TITLE: Prove the bounds for the matrix exponential
QUESTION [6 upvotes]: Prove that $ e^{-t\|A\|} \le \|e^{tA}\| \le e^{t\|A\|} $ for any matrix $A \in \mathbb{R}^{n \times n}, t > 0$, where $e^{tA} = I + tA + t^2 \dfrac{A^2}{2!} + t^3 \dfrac{A^3}{3!} + \cdots$
It is easy to prove the right part of the inequality $ \|e^{tA}\| \le e^{t\|A\|} $ for $t > 0$: $\|e^{tA}\| = \| I + tA + t^2 \dfrac{A^2}{2!} + t^3 \dfrac{A^3}{3!} + \cdots \| \le 1 + \|tA\| + \dfrac{\|tA\|^2}{2!} + \cdots = e^{\|tA\|}$
But I got stuck with the left part.
Great thanks for any help or ideas!

REPLY [7 votes]: The matrix exponential
satisfies the identity $e^a e^B = e^{A+B} $ if the matrices $A$ and $B$ commute $(AB=BA)$. In particular $e^{tA} e^{t(-A)} = I$ and therefore
$$
 1 = \| I \| = \| e^{tA} e^{t(-A)} \| \le \| e^{tA}\| \cdot \| e^{t(-A)}\|  \, .
$$
Using the already proven upper bound (for $-A$) one gets
$$
 1 \le  \| e^{tA}\| \cdot  e^{t\|-A\|} 
 = \| e^{tA}\| \cdot  e^{t\|A\|} \, ,
$$
from which the lower bound follows.