Results 1 to 4 of 4

Math Help - Cayley Transform

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    41

    Cayley Transform

    If U is a partial isometry with initial and final spaces M and N, respectively and such that (1-U)M is dense in H and A =i(1+U)(1-U)^-1 then A and U satisfy U = (A-i)(A+i)-1

    Algebraically manipulation shows U=(A-i)(A+i)^{-1}.

    The only thing left to check is (A+i) is indeed invertible.


    So by using
    Proposition 1.15:
    Let A: H-> K be a linear operator.
    (a) A is boundedly invertible if and only if ker A = (0), ran A = K, and the gra A is closed.
    (b) If A is bounded invertible, its inverse is unique and denoted by A^-1.

    How do I relate this proposition with what is known? I'm guessing is A is closed.

    We know that the deficiency subspaces are defined.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    The reason that A+i is invertible is that

    \begin{aligned}\|(A+i)x\|^2 &= \langle (A+i)x,(A+i)x\rangle \\ &= \langle Ax,Ax\rangle +i\langle x,Ax\rangle - i\langle Ax,x\rangle + \langle x,x\rangle = \|Ax\|^2 + \|x\|^2\end{aligned}

    (because A is symmetric). So if (A+i)x=0 then x=0. Thus (A+i)^{-1} is unambiguously defined on its domain, which is (A+i)D(A).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    41

    Question

    If A is a self-adjoint operator and U is its Cayley transform, then U is a unitary operator with ker(1-U)=(0). Conversely, if U is a unitary with 1 not in  sigma_p(U) , then the operator A defined by A=i(1+U)(1-U)^-1 is self-adjoint.

    I don't understand what the converse means with respect to 1 not being in the point spectrum.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Nusc View Post
    If A is a self-adjoint operator and U is its Cayley transform, then U is a unitary operator with ker(1-U)=(0). Conversely, if U is a unitary with 1 not in  sigma_p(U) , then the operator A defined by A=i(1+U)(1-U)^-1 is self-adjoint.

    I don't understand what the converse means with respect to 1 not being in the point spectrum.
    The point spectrum is the set of eigenvalues. If 1 is an eigenvalue of U then Ux = x for some nonzero vector x, so (IU)x = 0. This means that IU cannot have an inverse, so the definition A=i(1+U)(1-U)^{-1} can't be used in that case.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. cayley-Hamilton law
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 1st 2011, 06:30 AM
  2. Cayley Hamilton Theorem
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: September 6th 2011, 08:32 AM
  3. Cayley table
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 19th 2010, 03:37 AM
  4. cayley transform
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 31st 2010, 12:18 AM
  5. Cayley table and subgroup
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 22nd 2009, 08:52 PM

Search Tags


/mathhelpforum @mathhelpforum