Generalization to higher dimension of $e^r \not \in \mathbb Q$ The following is well known and not difficult to prove: 
$$\forall r \in \mathbb Q^*,  e^r \not \in \mathbb Q.$$
See for instance https://proofwiki.org/wiki/Exponential_of_Rational_Number_is_Irrational 
Could this be generalized with the following result, for $n\geq 2$:
$$\forall M \in \mathrm{GL}(n,\mathbb Q),  \exp(M) \in \mathrm{GL}(n,\mathbb R)\setminus \mathrm{GL}(n,\mathbb Q)?$$
If yes, do you have any proof or reference? 
 A: The answer seems to be yes.
The proof uses the Lindemann-Weierstrass theorem; specifically, we will use the following result:

If $a_1, \ldots, a_n$ are distinct algebraic numbers, then $e^{a_1}, \ldots, e^{a_n}$ are linearly independent over the algebraic numbers.

In particular (taking $a_1=0$), this means that no non-trivial linear combination of $e^{a_2}, \ldots, e^{a_n}$ over the algebraic numbers is an algebraic number when the powers are all non-zero.
Let us break down the proof into steps.  Let $\overline{\mathbb{Q}}$ be the field of algebraic numbers.  Recall that it is algebraically closed.  Let $M\in \operatorname{GL}(n, \mathbb{Q})$.
Step 1.
There exists a matrix $P\in \operatorname{GL}(n, \overline{\mathbb{Q}})$ and non-zero algebraic numbers $\lambda_1, \ldots, \lambda_r$ such that 
$$PMP^{-1} = J_{n_1}(\lambda_1) \oplus \ldots \oplus J_{n_r}(\lambda_r),$$  where
$$
 J_{n_i}(\lambda_i) = 
    \begin{pmatrix} 
      \lambda_i & 0 & 0 & \cdots & 0 \\
      1 & \lambda_i & 0 & \cdots & 0 \\
      0 & 1 & \lambda_i & \cdots & 0 \\
      \vdots & \vdots & \vdots & \ddots & \vdots \\
      0 & 0 & 0 & \cdots & \lambda_i        
    \end{pmatrix}
$$
is a Jordan block.  This is because $\overline{\mathbb{Q}}$ is algebraically closed.
Step 2. 
$$
\operatorname{exp}(M) = P^{-1} \Big(\operatorname{exp}\big(J_{n_1}(\lambda_1)\big) \oplus \ldots \oplus \operatorname{exp}\big(J_{n_r}(\lambda_r)\big) \Big)P.
$$
Step 3.
For each $i\in \{1,\ldots, r\}$, there is a nilpotent matrix $N_i$ such that $J_{n_i}(\lambda_i) = N_i + \lambda_i I_n$, where $I_n$ is the $n\times n$ identity matrix.
Step 4.
For each $i\in \{1,\ldots, r\}$,
$$
\operatorname{exp}\big(J_{n_1}(\lambda_1)\big) = e^{\lambda_i}\operatorname{exp}(N_i).
$$
This is because $N_i$ and $\lambda_i I_n$ commute.
Step 5.
We have that $\operatorname{exp}(N_i) \in \operatorname{GL}(n_i, \mathbb{Q})$.  This is because the sum $\sum_{j\geq 0} \frac{N_i^j}{j!}$ is finite when $N$ is nilpotent.
Step 6.
Putting all the previous steps together, we get
$$
\operatorname{exp}(M) = P^{-1} \Big(e^{\lambda_1}\operatorname{exp}(N_1) \oplus \ldots \oplus e^{\lambda_r}\operatorname{exp}(N_r)  \Big)P.
$$
The entries of this matrix are $\overline{\mathbb{Q}}$-linear combinations of $e^{\lambda_1}, \ldots, e^{\lambda_r}$.  Since $\operatorname{exp}(M)$ is non-zero, at least one of these linear combinations is non-zero.  By the Lindemann-Weierstrass theorem, it cannot be an algebraic number.
Thus $\operatorname{exp}(M)$ has an entry that is not an algebraic number.  In particular, is does not lie in $\operatorname{GL}(n, \mathbb{Q})$.
A: Actually, a stronger result is true: If $A$ is a non-singular matrix with algebraic (over $\mathbb{Q}$) entries, then $\exp A$ has at least one transcendental entry. To prove this, you can use the Jordan canonical form and assume that $A$ is upper triangular. Since $A$ is non-singular, we have $A_{ii}\neq 0$ for some $i$, and so the $(i,i)$ entry of $\exp A$ is transcendental by Lindemann--Weierstrass.
(The above argument turned out to be contained essentially in the earlier answer by Pierre-Guy.)
