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Chapter 7: Problem 14

Show that \(\|A\|_{F}=\left\|A^{T}\right\|_{F}\)

Short Answer

Expert verified
To show that \(\|A\|_{F}= \|A^{T}\|_{F}\), we can use the definition of the Frobenius norm and transpose of a matrix. The Frobenius norm of A is \(\|A\|_{F} = \sqrt{\sum_{i=1}^{m} \sum_{j=1}^{n} |a_{ij}|^{2}}\), and the transpose of A is \((A^{T})_{ij} = a_{ji}\). We can calculate the Frobenius norm of \(A^{T}\) as \(\|A^{T}\|_{F} = \sqrt{\sum_{i=1}^{n} \sum_{j=1}^{m} |a_{ji}|^{2}}\). Now, since the absolute value and squaring operations do not depend on the order of indices i and j, the sums in the Frobenius norms of A and A^T are equal, leading to the conclusion that \(\|A\|_{F} = \|A^{T}\|_{F}\).

Step by step solution

01

Write down the definition of the Frobenius norm

The Frobenius norm of a matrix A, denoted as \(\|A\|_{F}\), is defined as the square root of the sum of the squares of all matrix elements. In other words, \(\|A\|_{F} = \sqrt{\sum_{i=1}^{m} \sum_{j=1}^{n} |a_{ij}|^{2}}\) where A is an m x n matrix, and \(a_{ij}\) are the elements of the matrix A.
02

Write down the definition of the transpose of a matrix

The transpose of a matrix A, denoted as \(A^{T}\), is obtained by interchanging the rows and columns of A: \((A^{T})_{ij} = a_{ji}\)
03

Calculate the Frobenius norm of A^T using the definitions

Using the definition of the Frobenius norm, we have: \(\|A^{T}\|_{F} = \sqrt{\sum_{i=1}^{n} \sum_{j=1}^{m} |a_{ji}|^{2}}\)
04

Show that the Frobenius norm of A is equal to the Frobenius norm of A^T

To prove that \(\|A\|_{F} = \|A^{T}\|_{F}\), we will show that the sum of the squares of the elements in the original matrix A is equal to the sum of the squares of the elements in its transpose A^T. From Step 3, we see that the Frobenius norm of A^T is the sum of the squares of the elements of A^T, but with the indices i and j swapped. Since the absolute value and squaring operations do not depend on the order of indices i and j, the sum of the squares of the elements of A^T will be equal to the sum of the squares of the elements in A: \(\sum_{i=1}^{n} \sum_{j=1}^{m} |a_{ji}|^{2} = \sum_{i=1}^{m} \sum_{j=1}^{n} |a_{ij}|^{2}\) Therefore, the Frobenius norms of A and A^T are equal: \(\|A\|_{F} = \|A^{T}\|_{F}\)

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Most popular questions from this chapter

Let \(A_{1}=U \Sigma_{1} V^{T}\) and \(A_{2}=U \Sigma_{2} V^{T},\) where $$\Sigma_{1}=\left(\begin{array}{ccccc}\sigma_{1} & & & & \\ & \ddots & & \\\ & & \sigma_{r-1} & \\ & & & 0 \\ & & & & \ddots \\ & & & & \\ & & & \end{array}\right)$$ and $$\Sigma_{2}=\left(\begin{array}{cccccc} ^{0} 1 & & & & & \\ & \ddots & & & \\ & & \sigma_{r-1} & & \\ & & & & \sigma_{r} \\ & & & & 0 \\ & & & & & \ddots \\ & & & & & & 0 \end{array}\right)$$ and \(\sigma_{r}=\epsilon>0 .\) What are the values of \(\left\|A_{1}-A_{2}\right\|_{F}\) and \(\left\|A_{1}^{+}-A_{2}^{+}\right\|_{F} ?\) What happens to these values as we let \(\epsilon \rightarrow 0 ?\)

Let \(A\) be a nonsingular \(n \times n\) matrix, and let \(\|\cdot\|_{M}\) denote a matrix norm that is compatible with some vector norm on \(\mathbb{R}^{n}\). Show that \\[ \operatorname{cond}_{M}(A) \geq 1 \\]

If \(\mathbf{x} \in \mathbb{R}^{m},\) we can think of \(\mathbf{x}\) as an \(m \times 1\) matrix. If \(\mathbf{x} \neq \mathbf{0},\) we can then define a \(1 \times m\) matrix \(X\) by \\[ X=\frac{1}{\|\mathbf{x}\|_{2}^{2}} \mathbf{x}^{T} \\] Show that \(X\) and \(\mathbf{x}\) satisfy the four Penrose conditions and, consequently, that \\[ \mathbf{x}^{+}=X=\frac{1}{\|\mathbf{x}\|_{2}^{2}} \mathbf{x}^{T} \\]

Find the three-digit decimal floating-point representation of each of the following numbers: (a) 2312 (b) 32.56 (c) 0.01277 (d) 82,431

Let \\[ A=\left(\begin{array}{rr} 1 & -0.99 \\ -1 & 1 \end{array}\right) \\] Find \(A^{-1}\) and \(\operatorname{cond}_{\infty}(A)\)
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