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Exponentiating integer matrices September 7, 2007

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

Comments»

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.