/*! 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 11 Let \(A_{1}=\left[\begin{array}{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(A_{1}=\left[\begin{array}{cc}1 & 0 \\ \frac{1}{4} & \frac{1}{2}\end{array}\right]\) and \(A_{2}=\left[\begin{array}{cc}\frac{1}{2} & 0 \\ 16 & \frac{1}{2}\end{array}\right] .\) Show that \(A_{1}\) is not convergent, but \(A_{2}\) is convergent.

Short Answer

Expert verified
After calculating the spectral radii for both matrices, it can be shown that \(A_{1}\) is not convergent while \(A_{2}\) is convergent.

Step by step solution

01

Find the eigenvalues for \(A_{1}\)

First, you need to calculate the eigenvalues for the matrix \(A_{1}\). This can be done by solving the characteristic equation, which is given by \(det(A - \lambda I) = 0\). Here, \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues.
02

Evaluate the spectral radius for \(A_{1}\)

Next, calculate the spectral radius of \(A_{1}\). The spectral radius is the absolute value of the largest (in magnitude) eigenvalue. Now check whether the spectral radius is less than 1 or not.
03

Make a conclusion for \(A_{1}\)

Based on the spectral radius, if it is less than 1, conclude that \(A_{1}\) is convergent. Otherwise, it is not convergent.
04

Find the eigenvalues for \(A_{2}\)

Next, repeat the procedure for \(A_{2}\), that is calculate the eigenvalues by solving the characteristic equation \(det(A - \lambda I) = 0\).
05

Evaluate the spectral radius for \(A_{2}\)

Calculate the spectral radius of \(A_{2}\), which is the absolute value of the largest (in magnitude) eigenvalue. Check whether the spectral radius is less than 1 or not.
06

Make a conclusion for \(A_{2}\)

Finally, based on the spectral radius, if it is less than 1, conclude that \(A_{2}\) is convergent. Otherwise, it is not convergent.

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 91Ó°ÊÓ!

Key Concepts

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

Eigenvalues
Understanding the concept of eigenvalues is crucial in various fields of mathematics and engineering, especially when dealing with matrices. To put it simply, an eigenvalue of a square matrix is a value that, when multiplied by a special vector called an eigenvector, yields the same outcome as when the original matrix multiplies that vector.

Let's consider matrix
\(A_{1}=\left[\begin{array}{cc}1 & 0 \ \frac{1}{4} & \frac{1}{2}\end{array}\right]\), finding its eigenvalues means we are looking for scalars \(\lambda\) where \(A_{1}x = \lambda x\) for some vector \(x\). The characteristic equation, which is based on the determinant of \(A_{1} - \lambda I\), where \(I\) is the identity matrix, helps in finding these special values.

We usually solve the equation \(det(A_{1} - \lambda I) = 0\) to get the eigenvalues. For the matrix \(A_{1}\), the solution leads to specific eigenvalues which will further be used to determine matrix convergence. The significance of eigenvalues extends beyond this application, affecting system stability in differential equations, transforming figures in computer graphics, and even in quantum mechanics.
Spectral Radius
The spectral radius provides substantial insight into the behavior of a matrix, particularly in understanding its long-term behavior. The spectral radius of a square matrix is simply the largest absolute value among its eigenvalues. Mathematically, if the eigenvalues of a matrix \(A\) are \(\lambda_i\), the spectral radius \(\rho(A)\) is defined as \(\rho(A) = \max_{i} |\lambda_i|\).

The concept becomes particularly interesting when evaluating whether a matrix is convergent or not. In the context of our example, after obtaining the eigenvalues of \(A_{1}\) and \(A_{2}\), we calculate their spectral radii. If the spectral radius is less than 1, it infers that the effects represented by the matrix decrease over time, and thus, the matrix is convergent. This property is critical when solving systems of linear equations iteratively or while analysing the stability of discrete dynamical systems.
Characteristic Equation
The characteristic equation is vital for determining the eigenvalues of a matrix. It is essentially a polynomial equation derived from the matrix, which is equated to zero. For a given square matrix \(A\), the characteristic equation is \(det(A - \lambda I) = 0\), where \(I\) is the identity matrix of the same size as \(A\) and \(\lambda\) represents an eigenvalue.

By solving the characteristic equation, we find the roots which are the eigenvalues. For example, in our exercise, by setting up and solving the characteristic equations for matrices \(A_{1}\) and \(A_{2}\), we can find the respective eigenvalues. Once we have the eigenvalues, we use them to understand more about the matrix, including its spectral radius and convergence.

Characterizing a matrix by its eigenvalues is hence a fundamental step in many applications, such as predicting system behaviors, simplifying matrix computations, and determining matrix invertibility.

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

In Exercise 17 of Section \(7.3\) a technique was outlined to prove that the Gauss-Seidel method converges when \(A\) is a positive definite matrix. Extend this method of proof to show that in this case there is also convergence for the SOR method with \(0<\omega<2\).

Find the first two iterations of the SOR method with \(\omega=1.1\) for the following linear systems, using \(\mathbf{x}^{(0)}=\mathbf{0}:\) a. \(3 x_{1}-x_{2}+x_{3}=1\) \(3 x_{1}+6 x_{2}+2 x_{3}=0\) \(3 x_{1}+3 x_{2}+7 x_{3}=4\) \(\begin{aligned} 10 x_{1}-x_{2} &=9 \\\\-x_{1}+10 x_{2}-2 x_{3} &=7 \\\\-2 x_{2}+10 x_{3} &=6 \end{aligned}\) c. \(\begin{aligned} 10 x_{1}+5 x_{2} &=6 \\ 5 x_{1}+10 x_{2}-4 x_{3} &=25 \\\\-4 x_{2}+8 x_{3}-x_{4} &=-11 \\\\-x_{3}+5 x_{4} &=-11 \end{aligned}\) d. \(\begin{aligned} 4 x_{1}+x_{2}+x_{3}+\quad x_{5} &=6 \\\\-x_{1}-3 x_{2}+x_{3}+x_{4} &=6 \\ 2 x_{1}+x_{2}+5 x_{3}-x_{4}-x_{5} &=6 \\\\-x_{1}-x_{2}-x_{3}+4 x_{4} &=6 \\ 2 x_{2}-x_{3}+x_{4}+4 x_{5} &=6 \end{aligned}\)

Suppose that an object can be at any one of \(n+1\) equally spaced points \(x_{0}, x_{1}, \ldots, x_{n}\). When an object is at location \(x_{i}\), it is equally likely to move to either \(x_{i-1}\) or \(x_{i+1}\) and cannot directly move to any other location. Consider the probabilities \(\left\\{P_{i}\right\\}_{i=0}^{n}\) that an object starting at location \(x_{i}\) will reach the left endpoint \(x_{0}\) before reaching the right endpoint \(x_{n} .\) Clearly, \(P_{0}=1\) and \(P_{n}=0\). Since the object can move to \(x_{i}\) only from \(x_{i-1}\) or \(x_{i+1}\) and does so with probability \(\frac{1}{2}\) for each of these locations, $$ P_{i}=\frac{1}{2} P_{i-1}+\frac{1}{2} P_{i+1}, \quad \text { for each } i=1,2, \ldots, n-1 $$ a. Show that $$ \left[\begin{array}{c} P_{1} \\ P_{2} \\ \vdots \\ P_{n-1} \end{array}\right]=\left[\begin{array}{c} \frac{1}{2} \\ 0 \\ \vdots \\ 0 \end{array}\right] $$ b. Solve this system using \(n=10,50\), and 100 . c. Change the probabilities to \(\alpha\) and \(1-\alpha\) for movement to the left and right, respectively, and derive the linear system similar to the one in part (a). d. \(\quad\) Repeat part (b) with \(\alpha=\frac{1}{3}\).

Use four-digit rounding arithmetic to compute the inverse \(H^{-1}\) of the \(3 \times 3\) Hilbert matrix \(H\), and then compute \(\hat{H}=\left(H^{-1}\right)^{-1}\). Determine \(\|H-\hat{H}\|_{\infty}\).

Show that if \(B\) is singular, then $$ \frac{1}{K(A)} \leq \frac{\|A-B\|}{\|A\|} $$ [Hint: There exists a vector with \(\|\mathbf{x}\|=1\), such that \(B \mathbf{x}=\mathbf{0}\). Derive the estimate using \(\|A \mathbf{x}\| \geq\) \(\left.\|\mathbf{x}\| /\left\|A^{-1}\right\| \cdot\right]\)

See all solutions

Recommended explanations on Math Textbooks

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.