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Chapter 5: Problem 27

The result of Exercise 26 is not valid for norms other than the norm derived from the inner product. Give an example of this in \(\mathbb{R}^{2}\) using \(\|\cdot\|_{1}\)

Short Answer

Expert verified
In the \(\|\cdot\|_{1}\) norm, consider the vectors \(x = (1, 0)\) and \(y = (0, 1)\). We have \(\|x\|_{1} = 1\), \(\|y\|_{1} = 1\), and \(\|x + y\|_{1} = 2\). Since \(\|x + y\|_{1} = \|x\|_{1} + \|y\|_{1}\) but neither \(x\) nor \(y\) is a non-negative multiple of the other, this counterexample demonstrates that the result of Exercise 26 is not valid for the \(\|\cdot\|_{1}\) norm.

Step by step solution

01

1. Recall the definition of the \(\|\cdot\|_{1}\) norm

The \(\|\cdot\|_{1}\) norm of a vector in \(\mathbb{R}^{2}\), denoted as \(\|x\|_{1}\), is the sum of the absolute values of its components. Formally, given a vector \(x = (x_1, x_2) \in \mathbb{R}^{2}\), the \(\|\cdot\|_{1}\) norm is defined as follows: \[\|x\|_{1} = |x_1| + |x_2|\]
02

2. Recall the result of Exercise 26 and its conditions

The result of Exercise 26 is the following: for any vectors \(x, y \in \mathbb{R}^{2}\), if \(\|x + y\| = \|x\| + \|y\|\), then one of \(y\) is a non-negative multiple of the other. This result applies to the case when the norm is derived from the inner product, which is Euclidean norm. We will now show that this result is not valid for the \(\|\cdot\|_{1}\) norm.
03

3. Construct a counterexample

We will show a pair of vectors \(x, y \in \mathbb{R}^{2}\) for which \(\|x + y\|_{1} = \|x\|_{1} + \|y\|_{1}\), but neither of those vectors is a non-negative multiple of the other. Let's consider the following vectors: \[x = (1, 0)\] and \[y = (0, 1)\] Now, we have: \[\begin{aligned} \|x\|_{1} &= |1| + |0| = 1 \\ \|y\|_{1} &= |0| + |1| = 1 \\ x + y &= (1, 0) + (0, 1) = (1,1) \\ \|x + y\|_{1} &= |1| + |1| = 2 \end{aligned}\] Here, \(\|x + y\|_{1} = \|x\|_{1} + \|y\|_{1}\) but neither \(x\) nor \(y\) is a non-negative multiple of the other. This provides a counterexample, demonstrating that the result of Exercise 26 is not valid for the \(\|\cdot\|_{1}\) norm.

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Most popular questions from this chapter

Let \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) be an orthonormal basis for an inner product space \(V .\) If \(\mathbf{x}=c_{1} \mathbf{u}_{1}+c_{2} \mathbf{u}_{2}+c_{3} \mathbf{u}_{3}\) is a vector with the properties \(\|\mathbf{x}\|=5,\left\langle\mathbf{u}_{1}, \mathbf{x}\right\rangle=4\) and \(\mathbf{x} \perp \mathbf{u}_{2},\) then what are the possible values of \(c_{1}\) \(c_{2},\) and \(c_{3} ?\)

Let \(\theta\) be a fixed real number and let $$\mathbf{x}_{1}=\left(\begin{array}{c} \cos \theta \\ \sin \theta \end{array}\right) \quad \text { and } \quad \mathbf{x}_{2}=\left(\begin{array}{r} -\sin \theta \\ \cos \theta \end{array}\right)$$ (a) Show that \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}\right\\}\) is an orthonormal basis for \(\mathbb{R}^{2}\) (b) Given a vector \(\mathbf{y}\) in \(\mathbb{R}^{2},\) write it as a linear combination \(c_{1} \mathbf{x}_{1}+c_{2} \mathbf{x}_{2}\) (c) Verify that $$c_{1}^{2}+c_{2}^{2}=\|\mathbf{y}\|^{2}=y_{1}^{2}+y_{2}^{2}$$

Show that $$\|\mathbf{x}\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|$$ defines a norm on \(\mathbb{R}^{n}\)

Compute \(\|\mathbf{x}\|_{1},\|\mathbf{x}\|_{2},\) and \(\|\mathbf{x}\|_{\infty}\) for each of the following vectors in \(\mathbb{R}^{3}\) : (a) \(\mathbf{x}=(-3,4,0)^{T}\) (b) \(\mathbf{x}=(-1,-1,2)^{T}\) (c) \(\mathbf{x}=(1,1,1)^{T}\)

The trace of an \(n \times n\) matrix \(C,\) denoted \(\operatorname{tr}(C),\) is the sum of its diagonal entries; that is, $$\operatorname{tr}(C)=c_{11}+c_{22}+\cdots+c_{n n}$$ If \(A\) and \(B\) are \(m \times n\) matrices, show that (a) \(\|A\|_{F}^{2}=\operatorname{tr}\left(A^{T} A\right)\) (b) \(\|A+B\|_{F}^{2}=\|A\|_{F}^{2}+2 \operatorname{tr}\left(A^{T} B\right)+\|B\|_{F}^{2}\)
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