Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 36

If \(A_{1}, A_{2}, \ldots . A_{n}\) are operators on a dense domain such that $$ \sum_{i=1}^{n} A_{1}^{*} A_{1}=0 $$ show that \(A_{1}=A_{2}=\cdots=A_{n}=0 .\)

Short Answer

Expert verified
Applying the properties of operators and their adjoints, along with comparisons to the properties of inner products, demonstrates that each operator \( A_{i}\) must be zero.

Step by step solution

01

Understand the given condition

The problem begins with the condition \( \sum_{i=1}^{n} A_{1}^{*} A_{1} = 0 \). This can be understood as the sum of the product of each operator with its own adjoint being zero.
02

Apply the properties of inner product

The product of an operator and its adjoint closely resembles the concept of inner product in linear algebra. Recall that in this context, the inner product of a vector with itself equals zero if and only if the vector is zero. This implies that for every \(i\), we must have \( A_{i}^{*} A_{i} = 0\).
03

Conclude that each operator is zero

Recall that the only way for an operator's product with its own adjoint (or for an inner product of a vector with itself) to be zero is for the operator (or vector) to be zero. Therefore, each \(A_{i}\) must be zero.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Show that a vector subspace is a closed subset of \(\mathcal{H}\) with respect to the norm topology iff the limit of every sequence of vectors in \(V\) belongs to \(V\).

For every bounded operator \(A\) on a Hilbert space \(\mathcal{H}\) show that the exponential operator $$ \mathrm{e}^{A}=\sum_{n=0}^{\infty} \frac{A^{n}}{n !} $$ is well-defined and bounded on \(\mathcal{H}\). Show that (a) \(e^{0}=1\) (b) For all posituve integers \(n,\left(c^{A}\right)^{n}=e^{n A}\). (c) \(\mathrm{e}^{A}\) is invertuble for all bounded operators \(A\) (even if \(A\) is not mivertible) and \(e^{-A}=\left(e^{4}\right)^{-1} .\) (d) If \(A\) and \(B\) are commuting operators then \(e^{A+B}=\mathrm{e}^{A} \mathrm{e}^{\theta}\) (c) If \(A\) is hermitian then \(e^{i A}\) is unitary.

In the Hulbert space \(L^{2}([-1,1])\) let \(\left.\mid f_{n}(x)\right\\}\) be the sequence of functions \(1, x, x^{2}, \ldots, f_{n}(x)=x^{n} \ldots .\) (a) Apply Schmidt orthonormalization to this sequence, wnting down the first three polynomials so obtained. (b) The \(n\)th Legendre polynomial \(P_{n}(x)\) is defined as $$ P_{n}(x)=\frac{1}{2^{n} n !} \frac{d^{n}}{\mathrm{dx}^{n}}\left(x^{2}-1\right)^{n} $$ Prove that $$ \int_{-1}^{1} P_{m}(\mathrm{r}) P_{n}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{2}{2 n+1} \delta_{m \mathrm{~m}} $$ (c) Show that the \(n\)th member of the o.n. sequence obtained in (a) is \(\sqrt{n+\frac{1}{2}} P_{n}(x)\).

If \(A\) is a sclf-adjoint operator show that $$ \|(A+t) u\|^{2}=\|A u\|^{2}+I u \|^{2} $$ and that the operator \(A+1 I\) is invertible, Show that the operator \(U=\left(A-{ }_{1} I\right)(A+i I)^{-1}\) is unitary (called the Cayley transform of \(A\) ).

Let \(\ell_{0}\) be the subset of \(\ell^{2}\) consisting of sequences with only finutely many terms different from zero. Show that \(\ell_{0}\) is a vector subspace of \(\ell^{2}\), but that it is not closed. What is its closure \(\overline{\ell_{0}} ?\)
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

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.

Sign-up for free