/*! 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 19 Let \(\mathrm{V}\) be an inner p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(\mathrm{V}\) be an inner product space. Prove that (a) \(\|x \pm y\|^{2}=\|x\|^{2} \pm 2 \Re\langle x, y\rangle+\|y\|^{2}\) for all \(x, y \in \mathrm{V}\), where \(\Re\langle x, y\rangle\) denotes the real part of the complex number \(\langle x, y\rangle\). (b) \(|\|x\|-\|y\|| \leq\|x-y\|\) for all \(x, y \in \mathrm{V}\).

Short Answer

Expert verified
In an inner product space \(V\), we can prove that (a) \(\|x \pm y\|^{2}=\|x\|^{2} \pm 2 \Re\langle x, y\rangle+\|y\|^{2}\) for all \(x, y \in V\) by expanding the inner product and taking the real part; and (b) \(|\|x\|-\|y\|| \leq\|x-y\|\) for all \(x, y \in V\) using part (a) and the Cauchy-Schwarz inequality.

Step by step solution

01

Part (a): Prove the Equality

We are given an inner product space \(V\), and we want to prove that \(\|x \pm y\|^{2}=\|x\|^{2} \pm 2 \Re\langle x, y\rangle+\|y\|^{2}\) for all \(x, y \in V\). Let's start by considering the square of the norm of the sum or difference of two vectors: \(\|x \pm y\|^{2} = \langle x \pm y, x \pm y \rangle\) Now we can use the distributive property of the inner product: \(\langle x \pm y, x \pm y \rangle = \langle x, x \rangle \pm \langle x, y \rangle \pm \langle y, x \rangle + \langle y, y \rangle\) Now we know that \(\langle x, x \rangle = \|x\|^{2}\) and \(\langle y, y \rangle = \|y\|^{2}\). Also, the complex conjugate of \(\langle x, y \rangle\) is \(\langle y, x \rangle\): \(\|x \pm y\|^{2} = \|x\|^{2} \pm \langle x, y \rangle \pm \overline{\langle x, y \rangle} + \|y\|^{2}\) Taking the real part: \(\|x \pm y\|^{2} = \|x\|^{2} \pm 2\Re\langle x, y \rangle + \|y\|^{2}\)
02

Part (b): Prove the Inequality

We want to prove that \(|\|x\|-\|y\|| \leq\|x-y\|\) for all \(x, y \in V\). We can apply what we learned in part (a): \(\|x-y\|^{2} = \|x\|^{2} - 2\Re\langle x, y \rangle + \|y\|^{2}\) Now, using the Cauchy-Schwarz inequality, \(\langle x, y \rangle \leq \|x\| \|y\|\): \(-2\|x\|\|y\| \leq -2\Re\langle x, y \rangle \leq 2\|x\|\|y\|\) Shifting the terms: \( (\|x\| - \|y\|)^{2} \leq \|x\|^{2} - 2\Re\langle x, y \rangle + \|y\|^{2} \leq (\|x\| + \|y\|)^{2}\) Taking square roots: \(|\|x\| - \|y\|| \leq \|x - y\|\) Hence, the inequality is proven for all \(x, y \in V\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Norm of a Vector
The norm of a vector, often denoted as \( \|x\| \), measures the 'length' or 'magnitude' of a vector \( x \). In the context of an inner product space, it is defined using the inner product \( \langle x, x \rangle \) as \( \|x\| = \sqrt{\langle x, x \rangle} \).

The norm has several important properties that make it a fundamental tool in vector spaces. It is always non-negative, \( \|x\| \geq 0 \), and it is zero if and only if the vector itself is the zero vector. The norm also satisfies the triangle inequality, which states that the norm of the sum of two vectors is less than or equal to the sum of the norms of those vectors.

Understanding the norm is crucial to grasping concepts such as the distance between vectors and the Cauchy-Schwarz inequality. It's analogous to measuring the length of an object in physical space—except, in this case, we’re looking at the 'distance' from the origin to a point in a multi-dimensional space.
Cauchy-Schwarz Inequality
The Cauchy-Schwarz inequality is a powerful statement in mathematics that places a limit on the inner product of two vectors. For any vectors \( x \) and \( y \) in an inner product space, it asserts that \( |\langle x, y \rangle| \leq \|x\| \|y\| \), where \( |\langle x, y \rangle| \) denotes the absolute value of the inner product, and \( \|x\| \) and \( \|y\| \) represent the norms of \( x \) and \( y \) respectively.

This inequality implies that the absolute value of the inner product is at most the product of the magnitudes of the two vectors. It emphasizes the idea that the 'overlap' between vectors (as measured by the inner product) cannot exceed the product of their lengths. In the context of our exercise, it's used to show that the difference in lengths between two vectors is bounded by the norm of their difference. This foundational inequality is not only essential in abstract vector spaces but also has applications in subjects like statistics, physics, and economics.
Complex Inner Product
In the realms of vector spaces over the complex numbers, the complex inner product extends the notion of an inner product to include vectors with complex components. Unlike the real inner product, for complex vectors \( x \) and \( y \) the inner product \( \langle x, y \rangle \) can be a complex number, thus we often need to consider its real part, denoted as \( \Re\langle x, y \rangle \).

The complex inner product maintains similar properties to its real counterpart: it is conjugate symmetric, meaning \( \langle x, y \rangle = \overline{\langle y, x \rangle} \) where the overline denotes the complex conjugate, and it is linear in its first argument. An important usage of the complex inner product, as shown in the provided exercise, is in computing the square of the norm of the sum or difference of two vectors, taking into account that we need to consider the real part of the inner product when adding the contributions from \( \langle x, y \rangle \) and its conjugate \( \langle y, x \rangle \) to the formula.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\mathrm{W}=\operatorname{span}(\\{(i, 0,1)\\})\) in \(\mathrm{C}^{3}\). Find orthonormal bases for \(\mathrm{W}\) and \(\mathrm{W}^{\perp}\).

Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.

31\. Let \(\mathrm{H}_{u}\) be a Householder operator on a finite-dimensional inner product space V. Prove the following results. (a) \(\mathrm{H}_{u}\) is linear. (b) \(\mathrm{H}_{u}(x)=x\) if and only if \(x\) is orthogonal to \(u\). (c) \(\mathrm{H}_{u}(u)=-u\). (d) \(\mathrm{H}_{u}^{*}=\mathrm{H}_{u}\) and \(\mathrm{H}_{u}^{2}=\mathrm{I}\), and hence \(\mathrm{H}_{u}\) is a unitary [orthogonal] operator on \(\mathrm{V}\). (Note: If \(\mathrm{V}\) is a real inner product space, then in the language of Section \(6.11, \mathrm{H}_{u}\) is a reflection.) \({ }^{1}\) At one time, because of its great stability, this method for solving large systems of linear equations with a computer was being advocated as a better method than Gaussian elimination even though it requires about three times as much work. (Later, however, J. H. Wilkinson showed that if Gaussian elimination is done "properly," then it is nearly as stable as the orthogonalization method.)

Find the minimal solution to each of the following systems of linear equations. (a) \(x+2 y-z=12\) (b) \(2 x+3 y+z=2\) (c) \(x+y-z=0\) (d) \(\quad x+y+z-w=1\)

25\. Prove the converse to Exercise 24(a): Let \(V\) be a finite-dimensional real inner product space, and let \(H\) be a bilinear form on \(\mathrm{V}\). Then there exists a unique linear operator \(\mathrm{T}\) on \(\mathrm{V}\) such that \(H(x, y)=\langle x, \mathrm{~T}(y)\rangle\) for all \(x, y \in \mathrm{V}\). Hint: Choose an orthonormal basis \(\beta\) for \(\mathrm{V}\), let \(A=\) \(\psi_{\beta}(H)\), and let \(\mathrm{T}\) be the linear operator on \(\mathrm{V}\) such that \([\mathrm{T}]_{\beta}=A\). Visit goo.gl/bGAfSy for a solution.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.