Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

2 Answers

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.