Problem 11891
(American Mathematical Monthly, Vol.123, February 2016) Proposed by C. Chiser (Romania).
LetM be the set of all 3 × 3 matrices with complex entries. Suppose A, B ∈ M with AB = BA and Tr(Ak) = Tr(Bk) for k ∈ {1, 2, 3}. Suppose further that n ≥ 4 and that An− Bn is invertible.
Prove that there exist complex numbersα, β, γ such that
A2n+ B2n+ AnBn+ αAn+ βBn+ γI = 0.
Solution proposed by Moubinool Omarjee, Lyc´ee Henri IV, Paris, France, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.
We will show that the statement is true for any positive integer n.
Since A and B are commuting matrices, they are simultaneously triangularizable: there exist a invertible matrix P ∈ M such that the matrices P−1AP and P−1BP are both upper triangular.
Let
(a1, a2, a3) = diag(P−1AP) and (b1, b2, b3) = diag(P−1BP).
be the complex eigenvalues of A and B.
For any matrix M ∈ M, the characteristic polynomial is equal to det(xI − M ) = x3− e1x2+ e2x− e3
with
e1= Tr(M ), e2=1
2(Tr2(M ) − Tr(M2)), e3= 1
6(Tr3(M ) − 3Tr(M )Tr(M2) + 2Tr(M3)).
Thus, by Tr(Ak) = Tr(Bk) for k ∈ {1, 2, 3} , it follows that A and B have the same characteristic polynomial and therefore there is a permutation σ ∈ S3such that bk= aσ(k) for k ∈ {1, 2, 3}.
Now, since the matrix
(P−1AP)n− (P−1BP)n= P−1AnP− P−1BnP = P−1(An− Bn)P is upper triangular and An− Bn is invertible, we have that
Y
k∈{1,2,3}
(ank − anσ(k)) = Y
k∈{1,2,3}
(ank− bnk) = det((P−1AP)n− (P−1BP)n)
= det(P−1(An− Bn)P ) = det(An− Bn) 6= 0,
which imply that σ(k) 6= k for k ∈ {1, 2, 3}. So σ is a cyclic permutation in S3and the eigenvalues a1, a2, a3 are distinct complex numbers. Therefore A and B are simultaneously diagonalizable by an invertible matrix Q ∈ M and the equation to be solved is equivalent to
Q−1(A2n+ B2n+ AnBn+ αAn+ βBn+ γI)Q = 0, which yields the linear system,
a2n1 + a2nσ(1)+ an1anσ(1)+ αan1 + βanσ(1)+ γ = 0, a2n2 + a2nσ(2)+ an2anσ(2)+ αan2 + βanσ(2)+ γ = 0, a2n3 + a2nσ(3)+ an3anσ(3)+ αan3 + βanσ(3)+ γ = 0.
It is straightforward to verify that the above system is solved by
α= β = −(an1+ an2 + an3) = −t1, γ= an1an2+ an2an3 + an3an1 =1
2(t21− t2)
where t1:= Tr(An) = Tr(Bn) and t2:= Tr(A2n) = Tr(B2n).
