2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Right thank you so much! $\endgroup$ – IntegrateThis Jun 20 '19 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.