Question

Find an operator \(T \in \mathcal{B}\left(\ell_{2}\right)\) such that its spectrum \(\sigma(T)\) consists exactly of the points 0 and 1 and such that neither 0 nor 1 is an eigenvalue for \(T\).

Short answer

Use the operator \(T = \frac{1}{2}(S + S^*)\) where \(S\) is the forward shift and \(S^*\) is the backward shift on \(\u{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\textomanipulates}}}}}}}}}}}.

Step-by-step solution

Understand the Requirement

We need to find a bounded linear operator, denoted as \(T\), on the Hilbert space \(\ell_2\) such that its spectrum, \(\sigma(T)\), includes exactly the points 0 and 1. Furthermore, neither 0 nor 1 should be eigenvalues of \(T\).

Define a Suitable Operator

Consider a bilateral shift operator on \(\ell_2\). Specifically, we can use the forward shift operator \(S\), which is defined by \(S(e_n) = e_{n+1}\), where \(\{e_n\}\) is the standard orthonormal basis for \(\ell_2\).

Determine the Spectrum of the Operator

The spectrum \(\sigma(S)\) of the forward shift operator \(S\) on \(\ell_2\) is the closed unit disk in the complex plane, \(\{ \lambda \, : \, |\lambda| \leq 1 \}\).

Construct the Desired Operator

To restrict the spectrum to just 0 and 1, define the operator \(T = \frac{1}{2}(S + S^*)\), where \(S^*\) is the adjoint of \(S\). The operator \(S^*\) is the backward shift, with \(S^*(e_{n+1}) = e_n\) for \(n \geq 1\) and \(S^*(e_1) = 0\).

Verify the Spectrum

By construction, \(T = \frac{1}{2}(S + S^*)\) has a spectrum contained within \([-1, 1]\). Thus, scaling this operator appropriately, we can ensure that its spectrum is exactly \(\{0, 1\}\). We need to confirm that neither 0 nor 1 are eigenvalues.

Ensure No Eigenvalue Condition

For \(T\), no \(\lambda = 0 \) or \(1\) should be an eigenvalue if there does not exist any \(x\in \ell_2\) such that \(Tx = \lambda x\). This can be shown using the definition of \(T\) and the properties of shift operators.

Key concepts

Hilbert Space

A Hilbert space is a complete inner product space, meaning it has a well-defined inner product, and every Cauchy sequence converges within the space. In simpler terms, it's an environment where you can perform geometric and algebraic operations with infinite-dimensional vectors. One commonly referenced Hilbert space is \(\ell_2\), which consists of all infinite sequences of complex numbers that are square-summable. Imagine having vectors, but instead of having just a few components like in 3D space, these vectors have infinitely many components which must add up to a finite value when squared.

Key properties of Hilbert spaces include:

• Completeness: Any Cauchy sequence of vectors in the space will converge.
• Inner Product: A way to multiply vectors to get a scalar, typically denoted as \(\langle x,y \rangle\).
• Orthonormal Basis: An infinite set of vectors, which are mutually orthogonal and each of unit length, that span the entire space.

Understanding Hilbert spaces is crucial because operators, particularly bounded linear operators, act on these spaces, affecting the vectors within them in various structured ways.

Spectrum of an Operator

The spectrum of an operator in a Hilbert space is a critical concept in functional analysis. For a given bounded linear operator \(T\), its spectrum \(\sigma(T)\) consists of all the values \(\lambda\) in the complex plane for which the operator \(T - \lambda I\) is not invertible.

There are three types of spectra to consider:

• Point Spectrum: Values \(\lambda\) for which \(T - \lambda I\) is not injective, meaning \(\lambda\) is an eigenvalue.
• Continuous Spectrum: Values \(\lambda\) where \(T - \lambda I\) is injective and has dense range, but is not surjective.
• Residual Spectrum: Values \(\lambda\) for which \(T - \lambda I\) has surjective range, but the inverse is not bounded.

In our exercise, we looked for an operator whose spectrum is \[\{0, 1\}\], containing neither as an eigenvalue. This means we need a subset of the continuous or residual spectrum.

Forward Shift Operator

A forward shift operator, typically denoted as \(S\), is an operator that shifts each element of a sequence in \(\ell_2\) by one position to the right. Formally, if \(\{e_n\}\) is the standard orthonormal basis for \(\ell_2\), then \(S(e_n) = e_{n+1}\).

Consider the sequence \(x = (x_1, x_2, x_3, \dots)\). Applying the forward shift operator, \(S\) transforms it to \(Sx = (0, x_1, x_2, x_3, \dots)\). The forward shift operator extends to infinite-dimensional vectors smoothly, and its adjoint (the backward shift operator) complements this behavior by shifting elements left.

Key characteristics:

• The forward shift operator is bounded and linear.
• Its spectrum (\(\sigma(S)\) ) can be more complex than finite-dimensional matrices, often forming a disk in the complex plane.

In our context, \(S\) and its adjoint play a critical role in shaping the operator whose spectrum is confined to specific points.

Adjoint of an Operator

The adjoint of an operator is a concept tied closely to the properties of Hilbert spaces. If \(T\) is a linear operator on a Hilbert space, its adjoint, denoted as \(T^*\), is defined so that for all vectors \(x\) and \(y\) in the Hilbert space, the inner product \(\langle Tx, y \rangle = \langle x, T^*y \rangle\).

For example, consider the forward shift operator \(S\), defined by \(S(e_n) = e_{n+1}\). Its adjoint \(S^*\) is the backward shift operator, which moves elements one position to the left, i.e., \(S^*(e_{n+1}) = e_n\) and \(S^*(e_1) = 0\).

Properties include:

• Reversibility: The adjoint essentially