Is $\exp:\mathbb{M_n}\to\mathbb{M_n}$ injective?

Question

This is related to a personal exploration of isometries of directed graphs, motivated by my son's Lego Duplo train tracks and identifying "interesting" layouts. If $M$ is the adjacency matrix for a particular directed graph corresponding to a track layout, $e^M$ can aid in identifying a representative of the equivalence class under isometry. This is not the best approach to isometries of directed graphs, but it did raise the interesting question:

Let ${\mathbb{M}}_n$ be the space of square $n\times n$ matrices with real entries. For any $M\in{\mathbb{M}}_n$ we have $$e^M=\exp(M)=\sum_{k=0}^\infty \frac{1}{k!}M^k.$$

Is $e^M$ injective?

In other words, are there two distinct $M_0,M_1\in {\mathbb{M}}_n$ such that $e^{M_0}=e^{M_1}$?

For $n=1$ it is clearly injective. At $n=2$ I haven't been able to convince myself (never mind prove) it is injective.

Answer 1

Consider

$$M = \begin{pmatrix} 0 & -2\pi \\2\pi & 0\end{pmatrix}.$$

Basically, $M \hat{=} 2\pi i$, so $e^{M} = e^0 = I$.

We can embed $\mathbb{C}$ into $\mathbb{M}_2(\mathbb{R})$ as a subring via

$$x+iy \mapsto \begin{pmatrix}x & -y \\ y & x \end{pmatrix},$$

and thus for matrices of this form, the exponential function has period $2\pi i$, like the ordinary complex exponential function.

Answer 2

The answer is no and for example it's easy to see that

$$A\;\text{is diagonalizable and}\; \operatorname{sp}(A)\subset 2i\pi\mathbb Z\Rightarrow\exp(A)=I_n$$