# Matrix exponential: Bound $\|e^{X+Y} e^{-Y}-I\|$ in terms of only $X$.

Let $$X$$ and $$Y$$ denote symmetric real matrices of the same fixed size. Let $$\|\cdot\|$$ denote a submultiplicative matrix norm. (For concreteness, let $$\|\cdot\|$$ be the Frobenius norm, but I am interested in other norms too.)

Define $$f(X) = \sup_Y \|e^{X+Y}e^{-Y}-I\|$$ and $$g(X)=\inf_Y \|e^{X+Y}e^{-Y}-I\|.$$

Trivially, $$0 \le g(X) \le \|e^X-I\| \le f(X)$$. The question is to prove nontrivial bounds. Specific questions:

1. Is $$f(X)$$ finite?
2. Is $$g(X)$$ nonzero?
3. Is $$f(X) = \Theta(\|X\|)$$ as $$\|X\|\to0$$?
4. Is $$g(X) = \Theta(\|X\|)$$ as $$\|X\|\to0$$?
5. Is there a simple closed form expression for $$f(X)$$ or $$g(X)$$?

Note that $$\|e^X-I\|=\Theta(\|X\|)$$ as $$\|X\|\to0$$.

The reason I'm asking this question is that I want to have some understanding of how "smooth" the matrix exponential is.

A related bound is $$\|e^{X+Y}-e^Y\| \le \|X\|e^{\|X\|+\|Y\|}$$ and, more generally, $$\|e^X-e^Y\| \le \|X-Y\| e^{\max\{\|X\|,\|Y\|\}}$$.

• If you expand $e^{X+Y}e^{-Y}-I$ you get $X + (X^2 + XY - YX)/2 + \dots$. That seems to at least suggest the quantity can be taken to 0 for fixed $Y$. For non-fixed $Y$ I would guess $f(X)=\infty$, but I don't know. – Thomas Ahle May 14 '19 at 19:11