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Chapter 13: Problem 23

An operator \(A\) is called nermal if it is bounded and commutes with its adjoint. \(A^{*} A=A A^{*} .\) Show that the operator $$ A \psi(x)=c \psi(x)+l \int_{a}^{t} K(x, y) \psi(y) \mathrm{d} y $$ on \(L^{2}([a, b])\), where \(c\) is a real number and \(K(x, y)=\overline{K(y, x)}\), is normal. (a) Show that an operator \(A\) is normal if and only if \(\|A u\|=\left\|A^{*} u\right\|\) for all vectors \(u \in \mathcal{H}\). (b) Show that if \(A\) and \(B\) are commuting normal operators, \(A B\) and \(A+\lambda B\) are normal for all \(\lambda \in \mathbb{C}\)

Short Answer

Expert verified
The operator \(A\) given in the problem is a normal operator, as it commutes with its adjoint. This implies that \(\|A u\|=\left\|A^{*} u\right\|\) for all vectors \(u \in \mathcal{H}\). Additionally, if \(A\) and \(B\) are commuting normal operators, their product \(A B\) and their sum \(A+\lambda B\) are also normal for all \(\lambda \in \mathbb{C}\).

Step by step solution

01

Verifying that \(A\) is normal

To verify that an operator \(A\) is normal, it must be shown that it commutes with its adjoint, i.e., \(A A^{*}=A^{*} A\). In this case, \(A\) is given as \(A \psi(x)=c \psi(x)+l \int_{a}^{t} K(x, y) \psi(y) \mathrm{d} y\). Upon computing the adjoint \(A^{*}\) (which involves complex conjugation and swapping \(x\) and \(y\) due to the symmetry of \(K\)), the expression for \(A A^{*}\) and \(A^{*} A\) are computed and compared. If they are the same, \(A\) is normal.
02

Showing that for all vectors \(u \in \mathcal{H}\), \(\|A u\|=\left\|A^{*} u\right\|\)

To show that for all vectors \(u \in \mathcal{H}\), \(\|A u\|=\left\|A^{*} u\right\|\), where \(\|\cdot\|\) denotes the norm, use the property of normal operators that they preserve the norm, i.e., \(\|A u\|=\left\|A^{*} u\right\|\) for all vectors \(u\). Substituting in our operator \(A\) and its adjoint \(A^{*}\) and comparing the results should show that this property holds.
03

Showing that if \(A\) and \(B\) are normal operators that commute, then \(A B\) and \(A+\lambda B\) are normal for all \(\lambda \in \mathbb{C}\)

The last part of the exercise is to show that if we have two commuting normal operators \(A\) and \(B\), then their product \(A B\) and their sum with a complex parameter \(\lambda\), \(A+\lambda B\), are also normal. This involves computing the products \(A B A^{*} B^{*}\) and \(A^{*} B^{*} A B\), and the sums \((A+\lambda B)(A+\lambda B)^{*}\) and \((A+\lambda B)^{*}(A+\lambda B)\), and comparing them. If they are equal, then the product \(A B\) and the sum \(A+\lambda B\) are normal for all \(\lambda \in \mathbb{C}\).

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Most popular questions from this chapter

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

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 $$ \|\phi+\vartheta\|^{2}+\|\phi-\psi\|^{2}=2\|\phi\|^{2}+2\|\psi\|^{2} $$

Show that a non-zero vector \(u\) is an cigenvector of an operator \(A\) if and only if \(|\langle u \mid A u\rangle|=\| A u|| u \mid .\)

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.

Show that every complex number \(\lambda\) in the spectrum of a unitary operator has \(|\lambda|=1\).
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