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Chapter 5: Problem 34

Let \(\mathscr{A}\) be a unital commutative Banach algebra, and suppose \(A, B \in \mathscr{A}\). Show that \(r(A+B) \leq r(A)+r(B)\) and \(r(A B) \leq r(A) r(B)\). (Hint: Use the Gelfand transform.) Show the same result holds if \(\mathscr{A}\) is not assumed to be commutative, provided \(A B=B A\). Show the result fails in general (look in \(\left.M_{2}(\mathbb{C})\right)\).

Short Answer

Expert verified
In a commutative unital Banach algebra, \( r(A+B) \leq r(A)+r(B) \) and \( r(AB) \leq r(A)r(B) \). Without commutativity but with \( AB=BA \), the same holds. The inequality fails in \( M_{2}(\mathbb{C}) \) as shown by a specific counterexample.

Step by step solution

01

Define Spectral Radius

The spectral radius \( r(A) \) of an element \( A \) in a unital Banach algebra \( \mathscr{A} \) is defined as the maximum modulus of elements in its spectrum: \( r(A) = \max \{|\lambda| : \lambda \in \sigma(A)\} \). The spectrum \( \sigma(A) \) is the set of \( \lambda \in \mathbb{C} \) such that \( A - \lambda I \) is not invertible.
02

Use Gelfand Transform for \( A + B \)

The Gelfand transform \( \hat{A} \) of an element \( A \) is a homomorphism from a Banach algebra to the algebra of continuous functions on its maximal ideal space. For any \( \lambda \in \sigma(A) \) and an analogous element for \( B \), \( \sigma(A+B) \) satisfies \( |\lambda + \mu| \leq r(A) + r(B) \) by the properties of the complex modulus.
03

Spectral Radius Inequality for Sum

Given that \( \sigma(A+B) \subseteq \sigma(A) + \sigma(B) \), it follows that the spectral radius satisfies \( r(A+B) \leq r(A) + r(B) \). This uses the sub-additive property of the modulus and the operational rules of spectra in commutative Banach algebras.
04

Use Gelfand Transform for Product \( A B \)

By using the Gelfand transform on the product, we have \( \hat{AB}(\gamma) = \hat{A}(\gamma) \hat{B}(\gamma) \) \( \forall \gamma \) in the character space, demonstrating that the spectrum \( \sigma(AB) \) satisfies \( |\lambda| \leq r(A) r(B) \), so \( r(AB) \leq r(A) r(B) \).
05

Case for Non-Commutative \( \mathscr{A} \) and Commuting \( A \) and \( B \)

If \( \mathscr{A} \) is not commutative but \( AB = BA \), the function properties from the commutative case still hold; thus, \( r(A+B) \leq r(A)+r(B) \) and \( r(AB) \leq r(A)r(B) \) as the algebra generated by \( A \) and \( B \) is commutative.
06

Counterexample in \( M_{2}(\mathbb{C}) \)

In the matrix algebra \( M_{2}(\mathbb{C}) \), consider matrices \( A = \begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix} \) and \( B = \begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix} \). We have \( AB = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \) with \( r(AB) = 1 \), while \( r(A) = r(B) = 0 \) implies \( r(A)r(B) = 0 \), showing the inequality fails.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Banach Algebra
A Banach algebra is more than just a set and operation; it's a beautifully structured space where algebra meets the infinite. Essentially, a Banach algebra combines the properties of algebra and a complete normed space. This means every Cauchy sequence in this space converges within the space. Imagine it like a sturdy scaffold for complex numbers.
A unital Banach algebra, as its name suggests, includes a multiplicative identity element. This identity act as the neutral element for multiplication, just like the number 1 does for real numbers.
  • It has operations of addition and multiplication.
  • Each element has an associated norm, which intuitively measures its size.
  • Convergence in this space ensures the sequence stays within the algebra.
These features make Banach algebras essential in understanding functional analysis and addressing problems with spectral properties.
Spectral Radius
The spectral radius is a core concept when dealing with Banach algebras. Think of it as the size of the farthest point in the element's spectrum, similar to a radius of the largest circle that fits into a shape. Mathematically, the spectral radius of an element \( A \) is defined as \( r(A) = \max \{ |\lambda| : \lambda \in \sigma(A) \} \), where \( \sigma(A) \) is the set of complex numbers \( \lambda \) such that \( A - \lambda I \) is not invertible.
Essentially:
  • The spectrum \( \sigma(A) \) consists of all values that prevent \( A - \lambda I \) from having an inverse.
  • The spectral radius gives the maximum magnitude within this spectrum.
Understanding the spectral radius helps in analyzing stability and convergence issues within algebraic systems.
Gelfand Transform
The Gelfand transform is like a magical bridge from abstract algebra world to concrete function space. It takes elements of a Banach algebra and maps them into continuous functions.
  • Enables the study of algebra elements using the behavior of functions.
  • Transforms complex algebraic problems into more manageable analytical ones.
  • Reveals spectral properties by connecting elements' spectra to function values.
In the context of Banach algebras, it's instrumental for deriving properties like spectral inequalities, helping us show relationships such as \( r(A+B) \leq r(A) + r(B) \) and \( r(AB) \leq r(A)r(B) \). The Gelfand transform reinforces understanding by translating difficult algebraic structures into familiar functional scenarios.
Non-Commutative Algebra
Not every room is square, and not every algebra is commutative. Non-commutative algebras do not follow the order-agnostic property of multiplication seen in real numbers. Simply put, for elements \( A \) and \( B \), \( AB \) might not equal \( BA \). Non-commutativity introduces rich dynamics and complexity, essential for many mathematical and physical applications.
However:
  • Even in non-commutative Banach algebras, if two elements \( A \) and \( B \) commute (i.e., \( AB = BA \)), some commutative algebra results still hold.
  • These special conditions allow us to make comparisons like \( r(A+B) \leq r(A) + r(B) \) and \( r(AB) \leq r(A)r(B) \).
This balance between commutative and non-commutative properties is crucial for deeper insights into matrix and operator theory.
Matrix Algebra
Matrix algebra gives concrete examples of algebras in action. Each matrix can represent a linear transformation, facilitating complex computation and manipulation. These matrices aren't always straightforward! Elements can have unexpected properties, especially in higher-dimensional spaces.
  • Matrix algebra provides a context for discussing spectral properties and transformations.
  • As seen in \( M_{2}(\mathbb{C}) \), failing such as \( r(AB) eq r(A)r(B) \) showcases exceptions to general rules, illustrating matrix algebra's unique behavior.
Through examples like these, students can understand how calculation rules in finite-dimensional spaces differ from abstract algebras, paving the way for practical application of theoretical constructs.

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Most popular questions from this chapter

Let \(X=[0,1]\) and put a topology on \(X\) by declaring the open sets to be the empty set and those subsets of \(X\) whose complement is at most countable. Consider the set \(A=[0,1)\). Show that the closure of \(A\) is \([0,1]\), so that in particular 1 lies in the closure of \(A\). Show that there is no sequence \(\left\\{x_{n}\right\\}\) of points in \([0,1)\) that converges to 1 .

Suppose \(\mathscr{A}\) and \(\mathscr{B}\) are Banach algebras with common identity and \(\mathscr{B} \subseteq \mathscr{A}\). Show \(\sigma_{\mathscr{I}}(A) \subseteq \sigma_{\mathscr{B}}(A)\) and \(\partial \sigma_{\mathscr{D}}(A) \subseteq \partial \sigma_{\mathscr{I}}(A)\), for any \(A \in \mathscr{B}\). Hint for the second part: Since the first part implies that the interior of \(\sigma_{\sigma f}(A)\) is contained in the interior of \(\sigma_{\mathscr{B}}(A)\) for any \(A\) in \(\mathscr{B}\), argue first that it suffices to show that if \(\lambda \in \partial \sigma_{\mathscr{D}}(A)\), then \(\lambda \in \sigma_{\sigma d}(A)\).

Find the norm of the operator on \(\mathscr{B}(\mathscr{H})\) where \(\mathscr{H}=\mathbb{C}^{2}\) which is given by the matrix $$ \left[\begin{array}{ll} a & b \\ c & d \end{array}\right] $$ Give your answer in terms of the numbers \(S=|a|^{2}+|b|^{2}+|c|^{2}+|d|^{2}\) and \(D=\) \(a d-b c\).

Show that a Banach algebra \(\mathscr{A}\) with an involution satisfying $$ \left\|A^{*} A\right\| \geq\|A\|^{2} $$ is a \(C^{*}\)-algebra, meaning that equality holds in this inequality.

Suppose that \(A\) is a self-adjoint operator in \(\mathscr{B}(\mathscr{H})\) for some Hilbert space \(\mathscr{H}\). Show that if ker \((A-t I)=\\{0\\}\) and \(A-t I\) has closed range for some real number \(t\), then the range of \(A-t I\) is \(\mathscr{H}\). Conclude that a self-adjoint operator has no residual spectrum.
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