|c|c|c|}\hline \text { Species } & {\text { Wing length }} & {\text { Tarsus length }} & {\text { Beak length }} \\ \hline G . \text { difficilis } & {64} & {18.1} & {9.6} \\ {G . \text { fulliginosa }} & {62.1} & {17.9} & {9.6} \\ {\text {G. scandens}} & {73.1} & {21.1} & {14.5} \\\ \hline\end{array}$$ The proportion of time spent feeding on different types of food for these three species is given in the following table. $$\begin{array}{|c|c|c|c|}\hline \text { Species } & {\text { Seeds }} & {\text { Pollen }} & {\text { Other }} \\ \hline G . \text { difficilis } & {0.67} & {0.23} & {0.1} \\ {G . \text { fuliginosa }} & {0.7} & {0.28} & {0.02} \\ \hline G . \text { scandens } & {0.14} & {0} & {0.86} \\\ \hline\end{array}$$ (a) Thinking of the morphology of each species as a point in \(\mathbb{R}^{3},\) calculate the morphological distance between each pair of species. (b) Thinking of the diet of each species as a point in \(\mathbb{R}^{3}\) , calculate the diet distance between each pair of species. (c) Do species that are morphologically most similar also tend to have the most similar diets?

The change in population size of a species with juvenile and adult individuals is described by the matrix $$\left[ \begin{array}{cc}{0} & {2} \\ {\frac{1}{2}} & {\frac{1}{3}}\end{array}\right]$$ Find its eigenvalues and associated eigenvectors.

Methylation In Exercise 8.5 .20 we modeled DNA methylation with the recursion \(\mathbf{x}_{t+1}=A \mathbf{x}_{l},\) where $$A=\left[ \begin{array}{ll}{\frac{8}{10}} & {\frac{1}{2}} \\ {\frac{2}{10}} & {\frac{1}{2}}\end{array}\right]$$ (a) Calculate the eigenvalues of \(A\) . (b) The Perron-Frobenius Theorem can clearly be applied to the matrix A. Given your answer to part (a), what is the long-term behavior of \(\mathbf{x}_{r} ?\) (c) Using the initial condition \(\mathbf{x}_{0}=\left[ \begin{array}{c}{a} \\\ {1-a}\end{array}\right],\) express the solution to the recursion in terms of the eigenvectors and eigenvalues of \(A .\) (d) Verify your answer to part (b) using the solution you obtained in part (c).

Prove the Cauchy-Schwarz Inequality: $$|\mathbf{a} \cdot \mathbf{b}| \leq|\mathbf{a}||\mathbf{b}|$$

\(\begin{array}{c}{\text { Leslie matrices Consider the following model for the }} \\ {\text { population size } \mathbf{n}_{t} \text { of an age-structured population with two }} \\ {\text { age classes: }} \\\ {\mathbf{n}_{t+1}=\left[ \begin{array}{cc}{b} & {2} \\ {\frac{1}{2}} & {0}\end{array}\right] \mathbf{n}_{t}}\end{array}\) \(\begin{array}{c}{\text { An equilibrium is a value of the vector for which no change }} \\ {\text { occurs (that is, } \mathbf{n}_{t+1}=\mathbf{n}_{1} ) . \text { Denoting such values by } \hat{\mathbf{n}}, \text { they }} \\ {\text { must therefore satisfy the equation }} \\ {\hat{\mathbf{n}}=\left[ \begin{array}{cc}{b} & {2} \\ {\frac{1}{2}} & {0}\end{array}\right] \hat{\mathbf{n}}}\end{array}\) \(\begin{array}{l}{\text { (a) Suppose that } b \neq 0 . \text { Find all possible equilibrium }} \\ {\text { values. }} \\ {\text { (b) Suppose that } b=0 . \text { Find all possible equilibrium }} \\ {\text { values. }}\end{array}\)
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