It is well known that the Kronecker sum of matrices $A$,$B$, defined by
$A \oplus B = A \otimes I + I \otimes B$,
satisfies the nice identity
$ \text{exp}(A \oplus B) = \text{exp} A \otimes \text{exp} B . $
What does not seem to be often discussed is the behavior of $\text{exp}(A \otimes B)$.
I am interested mostly in the case when $A,B$ are Hermitian. I suppose this is easy in certain special cases (e.g. when $A^2 = I$ and $B^2 = I$, the infinite series can be simplified) but surely there must be something nice one can say more generally.
Best Answer
Here's one general approach: suppose that we can diagonalize $A$. That is, we have $A = VDV^*$ where $V$ is unitary and $$ D = \lambda_1 I_{m_1} \dot+ \cdots \dot+ \lambda_k I_{m_k} := \pmatrix{\lambda_1 I_{m_1}\\ & \ddots \\ && \lambda_k I_{m_k}}. $$ for some unitary $V$. Here, $I_k$ denotes the identity matrix of size $k$. Note that throughout, I use $\dot+$ to refer to the "diagonal sum" $$ A \dot+ B = \pmatrix{A&0\\0&B}. $$ We can now state that $$ A \otimes B = (VDV^*) \otimes B = (V \otimes I_n)(D \otimes B)(V \otimes I_n)^*. $$ where $n$ is the size of $B$. Now, assuming that we can compute $\exp(tB)$ easily, we can now compute $$ \exp(D\otimes B) = \exp[(I_{m_1} \otimes (\lambda_{m_1}B)) \dot{+} \cdots \dot + (I_{m_k} \otimes (\lambda_{m_k}B))]\\ = (I_{m_1} \otimes \exp(\lambda_{m_1}B)) \dot{+} \cdots \dot + (I_{m_k} \otimes \exp(\lambda_{m_k}B)). $$ With the above computed, we can now say that $$ \exp(A\otimes B) = (V \otimes I_n)^*\exp(D\otimes B)(V \otimes I_n). $$