Exponentiating integer matrices September 7, 2007

Posted by Ben Webster in crazy ideas, fun problems, Number theory.

So, another question that came up at tea you guys might like is the following:

Let $A$ be matrix with integer entries.  When does $e^A$ have integer entries?

One answer is not all that often: Since the eigenvalues of $e^A$ are the exponentials of the eigenvalues of $A$, the eigenvalues of $A$ must be algebraic integers (since $A$ has integer entries), whose exponentials are algebraic integers.  I have no idea whether this has been proved, but common sense suggests the only such algebraic integer is probably 0.  Assuming this, we obtain that $A$ must be nilpotent.  On the other hand, there is a large selection of nilpotent matrices which do have integer exponentials: for example, if all the entries are divisible by $(n-1)!$ where $A^n=0$ (for example, for all matrices whose square is 0).  Can any get these necessary and sufficient conditions a bit closer?

1. otogo - September 7, 2007

e^(non-zero algebraic number) = transcendental was proved by Lindemann in 1882.

http://mathworld.wolfram.com/Hermite-LindemannTheorem.html

2. ninguem - September 7, 2007

Lindemann, 1882: if a is a non-zero algebraic number, then exp(a) is transcendental. Hence, since exp(2i\pi)=1, \pi is transcendental. People don’t learn that these days?

3. David Speyer - September 8, 2007

If a matrix with integer entries is nilpotent then there is an integer change of basis which makes it strictly upper triangular (and still have integer entries). I’ll provide a proof on request. So it is enough to answer this question for strictly uper triangular integer matrices. I don’t think there is going to be a very nice answer, though, just a list of congruence conditions.

4. Ben Webster - September 8, 2007

I’ll provide a proof on request.

I, for one, would like to see a proof.

5. Ben Webster - September 8, 2007

Strike that. I figured it out.

6. Isabel - September 8, 2007

What about the matrix 2^A (or, more properly, exp(A ln 2)?) When does this have integer entries?

If A = [[a b] [b a]], then 2^A = [[2^k + 2^l, 2^k - 2^l] [2^k - 2^l, 2^k + 2^l]] with k = a+b-1, l = a-b-1. (Maple says so.) If a > b this gives an infinite family of solutions.

Thinking about upper triangular matrices, if A = [[a b] [0 d]], then 2^A = [[2^a, b(2^d - 2^a)/(d-a)] [0, 2^d]] and so for any a and d we can find b which makes 2^A have integer entries.

7. Sombra - November 12, 2007

somewhat along the same lines…If a matrix A is nilpotent how do you find A^p for large p? Lambda is 2 with multiplicity 3 and A is upper triangular with 2′s on the diagonal and 1′s above the diagonal.