8
$\begingroup$

Is there a basis-free formulation of Jordan normal form theorem?

From some search I did in Google, the answer is apparently yes. But I didn't find any article that I could understand. (I've only taken two semester course in linear algebra.)

My curiosity comes from the question whether the theorem can be generalized to infinite dimensional situation. If it's a separable Hilbert space, we can still represent the linear operator as a matrix, but does the theorem remain true?

In case of non-separable space, I think there's no way to put the linear operator in matrix form. So we need to find a basis-free formulation.

Wikipedia says that there is an analogue of Jordan normal form theorem for compact operators in Banach space. What is this analogous result?

  • 0
    I believe it is not an elementary problem. The Jordan canonical form is actually a statement about the existence of invariant subspaces for the operator. This is the approach of Herstein in his book on Algebra. In infinite dimension, you no longer have coordinates, but the idea of invariant subspaces survives. However, you need spectral theory and functional calculus, as stated in http://en.wikipedia.org/wiki/Spectral_theory_of_compact_operators#Invariant_subspaces2012-04-28
  • 0
    In non-separable Hilbert space, we can use matrices with uncountably many rows and columns.2012-04-28
  • 0
    Do such matrices have "diagonal"?2012-04-28

2 Answers 2