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Chapter 6: Problem 17

Prove: If \(\mathrm{A}\) has at least one nonzero entry, then \(\|\mathbf{A}\| \neq 0\).

Short Answer

Expert verified
Question: Show that if a matrix A has at least one nonzero entry, its norm is not equal to zero. Answer: Given that matrix A has at least one nonzero entry, we can calculate its Frobenius norm, which is a measure of the matrix's magnitude. In this case, the norm is positive due to the presence of the nonzero entry, proving that the norm of matrix A is not equal to zero.

Step by step solution

01

Define the matrix norm

The matrix norm, denoted as ||A||, is a measure of the magnitude of a matrix A, and can be defined in various ways. In this exercise, we will use the Frobenius norm, which for a matrix A is defined as: \(||A||_F = \sqrt{\sum_{i=1}^{m} \sum_{j=1}^{n} |a_{ij}|^2}\) Where m and n are the number of rows and columns in A, respectively, and \(a_{ij}\) denotes the entry in the ith row and jth column of A.
02

Consider the nonzero entry

Since matrix A has at least one nonzero entry, let \(a_{k \ell}\) be that entry (\(a_{k \ell} \neq 0\)).
03

Calculate the matrix norm

Now we will calculate the matrix norm for A using the Frobenius norm definition: \(||A||_F = \sqrt{\sum_{i=1}^{m} \sum_{j=1}^{n} |a_{ij}|^2}\) Since \(a_{k\ell} \neq 0\), it contributes a positive amount to the sum, so we can write: \(||A||_F = \sqrt{|a_{k \ell}|^2 + \sum_{\substack{i=1,j=1\\(i,j) \neq (k,\ell)}}^{m,n} |a_{ij}|^2}\) Since \(|a_{k \ell}|^2\) is positive (and the rest of the terms are non-negative), the sum inside the square root is also strictly positive: \(||A||_F = \sqrt{c}\), where \(c > 0\)
04

Conclusion

We have shown that \(||A||_F = \sqrt{c}\) for some positive constant c, and therefore \(||A||_F > 0\) and \(\|\mathbf{A}\| \neq 0\). This concludes the proof.

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Key Concepts

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

Frobenius norm
The Frobenius norm, sometimes referred to as the Hilbert-Schmidt norm, is a measure used to determine the magnitude of a matrix. This norm is one of the simplest and most commonly used among the many ways to calculate matrix norms. It is defined for a matrix \( A \) with elements \( a_{ij} \) as:
\[ ||A||_F = \sqrt{\sum_{i=1}^{m} \sum_{j=1}^{n} |a_{ij}|^2} \]
Think of it as the equivalent of the Euclidean norm for matrices, where we take each entry of the matrix, square it, sum all of these squares, and then take the square root of the total. This takes into account all the entries of the matrix, making it a comprehensive measure of the matrix's size. In the context of the problem provided, the Frobenius norm offers a direct way to show that a matrix with at least one nonzero entry must have a nonzero norm.
Matrix magnitude
When we talk about the magnitude of a matrix, we're essentially discussing a scalar that represents the 'size' of the elements within the matrix. In the real number system, magnitude is simply the absolute value of a number, telling us how far it is from zero. For matrices, the concept is extended to accommodate all the individual elements, using norms to calculate a single value that represents the entire matrix.
The Frobenius norm is a popular method for measuring the magnitude of a matrix because it is very straightforward to calculate and interprets the matrix elements in a geometric context, reminiscent of vectors in space. It captures the intuitive sense that a matrix with larger entries is 'bigger' in terms of its Frobenius magnitude.
Nonzero matrix entry
A nonzero matrix entry is any element within a matrix that is not equal to zero. This detail is critical when determining the characteristics of a matrix, such as its rank, determinant, and norms. For instance, if a matrix \( A \) has at least one nonzero entry, like \( a_{k \ell} eq 0 \), this single entry ensures that the Frobenius norm of the matrix is greater than zero.
Mathematically, a nonzero entry means that the entry contributes a positive amount when calculating squared values. Consequently, in the Frobenius norm, it guarantees that the sum of the squared elements is strictly positive, ensuring that the overall norm calculation yields a positive scalar value.
Real analysis
Real analysis is a branch of mathematics that deals with the behavior and properties of real numbers, sequences and series of real numbers, and real-valued functions. It includes theories such as limits, continuity, integration, and differentiation. In the context of the given proof about the Frobenius norm of a matrix, real analysis plays a crucial role, as the proof relies on concepts such as square roots and properties of real numbers (specifically, that the square of a nonzero real number is positive). Additionally, when considering the Frobenius norm in a broader scope, real analysis offers tools to study convergence, normed vector spaces, and the behavior of sequences of matrices, deeply enriching the understanding of how norms function within the matrix algebra framework.

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Most popular questions from this chapter

Calculate all first and second partial derivatives at \(\left(x_{0}, y_{0}\right)=(1,1)\) of the functions \(u\) and \(v\) that satisfy $$ \begin{array}{l} x^{2}+y^{2}+u^{2}+v^{2}=3 \\ x+y+u+v=3 \end{array} \quad u(1,1)=0, \quad v(1,1)=1 $$

Find \(\mathbf{F}^{-1}\) and \(\left(\mathbf{F}^{-1}\right)^{\prime}:\) (a) \(\left[\begin{array}{l}u \\ v\end{array}\right]=\mathbf{F}(x, y)=\left[\begin{array}{r}4 x+2 y \\ -3 x+y\end{array}\right]\) (b) \(\left[\begin{array}{c}u \\ v \\ w\end{array}\right]=\mathbf{F}(x, y, z)=\left[\begin{array}{r}-x+y+2 z \\ 3 x+y-4 z \\ -x-y+2 z\end{array}\right]\)

Suppose that \(\mathbf{F}\) and \(\mathbf{G}\) are transformations from \(\mathbb{R}^{n}\) to \(\mathbb{R}^{m}\) with common domain \(D\). Show that if \(\mathbf{F}\) and \(\mathbf{G}\) are continuous at \(\mathbf{X}_{0} \in D,\) then so are \(\mathbf{F}+\mathbf{G}\) and \(\mathbf{F}-\mathbf{G}\).

Suppose that \(\mathbf{F}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) and \(\overline{\mathbf{X}}\) is a limit point of \(D_{\mathbf{F}}\) contained in \(D_{\mathbf{F}} .\) Show that \(\mathbf{F}\) is continuous at \(\overline{\mathbf{X}}\) if and only if \(\lim _{k \rightarrow \infty} \mathbf{F}\left(\mathbf{X}_{k}\right)=\mathbf{F}(\overline{\mathbf{X}})\) whenever \(\left\\{\mathbf{X}_{k}\right\\}\) is a sequence of points in \(D_{\mathbf{F}}\) such that \(\lim _{k \rightarrow \infty} \mathbf{X}_{k}=\overline{\mathbf{X}}\). HINT: See Exercise 5.2.15.

Prove: If \(\mathbf{L}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) is defined by \(\mathbf{L}(\mathbf{X})=\mathbf{A}(\mathbf{X}),\) where \(\mathbf{A}\) is nonsingular, then $$ |\mathbf{L}(\mathbf{X})-\mathbf{L}(\mathbf{Y})| \geq \frac{1}{\left\|\mathbf{A}^{-1}\right\|}|\mathbf{X}-\mathbf{Y}| $$ for all \(\mathbf{X}\) and \(\mathbf{Y}\) in \(\mathbb{R}^{n}\).
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