# Proving convergence and bounding $e^A$ in terms of $|A|$ and $n$

My text wants me to prove that $e^A$ converges for any $n \times n$ matrix and put a bound on it in terms of $|A|$ and $n$ where $| |$ is the norm induced by identification of $M(n, n)$ with $\mathbb{R}^{n^2}$. I can prove convergence and it's obviously bounded by $e^{|A|}$ but I don't see how to get it in terms of $n$.

To prove convergence, Just note that $e^{|A|} = \sum_{i=0}^{\infty}\frac{1}{i!}|A|^{i}$ converges. This means that

$$\sum_{i=0}^{j}\frac{1}{i!}|A^{i}| \le \sum_{i=0}^{j}\frac{1}{i!}|A|^{i} \le e^{|A|}$$ This means that the partial sums of the norms are non-decreasing and bounded. They hence converge and the series is absolutely convergent which implies that it's convergent.

I can't get a bound on $|e^{A}|$ other than $e^{|A|}$ though. The problem says to bound it in terms of $|A|$ and $n$. I interpret this as proving that some function $f(|A|, n)$ exists with $|e^{A}| \le f(|A|, n)$ but I don't see how to do this. Any hints?

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I think you can look at the max norm and show that the remainder goes to zero (possibly by the Taylor Remainder Theorem?), and so it's Cauchy. – Michael Chen Apr 1 '11 at 22:11

Here's a sketch of the way I did this in class (using the norm $||A|| = \max\{|a_{ij}|\}$ but should still work):

The first step is to show that $||AB|| \leq n||A|||B||$. Then by induction, letting $B = A^{r-1}$, it can be proven that $||A^r|| \leq n^{r-1} ||A||^r$.

Consider the Taylor series $$T_n(x) = \sum_{k=0}^n \frac{A^k}{k!}$$ as a sequence of partial sums. Use Taylor's Remainder Theorem that $$|R_n| \le \frac{f^{(n+1)}(C)}{n!}(A-0)^n = \frac{e^C}{n!}A^n$$ for some $C$. Using the norm bound found above, since the factorial outgrows the exponential, you can show that this goes to zero for large $n$. Therefore it converges.

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I appreciate the help but I'm not sure how to get the bound I need out of this. I edited my question and expanded on how I'm currently proving convergence. – knucklebumpler Apr 1 '11 at 23:33

I ripped this from here: http://www.cis.upenn.edu/~cis610/diffgeom-n.pdf

If $A = (a_{ij})$ is an $n\times n$ matrix, and we let $\mu = max \{|a_{ij}|: 1\leq i,j\leq n\}$, then $|a_{i,j}^{(p)}| \leq (n\mu)^{p}$ (See link for a proof by induction), where $A^{p} = (a_{i,j}^{(p)})$. See the link for the rest. Only downside is that $|A|$ isn't involved.

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