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The problem is this:

Let A be a square $n \times n$ matrix, and define $$e^A=\sum_{k=0}^\infty \frac{1}{k!}A^k$$ Find a bound for $\lvert e^A \rvert$ in terms of $\lvert A \rvert$ and $n$.

I was able to show that this converges. I know that once $k \ge \lvert A \rvert$ , the values of $\left \lvert \frac{A^k}{k!}\right \rvert$ are going to start decreasing. So I'd imagine the strategy would be to take the partial sum up to $k \ge \lvert A \rvert$, and then then find some upper bound for the decreasing series that remains. I'm not sure where to go from here.

Additionally, I'm struggling to see the relevance of the dimension of the matrix. I mean, I could bound $\lvert A \rvert$ by $\sqrt{c^2n^2}$ if I take c = the max absolute value in A, but since I'm already expressing the result in the norm of A this doesn't seem like it would be a helpful thing to do.

Cookie
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JETM
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2 Answers2

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I'm also not sure what $n$ has to do with anything. The simplest thing that comes to mind is

$$\left | e^A \right | \leq \sum_{k=0}^\infty \left | \frac{A^k}{k!} \right | \leq \sum_{k=0}^\infty \frac{|A|^k}{k!} = e^{|A|}.$$

I'm not sure how much better a bound could possibly get, because for a diagonal matrix whose largest entry is positive (for example $A=I$), this bound is tight.

Ian
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  • Thanks, man. It's an early question in an introductory-style book, so that's probably it. Silly me, missing the obvious, easy solution. – JETM Dec 24 '14 at 22:44
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Actually, the above estimate $|e^A|\le e^{|A|}$ is only true for the operator norm. (In fact it holds for any sub-multiplicative norm satisfying $|I|=1$).

For the Frobenius norm, you have a problem since if the dimension is $d$ then $|I|_{F}=\sqrt{d} \neq 1$.

So,

$e^A=\sum_{k=0}^\infty \frac{A^k}{k!}=I+\sum_{k=1}^\infty \frac{A^k}{k!}$

Hence,

$\left | e^A \right | \leq |I|+\sum_{k=1}^\infty \left | \frac{A^k}{k!}\right | = \sqrt{d} + \sum_{k=1}^\infty \left | \frac{A^k}{k!}\right | \leq \sqrt{d} + \sum_{k=1}^\infty \frac{|A|^k}{k!} = (\sqrt{d}-1)+e^{|A|}$

Indeed, in the one-dimensional case, you recover (the rather non-interesting) estimate $e^a=\left | e^a \right | \leq e^{|a|}$ for any $a \in \mathbb{R}$ (where now $|\cdot|$ denotes the standard absolute value).

Asaf Shachar
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  • Do you get $|e^A| \leq |I|-1+e^{|A|}$ for a general norm, or do you need that the norm is submultiplicative in order to get the second inequality? (I was trying to use your answer to polish mine after my attention was drawn to this question by an upvote, but I got stuck on finding a counterexample to $|A^k| \leq |A|^k$.) – Ian Dec 28 '22 at 21:19
  • This is misleading. The proof in the accepted answer holds for all sub-multiplicative norms, since the only properties used are the triangle inequality and sub-multiplicativity. The Frobenius norm is sub-multiplicative, and $|I|=\sqrt{n}$ is no problem: Then the inequality states that $\sqrt{n}e\leq e^{\sqrt{n}}$, which is always true for $n\geq 1$. – SomeCallMeTim Feb 23 '24 at 14:05