/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Use the result from Exercise 23 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 24

Use the result from Exercise 23 to show that if \(\lambda\) is an eigenvalue of a stochastic matrix, then \(|\lambda| \leq 1\)

Short Answer

Expert verified
Using the result from Exercise 23, which states that "If 渭 is an eigenvalue of a stochastic matrix A, then 渭 = 1 or |渭| < 1," we can prove that if 位 is an eigenvalue of a stochastic matrix, then \(|\lambda| \leq 1\). Let x be an eigenvector of a stochastic matrix A with eigenvalue 位 such that \(Ax = \lambda x\). We can assume the sum of the elements of the eigenvector x is equal to 1, i.e., \(\sum_{i=1}^{n} x_i = 1\). It follows that: \(1 = \sum_{j=1}^{n} (Ax)_j = \sum_{j=1}^{n} \lambda x_j \Rightarrow \lambda = 1\) Thus, if 位 is an eigenvalue of a stochastic matrix, then \(|\lambda| \leq 1\).

Step by step solution

01

1. Definition of a Stochastic Matrix

A stochastic matrix is a square matrix where each of its columns sums up to 1. More formally, let A be an n x n matrix. A is a stochastic matrix if: 1. All its elements are non-negative i.e., \(A_{ij} \geq 0\) for all i, j. 2. The sum of each column is one i.e., \(\sum_{i=1}^{n} A_{ij} = 1\) for all j.
02

2. Reminder of Eigenvalues and Eigenvectors

Let A be an n x n matrix. 位 is an eigenvalue of A if there exists a non-zero vector x, called the eigenvector, such that \(Ax = \lambda x\).
03

3. Consider an Eigenvector x of A with Eigenvalue 位

Let x be an eigenvector of A with eigenvalue 位 such that \(Ax = \lambda x\). Now, note that for any scalar c, \(A(cx) = \lambda (cx)\). So, we can assume that the sum of the elements of the eigenvector x is equal to 1, i.e., \(\sum_{i=1}^{n} x_i = 1\). This is because if it doesn't, we can always scale the eigenvector so that its sum is equal to 1.
04

4. Compute the absolute value of 位

To compute the absolute value of 位, observe that: 1. \(Ax = \lambda x\), so the j-th component of Ax is given by: \( (Ax)_j = \lambda x_j\). 2. Summing over all j from 1 to n, we have \(\sum_{j=1}^{n} (Ax)_j = \sum_{j=1}^{n} \lambda x_j\). Now, we compute the left side of the equation: \(\sum_{j=1}^{n} (Ax)_j = \sum_{j=1}^{n} \sum_{i=1}^{n} A_{ij} x_i\) As A is a stochastic matrix, the sum of the elements in each column is 1, hence we have that: \(\sum_{j=1}^{n} \sum_{i=1}^{n} A_{ij} x_i = \sum_{i=1}^{n} x_i = 1\) Now, we compute the right side of the equation: \(\sum_{j=1}^{n} \lambda x_j = \lambda \sum_{j=1}^{n} x_j = \lambda\) Equating the left and right sides, we have the equation: \(\lambda = 1\)
05

5. Conclusion

Since the absolute value of 位 is equal to 1, we can conclude that if 位 is an eigenvalue of a stochastic matrix, then \(|\lambda| \leq 1\). This completes the proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Eigenvalue
In the world of linear algebra, an eigenvalue (\( \lambda \) ) is a special number associated with a matrix. When a matrix acts on a vector, the eigenvalue tells us how much the vector is scaled during that transformation. Imagine you have a matrix \( A \) and a vector \( x \), when you multiply them together (\( Ax \)), the result is often different from merely scaling \( x \). But, if there exists a scalar \( \lambda \) such that \( Ax = \lambda x \), then \( \lambda \) is called an eigenvalue of \( A \).Key aspects to remember:
  • An eigenvalue can be zero, positive, or negative.
  • It can also be complex, having both real and imaginary parts.
  • In special matrices, like stochastic matrices, eigenvalues have certain constraints, such as being within the unit circle (鈮� 1 in absolute value).
Eigenvector
Eigenvectors are fascinating objects in linear algebra. They help us understand the effect of multiplying a matrix by a vector. An eigenvector of a matrix \( A \) goes through a transformation without changing its direction, although its magnitude might be scaled.Suppose you have the equation \( Ax = \lambda x \), where \( \lambda \) is an eigenvalue. Here, \( x \) represents an eigenvector. This vector maintains its course even as you apply the matrix transformation.Characteristics of eigenvectors:
  • They are not unique; scalar multiples are also eigenvectors.
  • They reveal important structural properties of a matrix.
  • In practice, eigenvectors are often normalized for convenience.
Normalize here means scaling the vector so that its length is 1.
Linear Algebra
Linear algebra is the study of vectors, vector spaces, and linear transformations. It's a fundamental area of mathematics, critical in engineering, physics, computer science, and many other fields. At its core, linear algebra deals with things like matrices and operations you can perform on them. Concepts that are important in linear algebra:
  • Vectors: These are quantities with both magnitude and direction.
  • Matrices: Rectangular arrays of numbers that can represent linear transformations.
  • Determinants: Scalar values that provide information about the matrix, such as whether it is invertible.
  • Rank: The dimension of the vector space spanned by its rows or columns.
Linear algebra helps solve systems of linear equations, which have wide applications from simple problems to complex models.
Proof by Induction
Proof by induction is a mathematical technique used to prove statements about integers, particularly those that are true for all positive integers. It consists of two main steps: the base case and the inductive step.Here's an outline of the process:
  • Base Case: Start by proving the statement is true for an initial value, usually \( n = 1 \).
  • Inductive Step: Assume the statement is true for some arbitrary positive integer \( k \). Then, prove it is also true for \( k + 1 \).
Once these steps are successfully completed, the statement is considered proven for all natural numbers. This method is particularly useful when directly proving a statement for all elements is challenging, as it builds on established truths to create a logical progression.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \\[ A=\left(\begin{array}{rr} 1 & 2 \\ -1 & -2 \end{array}\right) \quad \text { and } \quad \mathbf{b}=\left(\begin{array}{r} 6 \\ -4 \end{array}\right) \\] (a) Compute the singular value decomposition of \(A\) and use it to determine \(A^{+}\) (b) Use \(A^{+}\) to find a least squares solution to the system \(A \mathbf{x}=\mathbf{b}\) (c) Find all solutions to the least squares problem \(A \mathbf{x}=\mathbf{b}\)

Let \\[ A=\left(\begin{array}{lll} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{array}\right) \quad \text { and } \quad \mathbf{u}_{0}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) \\] (a) Apply the power method to \(A\) to compute \(\mathbf{v}_{1}\) \(\mathbf{u}_{1}, \mathbf{v}_{2}, \mathbf{u}_{2},\) and \(\mathbf{v}_{3} .\) (Round off to two decimal places.) (b) Determine an approximation \(\lambda_{1}^{\prime}\) to the largest eigenvalue of \(A\) from the coordinates of \(\mathbf{v}_{3}\) Determine the exact value of \(\lambda_{1}\) and compare it with \(\lambda_{1}^{\prime} .\) What is the relative error?

Let \(A\) be a nonsingular \(n \times n\) matrix, and let \(\|\cdot\|_{M}\) denote a matrix norm that is compatible with some vector norm on \(\mathbb{R}^{n}\). Show that \\[ \operatorname{cond}_{M}(A) \geq 1 \\]

Let \\[ A=\left(\begin{array}{rr} 1 & 2 \\ 1 & 3 \\ 1 & 2 \\ 1 & -1 \end{array}\right), \quad \mathbf{b}=\left(\begin{array}{r} -3 \\ 10 \\ 3 \\ 6 \end{array}\right) \\] (a) Use Householder transformations to reduce \(A\) to the form \\[ \left(\begin{array}{c} R_{1} \\ O \end{array}\right)=\left(\begin{array}{cc} \times & \times \\ 0 & \times \\ 0 & 0 \\ 0 & 0 \end{array}\right) \\] and apply the same transformations to \(\mathbf{b}\) (b) Use the results from part (a) to find the least squares solution of \(A \mathbf{x}=\mathbf{b}\)

Let \(I\) denote the \(n \times n\) identity matrix. Determine the values of \(\|I\|_{1},\|I\|_{\infty},\) and \(\|I\|_{F}\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.