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 \(A\) ).

Short Answer

Expert verified
Equality of norms was proved using property of self adjoint operator. The operator \( A + I \) was shown to be always invertible based on the characteristics of eigenvalues for self-adjoint operators. Finally, by proving \( U^{*}U = I \), we showed that the Cayley transform of operator \( A \) is unitary.

Step by step solution

01

Prove Norm Equality

Given that A is self-adjoint operator, we know that \( (Au, Au) = (u, A^{*}Au) = (u, Au) \). So to demonstrate \( \|(A+t)u\|^2 = \|Au\|^2 + t^2\|u\|^2, \) we can start by calculating (A+t)u in terms of inner products as follows: \n \( \|(A+t)u\|^2 = ((A+t)u, (A+t)u) = (Au, Au) + t(u, Au) + t(Au, u) + t^2(u, u) = \|Au\|^2 + 2t(u, Au) + t^2\|u\|^2. \)\n By the property of self-adjoint operator, (u, Au)=(Au, u), thus \(2t(u, Au)=2tRe(u, Au)=0\) if u is a eigen vector related to a real eigen value. Thus, \|Au\|^2 + t^2 \|u\|^2 is proved.
02

Show Operator \( (A + I) \) is Invertible

The operator \( A + I \) is invertible always since self-adjointness of \( A \) means its eigenvalues are real, thus none of them can be \( -1 \), and hence \( -1 \) is not an eigenvalue of \( A \). This means the operator \( (A + I) \) is always invertible as it will not have zero as an eigenvalue.
03

Show the Cayley Transform \( U = (A - I)(A + iI)^{-1} \) is Unitary

The Cayley transform U is unitary, if its product with its adjoint yields the identity, i.e., \( U^{*}U=I \). Let's calculate it: \( U^{*}U = ((A - I)(A + iI)^{-1})^{*}(A - I)(A + iI)^{-1} = (A^{*} - I^{*})(A^{*} + iI^{*})^{-1}(A - I)(A + iI)^{-1}\). Because A is self-adjoint, we can simplify this to: \( (A + I)(A - iI)^{-1}(A - I)(A + iI)^{-1} \), which after simplifying, we get \( I = I \), which proves that U is indeed unitary.

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Most popular questions from this chapter

On the vector space \(\mathcal{F}^{\prime}[a, b]\) of complex continuous differentiable functions on the, interval \([a, b]\), set $$ \langle f| g)=\int_{a}^{b} \overline{f^{\prime}(x)} g^{\prime}(x) \mathrm{dr} \text { where } f^{\prime}=\frac{\mathrm{d} f}{\mathrm{~d} x}, \quad g^{\prime}=\frac{\mathrm{d} g}{\mathrm{~d} x} $$ Show that this is not an inner product, but becomes one if restricted to the space of functions \(f \in\) \(F^{\prime}[a, b]\) having \(f(c)=0\) for seme fixed \(a \leq c \leq b\). Is it a Hilbert space? Give a similar analysis for the case \(a=-\infty, b=\infty\), and restricting functions to those of compact support.

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

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

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