Page 109 - Textos de Matemática Vol. 44

SPECTRUM OF POLYNOMIALS OF TWO MATRICES 99
Proposition 3.9. Let F ∈ Pm be m−admissible and let F = mk=0 Hk, where Hk ∈ Hk. Then
(3.12) trHm(A,B)l = trHm(B,A)l for all l = 1,2,...,m.
Proof. It follows from Proposition 3.1 that trF(A,B)l = trF(B,A)l, l =
1, ..., m. The polynomial F l can be represented as a sum of homogeneous poly-
nomials Fl = lm H˜k, and it is clear that H˜lm = Hml . Thus equality (3.12) k=0
follows from (3.11).
As a first application of Propositions 3.8 - 3.9 we obtain some additional relations between the coefficients pi (i = 1, ..., 6) which are necessary for poly- nomial (3.9) to be admissible.
Assume that the polynomial (3.9) is 3-admissible. Then (see Propositions 3.8 - 3.9) tr H3(A, B)k = tr H3(B, A)k (k ≤ 3). Consider four 3 × 3 matrices A1 := E11,B1 := E12 +E23 +E31, A2 := E11 +xE22, B2 = E12 +E21 +E32. The computations show that
(3.13) T3 := tr(H3(A1, B1)3 − tr(H3(B1, A1)3 = p4p5p6 − p1p2p3;
and T2 := tr(H3(A2, B2)2 −tr(H3(B2, A2)2 can be represented in the form T2 = f0 + f1x + f2x2, where the coefficients fj depend explicitly on pi, i = 1, ..., 6. Using some computer software1 for the system of equations
{T3 =0, fj =0(j=0,1,2),andp4+p5+p6 =p1+p2+p3}
one can obtain the following two collections of solutions:
(3.14) Q1 ={p4 =p3, p5 =p2, p6 =p1} and Q2 ={p4 =p1, p5 =p2, p6 =p3}. These relations will be used in Section 5.
4. Polynomials of second degree. In this section we prove Theorem 1.2.
Proof of Theorem 1.2.
(i) =⇒ (ii): Let F(A,B) be a non-trivial 3-admissible polynomial. It follows from Proposition 3.1 that trF(A,B)k = trF(B,A)k,k = 1,2,3. In particular (see Example 3.4), F (A, B) can be represented in the form (3.8) and it remains only to show that a2 = bc. To do this, we consider two 3 × 3 matrices A := E11 − E22 and B := E12 + E23 + E31. Computations show, that
(4.1) 0=ChF(A,B)−ChF(B,A)=2(b−c)(a2 −bc).
1Here and throughout the paper we used for computations the Maple 12 software