Exponential of a matrix always converges

Question

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:

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

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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 $q^{n-1}\times a^n$

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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?

Answer 1

This topic is extraordinarily well explained in the book Naive Lie Theory. Here is an extract that will answer your question.

Answer 2

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 $$ \frac{d}{dt}\exp(tM)=M\exp(tM). $$

Answer 3

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.