Matrix exponentials, computing the product of $\exp(-iBt)$ and $\exp(-iB^{-1}t)$.

Question

Suppose I have some matrix exponential $U(t)=\exp(-iAt)$ where $t$ is some real valued number, $A$ is a hermitian matrix (so $U(t)$ is unitary) where $A=B+B^{-1}$ and $B$ is unitary. Because $B$ and $B^{-1}$ commute we have $$ U(t)=\exp(-iBt).\exp(-iB^{-1}t)$$ Can this expression for $U(t)$ then be simplified any further without diagonalising $B$ and $B^{-1}$?

Answer 1

One obvious simplification is \begin{align} U(t) &=\exp(-itB)\exp(-itB^{-1})\\ &=\exp(-it(B+B^{-1}))\ \text{ (because $B$ commutes with $B^{-1}$)}\\ &=\exp(-it(B+B^\ast)) \end{align} but you probably already know this.