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Sylvester's Criterion for Infinite-dimensional Matrices

  1. Apr 26, 2012 #1
    Does Sylvester's Criterion hold for infinite-dimensional matrices? Thanks!
     
  2. jcsd
  3. Apr 26, 2012 #2


    I can't see how unless one defines rationally the determinant of infinite matrices...

    DonAntonio
     
  4. Apr 26, 2012 #3
    Well, Sylvester's criterion only requires us to find determinants of the principal minors, which are all finite square matrices. There are just an infinite number of them.
     
  5. Apr 26, 2012 #4


    If I remember correctly S.C. requires to find out ALL the principal minors' determinants, up to and including the whole determinant's...

    DonAntonio
     
  6. Apr 26, 2012 #5
    Yes, but isn't that just the natural way to word the condition in the finite-dimensional case? Does the proof actually use the fact that the entire matrix's determinant is positive or that the determinant is finite-dimensional?
     
  7. Apr 26, 2012 #6
    I guess if all principal minors are positive, then the infinite matrix is the limit of positive definite matrices and is thus positive semi-definite.
     
  8. Apr 26, 2012 #7


    The limit...in what sense? I think you may need some topology to have any chance of defining limit in some meaningful way.

    DonAntonio
     
  9. Apr 26, 2012 #8
    I guess I haven't really told you about the specific problem I'm working on. My infinite matrix is mapping a Banach space into itself, so the matrix operator itself lives in a Banach space, which has a topology.
     
  10. Apr 26, 2012 #9
    But I think we've strayed from the initial question... Does anyone know if there's something similar to Sylvester's criterion for a countably infinite square matrix?
     
  11. Apr 28, 2012 #10
    The short answer is "it depends"

    There are 2 main interpretations of the quadratic forms with infinite matrices:

    The first one is that while we have infinite matrix, we only consider vectors with finitely many non-zero coordinates. The quadratic form is defined for all such vectors. In this case "positive definite" means that [itex](A x, x)>0[/itex] for all non-zero vectors [itex]x[/itex] with finitely many non-zero coordinates.

    In this case the Silvester's criterion works, i.e. the form is positive definite if and only if all finite principle minors are positive.

    The other interpretation of the quadratic form [itex](A x, x)[/itex] with infinite matrix [itex]A[/itex] is to assume that [itex]A[/itex] is a matrix of a bounded operator in [itex]\ell^2[/itex] (the space of sequences [itex]x=\{x_k\}_{k=1}^n[/itex] such that [itex]\sum_{k=1}^\infty|x_k|^2<\infty[/itex]).

    In this case if all finite principal minors are positive, we can only conclude that the matrix [itex]A[/itex] is positive-semidefinite, i.e. that [itex](Ax,x)\ge0[/itex] for all [itex]x\in\ell^2[/itex] (and "positive definite" means that [itex](Ax, x)>0[/itex] for all non-zero [itex]x\in\ell^2[/itex]). And it is not hard to construct an example where all finite principal minors are positive, but [itex](Ax, x)=0[/itex] for some non-zero [itex]x\in\ell^2[/itex].

    There can be more complicated interpretations of the quadratic forms with infinite matrices, but I am not going to discuss these right now.
     
  12. Apr 28, 2012 #11
    Thanks a lot.
     
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