/*! 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 22 Show that any operator \(T\) is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 22

Show that any operator \(T\) is the sum of a self-adjoint operator and a skew- adjoint operator.

Short Answer

Expert verified
To show that any operator \(T\) is the sum of a self-adjoint operator and a skew-adjoint operator, we can express \(T\) in terms of its adjoint \(T^*\). Let \(A\) be the self-adjoint operator and \(B\) be the skew-adjoint operator. We can then find the expressions for \(A\) and \(B\) by solving the following two equations: \(T = A + B\) and \(T^* = A - B\). Solving for \(A\) and \(B\), we obtain \(A = \frac{1}{2}(T + T^*)\) and \(B = \frac{1}{2}(T - T^*)\). Substituting these expressions back into the equation \(T = A + B\), we find that \(T = T\), confirming that any operator \(T\) can indeed be written as the sum of a self-adjoint operator and a skew-adjoint operator.

Step by step solution

01

Express T in terms of self-adjoint and skew-adjoint operators

Let's denote the self-adjoint operator as A and the skew-adjoint operator as B. We want to find A and B such that \(T = A + B\). Since A is self-adjoint and B is skew-adjoint, their respective adjoints will be equal to themselves and their negations, i.e., \(A^* = A\) and \(B^* = -B\). Now, let's add A and B and see if we can find a relationship with their adjoints: \(T = A + B\) Taking the adjoints on both sides, we have: \(T^* = A^* + B^*\) We know that \(A^* = A\) and \(B^* = -B\), so the equation becomes: \(T^* = A - B\)
02

Calculate the self-adjoint and skew-adjoint operators

From step 1, we have two equations: \(T = A + B\) and \(T^* = A - B\) Now we will solve these two equations for A and B. To find A, we can add the two equations: \(T + T^* = A + B + A - B\) Simplifying, we get: \(T + T^* = 2A\) So, dividing by 2: \(A = \frac{1}{2}(T + T^*)\) To find B, we can subtract the second equation from the first one: \(T - T^* = A + B - (A - B)\) Simplifying, we get: \(T - T^* = 2B\) So, dividing by 2: \(B = \frac{1}{2}(T - T^*)\)
03

Show that T is the sum of self-adjoint and skew-adjoint operators

Now that we have found the self-adjoint operator A and the skew-adjoint operator B, we can show that T is their sum: \(T = A + B\) Replacing A and B with the expressions found in step 2: \(T = \frac{1}{2}(T + T^*) + \frac{1}{2}(T - T^*)\) Simplifying the expression: \(T = T\) This confirms that any operator T can be written as the sum of a self-adjoint operator A and a skew-adjoint operator B, as desired.

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

Find the adjoint of \(G: \mathbf{C}^{3} \rightarrow \mathbf{C}^{3}\) defined by G(x, y, z)=[2 x+(1-i) y, \quad(3+2 i) x-4 i z, \quad 2 i x+(4-3 i) y-3 z]

Let \(V\) be an inner product space. Recall that each \(u \in V\) determines a linear functional \(\hat{u}\) in the dual space \(V^{*}\) by the definition \(\hat{u}(v)=\langle v, u\rangle\) for every \(v \in V\). (See the text immediately preceding Theorem 13.3 .) Show that the map \(u \mapsto \hat{u}\) is linear and nonsingular, and hence an isomorphism from \(V\) onto \(V^{*}\)

Find the adjoint of: (a) \(A=\left[\begin{array}{ll}5-2 i & 3+7 i \\ 4-6 i & 8+3 i\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{rr}3 & 5 i \\ i & -2 i\end{array}\right]\) (c) \(C=\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right]\)

Let \(T\) be a symmetric operator. Show that (a) The characteristic polynomial \(\Delta(t)\) of \(T\) is a product of linear polynomials (over \(\mathbf{R})\); (b) \(T\) has a nonzero eigenvector. (a) Let \(A\) be a matrix representing \(T\) relative to an orthonormal basis of \(V\); then \(A=A^{T}\). Let \(\Delta(t)\) be the characteristic polynomial of \(A\). Viewing \(A\) as a complex self-adjoint operator, \(A\) has only real eigenvalues by Theorem 13.4. Thus, $$ \Delta(t)=\left(t-\lambda_{1}\right)\left(t-\lambda_{2}\right) \cdots\left(t-\lambda_{n}\right) $$ where the \(\lambda_{i}\) are all real. In other words, \(\Delta(t)\) is a product of linear polynomials over \(\mathbf{R}\). (b) By (a), \(T\) has at least one (real) eigenvalue. Hence, \(T\) has a nonzero eigenvector.

\(\operatorname{Let} A=\left[\begin{array}{ll}2 & i \\ i & 2\end{array}\right] .\) Verify that \(A\) is normal. Find a unitary matrix \(P\) such that \(P^{*} A P\) is diagonal. Find \(P^{*} A P\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Calculus

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.