# A bound on $e^{A+B}$ for noncommutative $A$ and $B$

It is folklore that when two complex $n\times n$ matrices $A$ and $B$ commute, the equality $e^{A+B}=e^Ae^B$ holds. An elementary consequence of this is that for any submultiplicative norm, commuting matrices satisfy the inequality $$||e^{A+B}||\leq ||e^A||\cdot||e^B||.$$

Earlier this week, I was trying to prove a result (with $n=2$) that would have benefitted from the above inequality. However, my matrices did not commute. Testing a few thousand pairs of matrices in GNU Octave, not a single one failed to pass the test, which is reasonable evidence to suggest that the above inequality might hold generally or, if it fails, one might need to be clever in finding a counterexample.

I could not find a counterexample and did not get very far in proving it.

While I have moved on a proved the result without the inequality, it's still bothering me. Could somebody shed some light on this?

• I wonder whether you could parlay the Baker-Campbell-Hausdorff relation into a proof of this conjecture, or whether it could be used to construct a counterexample. Commented Mar 31, 2017 at 21:21
• Not sure how your test was done, but I found that the inequality is false for about 8% of all test cases: count=0;for k=1:10000,A=rand(2,2);B=rand(2,2);if norm(expm(A+B))>norm(expm(A))*norm(expm(B)),count=count+1;end;end;count Commented Apr 1, 2017 at 6:23
• I might have exaggerated the "thousands" part. I ran a few hundred with uniform sampling in (-1,1). Commented Apr 1, 2017 at 19:35

The inequality does not always hold. Counterexample: $$A=\pmatrix{0&-2\pi\\ 2\pi&0}, \ B=\pmatrix{0&-\pi\\ 4\pi&0}, \ A+B=\pmatrix{0&-3\pi\\ 6\pi&0}.$$ Then $e^A=e^B=I$ and $$e^{A+B}= \pmatrix{\cos\left(3\sqrt{2}\pi\right)&-\frac1{\sqrt{2}}\sin\left(3\sqrt{2}\pi\right)\\ \sqrt{2}\sin\left(3\sqrt{2}\pi\right)&\cos\left(3\sqrt{2}\pi\right)}.$$ When the operator norm is used, numerically we have $\|e^{A+B}\|\approx1.2735>1=\|e^A\|\|e^B\|$.