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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.

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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 unbounded operators, show that \(A^{*}+B^{*} \subseteq(A+B)^{\circ}\)

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 \(\lambda \in \mathbb{C}\)

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

For bounded linear operators \(A, B\) on a normed vector space \(V\) show that $$ \|\lambda A\|=|\lambda|\|A\|, \quad|A+B\|\leq\| A|+\|B\|, \quad \mid A B\|\leq\| A\|\| B \| $$ Hence show that \(|A|\) is a genuine norm on the set of bounded hnear operators on \(V\).
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