# Inequality: Norm of difference in exponential of matrices

Let $A$ and $B$ be two matrices. Then, $$\|e^{A+B}- e^A\| \leq \|B\|e^{\|A\|+\|B\|}.$$

Can anyone give me a proof or reference for this inequality, please? Thank you.

Yes, this is mostly just combinatorial and algebraic identities.

Letting $f(0)=A$ and $f(1)=B$, we can distribute terms to get $$(A+B)^n-A^n = \sum_{\sigma\in\{0,1\}^n \\ \;\;\sigma \not\equiv 0} \prod_{j=1}^n f(\sigma(j))$$ Thus \begin{align*}\|(A+B)^n-A^n\| &\leq \sum_{\sigma\in\{0,1\}^n \\ \;\;\sigma \not\equiv 0} \prod_{j=1}^n \|f(\sigma(j))\| \\&= \sum_{k=1}^n {{n}\choose {k}} \|A\|^k \|B\|^{n-k} \\ &= (\|A\|+\|B\|)^n - \|A\|^n \\ &= \|B\| \bigg( \sum_{j=0}^{n-1} (\|A\|+\|B\|)^{n-j-1}\|A\|^{j}\bigg) \\ &\leq n\cdot \|B\| \cdot (\|A\|+\|B\|)^{n-1}\end{align*} In the second line we just used basic combinatorics, in the third line we used the binomial theorem, in the fourth line we used the identity $x^n-y^n = (x-y)\big(\sum_{0}^{n-1} x^jy^{n-1-j}\big)$, and in the last line we just used the fact that $\|A\|^j \leq (\|A\|+\|B\|)^j$ for all $j$.

Therefore \begin{align*} \|e^{A+B}-e^{A}\| &\leq \sum_{n=1}^{\infty} \frac{\| (A+B)^n -A^n\|}{n!} \\ &\leq \sum_{n=1}^{\infty} \frac{n \cdot \|B\| \cdot (\|A\|+\|B\|)^{n-1}}{n!} \\ &= \|B\|\sum_{n=0}^{\infty} \frac{(\| A\|+\|B\|)^n}{n!} \\ &= \|B\|e^{\|A\|+\|B\|} \end{align*} This proves the claim. One could also use more advanced methods like Frechet calculus to prove even stronger bounds than this, like $\|e^{B}-e^{A}\| \leq \|B-A\|e^{\max \{\|A\|,\|B\|\}},$ or even better if $A,B$ are symmetric then $\|e^{B}-e^{A}\| \leq \|B-A\|e^{\max \Sigma(A) \cup \Sigma(B)}$.

• Where can I read about these inequalities? – Swapnil Tripathi May 22 '19 at 4:02