Matrix exponential: Formal notation for power series? Or, more?

Question

For a square matrix $A$, I'm already used to see and use: $$\sum_{n=0}^{\infty} \frac{A^n}{n!} = \lim_{n \to \infty} \left(I + \frac{A}{n}\right)^n = e^A$$

Which means a matrix $A$ is just like some complex number (apart from commutation) for which the above is factual.

My question is simply whether the last $=$ is "for real", or should it really be replaced with $\equiv$ (definition)?

(I remember that I have had a similar "doubt" regarding $e^{ix} = \cos x + i \sin x$ ($x \in \mathbb{R}$). But then I learned how to prove it.)

Answer 1

There are three equivalent (and most common) definitions of the exponential function of a matrix:

1) $\mathrm{e}^{tA}=\sum_{n=0}^\infty \dfrac{t^nA^n}{n!}$.

2) $\mathrm{e}^{tA}=\displaystyle\lim_{n\to\infty} \left(I+\dfrac{tA}{n}\right)^{\!n}$.

3) $\mathrm{e}^{tA}$ is the unique and globally defined solution of the initial value problem $$ X'=AX, \quad X(0)=I, $$ where $X\in\mathbb R^{n\times n}$ (or $X\in\mathbb C^{n\times n}$).

All of these definitions work equally well in $\mathbb R$ and $\mathbb C$.

Although the first one is the most common, I personally prefer the third one, as you can prove, through it, all the properties of the exponential much in a cleaner fashion, using primarily the uniqueness of the solutions of the IVP. (For example: $A\mathrm{e}^{tB}=\mathrm{e}^{tB}A$ for all $t$ iff $AB=BA$, as both $A\mathrm{e}^{tB}$ and $\mathrm{e}^{tB}A$ the same IVP: $X'=BX,\,X(0)=A$.)

4) Finally, there is a fourth one, using Functional Calculus (it works with operators in Banach spaces as well) $$ \mathrm{e}^{tA}=\frac{1}{2\pi i}\int_{\gamma}\frac{\mathrm{e}^{tz}\,dz}{zI-A}, $$ where $\gamma$ is a simple closed and positively oriented contour in $\mathbb C$, with all the eigenvalues of $A$ in its interior. This definition is also used for proofs. (If the real parts of all the eigenvalues of $A$ are negative, then $\mathrm{e}^{tA}\to 0$. To prove this, simply choose a $\gamma$ in $\{z\in\mathbb C:\mathrm{Re}\,z<0\}$.)

Answer 2

The matrix exponential is usually defined as $$ e^A = \sum_{n=0}^\infty \frac{A^n}{n!} = \lim_{m\to\infty} \sum_{n=0}^m \frac{A^n}{n!}. $$ (As part of the definition, you show that the limit actually exists.) You can then show that $\lim_{n\to\infty} (I+A/n)^n = e^A$. In both cases, you can choose any of common matrix norms for your definition of convergence, since they are all equivalent.

Answer 3

It should really be a definition, as it is in the complex case.

For complex numbers, it is not at all obvious what $e^{ix}$ should be. So, we take it to be defined one of the following two equivalent definitions: $$ e^{z} \equiv \sum_{n=0}^\infty \frac{z^n}{n!}\\ e^{z} \equiv \lim_{n \to \infty} \left( 1 + \frac{z}{n}\right)^n $$ The situation is similar for matrix exponentiation: we choose one of the two to be the definition, and stick to that story. The Taylor series is the more common definition, but the limit is particularly useful in Lie algebra and quantum mechanics, wherein we state that $A$ "generates" $e^A$.

Along these lines, you should ask yourself how exponentiation should be defined for irrational numbers, since there is no obvious connection between taking an irrational exponent and iterated multiplication.