Why does the matrix exponential $e^A$ always exist?

Question

Why does $e^A$ always exist for any given $n \times n$ matrix $A$?

I can't find anything discussing this question, which is quite suprising, since it is such a general question.

Answer 1

Consider the matrix norm $$\|A\|:=\max_{x\in\mathbb{R}^{n}}\frac{|Ax|}{|x|},$$ where $|\cdot |$ is the Euclidean metric on $\mathbb{R}^{n}$.

Let's prove that $\|AB\|\le\|A\|\|B\|$, for all $A,B\in M(n,\mathbb{R})$: \begin{align}\frac{|ABx|}{|x|}&=\frac{|ABx|}{|Bx|}\cdot\frac{|Bx|}{|x|}\le \max_{x\in\mathbb{R}^{n}}\frac{|ABx|}{|Bx|}\cdot \max_{x\in\mathbb{R}^{n}}\frac{|Ax|}{|x|}\le\\ &\le \max_{y\in\mathbb{R}^{n}}\frac{|Ay|}{|y|}\cdot \max_{x\in\mathbb{R}^{n}}\frac{|Ax|}{|x|}=\|A\|\|B\|.\end{align} In particular, $\|A^{k}\|\le\|A\|^{k}$, $\forall k\in\mathbb{N}$.

Pick $A\in M(n,\mathbb{R})$, hence you have \begin{align} \left\|\sum_{k=0}^{N}\frac{A^{k}}{k!}\right\|\le\sum_{k=1}^{N}\frac{\|A\|^{k}}{k!}\le\sum_{k=0}^{\infty}\frac{\|A\|^{k}}{k!}=e^{\|A\|}<+\infty, && \forall N\in\mathbb{N}.\end{align} It follows that the series converges absolutely, hence it is convergent, $\forall A\in M(n,\mathbb{R})$.

Answer 2

If maximal absolute value of element of $A$ is $x$, then maximal absolute value of element of $A^2$ is at most $n \cdot x^2$, of $A^3$ is $n^2 x^3$ and so on - maximal absolute value of element of $A^k$ is at most $n^{k - 1} x^k$, which growth slower than $k!$ - so $\sum\limits_{i=0}^\infty \frac{A^k}{k!}$ converges element-wise.

Answer 3

It is worth to treat the matrix $A$ as a linear bounded operator $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^n)$. Therefore defining the operator $\exp A:\mathbb{R}^n\to \mathbb{R}^n$ as a limit of the following sequence $$\bigg(\sum_{k=0}^n\frac{A^k}{k!}\bigg)_{n=0}^{\infty}\subset \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n)$$ in other to check convergence it suffices to check it is Cauchy sequence since $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^n)$ is Banach space (hence it is complete). However for fixed $\epsilon>0$ take $N\in\mathbb{N}$ such that $$\sum_{k=N}^{\infty}\frac{\|A\|^k}{k!}<\epsilon.$$ We can do this since this series converge. Such $N\in\mathbb{N}$ is a witness for being Cauchy since for any $n_1,n_2>N$ $$\bigg\| \sum_{k=0}^{n_1}\frac{A^k}{k!} - \sum_{k=0}^{n_2}\frac{A^k}{k!} \bigg\|=\bigg\| \sum_{k=n_1}^{n_2}\frac{A^k}{k!} \bigg\|\le \sum_{k=N}^{\infty}\frac{\|A\|^k}{k!} <\epsilon. $$ It is also worth to mention that the completeness of the space allow us to use generalised triangle inequality and conclude the sequence convergence from it.