Given $A$ is Hermitian, Show that a given $U$ is unitary

Question

Let $A\in\mathbb{C}^{n\times n}$ be Hermitian. Show that the matrix $U=(A+iI)^{-1}(A-iI)$ is a unitary matrix.

I have done so far: We want $U^*U=I$, so finding $U^*$

\begin{align}U^*&=((A+iI)^{-1}(A-iI))^*\\ &=(A-iI)^*((A+iI)^{-1})^*\\ &=(A^*+iI)((A+iI)^*)^{-1}\\ &=(A^*+iI)(A^*-iI)^{-1}\\ &=(A+iI)(A-iI)^{-1}\end{align}

Then we know that $U^*U=(A+iI)(A-iI)^{-1}(A+iI)^{-1}(A-iI).$

This is where I am stuck as im not sure how to manipulate $U^*U$ to show that it is $I$ given that matrices in general are not commutative.

Answer 1

$(A-iI)^{-1}$ commutes with $A$ and hence also with $A+iI$. Proof:

$(A-iI)^{-1}(A-iI)=I$ so $(A-iI)^{-1}A=I+i(A-iI)^{-1} $. Similarly,

$(A-iI)(A-iI)^{-1}=I$ so $A(A-iI)^{-1}=I+i(A-iI)^{-1}$. Since the right sides of these two equatins are the same we get $(A-iI)^{-1}A=A(A-iI)^{-1}$. So $(A-iI)^{-1}$ commutes with $A$. Hence it also commutes with $A+iI$.

Thus $(A+iI)(A-iI)^{-1}(A+iI)^{-1}(A-iI)=(A-iI)^{-1}(A+iI)((A+iI)^{-1}(A-iI)=I$.

Answer 2

$(A+iI)$ does commute with $(A-iI)^{-1}$ because \begin{aligned} (A+iI)(A-iI)^{-1} &=(A-iI)^{-1}\left[(A-iI)(A+iI)\right](A-iI)^{-1}\\ &=(A-iI)^{-1}\left[(A+iI)(A-iI)\right](A-iI)^{-1}\\ &=(A-iI)^{-1}(A+iI). \end{aligned} Alternatively, by Cayley-Hamilton theorem, $X^{-1}$ is a polynomial in $X$. Therefore $(A-iI)^{-1}$ is a polynomial in $A-iI$. In turn, it is a polynomial in $A$. Hence it commutes with every polynomial in $A$ and with $A+iI$ in particular.