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Let $A$, and $B$ be commuting $n\times n$ matrices, i.e., $A B = B A$. Let $$ \exp(A) := \sum_{i=0}^\infty\frac{1}{i!} A^i $$ Show that $\exp(A+B) = \exp(A) \exp(B)$.

Arctic Char
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user25004
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    Do you agree with my edits? – Rodrigo de Azevedo Mar 06 '23 at 13:43
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    @user25004 I do agree with your edit. I don’t understand why your post got closed after 7 years of original post. It’s like user are voting to close decade old post. Math SE community claim this site has evolved over time. PSQ post was justified in 2013. In this “model” every PSQ post exists on this site should be closed too. The major flaw of closing vote in my opinion is that once a post go in close vote review it comes out close. Reviewer don’t see date of publication of post. They just see PSQ, directly vote to close. – user264745 Mar 06 '23 at 15:44

1 Answers1

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By Cauchy product we have $$\exp(A)\exp(B)=\left(\sum_{i=0}^\infty\frac{1}{i!} A^i\right)\left(\sum_{i=0}^\infty\frac{1}{i!} B^i\right)=\sum_{i=0}^\infty c_i$$ where $$c_i=\sum_{k=0}^i\frac{1}{k!} A^k\frac{1}{(i-k)!} B^{i-k}=\frac{1}{i!}\sum_{k=0}^i{i\choose k}A^k B^{i-k}=\frac{1}{i!}(A+B)^i$$ hence $$\exp(A)\exp(B)=\sum_{i=0}^\infty\frac{1}{i!}(A+B)^i=\exp(A+B)$$

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    what was the use of the given A commutes with B? can you elaborate how did you implement the cauchy product (of scalars) into matrices? – Yarden Cohen Feb 13 '20 at 19:28
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    This doesn't invoke the commutativity of A and B. It is essentially identical to the proof for real numbers (not matrices). This is incomplete. – JoseOrtiz3 Feb 26 '20 at 07:08
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    It was used in the final step; the usual binomial expansion of $(A+B)^i$ doesn't work unless $A$ and $B$ commute. – Richard Jul 21 '20 at 22:40
  • To those who voted and closed this question and similar ones.... The timing of asking questions is actually important! When there are two similar questions, I think it makes highly more sense check timings and to vote to close the one which is asked later (here the other question which is asked 5 years later than mine). Note that if their research was done well or the duplicate detection worked well, they would not be able to post. Now unnecessarily closing well attended questions of earlier contributors (here me) I wonder what aspect of asking good questions is actually being rewarded?! – user25004 Sep 28 '20 at 17:23