04 - The Exponential Map
We explore the familiar exponential function in the context of groups of matrices.
The Trouble with Zero
As one-dimensional vector spaces, both $\mathbb{R}$ or $\mathbb{C}$ are abelian groups under the addition operation.
However, neither $\mathbb{R}$ nor $\mathbb{C}$ form a group under multiplication. There's a glaring inconsistency with the group axioms. Zero doesn't not have a multiplicative inverse. There is no number $x$ such that
$$ x\cdot 0 = 1.$$
Instead, we consider the multiplicative groups $\mathbb{R}^{\times}$ and $\mathbb{C}^{\times}$, which consist of all nonzero numbers. Specifically, for example,
$$\mathbb{C}^{\times} = \left\{ z \in \mathbb{C} \;\Big|\; z \neq 0\right\}.$$
It's not hard to check that both of these “punctured” fields form groups under multiplication.
This structure generalizes in a way we've already seen. Consider the vector space $\mathbb{C}^{n}$. The endomorphisms of $\mathbb{C}^{n}$ are just the $n\times n$ complex matrices,
$$\mathsf{End}(\mathbb{C}^{n}) = \mathsf{Hom}(\mathbb{C}^{n}).$$
These endomorphisms are not a group, but rather contain a subset
$$\mathsf{GL}(\mathbb{C}^{n}) = \left\{ M \in \mathsf{Hom}(\mathbb{C}^{n})\;\Big|\; \det M \neq 0\right\},$$
that is a group. The trouble once again involves zero.
The Exponential Map
The Real Numbers
Another way to deal with zero is to simply restrict ourselves to positive numbers. For $\mathbb{R}$, consider the map
$$\exp : t \mapsto e^{t} = \sum_{n=0}^{\infty} \frac{1}{n!}t^{n}.$$
The exponential function maps all of $\mathbb{R}$ to the positive, real line. In this context
$$0\mapsto 1,$$
and the limit
$$\lim_{t\rightarrow -\infty} e^{t} = 0.$$
In this sense, all of the negative, real numbers get mapped to the unit interval. Restricted to the strictly positive, real numbers, multiplication is once again a group operation.
Put another way, the image of the exponential map:
$$\exp[\mathbb{R}] = \left\{ \exp(x) \;\Big|\; x\in \mathbb{R}\right\},$$
is the set of strictly positive, real numbers,
$$\mathbb{R}^{+} = \left\{ x \in \mathbb{R} \;\Big|\; x > 0\right\}.$$
This forms a (connected) subgroup of $\mathbb{R}^{\times}$.
The Complex Numbers
This same construction carries over to the complex plane, with curious implications. The real numbers behave as they do with $\mathbb{R}$, but the entire imaginary axis is mapped to the unit circle,
$$ \exp : iy \mapsto e^{iy}.$$
Thus we find, for real parameters $x$ and $y$, a map to polar coordinates
\begin{equation}\label{polar}\exp : x + iy \mapsto e^{x + iy} = e^{x}e^{iy} \rightsquigarrow r e^{i\theta}.\end{equation}
The image of the exponential map, $\exp[\mathbb{C}]$ is $\mathbb{C}^{\times}$.
The Exponential of a Matrix
The definition of the exponential map ports directly to both $\mathsf{End}(\mathbb{R}^{n})$ and $\mathsf{End}(\mathbb{C}^{n})$. For either case, let $M$ be such an $n\times n$ matrix. Then
$$\exp: M \mapsto e^{M} = \sum_{n=0}^{\infty}\frac{1}{n!}M^{n}.$$
Using methods of multivariable calculus, one can show
\begin{equation}\label{jacobi}\det e^{M} = e^{\mathsf{Tr}\,M},\end{equation}
where $\mathsf{Tr}\,M$ represents the trace of the matrix $M$, that is to say, the sum of its diagonal elements. Therefore, given any matrix $M$, with finite elements, we find that the corresponding matrix $e^{M}$ is invertible.
TL;DR
To summarize, the exponential map took the fields $\mathbb{R}$ and $\mathbb{C}$ to the corresponding (multiplicative) groups, $\mathbb{R}^{+}$ and $\mathbb{C}^{\times}$. It also mapped endomorpisms to the groups,
$$ \exp : \mathsf{Hom}(\mathbb{F}^{n}) \rightarrow \mathsf{GL}(\mathbb{F}^{n}),$$
where $\mathbb{F}$ is either of $\mathbb{R}$ or $\mathbb{C}$.
The common thread here is that the exponential map takes vector spaces to groups by removing zero. More precisely, every member of the image of the exponential map is invertible.
Finally, notice that the exponential map itself is only invertible on its image. The so-called logarithm is undefined for zero, or for matrices with vanishing determinant.
