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.