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?

  • $\begingroup$ 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


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


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.)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.