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Chapter 7: Problem 15

Let \(A\) be a symmetric \(n \times n\) matrix. Show that \(\|A\|_{\infty}=\|A\|_{1}\)

Short Answer

Expert verified
For a symmetric \(n \times n\) matrix A, using the definitions of the 1-norm and infinity norm and the symmetry property \(a_{ij} = a_{ji}\), we can show that: \(\|A\|_{1} = \max_{1 \leq j \leq n} \sum_{i=1}^{n}|a_{ij}|\) \(\|A\|_{1} = \max_{1 \leq i \leq n} \sum_{j=1}^{n}|a_{ij}|\) \(\|A\|_{\infty} = \max_{1 \leq i \leq n} \sum_{j=1}^{n}|a_{ij}|\) Thus, for a symmetric \(n \times n\) matrix A, \(\|A\|_{\infty} = \|A\|_{1}\).

Step by step solution

01

Write the 1-norm of A as a function of row sums

Using the definition of the 1-norm: \(\|A\|_{1} = \max_{1 \leq j \leq n} \sum_{i=1}^{n}|a_{ij}|\)
02

Use the symmetry property of A, \(a_{ij} = a_{ji}\), to rewrite the 1-norm of A

As A is symmetric, we can substitute each occurrence of \(a_{ij}\) with \(a_{ji}\). So, the 1-norm becomes: \(\|A\|_{1} = \max_{1 \leq j \leq n} \sum_{i=1}^{n}|a_{ji}|\)
03

Rewrite the sum in terms of row sums instead of column sums

Notice that the sum inside the max function is now a sum of elements in a row instead of a column. So, we can rewrite the expression using the index variable i instead of j: \(\|A\|_{1} = \max_{1 \leq i \leq n} \sum_{j=1}^{n}|a_{ij}|\)
04

Use the definition of the infinity norm of A

The expression we obtained in step 3 is the definition of the infinity norm of A: \(\|A\|_{\infty} = \max_{1 \leq i \leq n} \sum_{j=1}^{n}|a_{ij}|\)
05

Confirm the equality

Comparing the expressions from steps 3 and 4, we can clearly see that: \(\|A\|_{\infty} = \|A\|_{1} \) Therefore, for a symmetric \(n \times n\) matrix A, it is true that \(\|A\|_{\infty} = \|A\|_{1}\).

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Most popular questions from this chapter

Let \(A\) be a \(3 \times 3\) matrix, and assume that \(A\) can be transformed into a lower triangular matrix \(L\) by using only column operations of type III; that is, \\[ A E_{1} E_{2} E_{3}=L \\] where \(E_{1}, E_{2}, E_{3}\) are elementary matrices of type III. Let $$ U=\left(E_{1} E_{2} E_{3}\right)^{-1} $$ Show that \(U\) is upper triangular with 1 's on the diagonal and \(A=L U .\) (This exercise illustrates a column version of Gaussian elimination.

Let \(A\) be an \(n \times n\) matrix and let \(Q\) and \(V\) be \(n \times n\) orthogonal matrices. Show that (a) \(\|Q A\|_{2}=\|A\|_{2}\) (b) \(\|A V\|_{2}=\|A\|_{2}\) (c) \(\|Q A V\|_{2}=\|A\|_{2}\)

Let \\[ A=\left(\begin{array}{rr} 1 & 2 \\ -1 & -1 \end{array}\right) \quad \text { and } \quad \mathbf{u}_{0}=\left(\begin{array}{l} 1 \\ 1 \end{array}\right) \\] (a) Compute \(\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3},\) and \(\mathbf{u}_{4},\) using the power method. (b) Explain why the power method will fail to converge in this case.

Let \\[ R=\left(\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\] Show that if \(\theta\) is not an integer multiple of \(\pi,\) then \(R\) can be factored into a product \(R=U L U\), where \(U=\left(\begin{array}{cc}1 & \frac{\cos \theta-1}{\sin \theta} \\ 0 & 1\end{array}\right) \quad\) and \(\quad L=\left(\begin{array}{cc}1 & 0 \\ \sin \theta & 1\end{array}\right)\) This type of factorization of a rotation matrix arises in applications involving wavelets and filter bases.

Let \(A=L U\), where \(L\) is lower triangular with 1 's on the diagonal and \(U\) is upper triangular. (a) How many scalar additions and multiplications are necessary to solve \(L \mathbf{y}=\mathbf{e}_{j}\) by forward substitution? (b) How many additions/subtractions and multiplications/divisions are necessary to solve \(A \mathbf{x}=\) \(\mathbf{e}_{j} ?\) The solution \(\mathbf{x}_{j}\) of \(A \mathbf{x}=\mathbf{e}_{j}\) will be the \(j\) th column of \(A^{-1}\) (c) Given the factorization \(A=L U\), how many additional multiplications/ divisions and additions/subtractions are needed to compute \(A^{-1} ?\)
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