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Chapter 5: Problem 45

Find the eigenvalues and eigenvectors for both of these Markov matrices \(A\) and \(A^{\infty}\) Explain why \(A^{100}\) is close to \(A^{\infty}\) : $$ A=\left[\begin{array}{ll} 6 & .2 \\ 4 & .8 \end{array}\right] \text { and } \quad A^{\infty}=\left[\begin{array}{ll} 1 / 3 & 1 / 3 \\ 2 / 3 & 2 / 3 \end{array}\right] $$

Short Answer

Expert verified
Eigenvalues for A: 6 and 0.8; Eigenvector for A: normalized vectors for each \( \lambda\). For \( A^{\infty} \), it's determined by the dominant eigenvalue 1, showing steady state.

Step by step solution

01

Understand Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are intrinsic properties of a matrix that describe its fundamental characteristics. Eigenvalues are scalars that, when multiplied by their corresponding eigenvectors, transform the matrix without changing its direction.
02

Find Eigenvalues of Matrix A

To find the eigenvalues of matrix \( A = \begin{bmatrix} 6 & 0.2 \ 4 & 0.8 \end{bmatrix} \), solve the characteristic equation \( \det(A - \lambda I) = 0 \). Substituting \( \lambda I = \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} \), the equation becomes:\[\begin{vmatrix} 6-\lambda & 0.2 \ 4 & 0.8-\lambda \end{vmatrix} = (6-\lambda)(0.8-\lambda) - (0.2)(4) = 0\]Simplifying the equation yields the eigenvalues, \( \lambda_1 = 6 \) and \( \lambda_2 = 0.8 \).
03

Find Eigenvectors of Matrix A

For each eigenvalue, \( \lambda_1 \) and \( \lambda_2 \), solve \( (A - \lambda I)\mathbf{v} = 0 \) for eigenvectors \( \mathbf{v} \).1. For \( \lambda_1 = 6 \):\[(A - 6I) = \begin{bmatrix} 0 & 0.2 \ 4 & -5.2 \end{bmatrix}\]Solve \( 0.2x + 0y = 0 \) and obtain the vector \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \), invalid, leading us to reconsider non-trivial solutions.2. For \( \lambda_2 = 0.8 \):\[(A - 0.8I) = \begin{bmatrix} 5.2 & 0.2 \ 4 & 0 \end{bmatrix}\]Setting \( 5.2x + 0.2y = 0 \) and \( 4x = 0 \), solving gives us the eigenvector \( \begin{bmatrix} -0.0385 \ 0.9992 \end{bmatrix} \), normalized.
04

Analyze A^{\infty}

Matrix \( A^{\infty} \) represents the limiting behavior of \( A^n \) as \( n \to \infty \). It shows the steady state of the system represented by \( A \), indicating the eigenvector associated with eigenvalue 1. \( A^{\infty} \) has eigenvalues \( \lambda_1 = 1 \) (dominant, steady state value) and \( \lambda_2 = 0 \), with eigenvectors \( \begin{bmatrix} \frac{1}{3} \ \frac{2}{3} \end{bmatrix} \) and \( \begin{bmatrix} \frac{1}{3} \ \frac{1}{3} \end{bmatrix} \).
05

Compare A^{100} with A^{\infty}

Calculate or conceptualize \( A^{100} \), which approaches \( A^{\infty} \) due to its reliance on its dominant eigenvalue. With small eigenvalues, high matrix powers diminish non-steady state contributions, leaving \( A^{\infty} \) primarily determined by \( \lambda_1 = 1 \). Decreasing influence of other eigenvalues makes \( A^{100} \) close to \( A^{\infty} \).

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Key Concepts

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

Markov Matrices
A Markov matrix, also known as a stochastic matrix, is a square matrix used in probability theory to describe the transitions of a Markov chain. Each of its rows sums to 1 and represents the transition probabilities from one state to another. This means that the matrix describes how probabilities are distributed across different states.
For example, if you have a simple system where a state can move to another state or stay in the same state based on certain probabilities, a Markov matrix captures this behavior.
  • The matrix can be used to represent all possibilities and doesn鈥檛 allow for any probability being greater than 1.
  • It helps in determining the future behavior of a system based on its current state, using repeated multiplication which represents subsequent transitions.
This is because by multiplying the Markov matrix by itself, you can predict where the system's probabilities will end up over time. More specifically, the eigenvectors and eigenvalues of a Markov matrix hold valuable information about the long-term behavior of the system.
Limiting Behavior
The limiting behavior of a Markov matrix refers to what the system does as time goes to infinity, i.e., as you multiply the system's transition matrix by itself numerous times. The intriguing aspect here is that, regardless of the starting point, a regular Markov matrix tends to stabilize in regards to its state transitions.
The matrix tends towards a stable matrix, often denoted as \( A^{\infty} \), which no longer changes with further multiplication by the original matrix.
  • Initially, the matrix might display varied behavior; however, over many iterations, it settles into this predictable pattern.
  • This behavior is tremendously reliant on its dominant eigenvalue, which for a Markov matrix in regular conditions is 1, dictating the system's steady state.
  • Higher powers of the matrix quash the contribution of other eigenvalues (usually less than 1), leading to exponential decay of those components.
Ultimately, it captures the idea that the system finds equilibrium, which provides critical insights into steady states.
Steady State System
In the realm of Markov matrices, the steady state of a system refers to a configuration where probabilities remain constant during transitions, rendering the system fully predictable.Resolved through eigenanalysis, it鈥檚 primarily revealed as the eigenvector associated with the eigenvalue of 1.
The presence of such a steady state suggests that no matter the initial distribution, the system will converge to this fixed state, given enough time.
  • This steady state provides a snapshot of the long-term behavior of the system, unaffected by initial disturbances or fluctuations.
  • Such behavior is typically desired in models that predict behavior over the long term, such as population models or financial predictions.
In practice, finding this steady state means computing powers of the system matrix until changes cease, but recognizing the pivotal eigenvectors directly offers a precise understanding without exhaustive computation.

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Most popular questions from this chapter

Write one significant fact about the eigenvalues of each of the following. (a) A real symmetric matrix. (b) A stable matrix: all solutions to \(d u / d t=A u\) approach zero. (c) An orthogonal matrix. (d) A Markov matrix. (e) A defective matrix (nondiagonalizable). (f) A singular matrix.

A door is opened between rooms that hold \(v(0)=30\) people and \(w(0)=10\) people. The movement between rooms is proportional to the difference \(v-w\) : $$ \frac{d v}{d t}=w-v \quad \text { and } \quad \frac{d w}{d t}=v-w $$ Show that the total \(v+w\) is constant ( 40 people). Find the matrix in \(d u / d t=A u\), and its eigenvalues and eigenvectors. What are \(v\) and \(w\) at \(t=1 ?\)

The powers \(A^{k}\) approach zero if all \(\left|\lambda_{i}\right|<1\), and they blow up if any \(\left|\lambda_{i}\right|>1\). Peter Lax gives four striking examples in his book Linear Algebra. $$ \begin{array}{ll} A=\left[\begin{array}{ll} 3 & 2 \\ 1 & 4 \end{array}\right] \quad B=\left[\begin{array}{rr} 3 & 2 \\ -5 & -3 \end{array}\right] \quad C=\left[\begin{array}{rr} 5 & 7 \\ -3 & -4 \end{array}\right] \quad D=\left[\begin{array}{rr} 5 & 6.9 \\ -3 & -4 \end{array}\right] \\ \left\|A^{1024}\right\|>10^{700} \quad B^{1024}=I \quad C^{1024}=-C \quad\left\|D^{1024}\right\|<10^{-78} \end{array} $$ Find the eigenvalues \(\lambda=e^{i \theta}\) of \(B\) and \(C\) to show that \(B^{4}=I\) and \(C^{3}=-1\).

For \(A=\left[\begin{array}{ll}0 & 2 \\ 0 & 5\end{array}\right]\), find the powers \(A^{k}\) (including \(A^{0}\) ) and show explicitly that their sum agrees with \((I-A)^{-1}\).

Suppose \(A\) and \(B\) have the same full set of eigenvectors, so that \(A=S \Lambda_{1} S^{-1}\) and \(B=S \Lambda_{2} S^{-1}\). Prove that \(A B=B A\).
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