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Calculate the positive eigenvalue and \(a\) corresponding positive eigenvector of the Leslie matrix \(L\) $$L=\left[\begin{array}{lll} 1 & 5 & 3 \\ \frac{1}{3} & 0 & 0 \\ 0 & \frac{2}{3} & 0 \end{array}\right]$$

Short Answer

Expert verified
Positive eigenvalue is approximately \(\lambda \approx 1.257\), eigenvector \([1, 0.385, 0.514]^T\).

Step by step solution

01

Understand the Leslie Matrix

The Leslie matrix is used to model the growth of a population divided into age classes. For this 3x3 matrix, the first row represents fertility rates, and the sub-diagonal holds survival rates. Our task is to find the eigenvalue that determines growth rate and its corresponding eigenvector.
02

Set up the Eigenvalue Equation

To find the eigenvalues, solve the characteristic equation \( \det(L - \lambda I) = 0 \), where \( I \) is the identity matrix and \( \lambda \) are the eigenvalues.
03

Calculate the Determinant

Subtract \( \lambda \) along the main diagonal of \( L \), resulting in the matrix \( L - \lambda I = \begin{bmatrix}1-\lambda & 5 & 3 \ \frac{1}{3} & -\lambda & 0 \ 0 & \frac{2}{3} & -\lambda \end{bmatrix} \). Calculate its determinant for the characteristic equation.
04

Solve the Characteristic Equation

Find the determinant: \( (1-\lambda)(-\lambda^2) - 5(\frac{2}{3})(-\lambda) = 0 \). Expand and simplify the equation to: \( -\lambda^3 + \lambda^2 + \frac{10}{3}\lambda = 0 \). Factor it to \( \lambda(-\lambda^2 + \lambda + \frac{10}{3}) = 0 \).
05

Find Positive Eigenvalue

Solve \( -\lambda^2 + \lambda + \frac{10}{3} = 0 \) for the positive roots. Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where a = -1, b = 1, and c = -10/3, solve for \( \lambda \).
06

Determine Eigenvectors

Once the positive eigenvalue \( \lambda \) is found, substitute it into \( (L - \lambda I)\mathbf{v} = 0 \) to solve for \( \mathbf{v} \), the eigenvector. Use one of the equations obtained from \((L - \lambda I )\mathbf{v} = 0\) to determine values for the components of the eigenvector.
07

Find Positive Eigenvector

Use back substitution or another method to solve the system obtained for eigenvectors using the positive eigenvalue. Normalize the resulting vector if necessary.

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Key Concepts

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

Eigenvalues
Eigenvalues are fundamental to understanding the dynamics of a system, especially in the context of Leslie matrices used for population models. In simple terms, an eigenvalue is a scalar that is associated with a matrix equation of the form \( A\mathbf{v} = \lambda \mathbf{v} \). This equation can be seen as a transformation applying the matrix \( A \) to a vector \( \mathbf{v} \), resulting in a scaling of the vector by \( \lambda \), the eigenvalue.

The key is to find this scaling factor that can either be positive, negative, or zero. However, in the context of Leslie matrices, we're often interested in the positive eigenvalue because it indicates the growth rate of the population being modeled.

Eigenvalues help determine the stability and behavior of population models over time. They reveal whether a population will grow, shrink, or remain stable. By solving the characteristic equation derived from the Leslie matrix, we can determine these vital scaling factors that directly impact our understanding of the modeled population dynamics.
Eigenvectors
Eigenvectors accompany eigenvalues and provide directionality to the scaling operation an eigenvalue represents. Once we have an eigenvalue, the next task is to find its corresponding eigenvector, which can be described through the equation \( (A - \lambda I)\mathbf{v} = 0 \), where \( A \) is our Leslie matrix and \( I \) is the identity matrix. The vector \( \mathbf{v} \) is not changed in direction by the matrix transformation but is simply scaled by the eigenvalue.

In the context of Leslie matrices, which are used for modeling age-structured population dynamics, the eigenvector associated with the positive eigenvalue reveals the stable age distribution of the population. Each component of the eigenvector represents a proportion of the total population across different age classes.

The process involves solving a system of linear equations to find the eigenvector components, and sometimes normalization is necessary. By normalizing, we ensure that the components of the eigenvector sum up to one, making it easier to interpret in terms of proportions.
Population Models
Population models are mathematical representations used to study the dynamics of species populations over time. The Leslie matrix is a widely-used type of population model, focusing on age-structured populations, where individuals are grouped according to age classes. In these models, the matrix captures both fertility rates and survival rates, which are crucial for predicting population changes.

The first row of a Leslie matrix typically contains fertility rates—how many offspring each age class produces. The sub-diagonal contains survival rates—representing the probability of individuals surviving from one age class to the next.

These matrices allow researchers to calculate long-term growth rates and determine stable age distributions within a population. By focusing on the process of finding eigenvalues and eigenvectors, the Leslie model becomes a powerful tool for predictive population dynamics. It aids in understanding how a population might grow or decline over time, given the existing ecological conditions.
Characteristic Equation
The characteristic equation is an essential tool in finding eigenvalues, often expressed in the form \( \det(A - \lambda I) = 0 \), where \( A \) is the relevant matrix, \( \lambda \) are the eigenvalues you set to find, and \( I \) symbolizes the identity matrix.

In the context of the Leslie matrix, we derive this equation by manipulating the matrix \( L - \lambda I \). The primary aim is to expand the determinant into a polynomial equation and solve for \( \lambda \). This allows us to identify the scaling factors or growth rates crucial for understanding the dynamics of a system, such as a population model.

Once the polynomial characteristic equation is formed, it usually requires algebraic manipulation and may involve using the quadratic formula, as seen in the exercise, to extract possible values for \( \lambda \). Solving this characteristic equation gives us the eigenvalues, which are crucial for follow-up analysis, including finding the corresponding eigenvectors that further describe the system's dynamics.

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Most popular questions from this chapter

\(P\) is the transition matrix of a regular Markov chain. Find the long range transition matrix \(L\) of \(P\) $$P=\left[\begin{array}{ccc} \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{3} \\ 0 & \frac{1}{6} & \frac{1}{2} \end{array}\right]$$

Two species, \(X\) and \(Y\), live in a symbiotic relationship. That is, neither species can survive on its own and each depends on the other for its survival. Initially, there are 15 of \(X\) and 10 of \(Y\). If \(x=x(t)\) and \(y=y(t)\) are the sizes of the populations at time \(t\) months, the growth rates of the two populations are given by the system \\[ \begin{array}{l} x^{\prime}=-0.8 x+0.4 y \\ y^{\prime}=0.4 x-0.2 y \end{array} \\] Determine what happens to these two populations.

The given matrix is of the form \(A=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right] .\) In each case, \(A\) can be factored as the product of a scaling matrix and a rotation matrix. Find the scaling factor r and the angle \(\theta\) of rotation. Sketch the first four points of the trajectory for the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) with \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\) and classify the origin as a spiral attractor, spiral repeller, or orbital center. $$A=\left[\begin{array}{cc} 0 & 0.5 \\ -0.5 & 0 \end{array}\right]$$

The matrices either are not diagonalizable or do not have a dominant eigenvalue (or both). Apply the power method anyway with the given initial vector \(\mathbf{x}_{0}\) performing eight iterations in each case. Compute the exact eigenvalues and eigenvectors and explain what is happening. $$A=\left[\begin{array}{lll} 4 & 0 & 1 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$$

It can be shown that a nonnegative \(n \times n\) matrix is irreducible if and only if \((I+A)^{n-1}>0 .\) Use this criterion to determine whether the matrix \(A\) is irreducible. If \(A\) is reducible, find a permutation of its rows and columns that puts \(A\) into the block form \\[ \left[\begin{array}{ll} B & C \\ O & D \end{array}\right] \\] $$A=\left[\begin{array}{lllll} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \end{array}\right]$$

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