Kronecker product: Is it possible to simplify this product $e^{-A} \otimes e^{A}$ where $A$ is an invertible and symmetric matrix [closed]

Question

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.

Answer 1

One can diagonalize your $A = VDV^{-1}$ explicitly; the closed formulas are here 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.