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Chapter 13: Problem 13

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

Short Answer

Expert verified
The operator \(A\) is the zero operator (\(A = 0\)) as it maps all vectors in the space \(\mathcal{H}\) to the zero vector.

Step by step solution

01

Recall Properties of Orthogonality and Dot Product

We must first remember the principle that if two vectors are orthogonal, their dot product is zero. This is denoted as \(u \cdot v = 0\) for \(u \perp v\). It follows that \(u \cdot A u = 0\) for all \(u \in \mathcal{H}\), under the condition \(A u \perp u\).
02

Expand the Dot Product

Now, expand the dot product using the linearity property of the dot product and distribute it over the operator \(A\): \(u \cdot A u = \| A u \|^2 = 0\). Since the square norm of a vector is zero, the vector itself must be the zero vector. This implies that the vector \(Au\) is the zero vector for every vector \(u\) in the space \(\mathcal{H}\).
03

Conclude Proof

We have shown that the resulting vector \(Au\) is a zero vector for any given vector \(u\). Consequently, the operator \(A\) must be the zero operator, denoted as \(A=0\), because it maps every vector in the space \(\mathcal{H}\) to zero.

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Most popular questions from this chapter

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

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

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

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