I'm struggling with the following question.
Let H be a Unitary matrix and define $H_c = (cI - H)(I-\bar{c}H)^{-1}$ for $c \in \mathbb{C}$. Show that $H_c$ is also unitary.
Using properties of the conjugate transpose we have that $$H_c^* H_c = (I - cH^*)^{-1}(\bar{c}I - H^*)(cI - H)(I - \bar{c}H)^{-1}.$$ But I don't see how to progress. The fact that polynomials of $H$ commutes might be useful. Does anyone know how to proceed?
Use the fact that $H$ is unitary twice to get \begin{align*} (\bar c I-H^\ast)(cI-H) &=(\bar c HH^\ast-H^\ast)(cHH^\ast-H)\\ &=(\bar cH-I) H^\ast H (cH^\ast-I)\\ &=(\bar cH-I) (cH^\ast-I)\\ \end{align*} Now since $H$, $H^\ast$ and $I$ all commute with each other, we can commute the above to get \begin{align*} H_c^\ast H_c &= (I-cH^\ast)^{-1}(\bar c I-H^\ast)(cI-H)(I-\bar cH)^{-1}\\ &= (I-cH^\ast)^{-1}(\bar cH-I) (cH^\ast-I)(I-\bar cH)^{-1}\\ &= (I-cH^\ast)^{-1}(cH^\ast-I)(\bar cH-I)(I-\bar cH)^{-1}\\ &=I \end{align*} The proof for $H_cH_c^\ast$ is almost identical.