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Chapter 13: Problem 35

If \(A\) is a symmetric operator, show that \(A^{*}\) is symmetric if and only if it is self-edjoint, \(A^{*}=A^{* *}\)

Short Answer

Expert verified
Given a symmetric operator \(A\), its adjoint can be said to be symmetric if and only if it's self-adjoint. This is proved by taking arbitrary vectors \(x, y\), applying the properties of symmetric and self-adjoint operators, and displaying that both resultant expressions can be equated due to the interchangeability of the inner product.

Step by step solution

01

- Define Given and Terms

A linear operator \(A\) on an inner product space is defined as symmetric if \(\langle Ax, y\rangle = \langle x, Ay\rangle\) for every vector \(x, y\). The adjoint of an operator \(A\), denoted as \(A^{*}\), is a unique operator such that \(\langle Ax, y\rangle = \langle x, A^{*}y\rangle\) for every vector \(x, y\). \(A^{*}\) is self-adjoint if and only if \(A^{*}=A^{* *}\).
02

- Prove Direction

Start by showing the direction that if \(A^{*}\) is symmetric, then it is self-adjoint. Here, let's take any vectors \(x, y\), then we can say \(\langle A^{*}x, y\rangle = \langle x, A^{*}y \rangle\). According to the definition of the adjoint operator, it can be rewritten as \(\langle x, A^{**}y \rangle\). Hence, proving that \(A^{*}\) is self-adjoint i.e., \(A^{*}=A^{**}\).
03

- Prove the Reverse Direction

Now, assume that the adjoint operator \(A^{*}\) is self-adjoint. Therefore, \(A^{*} = A^{* *}\). Then, \(\langle x, A^{*}y \rangle = \langle x, A^{**}y \rangle\) which according to the definition of the adjoint operator can be rewritten as \(\langle A^{*}x, y \rangle\). Hence proving that if \(A^{*}\) is self-adjoint, then it is symmetric.

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Most popular questions from this chapter

The norm \(\|\phi\|\) of a bounded linear operator \(\phi: \mathcal{H} \rightarrow \mathrm{C}\) is defined as the greatest lower bound of all \(M\) such that \(|\phi(u)| \leq M\|u\|\) for all \(u \in \mathcal{H}\). If \(\phi(u)=(v \mid u)\) show that \(\|\phi\|=\|v\|\). Hence show that ahe bounded lanear functional norm satisfies the parallelogram law $$ \|\phi+\vartheta\|^{2}+\|\phi-\psi\|^{2}=2\|\phi\|^{2}+2\|\psi\|^{2} $$

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 \(A_{1}=A_{2}=\cdots=A_{n}=0 .\)

Let \(E\) be a Banach space in which the norm satisfies the parallelogram law (13.2). Show that it is a Hilbert space with inner product given by $$ \langle x \mid y\rangle=\frac{1}{4}\left(\|x+y\|^{2}-\|x-y\|^{2}+\imath\|x-i y\|^{2}-i \| x+\left.1 y\right|^{2}\right) $$

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 \(\overline{\ell_{0}} ?\)

If \(A: \mathcal{H} \rightarrow \mathcal{H}\) is an operator such that \(A u \perp u\) for all \(u \in \mathcal{H}\), show that \(A=0\).
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