Let $A,B\in gl(n)$. Then prove $$e^{A+B} = \lim_{k\to\infty} \left (e^{\frac{A}{k}}e^{\frac{B}{k}}\right )^k$$

I found this theorem in this notes http://www4.ncsu.edu/~aalexan3/articles/liegroups.pdf but there is no proof. I am new to exponential of matrices and can't prove this by my own.

  • $\begingroup$ Would you be happy with a more general proof assuming that $A$,$B$ are bounded operators on a Banach space? $\endgroup$
    – user159517
    Nov 25, 2016 at 20:43
  • $\begingroup$ @user159517 Sorry I have no knowledge on Banach space. But I don't mind seeing a more general proof. $\endgroup$
    – Kenneth.K
    Nov 25, 2016 at 20:50
  • $\begingroup$ You can just read my answer thinking of $A,B$ as matrices and $\|.\|$ as some kind of matrix norm. $\endgroup$
    – user159517
    Nov 25, 2016 at 20:52

2 Answers 2


Let $A,B \in \mathcal{B}(H)$ be bounded operators on some Banach space $H$. Define $C := e^{(A+B)/k}$ and $D := e^{A/k}e^{B/k}$. We have the following estimates for the norm of $C,D$: $$\|C\|, \|D\| \leq \exp\left[\frac{\|A\| + \|B\|}{k}\right] = \exp(\|A\| + \|B\|)^{1/k} $$ Applying the Cauchy-Product formula on $D$ yields $$ D = e^{A/k}e^{B/k} = \sum_{i=0}^{\infty} \frac{(A/k)^{i}}{i!} \cdot \sum_{j=0}^{\infty} \frac{(B/k)^{j}}{j!} = \sum_{m=0}^{\infty} k^{-m} \sum_{i=0}^{m} \frac{A^{i}}{i!} \cdot \frac{B^{m-i}}{(m-i)!}$$ Which permits us to bound the norm of $C-D$ by \begin{align*} \|C-D\| &= \left\|\sum_{i=0}^{\infty} \frac{([A+B]/k)^{i}}{i!} - \sum_{m=0}^{\infty} k^{-m} \sum_{i=0}^{m} \frac{A^{i}}{i!} \cdot \frac{B^{m-i}}{(m-i)!} \right\| \\ &= \left\|\sum_{i=2}^{\infty} k^{-i} \frac{(A+B)^{i}}{i!} - \sum_{m=2}^{\infty} k^{-m} \sum_{i=0}^{m} \frac{A^{i}}{i!} \cdot \frac{B^{m-i}}{(m-i)!} \right\| \\ &\leq \frac{1}{k^2} \cdot\left[ \exp(\|A\| +\|B\|) + \sum_{m=2}^{\infty} \sum_{i=0}^{m} \frac{\|A\|^{i}}{i!} \cdot \frac{\|B\|^{m-i}}{(m-i)!}\right] \\ &= \frac{1}{k^2} \cdot\left[ \exp(\|A\| +\|B\|) + \sum_{m=2}^{\infty} \frac{1}{m!}\sum_{i=0}^{m} \binom{m}{i}\|A\|^{i} \cdot \|B\|^{m-i}\right]\\ &= \frac{1}{k^2} \cdot\left[ \exp(\|A\| +\|B\|) + \sum_{m=2}^{\infty} \frac{(\|A\|+\|B\|)^m}{m!}\right] \\ &\leq \frac{2}{k^2} \cdot \exp(\|A\| +\|B\|) \end{align*} We obtain $$\| C^{k} - D^{k} \| = \left\| \sum_{m=0}^{k-1} C^{m} (C-D) D^{k-m-1}\right\| \leq \exp(\|A\|+\|B\|)\cdot k \cdot \|C-D\|$$ where I used that $\|C\|^m\cdot \|D\|^{k-m-1} \leq \exp(\|A\|+\|B\|)^{\frac{k-1}{k}} \leq \exp(\|A\|+\|B\|).$ Together with the estimate for $\|C-D\|$, this yields $$\| C^{k} - D^{k} \| \leq \frac{2\exp(2\|A\| + 2\|B\|)}{k} $$ which proves the claim.

  • $\begingroup$ Thanks. This is a nice proof. $\endgroup$
    – Kenneth.K
    Nov 25, 2016 at 20:54
  • $\begingroup$ Can you explain more on the last equation that $||C^k-D^k||=||\dots||\leq exp(||A||+||B||) k ||C-D||$? $\endgroup$
    – Kenneth.K
    Nov 25, 2016 at 21:38
  • $\begingroup$ I supplied more details. Is it clear now? $\endgroup$
    – user159517
    Nov 25, 2016 at 21:47
  • $\begingroup$ That's clearer. And for the last inequality, should it be $||C^k-D^k||\leq \frac{2\exp(2||A||+2||B||)}{k}$? $\endgroup$
    – Kenneth.K
    Nov 25, 2016 at 21:48
  • $\begingroup$ @Kenneth.K Yes, it should. $\endgroup$
    – user159517
    Nov 25, 2016 at 21:58

For any matrix $P$, $$ e^P = I + P + {P^2 \over 2!} + {P^3 \over 3!} + \cdots + {P^n \over n!} + \cdots $$

Thus, it follows that $$ e^{A \over m} e^{B \over m} = I + {A \over m} + {B \over m} + O\left({1 \over m^2} \right) $$

Since $e^{A \over m} e^{B \over m} \to I$ as $m \to \infty$, we see that $e^{A \over m} e^{B \over m}$ is in the domain of the logarithm for all $m$ sufficiently large.

We also know that for all $n \times n$ matrices with $\Vert P \Vert < {1 \over 2}$, $$ \log(I + P) = P + O(\Vert P \Vert^2) $$

Hence, for large values of $m$, we have $$ \log\left( e^{A \over m} e^{B \over m} \right) = \log\left( I + {A \over m} + {B \over m} + O\left({1 \over m^2} \right) \right) $$ i.e. $$ \log\left( e^{A \over m} e^{B \over m} \right) = {A \over m} + {B \over m} + O\left( \left\Vert {A \over m} + {B \over m} + O\left({1 \over m^2}\right) \right\Vert^2 \right) $$

Thus, $$ \log\left( e^{A \over m} e^{B \over m} \right) = {A \over m} + {B \over m} + O\left( {1 \over m^2} \right) $$

Exponentiating the logarithm, we get $$ e^{A \over m} e^{B \over m} = \exp\left[ {A \over m} + {B \over m} + O\left( {1 \over m^2} \right) \right] $$

Therefore, $$ \left( e^{A \over m} e^{B \over m} \right)^m = \exp\left[ A + B + O\left( {1 \over m} \right) \right] $$

By the continuity of the exponential, we conclude that $$ \lim\limits_{m \to \infty} \ \left( e^{A \over m} e^{B \over m} \right)^m = exp(A + B) $$ which proves the Lie product formula.


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