/*! 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 21 Prove that if \(\|A B-I\|<1\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Prove that if \(\|A B-I\|<1\), then \(2 B-B A B\) is a better approximate inverse for \(A\) than \(B\), in the sense that \(A(2 B-B A B)\) is closer to \(I\).

Short Answer

Expert verified
The approximation \(2B - BAB\) is closer to the inverse as \(\|I - C^2\| < \|C\|\).

Step by step solution

01

Understand the Problem

The exercise involves proving that one approximation of the inverse is better than another under a specific condition. We are given a matrix condition \(\|A B - I\| < 1\). We need to compare \(A(2B - BAB)\) and \(AB\) in terms of how close they are to the identity matrix \(I\).
02

Analyze the Expression \(A(2B - BAB)\)

We need to expand the expression \(A(2B - BAB) = 2AB - ABAB\). This will allow us to compare the result directly with \(I\).
03

Derive the Proximity to Identity Matrix

Subtract \(I\) from \(A(2B - BAB)\), which gives us \(A(2B - BAB) - I = 2AB - ABAB - I\). We need to show this expression has a smaller norm than \(AB - I\).
04

Simplify the Expression \(2AB - ABAB - I\)

Start by acknowledging \(AB - I = C\) with \(\|C\| < 1\). Then substitute \(AB = I + C\) into \(2AB - ABAB\), resulting in \(2(I+C) - (I+C)(I+C) = 2I + 2C - I - 2C - C^2\). Simplifying gives \(I - C^2\).
05

Evaluate \(\|I - C^2\|\) vs. \(\|C\|\)

Since \(\|C\| < 1\), we have \(\|C^2\| = \|C\|^2 < \|C\|\). The expression \(I - C^2\) indicates \(\|I - C^2\|\) should indeed be less than \(\|C\|\), since subtracting \(C^2\) from \(I\) moves it closer to \(I\) itself.
06

Conclude that \(2B - BAB\) is a Better Approximate Inverse

Since \(\|A(2B - BAB) - I\| = \|I - C^2\|\) is smaller than \(\|C\|\), \(2B - BAB\) provides a closer approximation to the inverse of \(A\) compared to \(B\).

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.

Matrix Norms
When working with matrices, it's essential to have tools to measure how large or small a matrix is. This is where matrix norms come into play. A matrix norm is a number that provides a measure of the "size" or "length" of a matrix. In mathematical terms, it assigns a non-negative value to each matrix.

Some commonly used matrix norms include:
  • The Frobenius norm, which treats a matrix in a similar way to a multi-dimensional vector.
  • The 1-norm, calculated as the maximum absolute column sum of the matrix.
  • The infinity norm, which is the maximum absolute row sum.
Matrix norms have various applications, one of which is in approximating how close certain matrices come to behaving like the identity matrix. This is crucial in verifying if a matrix is an approximate inverse of another. When analyzing the condition \( \| AB - I \| < 1 \), the norm tells us how close the product of matrices \( A \) and \( B \) is to the identity matrix \( I \). A smaller norm indicates closer proximity. This information is key when confirming a matrix as a better approximation of an inverse.
Approximate Inverse
In linear algebra, finding the exact inverse of a matrix can be both difficult and computationally expensive. That's where approximate inverses become incredibly useful. An approximate inverse refers to a matrix that is not exactly the inverse, but achieves a close approximation.

In our original exercise, we're examining two potential approximate inverses for a matrix \( A \): \( B \) and \( 2B - BAB \). The goal is to determine which of these is closer to being the true inverse. If \( A(2B - BAB) \) is closer to the identity matrix \( I \) than \( AB \) is, then \( 2B - BAB \) is a better approximation.

The task becomes analyzing and calculating expressions to see how well they approach the behavior of an inverse. Practically, finding a better approximate inverse can help simplify computations in larger systems, reducing error margins and optimizing performance.
Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra, where two matrices are combined to produce another matrix. This operation is not as straightforward as multiplying numbers because the order of the matrices matters—meaning \( AB \) may not be the same as \( BA \).

When multiplying matrices, the number of columns in the first matrix must match the number of rows in the second. The resulting matrix will have the number of rows of the first and columns of the second.
  • The entry in the \( i^{th} \) row and \( j^{th} \) column of the resultant matrix is the dot product of the \( i^{th} \) row of the first matrix and the \( j^{th} \) column of the second matrix.
Matrix multiplication is essential in discussing matrix inversion, as both exact and approximate inverses are demonstrated using multiplication. In our exercise, the expression \( A(2B - BAB) \) required expanding and multiplying matrices to evaluate how close the product is to the identity matrix \( I \). This multiplication gives insight into the effectiveness of an approximate inverse in replicating the identity matrix when used with \( A \).

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 a normed linear space, prove that if a sequence of vectors converges, then it must satisfy the Cauchy criterion.

Consider the symmetric tridiagonal positive definite matrix $$A=\left[\begin{array}{rrrr} 136.01 & 90.860 & 0.0 & 0.0 \\ 90.860 & 98.810 & -67.590 & 0.0 \\ 0.0 & -67.590 & 132.01 & 46.260 \\ 0.0 & 0.0 & 46.260 & 177.17 \end{array}\right]$$ Using five significant figures, factor \(A\) in the following ways: a. \(A=L U\), where \(L\) is unit lower triangular and \(U\) is upper triangular. b. \(A=L D U\), where \(L\) is unit lower triangular, \(D\) is diagonal, and \(U\) is unit upper triangular. c. \(A=L U\), where \(L\) is lower triangular and \(U\) is unit upper triangular. d. \(A=L L^{T}\), where \(L\) is lower triangular.

For any real number \(p \geq 1\), the formula $$ \|x\|_{p}=\left(\sum_{i=1}^{n}\left|x_{i}\right|^{p}\right)^{1 / P} $$ defines a norm. (For the proof, consult Bartle [1976, p. 61].) Prove that for each \(x \in \mathbb{R}^{n}\). $$ \lim _{p \rightarrow \infty}\|x\|_{p}=\|x\|_{\infty} $$ This explains why the notation \(\|\cdot\|_{\infty}\) is used.

(Continuation) Write an algorithm to compute the \(L D L^{T}\) -factorization of a symmetric matrix \(A\). Your algorithm should do approximately half as much work as the standard Gaussian algorithm. Note: This algorithm can fail if some principal minors of \(A\) are singular. (This modification of the Cholesky algorithm does not involve square root calculations.)

Determine directly the \(L L^{T}\) -factorization, in which \(L\) is a lower triangular matrix with oositive diagonal elements, for the matrix $$ A=\left[\begin{array}{ccc} 4 & \frac{1}{2} & 1 \\ \frac{1}{2} & \frac{17}{16} & \frac{1}{4} \\ 1 & \frac{1}{4} & \frac{33}{64} \end{array}\right] $$

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.