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Chapter 5: Problem 24

Show that, in any vector space with a norm, $$\|-\mathbf{v}\|=\|\mathbf{v}\|$$

Short Answer

Expert verified
To show that \(\|-\mathbf{v}\|=\|\mathbf{v}\|\) in any vector space with a norm, we use the homogeneity property of norms which states \(\|cv\| = |c|\|v\|\). For \(c=-1\), we get \(\|(-1)\mathbf{v}\| = |-1|\|\mathbf{v}\|\), which simplifies to \(\|-\mathbf{v}\| = 1\|\mathbf{v}\|\). Since multiplying by 1 does not change the value, we conclude that \(\|-\mathbf{v}\| = \|\mathbf{v}\|\).

Step by step solution

01

Definition of a norm

A norm is a function \(\|\cdot\|: V \rightarrow \mathbb{R}\) on a vector space \(V\) that satisfies the following properties for all \(v, u \in V\) and \(c \in \mathbb{R}\): 1. \(\|v\| \geq 0\), and \(\|v\| = 0\) if and only if \(v = 0\), 2. \(\|v + u\| \leq \|v\| + \|u\|\) (Triangle Inequality), 3. \(\|cv\| = |c|\|v\|\) (Homogeneity). We are given a vector space with a norm, so these properties hold. Our goal is to use these properties to show that \(\|-\mathbf{v}\|=\|\mathbf{v}\|\).
02

Using the Homogeneity Property

To show that \(\|-\mathbf{v}\|=\|\mathbf{v}\|\), let's use the property of homogeneity for norms that states that \(\|cv\| = |c|\|v\|\). Here, let \(c=-1\): \[\|(-1)\mathbf{v}\| = |-1|\|\mathbf{v}\|\] Since \(|-1| = 1\), we have \[\|-1\mathbf{v}\| = 1\|\mathbf{v}\|\]
03

Simplifying and Concluding the Proof

Knowing that the additive inverse (negation) of the vector \(\mathbf{v}\) is \(-\mathbf{v} = (-1) \mathbf{v}\), we can replace the left side of the equation: \[\|-\mathbf{v}\| = 1\|\mathbf{v}\|\] Since multiplying by 1 does not change the value, we can simplify the equation: \[\|-\mathbf{v}\| = \|\mathbf{v}\|\] Thus, we have shown that in any vector space with a norm, the norm of a vector and its additive inverse (negation) are the same.

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Key Concepts

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

Norm Properties
A norm is a fundamental concept in mathematics, especially in linear algebra and functional analysis. It is a function that assigns a non-negative length or size to each vector in a vector space.
  • **Non-negativity**: For any vector \( v \), \( \|v\| \geq 0 \). Only the zero vector has a norm of zero, meaning \( \|v\| = 0 \) if and only if \( v = 0 \).
  • **Triangle Inequality**: This property asserts that for any vectors \( v \) and \( u \), the norm of their sum is at most the sum of their norms; mathematically, \( \|v + u\| \leq \|v\| + \|u\| \).
  • **Homogeneity**: Also known as scalability, this property means the norm of a scalar multiplied by a vector equals the absolute value of the scalar times the norm of the vector, \( \|cv\| = |c|\|v\| \) for any scalar \( c \).
A norm is a way to measure vectors' sizes, and these properties ensure that norms behave intuitively like lengths in the space.
Vector Space
A vector space is a collection of objects called vectors. It is an abstract structure that allows the operations of vector addition and scalar multiplication to satisfy certain axioms.
  • **Vectors**: The components of a vector space. They can be added together and multiplied by scalars (numbers).
  • **Scalars**: Elements that can stretch or shrink vectors when multiplied with them.
For a set to qualify as a vector space, it must adhere to rules such as closure under addition and scalar multiplication, existence of an additive identity (the zero vector), and every vector having an additive inverse. These qualities provide a framework for various mathematical operations and proofs. The exercise, for example, deals with the nature of norms within a vector space, showing how norms relate to the properties of the space.
Homogeneity
Homogeneity, in the context of norms, refers to how norms behave when vectors are scaled by a scalar. It is one of the defining characteristics of a norm. Homogeneity ensures that if a vector is multiplied by a scalar, its norm is affected linearly. Mathematically, this is expressed as:\[ \|cv\| = |c|\|v\| \]where \( c \) is a scalar and \( v \) is a vector.This principle is crucial because it maintains the proportionality of the vector size under scaling, which is why it plays a vital role in the proof provided in the exercise. When you multiply a vector by \(-1\), homogeneity guarantees that its norm remains unchanged, after simplifying absolute values.
Additive Inverse
The additive inverse of a vector is another vector that, when added to the original, yields the zero vector. In simple terms, if you have a vector \( \mathbf{v} \), its additive inverse is \(-\mathbf{v}\).
  • The notation for the additive inverse is consistent: \( -\mathbf{v} \) implies the vector \( \mathbf{v} \) with its direction reversed.
  • The key property of the additive inverse is: \( \mathbf{v} + (-\mathbf{v}) = 0 \).
In the context of norms in a vector space, the additive inverse plays a significant role. The exercise demonstrates that the norm of a vector and its additive inverse are equal, \( \|\mathbf{v}\| = \|-\mathbf{v}\| \), using the homogeneity property. This result is intuitive because flipping the direction of a vector does not change its length or magnitude.

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Most popular questions from this chapter

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 $$c_{1}^{2}+c_{2}^{2}=\|\mathbf{y}\|^{2}=y_{1}^{2}+y_{2}^{2}$$

Let $$A=\left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array}\right)$$ (a) Show that the column vectors of \(A\) form an orthonormal set in \(\mathbb{R}^{4}\) (b) Solve the least squares problem \(A \mathbf{x}=\mathbf{b}\) for each of the following choices of \(\mathbf{b}\) (i) \(\mathbf{b}=(4,0,0,0)^{T}\) (ii) \(\mathbf{b}=(1,2,3,4)^{T}\) (iii) \(\mathbf{b}=(1,1,2,2)^{T}\)

The functions \(\cos x\) and \(\sin x\) form an orthonormal \(\operatorname{set} \operatorname{in} C[-\pi, \pi] .\) If \(f(x)=3 \cos x+2 \sin x\) and \(g(x)=\cos x-\sin x\) use Corollary 5.5 .3 to determine the value of $$\langle f, g\rangle=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) g(x) d x$$

For each of the following, use the Gram-Schmidt process to find an orthonormal basis for \(R(A)\) (a) \(A=\left(\begin{array}{rr}-1 & 3 \\ 1 & 5\end{array}\right)\) (b) \(A=\left(\begin{array}{rr}2 & 5 \\ 1 & 10\end{array}\right)\)

Show that the Chebyshev polynomials have the following properties: (a) \(2 T_{m}(x) T_{n}(x)=T_{m+n}(x)+T_{m-n}(x),\) for \(m>n\) (b) \(T_{m}\left(T_{n}(x)\right)=T_{m n}(x)\)
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