Cayley Transform of self-adjoint operator is unitary

Question

If $A$ is a self-adjoint operator, I want to show that the Cayley transform of $A$ defined as the operator $U=(A-iI)(A+iI)^{-1}$ is unitary. Here's my trial: For all $x,y \in \mathcal{H}$, keeping in mind that $A$ and $I$ have adjoints:

$$\langle Ux,Uy \rangle = \langle U^*Ux,y \rangle$$

All i need to show now is that $U$ is an Isometry, i.e, $U^*=U^{-1}$ and I'm all set (right?). Therefore: $$ U^*=[(A-iI)(A+iI)^{-1}]^* = [(A+iI)^{-1}]^*(A-iI)^*$$ But i cannot go further because $(A+iI)$ is not self-adjoint, so I can't switch the inverse and the adjoint operations.

Does somenone have some tip on how to get the inverse of this operator or an alternative way to show that $U^*=U^{-1}$? all the help will be appreciated.

Answer 1

I was just tutoring a graduate lin alg student and a problem very similar to this came up. (See Bau Trefethan Numerical Linear Algebra Problem 2.5 part c)

Here's how I solved that problem in 5 steps using definition of unitary ($U^{H} = U^{-1}$) and definition of self adjoint ($A^H = A^{*,T} = A$).

Given $U = (A - iI)(A+iI)^{-1}$,

$$U^{H} = \Big( (A - iI)(A+iI)^{-1} \Big)^{H}$$

1. Act the Hermitian operator $$ = \Big((A+iI)^{-1} \Big)^{H} (A - iI)^H $$
2. Switch Hermitian and inverse operations $$ = \Big((A+iI)^{H} \Big)^{-1} (A - iI)^H $$
3. Take Hermitian, A and I are both self adjoint so only $i$ flips sign. $$ = (A-iI)^{-1} (A + iI) $$
4. Commute matrices $$ = (A + iI) (A-iI)^{-1}$$
5. Definition of $U^{-1}$ $$ = U^{-1} $$ DONE!! Note that we have NOT proven the commutation of these matrices. If you're interested in the proof it's below.

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We'll use the eigendecomposition of $A = Q^{-1} \Lambda Q $

$$A + iI = Q^{-1} (\Lambda + iI) Q $$ $$(A - iI)^{-1} = \Big( Q^{-1} (\Lambda - iI) Q \Big)^{-1} = Q^{-1} (\Lambda^{-1} - iI) Q $$ Thus $$(A + iI)(A - iI)^{-1} = Q^{-1} (\Lambda + iI) Q Q^{-1} (\Lambda^{-1} - iI) Q $$ $$ = Q^{-1} (\Lambda + iI) (\Lambda^{-1} - iI) Q $$ $$ = Q^{-1} (2 I) Q $$ $$ = Q^{-1} (\Lambda^{-1} - iI) (\Lambda + iI) Q $$ $$ = Q^{-1} (\Lambda^{-1} - iI) Q Q^{-1} (\Lambda + iI) Q $$ $$ = (A - iI)^{-1} (A + iI)$$

Answer 2

Note that \begin{align} \|(A\pm iI)x\|^2&=\|Ax\|^2+\langle Ax,ix\rangle+\langle Ax,-ix\rangle+\|x\|^2 \\ &= \|Ax\|^2+\|x\|^2,\;\;\; x\in\mathcal{D}(A). \end{align} Therefore, $\|(A\pm iI)x\| \ge \|x\|$ for all $x\in\mathcal{D}(A)$, which is enough to show that both $A\pm iI$ have closed ranges and bounded inverses $(A\pm iI)^{-1}$ on their respective ranges. These ranges are dense because $$ \langle (A+iI)x,y\rangle=0,\; \forall x\in\mathcal{D}(A)\implies y\in\mathcal{D}(A)=\mathcal{D}(A-iI) \mbox{ and } (A-iI)y=0, $$ which is possible iff $y=0$. Therefore $(A+iI)(A-iI)^{-1}$ and $(A-iI)(A+iI)^{-1}$ are bounded, isometric inverses of each other, making both unitary.