Page 113 - Textos de Matemática Vol. 44

SPECTRUM OF POLYNOMIALS OF TWO MATRICES 103
Proof. We represent polynomials F1, and F2 in the form (3.3): (6.3) F1(A,B)= apCp, F2(A,B)= bpCp.
There exists a vector w, such that aw ̸= 0. Let us compare the coefficients of monomials CwCp in the left and right hand sides of (6.1). We have
(6.4) awbp = bwap.
Recallthatwisfixedandaw ̸=0.Ifbw =0,then(see(6.4))allbp =0,which
is not true. Thus, bw ̸= 0 and we can rewrite (6.4) in the form
(6.5)
hence it implies that (6.6)
Denote
(6.7)
and we obtain (6.2).
ap = bp , aw bw
F1(A, B) = F2(B, A) aw bw
:= F0(A, B). G(A,B)= aw bw F0(A,B)
Remark 6.7. Note that the the polynomial F(A,B) from Theorem 6.6 is always admissible, but may be trivial or non-trivial. For the trivial case we can take F1(A, B) = A + B = F2(A, B). The non-trivial admissible polynomial F (A, B) = (AB)m (see Example 6.1) satisfies the conditions of Theorem 6.6 if and only if m = 2l + 1. If this is the case, then F(A,B) = G(A,B)G(B,A), where G(A, B) = (AB)lA.
Remark 6.8. Example 6.1 can be also used to show that for any odd num- bers p, q ∈ N, there exists a non-trivial homogeneous admissible polynomial of order p + q, which satisfies the conditions of Theorem 2.1 with deg F1 = p and degF2 =q.
Proof. Let p = 2r+1,q = 2s+1 and F(A,B) = (AB)r+s+1. We set F1(A,B) = (AB)rA and F2(A,B) = B(AB)s. The rest is plane.
Conjecture 6.9. Let a general (not necessary homogeneous) polynomial be represented in the form F (A, B) = F1(A, B)F2(A, B), where deg F1 = deg F2, and let the polynomial F(A,B) satisfy the conditions of Theorem 2.1. Then the polynomial F(A,B) can be represented in the form
(6.8) F(A,B)=G(A,B)G(B,A)+δE, δ∈C.
For illustrations consider two particular cases (see Proposition 6.10 and Theo- rem 6.11 below) in which this conjecture is true.