Infinite Dimensional Vector Space

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SUMMARY

The statement that every linear operator T: V → V on an infinite-dimensional vector space V over the complex numbers C has an eigenvector is false. A standard counterexample is the bilateral shift operator defined on the space ℓ(ℤ), which maps a sequence (a_i)_{i∈ℤ} to (a_{i+1})_{i∈ℤ}. This operator does not possess any eigenvectors, illustrating the existence of a spectrum in infinite-dimensional spaces.

PREREQUISITES
  • Understanding of linear operators in vector spaces
  • Familiarity with eigenvectors and eigenvalues
  • Knowledge of infinite-dimensional vector spaces
  • Basic concepts of functional analysis, particularly spectrum theory
NEXT STEPS
  • Study the properties of the bilateral shift operator in functional analysis
  • Explore the concept of spectrum in infinite-dimensional spaces
  • Learn about other types of operators in Hilbert spaces
  • Investigate the implications of the lack of eigenvectors in infinite-dimensional vector spaces
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Mathematicians, students of functional analysis, and anyone studying linear operators in infinite-dimensional vector spaces.

Bachelier
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Can you guys provide a counter example of why this statement is False.

If T: V-->V with V a VS over C then T has an eigenvector?

This is not always true as if V is infinite dim., it'll have a Spectrum.

Any counter examples?

Thanks
 
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The standard example is the bilateral shift on \ell(\mathbb{Z}):

T:\ell(\mathbb{Z})\to \ell(\mathbb{Z})
(a_i)_{i\in\mathbb{Z}}\mapsto (a_{i+1})_{i\in\mathbb{Z}}.
 

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