I want to ask General strategy of the error bound of the matrix exponential. For example, suppose, $A, B$ are finite dimension $n \times n$ matrices with complex coefficients. Using Baker–Campbell–Hausdorff formula or simple calculation. One can show that ($x \geq 0$)


But the thing I want is that the clear bounds. I want to find the $c$ such that $||e^{Ax}e^{Bx}-e^{Bx}e^{Ax}-[A,B]x^2|| \leq c|x^3|.$ Here $c$ should be function of $A,B.$

The first method I want to try is to use Taylor's theorem. But the computation is so tedious. Is there any way that I can find the clear expression for $c$?

More generally, given any matrix expression in terms of parameters $x$, one can easily see the first few terms, but is there a more elegant way to control the error bounds? Or in the simplest case, what is the error bounds for Baker–Campbell–Hausdorff formula $e^{Ax}e^{Bx}$ to $n$ order. I mean the clear expression of the constant before $|x|^n$ not just the big O notation.

  • $\begingroup$ One can get reasonable error bounds by upper bounding the third derivative of $x \mapsto e^{Ax} e^{Bx} - e^{Bx} e^{Ax}$ (i.e., $e^{Ax} (A^3+3A^2B + 3AB^2+B^3)e^{Bx} - e^{Bx} (B^3+3B^2A+3BA^2+A^3) e^{Ax}$) in, say, the operator norm, and applying Taylor's theorem with remainder. $\endgroup$
    – Terry Tao
    Nov 30, 2021 at 22:38
  • $\begingroup$ Thanks, professor, But I would guess that c should involve a term like $[A, B]$ commutator. So then when $[A, B]=AB-BA=0$. I will get zero left-hand sides. The problem with using Taylor remainder is that the bound will become $(|A|+|B|)^3 e^{|Ax|+|Bx|}$ somehow do not reflect the fact that when $[A, B]=0$, I should get zero. $\endgroup$
    – En-Jui Kuo
    Dec 1, 2021 at 4:27
  • $\begingroup$ Estimates of this type can be obtained with more work. Starting with the Duhamel formula $[e^{Ax}, B] = \int_0^x e^{Ay} [A,B] e^{A(x-y)}\ dy$ (arising from applying the fundamental theorem of calculus to $y \mapsto e^{Ay} B e^{A(x-y)}$) we obtain also $[e^{Ax}, e^{Bx}] = \int_0^x \int_0^x e^{Ay} e^{Bz} [A,B] e^{B(x-z)} e^{A(x-y)}\ dz dy$. Taylor expansion of the $e^{Ay}, e^{Bz}, e^{B(x-z)}, e^{A(x-y)}$ terms will then give a satisfactory approximation. $\endgroup$
    – Terry Tao
    Dec 1, 2021 at 5:21
  • $\begingroup$ Professor, thanks a lot. I will investigate more $\endgroup$
    – En-Jui Kuo
    Dec 1, 2021 at 14:23

1 Answer 1


Here is a method that gives a simple explicit formula in this case and should give similar ones in more general cases.

Fix a matrix norm that is submultiplicative, such as the largest singular value. Assume $x \leq 1$.

We have

$$ e^{Ax} e^{Bx} - e^{Bx} e^{Ax} - [A,B]x^2 = \sum_{n,m=0}^{\infty} \frac{A^n B^m }{ n!m!} x^{n+m} - \sum_{n,m=0}^{\infty} \frac{B^m. A^n }{ n!m!} x^{n+m} - (AB-BA) x^2 = \sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{A^n B^m }{ n!m!} x^{n+m} -\sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{ B^mA^n }{ n!m!} x^{n+m} $$

so $$ \left| e^{Ax} e^{Bx} - e^{Bx} e^{Ax} - [A,B]x^2\right| = \left| \sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{A^n B^m }{ n!m!} x^{n+m} -\sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{ B^mA^n }{ n!m!} x^{n+m}\right| \leq \sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{|A|^n |B|^m }{ n!m!} x^{n+m} +\sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{ |B|^m |A|^n }{ n!m!} x^{n+m} \leq 2\sum_{0 \leq n, m \leq \infty}\frac{ |B|^m|A|^n }{ n!m!}x^3 = 2e^{ |A| + |B|} x^3. $$


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