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The spectrum of a bounded differential equation

  1. Mar 26, 2014 #1
    is it possible to work backwards from a spectrum to which operator?
  2. jcsd
  3. Mar 26, 2014 #2
    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.
  4. Mar 26, 2014 #3
    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.
  5. Mar 26, 2014 #4
    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.
  6. Mar 26, 2014 #5
    Yes, I thought that was given, but apparently I was mistaken.
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