Is $\exp:\overline{\mathbb{M}}_n\to\mathbb{M}_n$ injective?

Question

More specific to my problem, this is a variation on Is $\exp:\mathbb{M_n}\to\mathbb{M_n}$ injective? which was promptly answered with a counterexample.

Let $\mathbb{M}_n$ be the space of $n\times n$ matrices with real entries.

Let $\overline{\mathbb{M}}_n$ be the space of square $n\times n$ matrices with entries $0$ and $1$.

For any $M\in\overline{\mathbb{M}}_n$ we have $$e^M=\exp(M)=\sum_{k=0}^\infty \frac{1}{k!}M^k.$$

Is $\exp:\overline{\mathbb{M}}_n\to\mathbb{M}_n$ injective?

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

Since for fixed $n$ the space $\overline{\mathbb{M}}_n$ is finite we can manually check -- it is injective for $n\le4$, but I haven't been able to convince myself for arbitrary $n$.

For $n=4$ there are $2^{n^2}=65536$ matrices to check, and MATLAB's expm and my hastily written matrix building code took half an hour to complete. For $n=5$ we have $2^{25}>3.3\cdot 10^7$ matrices to test which would take ten days to run. For larger $n$ a direct test is out of the question.

The smallest nontrivial examples in my problem are $n=6$, and the interesting ones are $n=12$ and larger.

Answer 1

Proposition: Let $E=\{A\in M_n(\mathbb{C})|A\text{ has algebraic entries }\}$. Then the exponential map is injective on $E$.

Proof: Assume that $e^A=e^B$. Here $A,B$ have algebraic entries ; according to "Wermuth, 2 remarks on matrix exponentials" (in free access) http://www.sciencedirect.com/science/article/pii/0024379589905545 $AB=BA$. Thus $A,B$ are simultaneously triangularizable over $\mathbb{C}$ with diagonals $(\lambda_j)_j,(\mu_j)_j$. Necessarily $e^{\lambda_j}=e^{\mu_j}$, that is $\lambda_j=\mu_j+2k_j\pi$. $\lambda_j$ and $\mu_j$ are algebraic numbers, that implies $k_j=0$ and therefore $\lambda_j=\mu_j$.

$A$ is similar over $\mathbb{C}$ to a matrix in the form $A_1=diag(\lambda_1I_{i_1}+N_1,\cdots)$ where the $\lambda_j$ are distinct and the $N_j$ are nilpotent.

EDIT: Since $AB=BA$, $B_1$ is in the form $B_1=diag(U_1,\cdots)$ with $U_jN_j=N_jU_j$. The previous reasoning and $e^{U_j}=e^{\lambda_j}e^{N_j}$ imply that $U_j$ has a sole eigenvalue $\lambda_j$. Then $B_1=diag(\lambda_1 I_{i_1}+ M_1,\cdots)$ with $M_j$ nilpotent and $N_jM_j=M_jN_j$ ;

in particular, $M_j-N_j$ is nilpotent. Moreover $e^{\lambda_jI_{i_j}+N_j}=e^{\lambda_jI_{i_j}+M_j}$, that is $e^{N_j-M_j}=I_{i_j}$. The exponential map is an isomorphism between nilpotent matrices and unipotent ones ; then $N_j-M_j=0$, $A_1=B_1$ and finally $A=B$.