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As an example that convergent iteration methods can behave in unusual ways, consider $$ x^{(k+1)}=b+A x^{(k)}, k \geq 0 $$ with $$ A=\left[\begin{array}{cc} \lambda & c \\ 0 & -\lambda \end{array}\right], \quad b \in \mathbf{R}^{2} $$ Assuming \(|\lambda|<1\), we have \((I-A)^{-1}\) exists and \(x^{(k)} \rightarrow x^{*}=(I-A)^{-1} b\) for all initial guesses \(x^{0} .\) Find explicit formulas for \(A^{k}, x^{*}-x^{(k)}\), and \(x^{(k+1)}-x^{(k)} .\) By suitably adjusting \(c\) relative to \(\lambda\), show that it is possible for \(\left\|x^{*}-x^{(k)}\right\|_{\infty}\) to alternately increase and decrease as it converges to zero. Look at the corresponding values for \(\left\|x^{(k+1)}-x^{(k)}\right\|_{\infty} .\) For simplicity, use \(x^{(0)}=0\) in all calculations.

Short Answer

Expert verified
Use the formulas for \( A^k \) and \( (I-A)^{-1} \) to show \( x^{*} - x^{(k)} \) behavior depends on c. Adjusting c can cause alternating increases and decreases in the infinity norm.

Step by step solution

01

Understanding the Problem

We are given an iterative method and need to find the explicit formulas for several terms, and analyze the behavior of the method under specific conditions.
02

Given Matrix A

The matrix A is given as: \[ A = \begin{pmatrix} \lambda & c \ 0 & -\lambda \end{pmatrix} \] where \lambda and c are constants.
03

Iteration Formula

The iteration formula is given by: \[ x^{(k+1)} = b + A x^{(k)} \] with initial guess \[ x^{(0)} = 0 \].
04

Find A^k

Using induction, we can prove that: \[ A^k = \begin{pmatrix} \lambda^k & c \sum_{i=0}^{k-1}\lambda^i \ 0 & (-\lambda)^k \end{pmatrix} \].
05

Stationary Point x^*

Given that \( (I-A)^{-1} \) exists, we can find the stationary point by solving: \[ x^{*} = (I - A)^{-1} b \].
06

Formula for (I-A)^{-1}

To find \( (I-A)^{-1} \), we start by calculating \( I - A \): \[ I - A = \begin{pmatrix} \1-\lambda & -c \ 0 & 1+\lambda \end{pmatrix} \]. Thus, \( (I - A)^{-1} \) is: \[ (I - A)^{-1} = \begin{pmatrix} \frac{1}{1-\lambda} & \frac{-c}{(1-\lambda)(1+\lambda)} \ 0 & \frac{1}{1+\lambda} \end{pmatrix} \].
07

Explicit Formula for x^*

Now calculate \( x^* \) by plugging in b and \( (I - A)^{-1} \): \[ x^{*} = (I - A)^{-1} b \].
08

Find x^{(k)} - x^*

We need to find the difference between the iterated value and the stationary point: \[ x^{*} - x^{(k)} = (I - A)^{-1} b - x^{(k)} \].
09

Compare Infinity Norms

Calculate \( \| x^{*} - x^{(k)} \|_{\infty} \) and analyse its behavior under different values of c and \lambda, showing that adjusting c can cause the norm to alternately increase and decrease as it converges to zero.
10

Convergence Behavior

Determine \( \| x^{(k+1)} - x^{(k)} \|_{\infty} \) and analyze its values under different iterations.

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Key Concepts

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

Iterative Methods
Iterative methods are mathematical procedures used to generate a sequence of improving approximate solutions for a class of problems. In the given exercise, we use an iterative method defined by the formula: \[ x^{(k+1)} = b + A x^{(k)} \] where \( A \) is a matrix and \( b \) is a vector.

The main idea here is that starting from an initial guess \( x^{(0)} \), we repeatedly apply the formula to get closer to the true solution, \( x^* \).

Because \( | \lambda | < 1 \) in our exercise, convergence is guaranteed, meaning our sequence will ultimately reach (or get very close to) the stationary point \( x^* \).

Iterative methods are useful in many areas such as solving large linear systems, finding roots of nonlinear equations, and even for optimizations in machine learning.
Matrix Analysis
Matrix analysis plays a critical role in understanding the behavior of iterative methods. In this exercise, we are given:

\[ A = \begin{pmatrix} \lambda & c \ 0 & -\lambda \end{pmatrix} \]

To analyze how iterations evolve, we must find powers of this matrix, \( A^k \). By induction, it's proven that: \[ A^k = \begin{pmatrix} \lambda^k & c \sum_{i=0}^{k-1} \lambda^i \ 0 & (-\lambda)^k \end{pmatrix} \]

Knowing \( A \) and its behavior as \( k \) increases is fundamental. It allows us to predict how the iterations behave, showing us the impact of the matrix's properties on convergence speed and pattern.
Numerical Analysis
Numerical analysis helps us understand and control the behavior of iterative methods in practice.

For our iterative method: \[ x^{(k+1)} = b + A x^{(k)} \]

we can analyze the stationary point \( x^* \) by solving: \[ x^* = (I - A)^{-1} b \]

Using numerical techniques, we calculate: \[ (I - A)^{-1} = \begin{pmatrix} \frac{1}{1-\lambda} & \frac{-c}{(1-\lambda)(1+\lambda)} \ 0 & \frac{1}{1+\lambda} \end{pmatrix} \]

Plugging in \( b \) gives us the explicit stationary point \( x^* \).

Numerical analysis also shows us how to calculate \( x^{(k)} - x^* \) and how to measure the convergence quality by evaluating the infinity norm \( \| x^* - x^{(k)} \|_{\infty} \).
Convergence Behavior
Understanding the convergence behavior is crucial for iterative methods.

The convergence of our iterative method depends heavily on the values of \( \lambda \) and \( c \). By adjusting \( c \) with respect to \( \lambda \), we can influence the convergence pattern.

Specifically, it is possible for: \[ \left\| x^* - x^{(k)} \right\|_{\infty} \] to alternately increase and decrease as it converges to zero. This oscillating behavior can be analyzed by looking at the change in terms of: \[ \left\| x^{(k+1)} - x^{(k)} \right\|_{\infty} \]

By examining these differences, we can gain insights into how quickly and smoothly our solution converges to \( x^* \). Studying these norms under different parameter sets (\( \lambda \) and \( c \)) allows us to better control and understand convergence in practical applications.

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Most popular questions from this chapter

Let \(A\) be nonsingular. Let. \(A=L U=L D M\), with all \(l_{i i}, m_{i i}=1\), and \(D\) diagonal. Further assume \(A\) is symmetric. Show that \(M=L^{T}\), and thus \(A=L D L^{T} .\) Show \(A\) is positive definite if and only if all \(d_{i i}>0 .\)

Let \(A\) be real, symmetric, positive definite, and of order \(n\). Consider solving \(A x=b\) using Gaussian elimination without pivoting. The purpose of this problem is to justify that the pivots will be nonzero. (a) Show that all of the diagonal elements satisfy \(a_{i i}>0\). This shows that \(a_{11}\) can be used as a pivot element. (b) After elimination of \(x_{1}\) from equations 2 through \(n\), let the resulting matrix \(A^{(2)}\) be written as $$ A^{(2)}=\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ 0 & & & \\ \vdots & & \hat{A}^{(2)} & \\ 0 & & & \end{array}\right] $$ Show that \(\hat{A}^{(2)}\). is symmetric and positive definite. This procedure can be continued inductively to each stage of the elimination process, thus justifying the existence of nonzero pivots at evéry step. Hint: To prove \(\hat{A}^{(2)}\) is positive definite, first prove the identity $$ \sum_{i, j=2}^{n} a_{i j}^{(2)} x_{i} x_{j}=\sum_{i, j=1}^{n} a_{i j} x_{i} x_{j}-a_{11}\left[x_{1}+\sum_{j=2}^{n} \frac{a_{j 1}}{a_{11}} x_{j}\right]^{2} $$ for any choice of \(x_{1}, x_{2}, \ldots, x_{n} .\) Then choose \(x_{1}\) suitably.

The system \(A x=b\), $$ A=\left[\begin{array}{rrrrrr} 4 & -1 & 0 & -1 & 0 & 0 \\ -1 & 4 & -1 & 0 & -1 & 0 \\ 0 & -1 & 4 & 0 & 0 & -1 \\ -1 & 0 & 0 & 4 & -1 & 0 \\ 0 & -1 & 0 & -1 & 4 & -1 \\ 0 & 0 & -1 & 0 & -1 & 4 \end{array}\right] \quad b=\left[\begin{array}{l} 2 \\ 1 \\ 2 \\ 2 \\ 1 \\ 2 \end{array}\right] $$ has the solution \(x=[1,1,1,1,1,1]^{T} .\) Solve the system using the Gauss- Jacobi iteration method, and then solve it again using the Gauss-Seidel method. Use the initial guess \(x^{(0)}=0\). Note the rate at which the iteration error decreases. Find the answers with an accuracy of \(\epsilon=.0001\).

Solve the following systems \(A x=b\) by Gaussian elimination without pivoting. Check that \(A=L U\), as in \((8.1 .5)\). (a) \(A=\left[\begin{array}{rrr}1 & 1 & -1 \\ 1 & 2 & -2 \\ -2 & 1 & 1\end{array}\right] \quad b=\left[\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right]\) (b) \(A=\left[\begin{array}{llll}4 & 3 & 2 & 1 \\ 3 & 4 & 3 & 2 \\ 2 & 3 & 4 & 3 \\ 1 & 2 & 3 & 4\end{array}\right] \quad b=\left[\begin{array}{r}1 \\ 1 \\\ -1 \\ -1\end{array}\right]\) (c) \(A=\left[\begin{array}{rrrr}1 & -1 & 1 & -1 \\ -1 & 3 & -3 & 3 \\ 2 & -4 & 7 & -7 \\ -3 & 7 & -10 & 14\end{array}\right] \quad b=\left[\begin{array}{r}0 \\ 2 \\ -2 \\ 8\end{array}\right]\)

Define the order \(n\) tridiagonal matrix $$ A_{n}=\left[\begin{array}{rrrrrr} 2 & -1 & 0 & & \cdots & 0 \\ -1 & 2 & -1 & 0 & & \\ 0 & -1 & 2 & -1 & & \vdots \\ \vdots & & & \ddots & \cdot & \\ 0 & & \cdots & & -1 & 2 \end{array}\right] $$ Find a general formula for \(A_{n}=L U .\) Hint: Consider the cases \(n=3,4,5\), and then guess the general pattern and verify it.

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