Thread: 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.

The reason that A+i is invertible is that



(because A is symmetric). So if (A+i)x=0 then x=0. Thus is unambiguously defined on its domain, which is .

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 , 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.

Originally Posted by Nusc

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 , 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 (I–U)x = 0. This means that I–U cannot have an inverse, so the definition can't be used in that case.