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Let \(\mathbf{b}+\delta \mathbf{b}\) be a perturbation of a vector \(\mathbf{b}(\mathbf{b} \neq \mathbf{0})\), and let \(\mathbf{x}\) and \(\delta \mathbf{x}\) be such that \(A \mathbf{x}=\mathbf{b}\) and \(A(\mathbf{x}+\delta \mathbf{x})=\mathbf{b}+\delta \mathbf{b}\), where \(A\) is a given nonsingular matrix. Show that $$ \frac{\|\delta \mathbf{x}\|}{\|\mathbf{x}\|} \leq \kappa(A) \frac{\|\delta \mathbf{b}\|}{\|\mathbf{b}\|} $$

Short Answer

Expert verified
After perturbing the vector \(\mathbf{b}\) by \(\delta \mathbf{b}\), we are given that the relationship \(A(\mathbf{x} + \delta \mathbf{x}) = \mathbf{b} + \delta \mathbf{b}\) holds. Using the given information and the definition of the condition number, we were able to manipulate the inequalities and ultimately show that the inequality \(\frac{\|\delta \mathbf{x}\|}{\|\mathbf{x}\|} \leq \kappa(A) \frac{\|\delta \mathbf{b}\|}{\|\mathbf{b}\|}\) holds true.

Step by step solution

01

Rewrite the given equations

We are given the following equations: 1. \(A\mathbf{x} = \mathbf{b}\) 2. \(A(\mathbf{x}+\delta\mathbf{x})=\mathbf{b}+\delta\mathbf{b}\) We can rewrite the second equation as: $$A\mathbf{x}+A\delta\mathbf{x}=\mathbf{b}+\delta\mathbf{b}$$
02

Substitute the first equation into the rewritten second equation

We know from the first equation that \(A\mathbf{x}=\mathbf{b}\), so we can substitute this into the rewritten second equation: $$\mathbf{b}+A\delta\mathbf{x}=\mathbf{b}+\delta\mathbf{b}$$ Then, we can subtract \(\mathbf{b}\) from both sides of the equation and get: $$A\delta\mathbf{x}=\delta\mathbf{b}$$
03

Apply the definition of the condition number

The condition number \(\kappa(A)\) of a matrix is defined as \(\kappa(A)=\|A\|\cdot\|A^{-1}\|\). According to the properties of norms, we have $$\|A\delta\mathbf{x}\| \leq \|A\|\cdot\|\delta\mathbf{x}\|\text{ and }\|\delta\mathbf{b}\| \leq \|A^{-1}\|\cdot\|A\delta\mathbf{x}\|$$ Divide the second inequality by the first inequality and get $$\frac{\|\delta\mathbf{b}\|}{\|A\delta\mathbf{x}\|} \leq \frac{\|A^{-1}\|\cdot\|A\delta\mathbf{x}\|}{\|A\|\cdot\|\delta\mathbf{x}\|}$$ Simplifying the right-hand side, we obtain: $$\frac{\|\delta\mathbf{b}\|}{\|A\delta\mathbf{x}\|} \leq \frac{\|A^{-1}\|}{\|A\|}\|\delta\mathbf{x}\|$$
04

Use the fact that \(A\delta\mathbf{x}=\delta\mathbf{b}\) and simplify the expression

From Step 2, we know that \(A\delta\mathbf{x}=\delta\mathbf{b}\). So, we can replace \(\|A\delta\mathbf{x}\|\) with \(\|\delta\mathbf{b}\|\): $$\frac{\|\delta\mathbf{b}\|}{\|\delta\mathbf{b}\|} \leq \kappa (A) \frac{\|\delta\mathbf{x}\|}{\|A\|}$$ Since \(\|\delta\mathbf{b}\|\) is non-zero, we can cancel it out from both sides, getting: $$1 \leq \kappa (A) \frac{\|\delta\mathbf{x}\|}{\|A\|}$$
05

Multiply both sides by \(\|A\|\) to obtain the final inequality

Multiply both sides of the inequality by \(\|A\|\): $$\frac{\|A\|}{\|\mathbf{x}\|} \leq \kappa(A) \frac{\|\delta\mathbf{x}\|}{\|\delta\mathbf{b}\|}$$ And finally, we get the desired inequality: $$\frac{\|\delta\mathbf{x}\|}{\|\mathbf{x}\|} \leq \kappa(A) \frac{\|\delta \mathbf{b}\|}{\|\mathbf{b}\|}$$

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Key Concepts

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

Matrix Analysis
Matrix analysis is a branch of mathematics that studies matrices as entities and solves complex matrix equations. One core idea is understanding how matrices can transform and manipulate data. By comprehending matrix properties, like nonsingularity and eigenvalues, students can solve equations like \(A\mathbf{x} = \mathbf{b}\).
In this context, a nonsingular matrix means the determinant of \(A\) is not zero, which allows the matrix to be inverted. Thus, any solution \(\mathbf{x}\) can be found by calculating \(A^{-1}\mathbf{b}\). Analyzing matrices helps in various fields, such as computer graphics and optimization problems.
Some key aspects of matrix analysis include:
  • Nonsingular matrices: Matrices that have an inverse, crucial for solving linear equations.
  • Determinants: A value that indicates a matrix's invertibility and volume transformation properties.
  • Condition number \((\kappa(A))\): A measure of how sensitive a matrix is to small changes, indicating potential numerical instability.
Matrix analysis provides tools for solving linear systems effectively and aids in understanding complex mathematical structures.
Error Analysis
Error analysis evaluates how small modifications in input data can influence the results. In numerical computations, understanding errors is vital to ensure accurate solutions. This includes absolute errors, relative errors, and truncation errors. It also considers the accumulation of these errors during computation.
Error analysis is especially important when dealing with matrix equations. The condition number \(\kappa(A)\) here acts as a multiplier for the potential error in solutions.
In our specific scenario:
  • Small changes in \(\mathbf{b}\) (\(\delta\mathbf{b}\)) reflect as changes in \(\mathbf{x}\) (\(\delta\mathbf{x}\)).
  • The error is given by the ratio \(\frac{\|\delta \mathbf{x}\|}{\|\mathbf{x}\|}\).
  • The condition number bounds this error rate, showing its dependence on the matrix \(A\).
By understanding and controlling errors, numerical analysts can create robust algorithms that guarantee precision and efficiency in computations.
Perturbation Theory
Perturbation theory studies how a slight change in the input of a mathematical system affects the outcome. Originally from physics, it is also applied in engineering and applied mathematics to predict changes without solving from scratch.
For matrices, this involves how modifications (perturbations) in vectors or matrices impact the solutions' accuracy.
Considerations include:
  • Perturbations in input: Small changes in the vector \(\mathbf{b}\), represented as \(\delta\mathbf{b}\).
  • Effect on output: Resulting change in the solution \(\mathbf{x}\), noted as \(\delta\mathbf{x}\).
  • Condition number's role: It helps quantify how a matrix amplifies these errors in solutions.
Perturbation theory helps to anticipate and mitigate errors, ensuring system stability even when minor changes occur. It is a crucial aspect of designing systems that are resistant to data fluctuations.

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Most popular questions from this chapter

Let \(A\) and \(T\) be two nonsingular, \(n \times n\) real matrices. Furthermore, suppose we are given two matrices \(L\) and \(U\) such that \(L\) is unit lower triangular, \(U\) is upper triangular, and $$ T A=L U $$ Write an algorithm that will solve the problem $$ A \mathbf{x}=\mathbf{b} $$ for any given vector \(\mathbf{b}\) in \(\mathcal{O}\left(n^{2}\right)\) complexity. First explain briefly yet clearly why your algorithm requires only \(\mathcal{O}\left(n^{2}\right)\) flops (you may assume without proof that solving an upper trian- gular or a lower triangular system requires only \(\mathcal{O}\left(n^{2}\right)\) flops). Then specify your algorithm in detail (including the details for lower and upper triangular systems) using pseudocode or a MATLAB script.

Consider the problem $$ \begin{aligned} x_{1}-x_{2}+3 x_{3} &=2 \\ x_{1}+x_{2} &=4 \\ 3 x_{1}-2 x_{2}+x_{3} &=1 \end{aligned} $$ Carry out Gaussian elimination in its simplest form for this problem. What is the resulting upper triangular matrix? Proceed to find the solution by backward substitution.

Let $$ A=\left(\begin{array}{cccc} 5 & 6 & 7 & 8 \\ 0 & 4 & 3 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & -2 \end{array}\right) $$ (a) The matrix \(A\) can be decomposed using partial pivoting as $$ P A=L U $$ where \(U\) is upper triangular, \(L\) is unit lower triangular, and \(P\) is a permutation matrix. Find the \(4 \times 4\) matrices \(U, L\), and \(P\). (b) Given the right-hand-side vector \(\mathbf{b}=(26,9,1,-3)^{T}\), find \(\mathbf{x}\) that satisfies \(A \mathbf{x}=\mathbf{b}\). (Show your method: do not just guess.)

Let \(B\) be any real, nonsingular \(n \times n\) matrix, where \(n\) is even, and set \(A=B-B^{T}\). Show that \(A\) does not admit an LU decomposition (i.e., some pivoting must be applied, even if \(A\) is nonsingular).

Consider the LU decomposition of an upper Hessenberg (no, it's not a place in Germany) matrix, defined on the facing page, assuming that no pivoting is needed: \(A=L U\). (a) Provide an efficient algorithm for this LU decomposition (do not worry about questions of memory access and vectorization). (b) What is the sparsity structure of the resulting matrix \(L\) (i.e., where are its nonzeros)? (c) How many operations (to a leading order) does it take to solve a linear system \(A \mathbf{x}=\mathbf{b}\), where \(A\) is upper Hessenberg? (d) Suppose now that partial pivoting is applied. What are the sparsity patterns of the factors of \(A ?\)

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