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Chapter 7: Problem 9

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 \(\|A \mathbf{x}\| \geq\) \(\left.\|\mathbf{x}\| /\left\|A^{-1}\right\| \cdot\right]\)

Short Answer

Expert verified
Showing the inequality involves constructing a normalized vector \(x\) such that \(Bx = 0\), exploring the implications on \(Ax\) and \(A - B\), and then applying an inequality related to the condition number of \(A\). The work leads to the desired result of \(1/ K(A) \leq \|A-B\|\) when multiplied by \|A\|.

Step by step solution

01

Identify a Vector x With Norm 1 Such That Bx = 0

Since \(B\) is singular, there exists a non-zero vector \(x\) such that \(Bx = 0\). We can normalize \(x\) to create a new vector with length 1: \(x = x / \|x\|\). This way, we maintain the property that \(Bx = 0\), and also now have \(\|x\| = 1\).
02

Evaluate Norm of (A - B)x

The difference between \(A\) and \(B\) when multiplied by vector \(x\) can be written as \((A - B)x = Ax - Bx\). Given that \(Bx = 0\), we simplify to \((A - B)x = Ax\). From this, taking norms on both sides results in \(\|(A - B)x\| = \|Ax\|\).
03

Apply the Inequality Involving Inverse of A

As the hint suggests, apply the inequality \(\|Ax\| \geq \|\mathbf{x}\| /\|A^{-1}\|\). As \(\|x\| = 1\), this simplifies to \(\|Ax\| \geq 1/\|A^{-1}\|\). Now, as \(\|(A - B)x\| = \|Ax\|\), we have \(\|(A - B)x\| \geq 1/\|A^{-1}\|\). Taking reciprocals of both sides, and remembering that \(K(A) = \|A\| \|A^{-1}\|\), gives \(1/ K(A) \leq \|A-B\|\) when multiplied by \|A\|.

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Key Concepts

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

Condition Number
The condition number of a matrix is a crucial concept in numerical analysis, which provides a measure of how sensitive a matrix is to numerical errors during computations, such as finding its inverse or solving linear equations. It is defined for a non-singular matrix and is a key indicator of the numerical stability of matrix operations.

The condition number, denoted as \(K(A)\), is calculated using the matrix norm and the norm of its inverse. It is expressed as \(K(A) = \|A\| \cdot \|A^{-1}\|\), where \(\|A\|\) is the norm of matrix \(A\), and \(\|A^{-1}\|\) is the norm of its inverse. When the condition number is high, small changes or errors in the matrix or in the input can lead to large variations in the results, which means the matrix is ill-conditioned. Conversely, a small condition number indicates that the matrix is well-conditioned, and the computations are expected to be more accurate.
Matrix Norm
The matrix norm is a tool used to quantify the size of elements in a matrix. For a given matrix \(A\), its norm \(\|A\|\) is a non-negative number that provides a sense of the 'magnitude' of the matrix. There are several types of matrix norms, but a common one used in numerical analysis is the Euclidean norm (or \(L_2\) norm), which can be thought of as an extension of the Euclidean distance from vectors to matrices.

In the given exercise, the matrix norm plays an essential role in deriving the inequality. It bounds the difference between two matrices \(A\) and \(B\), giving us an understanding of how 'far' they are from each other. Importantly, when we apply the matrix norm to the operation \(A - B\), it gives us information about the sensitivity of the matrix with respect to changes, which is invaluable when analyzing numerical stability.
Inverse of a Matrix
The inverse of a matrix \(A\), denoted by \(A^{-1}\), is a fundamental concept in linear algebra that refers to a matrix which, when multiplied by \(A\), results in the identity matrix, symbolizing the notion of a reciprocal in matrix algebra. Not all matrices have inverses; only non-singular (invertible) matrices possess an inverse that is unique to them.

In the context of the given exercise, the existence of an inverse is taken into account to derive an inequality that is reliant on the norm of the inverse. The notion that \(\|Ax\| \geq 1/\|A^{-1}\|\) for a vector \(x\) with norm 1 is derived from the properties of the inverse. If the matrix \(B\) is singular (does not have an inverse), it becomes important to understand the effects of small perturbations or errors on solutions to equations involving \(A\) and \(B\), which is where the concept of the inverse plays a significant analytical role.

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Most popular questions from this chapter

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.

The linear system \(A \mathbf{x}=\mathbf{b}\) given by $$ \left[\begin{array}{cc} 1 & 2 \\ 1.00001 & 2 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]=\left[\begin{array}{c} 3 \\ 3.00001 \end{array}\right] $$ has solution \((1,1)^{t}\). Use seven-digit rounding arithmetic to find the solution of the perturbed system $$ \left[\begin{array}{cc} 1 & 2 \\ 1.000011 & 2 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]=\left[\begin{array}{l} 3.00001 \\ 3.00003 \end{array}\right] $$ and compare the actual error to the estimate (7.25). Is \(A\) ill-conditioned?

Find the complex eigenvalues and associated eigenvectors for the following matrices. a. \(\left[\begin{array}{rrr}1 & 0 & 2 \\ 0 & 1 & -1 \\ -1 & 1 & 1\end{array}\right]\) b. \(\left[\begin{array}{rrr}0 & 1 & -2 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}\right]\)

The \(n \times n\) Hilbert matrix \(H^{(n)}\) (see page 512) defined by $$ H_{i j}^{(n)}=\frac{1}{i+j-1}, \quad 1 \leq i, j \leq n $$ is an ill-conditioned matrix that arises in solving the normal equations for the coefficients of the least-squares polynomial (see Example 1 of Section 8.2). a. Show that $$ \left[H^{(4)}\right]^{-1}=\left[\begin{array}{rrrr} 16 & -120 & 240 & -140 \\ -120 & 1200 & -2700 & 1680 \\ 240 & -2700 & 6480 & -4200 \\ -140 & 1680 & -4200 & 2800 \end{array}\right], $$ and compute \(K_{\infty}\left(H^{(4)}\right)\). b. Show that $$ \left[H^{(5)}\right]^{-1}=\left[\begin{array}{rrrrr} 25 & -300 & 1050 & -1400 & 630 \\ -300 & 4800 & -18900 & 26880 & -12600 \\ 1050 & -18900 & 79380 & -117600 & 56700 \\ -1400 & 26880 & -117600 & 179200 & -88200 \\ 630 & -12600 & 56700 & -88200 & 44100 \end{array}\right], $$ and compute \(K_{\infty}\left(H^{(5)}\right)\). c. Solve the linear system $$ H^{(4)}\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \\ 1 \end{array}\right] $$ using five-digit rounding arithmetic, and compare the actual error to that estimated in (7.25).

The following excerpt from the Mathematics Magazine [Sz] gives an alternative way to prove the Cauchy-Buniakowsky-Schwarz Inequality. a. Show that when \(\mathbf{x} \neq \mathbf{0}\) and \(\mathbf{y} \neq \mathbf{0}\), we have $$ \frac{\sum_{i=1}^{n} x_{i} y_{i}}{\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{1 / 2}\left(\sum_{i=1}^{n} y_{i}^{2}\right)^{1 / 2}}=1-\frac{1}{2} \sum_{i=1}^{n}\left(\frac{x_{i}}{\left(\sum_{j=1}^{n} x_{j}^{2}\right)^{1 / 2}}-\frac{y_{i}}{\left(\sum_{j=1}^{n} y_{j}^{2}\right)^{1 / 2}}\right)^{2}. $$ b. Use the result in part (a) to show that $$ \sum_{i=1}^{n} x_{i} y_{i} \leq\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{1 / 2}\left(\sum_{i=1}^{n} y_{i}^{2}\right)^{1 / 2} $$
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