Problem 11917
(American Mathematical Monthly, Vol.123, June-July 2016) Proposed by O. Furdui (Romania).
LetA be a 2 × 2 matrix with rational entries and both eigenvalues less than one in absolute value.
Prove that
log(I − A) := −
∞
X
k=1
Ak k has rational entries if and only ifA2= 0.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let L = log(I − A). If A2= 0 then L = −A and therefore L has rational entries.
Now we assume that L has rational entries. It is known (see K. S. Williams, The nth power of a 2 × 2 matrix, Math. Mag. 65(1992), 336) that for k ≥ 1,
Ak =
αk A − βI α − β
− βk A − αI α − β
, if α 6= β
αkI + kαk−1(A − αI), if α = β.
where α and β are the eigenvalues of the matrix A.
Hence, since |α| < 1 and |β| < 1, we have
L =
log(1 − α) A − βI α − β
− log(1 − β) A − αI α − β
, if α 6= β
log(1 − α)I − A − αI
1 − α , if α = β.
If α = β then,
l11= log(1 − α) −a11− α
1 − α ⇒ log(1 − α) = l11+a11− α 1 − α
Since l11, a11∈ Q and α is an algebraic number, it follows that 1 − α and l11+a1−α11−α are algebraic numbers. By Hermite-Lindemann theorem, the natural logarithm of a positive algebraic number different from 1 is transcendental, therefore 1 − α = 1, that is α = 0. Therefore tr(A) = 2α = 0, det(A) = α2= 0 which imply, by Cayley-Hamilton theorem, that A2= 0.
Now, it suffices to show that the condition α 6= β leads always to a contradiction. If a216= 0 then
(α − β)l21= log(1 − α)a21− log(1 − β)a21 ⇒ log 1 − α 1 − β
=(α − β)l21 a21 .
Hence, by using the same argument as above, we have that (1 − α)/(1 − β) = 1, that is α = β.
Contradiction.
So a21= 0 and, in a similar way, we get a12= 0. Thus, a11= α, a22= β and
(α − β)l11= log(1 − α)(a11− β) − log(1 − β)(a11− α) = log(1 − α)(α − β) ⇒ log(1 − α) = l11, (α − β)l22= log(1 − α)(a22− β) − log(1 − β)(a22− α) = − log(1 − β)(β − α) ⇒ log(1 − β) = l22. Therefore, again by using the same argument, we conclude that α = 0 and β = 0. Contradiction. Remark. If A is a 2 × 2 matrix with integer entries and both eigenvalues less than one in absolute value then it follows immediately that tr(A) = det(A) = 0, and therefore A2= 0 by Cayley-Hamilton theorem.
