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Chapter 7: Problem 39

Let \(A\) and \(B\) be nonsingular \(n \times n\) matrices. Show that \\[ \operatorname{cond}(A B) \leq \operatorname{cond}(A) \operatorname{cond}(B) \\]

Short Answer

Expert verified
The desired inequality is proved using the definition of the condition number, properties of matrix norms, and the sub-multiplicative property: \( \operatorname{cond}(AB) \leq \operatorname{cond}(A) \operatorname{cond}(B) \).

Step by step solution

01

Recall the definition of the condition number

The condition number of a matrix is defined as the product of its norm and its inverse's norm: \( \operatorname{cond}(M) = \|M\| \|M^{-1}\| \) for any nonsingular matrix M.
02

Compute the norm of the product of matrices

Recall the property, for any matrices M and N, that \( \|M N\| \leq \|M\| \|N\| \). This is known as sub-multiplicative property of matrix norms.
03

Prove the inequality

We want to prove that \( \operatorname{cond}(AB) \leq \operatorname{cond}(A) \operatorname{cond}(B) \). Let's start with calculating the condition number of the product AB: \( \operatorname{cond}(AB) = \|AB\| \| (AB)^{-1} \| \) Now, we know that the inverse of the product of matrices is the product of their inverses in reverse order: \((AB)^{-1} = B^{-1} A^{-1}\). So, we have: \( \operatorname{cond}(AB) = \|AB\| \| B^{-1} A^{-1} \| \) Using the sub-multiplicative property from Step 2, we can write: \( \|AB\| \leq \|A\| \|B\| \) and \( \| B^{-1} A^{-1} \| \leq \|B^{-1}\| \|A^{-1}\| \) Thus, we obtain: \( \operatorname{cond}(AB) = \|AB\| \| B^{-1} A^{-1} \| \leq \|A\| \|B\| \|B^{-1}\| \|A^{-1}\| \) Now, from Step 1, we know that \( \operatorname{cond}(A) = \|A\| \|A^{-1}\| \) and \( \operatorname{cond}(B) = \|B\| \|B^{-1}\| \). So, we can rewrite the above inequality as: \( \operatorname{cond}(AB) \leq \operatorname{cond}(A) \operatorname{cond}(B) \) Hence, the desired inequality is proved.

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Most popular questions from this chapter

Let \\[ A=\left(\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right) \quad \text { and } \quad \mathbf{b}=\left(\begin{array}{l} 5 \\ 2 \end{array}\right) \\] The solution computed using two-digit decimal floating-point arithmetic is \(\mathbf{x}=(1.1,0.88)^{T}\) (a) Determine the residual vector \(\mathbf{r}\) and the value of the relative residual \(\|\mathbf{r}\|_{\infty} /\|\mathbf{b}\|_{\infty}\) (b) Find the value of \(\operatorname{cond}_{\infty}(A)\) (c) Without computing the exact solution, use the results from parts (a) and (b) to obtain bounds for the relative error in the computed solution. (d) Compute the exact solution \(\mathbf{x}\) and determine the actual relative error. Compare your results with the bounds derived in part (c)

Let \(A\) be a nonsingular \(n \times n\) matrix and let \(Q\) be an \(n \times n\) orthogonal matrix. Show that (a) \(\operatorname{cond}_{2}(Q A)=\operatorname{cond}_{2}(A Q)=\operatorname{cond}_{2}(A)\) (b) if \(B=Q^{T} A Q,\) then \(\operatorname{cond}_{2}(B)=\operatorname{cond}_{2}(A)\)

Let \(A\) be an \(m \times n\) matrix. Show that \(\|A\|_{1,2} \leq\|A\|_{2}\)

Given \(\mathbf{x} \in \mathbb{R}^{3},\) define \\[ r_{i j}=\left(x_{i}^{2}+x_{j}^{2}\right)^{1 / 2} \quad i, j=1,2,3 \\] For each of the following, determine a Givens transformation \(G_{i j}\) such that the \(i\) th and \(j\) th coordinates of \(G_{i j} \mathbf{x}\) are \(r_{i j}\) and \(0,\) respectively: (a) \(\mathbf{x}=(3,1,4)^{T}, i=1, j=3\) (b) \(\mathbf{x}=(1,-1,2)^{T}, i=1, j=2\) (c) \(\mathbf{x}=(4,1, \sqrt{3})^{T}, i=2, j=3\) (d) \(\mathbf{x}=(4,1, \sqrt{3})^{T}, i=3, j=2\)

Let \(A=\left(\begin{array}{rrr}1 & 2 & -4 \\ 2 & 6 & 7 \\ -2 & 1 & 8\end{array}\right) \quad\) and \(\quad \mathbf{b}=\left(\begin{array}{r}9 \\ 9 \\\ -3\end{array}\right)\) (a) Use Householder transformations to transform \(A\) into an upper triangular matrix \(R\). Also transform the vector \(\mathbf{b} ;\) that is, compute \(\mathbf{b}^{(1)}=\) \(H_{2} H_{1} \mathbf{b}\) (b) Solve \(R \mathbf{x}=\mathbf{b}^{(1)}\) for \(\mathbf{x}\) and check your answer by computing the residual \(\mathbf{b}-A \mathbf{x}\)
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