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If $||\cdot||$ is a matrix norm and $A$ is a matrix, is there any relation between the two quantities $||A||$ and $||e^A - I||$, where $I$ is the identity matrix?

I am tempted to say the former is always smaller than or equal to the latter, analogous to $x \leq e^x - 1$ for all $x \in \mathbb{R}$, but my numerical experiment suggests that it is the other way: $||A|| \geq ||e^A - I||$. Can some one please provide some idea?

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  • $\begingroup$ There is no such relationship for the Euclidean norm. Consider $A=I$ and $A=-I$ which give $\le$ and $\ge$ respectively. $\endgroup$ – Peter Foreman Jan 15 at 22:40
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If the norm is submultiplicative (i.e. $\ \|AB\,\|\le \|A\,\| \|B\,\|\ $ for all $\ A,B\ $), then you do get \begin{align} \left\|e^A-I\right\|&= \left\|\sum_{n=1}^\infty\frac{A^n}{n!} \right\|\\ &\le \sum_{n=1}^\infty\frac{\|A\|^n}{n!}\\ &=e^{\|A\|}-1\ , \end{align} but since $\ e^x\ $ is not bounded above by any polynomial in $\ x\ $, neither is $\ e^{\|A\|}-1\ $ bounded above by any polynomial in $\ \|A\|\ $.

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