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I am trying to show that the exponential of a matrix converges for any given square matrix of size $n\times n$:

$M\mapsto e^M$ e.g. $\displaystyle e^M = \sum_{n=0}^\infty \frac{M^n}{n!}$

Can I argue that: Since $n!$ necessarily grows faster than $k^n$ will, that this converges. This seems to be an obvious fact, since:

$$n!=1\times 2\times 3\times \cdots \times k\times (k+1)\times (k+2)\times \cdots$$ $$k^n=k\times k\times k \times\cdots\times k \times k\times \cdots$$


If we have some $q\times q$ matrix, with $a$'s in each position(which will grow as fast as we make our $a$ and $q$ large) we still only get increasing at a rate of $q^{n-1}\times a^n$


In light of the comments, I know that in this banach space, I need only show that $\displaystyle e^M = \sum_{n=0}^\infty \frac{||M||^n}{n!}$ converges. Now I have many matrix norms to choose from, and I can't seem to get a good argument going rigorously. Any ideas?

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  • $\begingroup$ That's not quite valid since you're dealing with matrices. $\endgroup$ – Qudit Apr 9 '15 at 3:18
  • $\begingroup$ @Qudit But the maximum my matrix $q\times q$ can in increase if I fill all entries with $a$ where $a$ is some number is $q a^2, q^2 a^3,q^3 a^4$ in each position, and the same applies I'll edit this in $\endgroup$ – Oceans Bleed Apr 9 '15 at 3:19
  • $\begingroup$ @CameronWilliams Meaning that it converges in magnitude? $\endgroup$ – Oceans Bleed Apr 9 '15 at 3:26
  • $\begingroup$ @CameronWilliams A banach space is a space where all cauchy sequences converge(due to completeness) and we have a norm( standard euclidean norm here?)? $\endgroup$ – Oceans Bleed Apr 9 '15 at 3:28
  • $\begingroup$ Let me revise my comment. I meant to say something else. In a Banach space (a complete normed vector space), like the space of $n\times n$ matrices, if a series converges absolutely, then it converges. So if you can show that $\sum_n \frac{\|M\|^n}{n!}$ converges, then you know that your series above converges. $\endgroup$ – Cameron Williams Apr 9 '15 at 3:29
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This topic is extraordinarily well explained in the book Naive Lie Theory. Here is an extract that will answer your question.

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  • $\begingroup$ Thanks. I will read this. What is the goal of your website btw? $\endgroup$ – Oceans Bleed Apr 15 '15 at 7:04
  • $\begingroup$ @OceansBleed It is explained in the website itself, in the About section. It is in its early steps right now, but a big growth is scheduled for the next months. Contact us for more info! $\endgroup$ – mathifold.org Apr 15 '15 at 7:08
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A very elegant, and a bit more advanced way of showing it is the following. Consider the Initial Value Problem $$ \left\{\begin{array}{cc} X'=MX, \\ X(0)=I, \end{array} \right. $$ where $X$ is the unknown, an $N\times N$ matrix and $I$ the identity matrix. Picard-Lindelöf Theorem, for linear systems, implies that the recursive sequence $$ X_{n+1}(t)=I+\int_0^t MX_n(s)\,ds, $$ converges uniformly in every closed interval. It is readily seen that $$ X_n=\sum_{k=0}^n\frac{t^kM^k}{k!}. $$ Hence, the Neuman series $\sum_{n=0}^\infty\frac{t^nM^n}{n!}=\exp(tM)$ is has infinite radius of convergence and satisfies the equation $$ \frac{d}{dt}\exp(tM)=M\exp(tM). $$

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For any nxn matrix A, the sequence

$I+A+\frac{A^2}{2!}+...$

converges.

Proof

Let m be the largest $|a_{ij}|$ in A. Then

The biggest element in first term is 1.

The biggest element in second term is m.

The biggest element in third term is $\leq\frac{nm^2}{2!}$.

The biggest element in fourth term is $\leq\frac{n^2m^3}{3!}$. etc.

Any ij sequence is dominated by 1, $m$, $\frac{nm^2}{2!}$, $\frac{n^2m^3}{3!}$......$\frac{n^{k-2}m^{k-1}}{{k-1}!}$.....

Applying the ratio test to this maximal sequence gives

$\frac{n^{k-1}m^{k}}{{k}!} \frac{{(k-1)}!}{n^{k-2}m^{k-1}} = \frac{nm}{k}$ Since n and m are fixed, the ratio goes to 0 as $k \to \infty $, proving (absolute) convergence.

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    $\begingroup$ Link only answers are discouraged because links can break. It is better for you to include the main points of the proof, or reproduce it completely, here (with full attribution of course). $\endgroup$ – Michael Albanese Jan 14 '16 at 11:56

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