I'm trying to prove this inequality:

$\|e^A\|\le e^{\|A\|}$, where $A$ is a matrix and $\|A\|:=\sup_{|x|=1} |Ax|$.

My attempt of solution:

Since $e^A:=I+A+A^2/2!+A^3/3!+\ldots$

we have


$$=\sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\| $$ $$\leq \sup_{|x|}\|Ix\| +\sup_{|x|=1}\|Ax\|+\frac{\sup_{|x|}\|A^2x\|}{2!}+\frac{\sup_{|x|}\|A^3x\|}{3!}+\ldots$$

Am I right so far? I couldn't go further

I need help!

Thanks a lot.

  • $\begingroup$ Use the fact that $\|A+B\|\le \|A\|+\|B\|$. $\endgroup$ Jun 14, 2013 at 22:35
  • 1
    $\begingroup$ That's true in any unital Banach algebra by triangular inequality and submultiplicativity of the norm. $\endgroup$
    – Julien
    Jun 14, 2013 at 22:56
  • $\begingroup$ Here is the technique you need. $\endgroup$ Jun 15, 2013 at 6:17

1 Answer 1


If you've established a few basic properties (subadditivity, submultiplicativity, and continuity) of the norm, you can do it without having to bring in the $\sup$ manipulation. Namely, for any $n$, we have by subadditivity of the norm that $$ \| I+A+\frac{A^2}{2!}+\cdots +\frac{A^n}{n!} \| \leq \| I\| + \|A\|+\|\frac{A^2}{2!}\|+\cdots +\|\frac{A^n}{n!} \| $$ Then since we know $\| A^2\| \leq \| A\|^2$, we can pull all the exponents out:

$$ \| I\| + \|A\|+\|\frac{A^2}{2!}\|+\cdots +\|\frac{A^n}{n!} \| \leq \| I\| + \|A\|+\frac{1}{2!} \|A\|^2 +\cdots +\frac{1}{n!} \|A \|^n $$ Then since the norm is a continuous function, we can actually pass onto the limit as $n\to \infty$ of the above inequality, and obtain $$ \|e^A \| = \|I+A+\cdots \| \leq \| I\| + \|A\|+\frac{1}{2!} \|A\|^2 +\cdots +\frac{1}{n!} \|A \|^n + \cdots = e^{\| A\|} $$

  • $\begingroup$ Ah, good point! $\endgroup$ Jun 14, 2013 at 22:47
  • $\begingroup$ Great answer, thank you very much $\endgroup$
    – user42912
    Jun 15, 2013 at 2:26

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