Showing that $e^{iB \otimes e_1 \otimes e_1^t} = e^{iB} \otimes e_1 \otimes e_1^t + I_n \otimes(I_n - e_1 \otimes e_1^t)$ for a matrix $B$

Question

Suppose I have a square matrix $B$ of dimension $n$, a paper I am reading states without proof that $e^{iB \otimes e_1 \otimes e_1^t} = e^{iB} \otimes e_1 \otimes e_1^t + I_n \otimes(I_n - e_1 \otimes e_1^t)$ where this last equality follows from the series expansion of the exponential, and $e_1$ is the first basis element of the canonical basis of $\mathbb{C}^n$.

Following this statement and using the fact that $A^n \otimes B^n = (A \otimes B)^n$, then : $e^{iB \otimes e_1 \otimes e_1^t} = \sum_{k=0}^\infty \frac {1} {k!} (iB \otimes e_1 \otimes e_1^t)^k$, and $(e_1 \otimes e_1^t)^k = (e_1 \otimes e_1^t)$ for any positive integer $k$. , and so it seems to me that : $e^{iB \otimes e_1 \otimes e_1^t} =\sum_{k=0}^\infty \frac {1} {k!} (iB)^k \otimes (e_1 \otimes e_1^t) = (\sum_{k=0}^\infty \frac {1} {k!} (iB)^k) \otimes (e_1 \otimes e_1^t) = e^{iB} \otimes (e_1 \otimes e_1^t)$, so that I am not sure how this identity was obtained.

Answer 1

You almost got it right. You correctly noted that $(e_1 \otimes e_1^t)^k = (e_1 \otimes e_1^t)$ for any positive (!) integer $k$, but then later in your derivation you implicitly used (incorrectly) also that $(e_1 \otimes e_1^t)^0 = (e_1 \otimes e_1^t)$. Now let us reconsider the expansion of $e^{iB \otimes e_1 \otimes e_1^t}$ by first separating the $0$th (identity) term: $$e^{iB \otimes e_1 \otimes e_1^t} = I_n \otimes I_n + \sum_{k=1}^\infty \frac {1} {k!} (iB)^k \otimes (e_1 \otimes e_1^t)^k = I_n \otimes I_n +(\sum_{k=1}^\infty \frac {1} {k!} (iB)^k) \otimes (e_1 \otimes e_1^t)\\ =I_n \otimes (I_n -e_1 \otimes e_1^t) + I_n \otimes e_1 \otimes e_1^t +(\sum_{k=1}^\infty \frac {1} {k!} (iB)^k) \otimes (e_1 \otimes e_1^t) \\ = I_n\otimes (I_n -e_1 \otimes e_1^t) +(\sum_{k=0}^\infty \frac {1} {k!} (iB)^k) \otimes (e_1 \otimes e_1^t)= e^{iB} \otimes (e_1 \otimes e_1^t),$$ which is exactly what we wanted to prove.