/*! 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 8 Prove that there does not exist ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 8

Prove that there does not exist a self-adjoint operator \(T \in \mathcal{L}\left(\mathbf{R}^{3}\right)\) such that \(T(1,2,3)=(0,0,0)\) and \(T(2,5,7)=(2,5,7)\).

Short Answer

Expert verified
To prove that there does not exist a self-adjoint operator \(T \in \mathcal{L}\left(\mathbf{R}^{3}\right) \) such that \(T(1,2,3)=(0,0,0) \) and \(T(2,5,7)=(2,5,7)\), we consider the inner products: \( \langle T(u), v \rangle = \langle (0,0,0),(2,5,7) \rangle = 0 \) \( \langle u, T(v) \rangle = \langle (1,2,3), (2,5,7) \rangle = 33 \) Since 0 鈮� 33, the property of a self-adjoint operator (\( \langle T(x), y \rangle = \langle x, T(y) \rangle \)) is contradicted. Therefore, such a self-adjoint operator T does not exist.

Step by step solution

01

Understand Self-Adjoint Operators

A linear operator T is self-adjoint if, for all x and y in the linear space \( \mathcal{L}\left(\mathbf{R}^{3}\right) \), we have: \( \langle T(x), y \rangle = \langle x, T(y) \rangle \) Now, let's apply this property to the given vectors.
02

Check Whether the Given Property Holds for the Vectors

Let's specifically look at the given vectors (1,2,3) and (2,5,7). We will denote them as u and v respectively: u = (1,2,3) v = (2,5,7) According to property 2, T(u) = (0,0,0). Now, for T to be self-adjoint, we must have: \( \langle T(u), v \rangle = \langle u, T(v) \rangle \) Substituting the given values for T(u) and T(v): \( \langle (0,0,0), v \rangle = \langle u, (2,5,7) \rangle \)
03

Compute the Inner Products

Now, we compute the inner products on both sides of the equation: Left side: \( \langle (0,0,0), (2,5,7) \rangle = (0)(2) + (0)(5) + (0)(7) = 0 \) Right side: \( \langle (1,2,3), (2,5,7) \rangle = (1)(2) + (2)(5) + (3)(7) = 2 + 10 + 21 = 33 \)
04

Compare the Results

After calculating the inner products, we see that: 0 鈮� 33 This inequality contradicts the property of being self-adjoint: \( \langle T(u), v \rangle = \langle u, T(v) \rangle \) Therefore, there does not exist a self-adjoint operator T in \( \mathcal{L}\left(\mathbf{R}^{3}\right) \) such that T(1,2,3) = (0,0,0) and T(2,5,7) = (2,5,7).

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 91影视!

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

Give an example of an operator \(T\) on an inner product space such that \(T\) has an invariant subspace whose orthogonal complement is not invariant under \(T\)

Prove or disprove: the identity operator on \(\mathbf{F}^{2}\) has infinitely many self-adjoint square roots.

Suppose \(T \in \mathcal{L}(V)\) has singular-value decomposition given by $$T v=s_{1}\left\langle v, e_{1}\right\rangle f_{1}+\cdots+s_{n}\left\langle v, e_{n}\right\rangle f_{n}$$ for every \(v \in V\), where \(s_{1}, \ldots, s_{n}\) are the singular values of \(T\) and \(\left(e_{1}, \ldots, e_{n}\right)\) and \(\left(f_{1}, \ldots, f_{n}\right)\) are orthonormal bases of \(V\) (a) Prove that $$T^{*} v=s_{1}\left\langle v, f_{1}\right\rangle e_{1}+\cdots+s_{n}\left\langle v, f_{n}\right\rangle e_{n}$$ for every \(v \in V\) (b) Prove that if \(T\) is invertible, then $$T^{-1} v=\frac{\left\langle v, f_{1}\right\rangle e_{1}}{s_{1}}+\dots+\frac{\left\langle v, f_{n}\right\rangle e_{n}}{s_{n}}$$ for every \(v \in V\)

Prove that a normal operator on a complex inner-product space is self-adjoint if and only if all its eigenvalues are real.

Define \(T \in \mathcal{L}\left(\mathbf{F}^{3}\right)\) by $$T\left(z_{1}, z_{2}, z_{3}\right)=\left(z_{3}, 2 z_{1}, 3 z_{2}\right)$$ Find (explicitly) an isometry \(S \in \mathcal{L}\left(\mathbf{F}^{3}\right)\) such that \(T=S \sqrt{T^{*} T}\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation
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.