/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 The following definition is used... [FREE SOLUTION] | 91Ó°ÊÓ

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The following definition is used in Exercises 26-30. Definition. Let \(\mathrm{V}\) be a vector space over \(F\), where \(F\) is either \(R\) or \(C\). Regardless of whether \(\mathrm{V}\) is or is not an inner product space, we may still define a norm \(\|\cdot\|_{V}\) as a real-valued function on \(\mathrm{V}\) satisfying the following three conditions for all \(x, y \in \mathrm{V}\) and \(a \in F\) : (1) \(\|x\|_{v} \geq 0\), and \(\|x\|_{v}=0\) if and only if \(x=0\). (2) \(\|a x\|_{V}=|a| \cdot\|x\|_{V}\). (3) \(\|x+y\|_{v} \leq\|x\|_{v}+\|y\|_{v}\). Let \(\|\cdot\|_{v}\) be a norm on a vector space \(\mathrm{V}\), and define, for each ordered pair of vectors, the scalar \(d(x, y)=\|x-y\|_{v}\), called the distance between \(x\) and \(y\). Prove the following results for all \(x, y, z \in \mathrm{V}\). (a) \(d(x, y) \geq 0\). (b) \(d(x, y)=d(y, x)\). (c) \(d(x, y) \leq d(x, z)+d(z, y)\). (d) \(d(x, x)=0\) if and only if \(x=0\). (e) \(d(x, y) \neq 0\) if \(x \neq y\).

Short Answer

Expert verified
In summary, given a vector space \(V\) and a norm \(\|\cdot\|_{V}\) with certain properties, we've established the following results regarding the distance function \(d(x, y) = \|x - y\|_{V}\) for all \(x, y \in V\): (a) Non-negativity: \(d(x, y) \geq 0\) (b) Symmetry: \(d(x, y) = d(y, x)\) (c) Triangle inequality: \(d(x, y) \leq d(x, z) + d(z, y)\) (d) Zero-distance property: \(d(x, x) = 0\) if and only if \(x = 0\) (e) Distinct elements property: \(d(x, y) \neq 0\) if \(x \neq y\)

Step by step solution

01

(a) Prove non-negativity property of d(x, y)

To show that \(d(x, y) \geq 0\), we use property (1) of the norm, which states that \(\|z\|_{V} \geq 0\) for all \(z \in V\). By definition, \(d(x, y) = \|x - y\|_{V}\). Since \(x - y \in V\), we have \(\|x - y\|_{V} \geq 0\). Therefore, \(d(x, y) \geq 0\).
02

(b) Prove symmetry property of d(x, y)

To prove that \(d(x, y) = d(y, x)\), we will use property (1) and (2) of the norm. First, we have: \[d(x, y) = \|x - y\|_{V}\] Now, consider the term inside the norm: \(x - y = -(y - x)\). By property (2), this means that: \[\|x - y\|_{V} = \|-(y - x)\|_{V} = |-1| \cdot \|(y - x)\|_{V} = \|y - x\|_{V}\] Thus, we have \(d(x, y) = d(y, x)\).
03

(c) Prove triangle inequality property of d(x, y)

To prove the triangle inequality, we will use property (3) of the norm. We have: \(d(x, y) = \|x - y\|_{V}\) Now, we need to compare it with \(d(x, z) + d(z, y)\). Using the definition of distance, we have: \(d(x, z) + d(z, y) = \|x - z\|_{V} + \|z - y\|_{V}\) Using property (3) of the norm, the triangle inequality, we have: \(\|x - y\|_{V} \leq \|x - z\|_{V} + \|z - y\|_{V}\) Rewriting this in terms of the distance function, we get: \(d(x, y) \leq d(x, z) + d(z, y)\)
04

(d) Prove zero-distance property of d(x, x)

For any vector \(x \in V\), we have: \(d(x, x) = \|x - x\|_{V} = \|0\|_{V}\) Using property (1) of the norm, we see that if \(x = 0\), then \(\|0\|_{V} = 0\), and therefore \(d(x, x) = 0\). Conversely, if \(d(x, x) = 0\), then \(\|x - x\|_{V} = 0\), which implies \(x - x = 0\), and thus \(x = 0\).
05

(e) Prove distinct elements property of d(x, y)

If \(x \neq y\), then \(x - y \neq 0\), since otherwise \(x = y\). By property (1) of the norm, we have \(\|x - y\|_{V} > 0\) because \(x - y \neq 0\). Therefore, we get: \(d(x, y) = \|x - y\|_{V} > 0\) So, \(d(x, y) \neq 0\) if \(x \neq y\). In conclusion, we have proved all the desired properties of the distance function using the properties of the given norm on the vector space \(V\).

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Key Concepts

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

Vector Space
A vector space is a fundamental concept in linear algebra that plays a crucial role in various areas of mathematics and physics. It is a collection of objects known as vectors, which can be added together and multiplied by scalars (numbers) to produce another vector within the same space.

In a vector space, two operations are defined: vector addition and scalar multiplication. These operations must adhere to certain rules, called axioms, such as associativity, commutativity of addition, existence of an additive identity (zero vector), and others. The concept of a vector space extends beyond mere three-dimensional space to encompass an infinite number of dimensions, and the vectors are not restricted just to physical quantities like force or velocity; they can be functions, symbols, or sequences, as long as they satisfy the vector space axioms.
Triangle Inequality
One of the fundamental properties of vector spaces equipped with a norm is the triangle inequality. This property is important as it provides a way to gauge the distance between vectors not directly related.

The triangle inequality states that for any vectors \(x\), \(y\), and \(z\) in a vector space \(V\), the sum of the lengths (norms) of any two sides of a triangle is always greater than or equal to the length of the third side. In mathematical terms, for any vectors \(x\) and \(y\) in \(V\), \(\|x+y\|_V \leq \|x\|_V + \|y\|_V\). This property mirrors the geometric intuition behind the triangle inequality in Euclidean space and is central to the concept of metric spaces, where distance measurements are paramount.
Real-valued Function
A real-valued function is a function that assigns a real number to each element in its domain. In the context of vector spaces, norms are examples of real-valued functions. As demonstrated in the exercise, a norm \(\|\cdot\|_V\) assigns a non-negative real number to each vector in the vector space \(V\), which represents the size or length of the vector.

Real-valued functions are omnipresent in mathematics and are used to describe diverse phenomena in real-world applications, from physics to economics. These functions have varied operations and properties, such as continuity, differentiability, and integrability, which are studied extensively in analysis and calculus.
Inner Product Space
An inner product space is a special type of vector space equipped with an additional structure called an inner product. The inner product is a function that takes two vectors and returns a scalar, which intuitively reflects the 'dot product' in Euclidean vector spaces. It is denoted as \(\langle x, y \rangle\) for two vectors \(x\) and \(y\).

An inner product allows for the definition of geometric concepts like angles and lengths within vector spaces. In an inner product space, the norm of a vector \(x\) can be defined using the inner product as \(\|x\| = \sqrt{\langle x, x \rangle}\). This enables one to establish a true sense of distance and orthogonality, similar to what we are accustomed to in Euclidean geometry, but in a more abstract setting. Inner product spaces form the backbone of functional analysis and quantum mechanics, where they are used to describe the state space of quantum systems.

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Most popular questions from this chapter

Let \(\mathrm{T}\) and \(\mathrm{U}\) be positive definite operators on an inner product space V. Prove the following results. (a) \(\mathrm{T}+\mathrm{U}\) is positive definite. (b) If \(c>0\), then \(c \mathrm{~T}\) is positive definite. (c) \(\mathrm{T}^{-1}\) is positive definite. Visit goo.gl/cQch7i for a solution.

Let \(U\) be a unitary operator on an inner product space \(V\), and let \(W\) be a finite-dimensional U-invariant subspace of V. Prove that (a) \(U(W)=W\); (b) \(\mathrm{W}^{\perp}\) is U-invariant. Contrast (b) with Exercise 16 .

Consider three coordinate systems \(S, S^{\prime}\), and \(S^{\prime \prime}\) with the corresponding axes \(\left(x, x^{\prime}, x^{\prime \prime} ; y, y^{\prime}, y^{\prime \prime} ;\right.\) and \(\left.z, z^{\prime}, z^{\prime \prime}\right)\) parallel and such that the \(x-, x^{\prime}-\), and \(x^{\prime \prime}\)-axes coincide. Suppose that \(S^{\prime}\) is moving past \(S\) at a velocity \(v_{1}>0\) (as measured on \(S\) ), \(S^{\prime \prime}\) is moving past \(S^{\prime}\) at a velocity \(v_{2}>0\) (as measured on \(S^{\prime}\) ), and \(S^{\prime \prime}\) is moving past \(S\) at a velocity \(v_{3}>0\) (as measured on \(S\) ), and that there are three clocks \(C, C^{\prime}\), and \(C^{\prime \prime}\) such that \(C\) is stationary relative to \(S, C^{\prime}\) is stationary relative to \(S^{\prime}\), and \(C^{\prime \prime}\) is stationary relative to \(S^{\prime \prime}\). Suppose that when measured on any of the three clocks, all the origins of \(S, S^{\prime}\), and \(S^{\prime \prime}\) coincide at time 0 . Assuming that \(\mathrm{T}_{v_{3}}=\mathrm{T}_{v_{2}} \mathrm{~T}_{v_{1}}\) (i.e., \(B_{v_{3}}=B_{v_{2}} B_{v_{1}}\) ), prove that $$ v_{3}=\frac{v_{1}+v_{2}}{1+v_{1} v_{2}} . $$ Note that substituting \(v_{2}=1\) in this equation yields \(v_{3}=1\). This tells us that the speed of light as measured in \(S\) or \(S^{\prime}\) is the same. Why would we be surprised if this were not the case?

Determine which of the mappings that follow are bilinear forms. Justify your answers. (a) Let \(\mathrm{V}=\mathrm{C}[0,1]\) be the space of continuous real-valued functions on the closed interval \([0,1]\). For \(f, g \in \mathrm{V}\), define $$ H(f, g)=\int_{0}^{1} f(t) g(t) d t . $$ (b) Let \(\mathrm{V}\) be a vector space over \(F\), and let \(J \in \mathcal{B}(\mathrm{V})\) be nonzero. Define \(H: \mathrm{V} \times \mathrm{V} \rightarrow F\) by $$ H(x, y)=[J(x, y)]^{2} \quad \text { for all } x, y \in \mathrm{V} . $$ (c) Define \(H: R \times R \rightarrow R\) by \(H\left(t_{1}, t_{2}\right)=t_{1}+2 t_{2}\). (d) Consider the vectors of \(\mathrm{R}^{2}\) as column vectors, and let \(H: \mathrm{R}^{2} \rightarrow R\) be the function defined by \(H(x, y)=\operatorname{det}(x, y)\), the determinant of the \(2 \times 2\) matrix with columns \(x\) and \(y\). (e) Let \(\mathrm{V}\) be a real inner product space, and let \(H: \mathrm{V} \times \mathrm{V} \rightarrow R\) be the function defined by \(H(x, y)=\langle x, y\rangle\) for \(x, y \in \mathrm{V}\). (f) Let \(\mathrm{V}\) be a complex inner product space, and let \(H: \mathrm{V} \times \mathrm{V} \rightarrow C\) be the function defined by \(H(x, y)=\langle x, y\rangle\) for \(x, y \in \mathrm{V}\).

Let \(\mathrm{V}=\mathrm{C}([-1,1]) .\) Suppose that \(\mathrm{W}_{e}\) and \(\mathrm{W}_{o}\) denote the subspaces of \(\mathrm{V}\) consisting of the even and odd functions, respectively. (See Exercise 22 of Section 1.3.) Prove that \(\mathrm{W}_{e}^{\perp}=\mathrm{W}_{o}\), where the inner product on \(\mathrm{V}\) is defined by $$ \langle f, g\rangle=\int_{-1}^{1} f(t) g(t) d t . $$

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