I am considering an exponential on the following form:
$$\exp(X + A \otimes I_B),$$
where $X$ is a Hermitian operator on a tensor Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$, $A$ is a Hermitian operator on $\mathcal{H}_A$, and $I_B$ is the identity opeartor on $\mathcal{H}_B%$. Furthermore, all Hilbert spaces are finite dimensional.
My question is now if there exists an operator $Z_A$ on $\mathcal{H}_A$, such that
$$\exp(X + A \otimes I_B) = \exp(X)Z_A\otimes I_B.$$
I am hoping that this might be the case, since I'm trying to factor out something that acts nontrivially on $\mathcal{H}_A$ only. I think a place to start might be the Zassenhaus formula, but I'm not sure. I only need to know about existence.
Update:
So here is my attempt at a proof. Since $[X,A\otimes I_B]$ should be an operator acting non-trivially on $\mathcal{H}_A$ only(?), the Zassenahus formula (linked to above), means we can write
$$\exp(X + A \otimes I_B) = \exp(X)\exp(A\otimes I_B)C_A\otimes I_B,$$
Where $C_A\otimes I_B$ is given by the infinite product of exponentials of nested commutators in the Zassenhaus formula. My biggest worry now is about the existence of this infinite product. I am guaranteed that the operator $C_A$ here is well-defined?
