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][1] 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][2] 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. 


  [1]: https://mathworld.wolfram.com/MatrixExponential.html
  [2]: https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#:~:text=Zassenhaus%20formula,-A%20related%20combinatoric&text=where%20the%20exponents%20of%20higher,of%20the%20above%20BCH%20expansion.