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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

Show that if \(\left(A, D_{A}\right)\) and \(\left(B, D_{B}\right)\) arc operators on dense domains in \(H\) then \(B^{*} A^{*} \subseteq\) \((A B)^{\circ}\)

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 $$ \langle x \mid y\rangle=\frac{1}{4}\left(\|x+y\|^{2}-\|x-y\|^{2}+\imath\|x-i y\|^{2}-i \| x+\left.1 y\right|^{2}\right) $$

Which of the following is a vector subspace of \(\ell^{2}\), and which are closed? In each case find the space of vectors orthogonal to the set. (a) \(V_{N}=\left\\{\left(x_{1}, x_{2} \ldots\right) \in \ell^{2} \mid x_{1}=0\right.\) for \(\left.i>N\right]\) (b) \(V=\bigcup_{N=1}^{\infty} V_{N}=\left|\left(x_{1}, x_{2}, \ldots\right) \in \ell^{2}\right| x_{t}=0\) for \(i>\) some \(\left.N\right]\). (c) \(U=\left|\left(x_{1}, x_{2}, \ldots\right) \in \ell^{2}\right| x_{1}=0\) for \(\left.t=2 n\right\\} .\) (d) \(W=\left\\{\left(x_{1}, x_{2}, \ldots\right) \in \ell^{2} \mid x_{1}=0\right.\) for some \(\left.i\right\\}\)

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.

If \(\left(A, D_{A}\right)\) is a densely defined openator and \(D_{A}\) is dense in \(\mathcal{H}_{1}\) show that \(A \subseteq A^{* *}\).
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