Let $A$ be an invertible, symmetric and tridiagonal matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the elements in the sub- and super-diagonal of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.

I know that, given the Kronecker sum property of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n +I_n \otimes A}. \end{equation}

Since $A \otimes I_n$ and $I_n \otimes A$ commutes, using Zassenhaus formula,

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n} e^{I_n \otimes A}=(e^{-A}\otimes I_n)(I_n \otimes e^{A}) \end{equation}

Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.