# 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 $B$

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.

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.

• Right thank you so much! – IntegrateThis Jun 20 '19 at 22:55