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Problem of the Week #92 - December 30th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's the last University POTW of 2013!

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Problem: Suppose that $A$ and $B$ are $n\times n$ matrices and that $AB=BA$. Prove that $e^{A+B} = e^{A}e^{B}$, where $e^A$ denotes the matrix exponential.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
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Jan 26, 2012
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This week's problem was correctly answered by Deveno. You can find his solution below.

Since the matrix exponential converges everywhere, the LHS and RHS factors are well-defined, and thus so is the product series of the RHS. Now the $k$-th term of $e^Ae^B$ is:

$\displaystyle \sum_{i=0}^k \left(\frac{A^i}{i!}\right)\left(\frac{B^{k-i}}{(k-i)!}\right)$

$\displaystyle = \frac{1}{k!}\sum_{i = 0}^k \frac{k!}{i!(k-i)!}A^iB^{k-i}$

$\displaystyle = \frac{1}{k!}\sum_{i = 0}^k \binom k i A^iB^{k-i}$

$\displaystyle = \frac{1}{k!} (A+B)^k$

(we can only justify using this binomial expansion BECAUSE $A$ and $B$ commute: for example, in general we have

$(A+B)^2 = A^2 + AB + BA + B^2 = A^2 + 2AB + B^2$ only if $AB = BA$ and similarly with higher powers)

which is precisely the $k$-th term of $e^{A+B}$.
 
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