Page 115 - Textos de Matemática Vol. 44

(6.17)
Fˆ1(A, B) = H0(A, B) + N1(A, B) + m1;
(6.20)
and it follows from (6.18) that
SPECTRUM OF POLYNOMIALS OF TWO MATRICES 105
Using Equality (6.6) we may assume that H1(A, B) = αHˆ (A, B) and
H2(A, B) = βHˆ (B, A), where α, β ∈ C. Thus the polynomial F can be repre-
sented in the form F = Fˆ Fˆ and the polynomials Fˆ (A, B), Fˆ (A, B) can be 1212
represented as the sums
F2(A, B) = H0(B, A) + N2(B, A) + m2,
where H0 is a homogeneous polynomials of second degree and Nj are linear
homogeneous polynomials. It follows from (6.16) that
(6.18) Fˆ1(A,B)Fˆ2(A,B) = Fˆ2(B,A)Fˆ1(B,A).
Comparing the terms of third degree in (6.18), we obtain
(6.19) [N1(A,B)−N2(A,B)]H0(B,A)=H0(A,B)[N1(B,A)−N2(B,A)]. Using Lemma 6.12 for the polynomial
L(A, B) = H0(A, B) [N1(B, A) − N2(B, A)]
of third degree, we conclude that there are two possibilities. Either N1(B, A) = N2(B, A) or, up to a constant factors, H0(A, B) = (A + B)2 and N2(B, A) − N1(B, A) = B + A.
Let first N2(B, A) = N1(A, B). Then
Fˆ(A,B)=H (A,B)+N (A,B)+m E, 1011
Fˆ(A,B)=H (B,A)+N (B,A)+m E, 2012
(6.21) (H0(A, B) + N1(A, B) + m1E) (H0(B, A) + N2(B, A) + m2E)
= (H0(A, B) + N2(A, B) + m2E) (H0(B, A) + N1(B, A) + m1E) .
Using this equality and Proposition 6.10 the polynomial F(A,B) can be rep- resented in the form (6.13). Thus, in this case the theorem is proved.
Consider the second case: N1(A, B) = aA+bB, N2(A, B) = (A+B)+(aA+ bB) = (a + 1)B + (b + 1)A. Comparing the terms of degree two in (6.20) we obtain
(6.22) (aA+bB) [(a+1)B+(b+1)A] = [(a+1)A+(b+1)B] (aB+bA) .
It follows from here that a = b. Thus, F(A,B) = f(A+B) = F(B,A), where f(X) is a polynomial of the matrix X. But this contradicts the condition that F (A, B) is non-trivial.