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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?

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  • $\begingroup$ What if $M^k=0\ne M$ for some $k>1$? $\endgroup$ Jan 31, 2017 at 5:30
  • $\begingroup$ If $M^k=0$, then $M\not \in \mathrm{GL}(n,\mathbb Q)$. $\endgroup$ Jan 31, 2017 at 5:57
  • $\begingroup$ If $M$ is a nonsingular matrix with algebraic entries, it has a nonzero algebraic eigenvalue $\lambda$ and a corresponding eigenvector $v$ with algebraic entries. Hence (Lindemann-Weierstrass) all nonzero entries of $e^\lambda v$ are transcendental. In turn, $e^M$ contains a transcendental entry, because $e^Mv=e^\lambda v$. $\endgroup$
    – user1551
    Feb 16, 2017 at 20:54

2 Answers 2

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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 $\operatorname{GL}(n, \mathbb{Q})$.

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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.)

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  • $\begingroup$ That is the proof that I've detailed in my answer, and it does indeed give the result with this level of generality. $\endgroup$ Feb 16, 2017 at 19:39
  • $\begingroup$ @Pierre-GuyPlamondon I added a comment on this to the answer, thanks! $\endgroup$
    – heptagon
    Feb 16, 2017 at 19:47

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