Page 95 - Textos de Matemática Vol. 39

SPECTRALLY ARBITRARY FACTORIZATION: THE NONDEROGATORY CASE 85
or
x3+αx1 (Σ1−(2+α+γ+x1))=(1+α)(1+γ)+x1(1+γ)−Σ2−α(Σ1−(2+α+γ+x1)).
x3
This is a monic quadratic in x3, which with one more restriction on x1 (since α ̸= 0) may be made to have nonzero linear and constant terms, ensuring the existence of a nonzero x3 and the cospectrality of B with A.
For the nonderogatory case, we need that whenever λ ∈ σ(A), rank(B − λI) = 2. This is ensured with a few final linear restrictions upon x1, by choosing x1 so that det(B[1, 2] − λI) ̸= 0 when λ ∈ σ(A). In the end, only finitely many values of x1, are excluded, ensuring existence of the desired B in the nonderogatory case.
In case A is not nonderogatory, there are two possible Jordan structures for A, as A is not scalar:
⎡⎣ a 0 0 ⎤⎦ ⎡⎣ a 1 0 ⎤⎦ (1) 0a0 ,b̸=aand(2) 0a0 .
00b 00a
We may assume, without loss of generality, that a = 1, and then det A = b in case (1) and det(A) = 1 in case (2). In either event, because of the hypotheses, noαi =1,andαcannotbebincase(1)or1incase(2).
Now suppose
⎡⎣ 1 0 0 ⎤⎦ ⎡⎣ γ 1 1 y 3 ⎤⎦ ⎡⎣ γ 1 1
B=LU= x1 1 0 0 γ2 1 = x1γ1 x1+γ2
y 3 ⎤⎦ x1y3+1 .
1+x3y3 +γ3
x3 1 1 0
0 γ3 x3γ1 x3 +γ2
Our requirements on B are now that rank(B−I) = 1, γ1γ2 = α and γ1γ2γ3 = b in case (1) and = 1 in case (2). Then rank(B−I) = 1 (assuming B is nonscalar) if
det(B[1,2]−I)=det(B−I)[1,2;1,3]=det(B−I)[2,3;1,2]=det(B[2,3]−I)=0. From these, we obtain that x1, x3, y3 and γ1 must be chosen so that
x1 =(γ1 −1)(γ2 −1) 1=(γ2 −1)(γ3 −1)
and
This yields the solution for B,
x3 = γ2(1 − γ3)x1 y3= 1 ,γ2̸=1.
γ2 − 1
⎡ −1 −bα+α2 1 1 (b−α) ⎤
bα
⎣ −1 −bα+α2 bα+α3−bα2 −b α + 1 bα+α3−bα2 1 b−α bα+α3−bα2+1⎦
b bα−b2 −bα+α2 bα−b2 α bα−b2
− 1 −bα+α2 b−bα+α2 −b α + 1 b−bα+α2 b + 1 b−α b−bα+α2+1 b(−b+α) −bα+α2 −b+α α α −b+α