Page 110 - Textos de Matemática Vol. 44

100 N. KRUPNIK AND Y. SPIGEL
Since F (A, B) is nontrivial it follows that b ̸= c, and hence, a2 = bc, i.e.,
(4.2) F(A,B) = a(A2+B2)+bAB+cBA+d(A+B)+qI,
where a,b,c,d,q ∈ C and a2 = bc. √ √
(ii) =⇒ (iii): Since in (4.2) a2 = bc, it follows that a = b c with some
√
√
b, r =
a=pr, b=p2, c=r2.Notethatp+r̸=0,otherwiseb=p2 =(−p)2 =r2 =c. Denote by s and γ the following numbers: s := d/(p+r) and γ = q−s2. Then
we come to representations (1.3), Namely,
(4.3) F(A,B)=a(A2 +B2)+bAB+cBA+d(A+B)+qE
=pr(A2 +B2)+p2AB+r2BA+s(p+r)(A+B)+qE = (pA + rB + sI)(rA + pB + sI) + γE.
The relation (iii) =⇒ (iv) follows from Theorem 2.1 and the relation (iii) =⇒
appropriate choice of the roots of b and c. We set p =
c. Then
(i) is evident.
5. Polynomials of third degree. In this section we prove Theorem 1.3.

Proof of Theorem 1.3. Let F(A,B) be an admissible non-trivial polyno- mial of the third degree. Using equality (3.9) and relations (3.14) the polyno- mial F(A,B) can be represented in one of the following two forms:
(5.1) F1(A,B) =m(A3+B3)+k(ABA+BAB)+r(A2B+B2A) +n(BA2+AB2)+a(A2+B2)+bAB+cBA+d(A+B)+qE
or
(5.2) F2(A,B) =m(A3+B3)+k(ABA+BAB)+r(A2B+AB2)
+n(BA2+B2A)+a(A2+B2)+bAB+cBA+d(A+B)+qE.
Now we are going to add some additional necessary conditions in order to be able to prove the relation (i) =⇒ (ii) in Theorem 1.3. We start with the polynomial F1. One may assume in (5.1) that b ̸= c, otherwise the polynomial (5.1) is trivial. Consider two 3 × 3 matrices A := xE11 + E23 + x2E33 and B := E12 +E23 +E31. Denote fi(x) = trFi(A,B)2 −trFi(B,A)2 and gi(x) = tr Fi(A, B)3 − tr Fi(B, A)3 (i = 1, 2) Computations show that f1(1) = 4(r − n)(a − b). Since b ̸= c, and F1 is admissible, it follows from Proposition 3.1 that f1(1) = 0, i.e. n = r. We take this into consideration and compute the function g1(x) which turns out to be of the form g1(x) = h(x) 4k=0ckxk, where h(x) depends only on x and the coefficients ck only on the coefficients of