The spectrum of a bounded differential equation

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Discussion Overview

The discussion revolves around the relationship between the spectrum of an operator and the operator itself, particularly in the context of bounded differential equations. Participants explore the implications of knowing the spectrum in both finite and infinite dimensions, and the conditions under which one can deduce the operator from its spectrum.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that in finite dimensions, knowing all eigenvalues allows one to determine the linear transformation up to unitary equivalence, provided the multiplicities of the eigenvalues are also known.
  • Others argue that this relationship does not hold in infinite dimensions, citing examples such as multiplication operators on function spaces that share the same spectrum but are not equivalent.
  • A later reply questions the assumption of Hermitian operators, noting that while eigenvectors are orthogonal in this case, the context of differential equations may not always guarantee self-adjointness.
  • Some participants suggest that additional information about how the operator acts on a dense subset of the solution space could allow for the reconstruction of the operator from its spectrum.

Areas of Agreement / Disagreement

Participants express differing views on the implications of knowing the spectrum in finite versus infinite dimensions, with no consensus reached on the ability to reconstruct the operator solely from its spectrum in the infinite-dimensional case.

Contextual Notes

Limitations include the dependence on the assumptions of self-adjointness and the specific nature of the operators discussed, which may affect the conclusions drawn about the relationship between spectrum and operator.

greentea28a
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is it possible to work backwards from a spectrum to which operator?
 
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Think about finite dimensions. If I give you all the eigenvalues, you still need to know what the eigenvectors are before you know the linear transformation.
 
homeomorphic said:
Think about finite dimensions. If I give you all the eigenvalues, you still need to know what the eigenvectors are before you know the linear transformation.

But in that case you do know the transformation up to unitary equivalence (well, if you also know the multiplicities of the eigenvalues)! So the spectrum forms a very nice invariant in finite dimensions (called a unitary invariant).

Sadly enough, the same is not true for infinite dimensions, not even for bounded operators. For example, multiplication by ##x## on ##L^2([0,1])## and an operator with a complete set of eigenfunctions whose eigenvalues are dense in ##[0,1]##. These two operators can't be much different, but they do share the same spectrum.
Finding a stronger unitary invariant can be done and yields the commutative multiplicity theorem. I refer to the beautiful book by Reed and Simon for a discussion.
 
But in that case you do know the transformation up to unitary equivalence (well, if you also know the multiplicities of the eigenvalues)! So the spectrum forms a very nice invariant in finite dimensions (called a unitary invariant).

Well, if you assume it's Hermitian, right? Then all your eigenvectors are orthogonal.

The context was that of differential equations. That means the operator will be self-adjoint if your boundary conditions make it so, but not always.

I assume with enough extra information, you can eventually get back the operator. For example, if you knew how the operator acted on dense subset of the solution space, you should be good to go.
 
homeomorphic said:
Well, if you assume it's Hermitian, right? Then all your eigenvectors are orthogonal.

Yes, I thought that was given, but apparently I was mistaken.
 

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