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

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#### Chris L T521

##### Well-known member
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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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#### Chris L T521

##### Well-known member
Staff member
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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