MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A,B$ be full rank $n \times n$ matrices. If $AB = BA$, then $\exp(\log(A)+\log(B))=AB$.

Supposing $A = USL$ and $B = VSL$ where $U,V,S,L$ are integer valued matrices, $det(L)=1$ and $U = LVL^{-1}$. If $AB = (USL)(VSL) = (L(VSL)L^{-1})(L^{-1}(USL)L) = (LBL^{-1})(L^{-1}AL)$, is there a possibility to use exponentials to calculate $AB$?

share|cite|improve this question
Yes, the magic words are "Cambpell-Baker-Hausdorff", but why? – Igor Rivin Aug 6 '12 at 17:34
Hi Igor: I thought the case here may be easier. I am trying to represent something that is commutative. But this is the best I am getting. – Turbo Aug 7 '12 at 3:52
Also would you happen to know the complexity of CBH. I have a long iterative chain of such matrices. – Turbo Aug 7 '12 at 3:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.