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](https://mathworld.wolfram.com/MatrixExponential.html) 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](https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#:%7E:text=Zassenhaus%20formula,-A%20related%20combinatoric&text=where%20the%20exponents%20of%20higher,of%20the%20above%20BCH%20expansion.) 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.



Kronecker product: Is it possible to simplify this product $e^{-A} \otimes e^{A}$ where $A$ is an invertible and symmetric matrix
One can diagonalize your $A = VDV^{-1}$ explicitly; the closed formulas are [here](https://de.wikipedia.org/wiki/Tridiagonal-Toeplitz-Matrix) for instance.


Once you have those matrices, you can write the orthogonal eigendecomposition
$$
\exp(-A) \otimes \exp(A) = (V\otimes V) \,\, (\exp(-D)\otimes \exp(D)) \, \,(V\otimes V)^{-1}.
$$
The diagonal matrix $\exp(-D)\otimes \exp(D)$ has elements $\exp(-\lambda\_i + \lambda\_j)$, $i,j=1,\dots,n$; when you replace it with the formulas for the eigenvalues, the $a$'s simplify out and you get a difference of two cosines that you can further manipulate using the sum-to-product formulas. I don't think it gets any simpler than that.


Alternatively, you can write $A = aI + bZ$, where $Z$ is the matrix with ones on the super- and subdiagonal, and write $\exp(A) = e^a \exp(bZ)$ (since the two summands commute) and
$$
\exp(-A) \otimes \exp(A) = \exp(-bZ) \otimes \exp(bZ).
$$
This shows in a simpler way that $a$ simplifies out from your expression.


