Solve *-Algebra Problem: $\sigma(\lambda{e}-x)=\lambda-\sigma(x)$

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SUMMARY

The discussion focuses on the relationship between the spectrum of an element in a *-algebra and a linear transformation involving a complex scalar. Specifically, it establishes that for a *-algebra \(X\) with identity \(e\) and a complex number \(\lambda\), the equation \(\sigma(\lambda{e}-x)=\lambda-\sigma(x)\) holds true. The proof is based on the invertibility condition, where \(v\in\sigma(\lambda e-x)\) if and only if \((\lambda-v)e-x\) is invertible, leading to the conclusion that \(\lambda-v\in\sigma(x)\).

PREREQUISITES
  • Understanding of *-algebras and their properties
  • Familiarity with the concept of spectrum in functional analysis
  • Knowledge of invertibility in the context of operators
  • Basic grasp of complex numbers and their operations
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  • Study the properties of *-algebras and their applications in functional analysis
  • Learn about the spectrum of operators in Hilbert spaces
  • Explore the concept of invertibility in operator theory
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Mathematicians, particularly those specializing in functional analysis, theoretical physicists working with quantum mechanics, and students studying operator theory will benefit from this discussion.

Cairo
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Let $X$ be a *-algebra with identity $e$, and let $e\in{X}$, $\lambda\in\mathbb{C}$. Can somebody show me how $\sigma(\lambda{e}-x)=\lambda-\sigma(x)$, where $\sigma(x)$ is the spectrum of an element.

Thanks in advance.
 
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$v\in\sigma(\lambda e-x)$ if and only if $(\lambda-v)e-x$ is invertible, that is if and only if $\lambda-v\in\sigma(x)$, which gives the result.
 

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