Page 138 - Textos de Matemática Vol. 39

128 ALICIA ROCA AND FERNANDO C. SILVA
Proof. Suppose that there exists a matrix pencil strictly equivalent to P(x) containing Q(x) as a subpencil.
Firstly, suppose that P (x) and Q(x) do not have infinite elementary divi- sors.Letβ1 |···|βr betheinvariantfactorsofP(x)andletα1 |···|αs bethe invariant factors of Q(x). According to Theorem 1.2, s ≤ r ≤ s+m+n−p−q, and (1.3) and (1.4) are satisfied. Then
β˜ i | α˜ i , for every i ∈ {1,...,s}, and
α˜ i | β˜ i + m + n − p − q ,
for every i ∈ {1,...,s} such that i+m+n−p−q ≤ r. That is, (1.5) and (1.6) are satisfied.
Now consider the general case. Let G be an infinite extension of F . Then there exists a nonsingular matrix S ∈ G2×2 such that the rank of the coefficient of x of the pencil PS(P(x)) is equal to rank(PS(P(x))) and the rank of the coefficient of x of the pencil PS(Q(x)) is equal to rank(PS(Q(x))). That is, PS(P(x)) and PS(Q(x)) do not have infinite elementary divisors. Moreover, PS(P(x)) is strictly equivalent to a pencil containing PS(Q(x)) as a subpencil. According to the previous case, s ≤ r ≤ s + m + n − p − q,
(1.7) ΠS(δi) | ΠS(γi),
for every i ∈ {1,...,s}, and
(1.8) ΠS(γi) | ΠS(δi+m+n−p−q),
for every i ∈ {1,...,s} such that i+m+n−p−q ≤ r. From (1.7) and (1.8) follow (1.5) and (1.6).
Using Theorem 1.1, with the same technique, the converse of Theorem 1.3 can be proved for regular pencils. The resulting theorem is due to Baragan˜a, but Baragan˜a’s proof is different from the one given here. Recall that the regular pencils are the square pencils with nonzero determinant.
Theorem 1.4. [1] Suppose that F is infinite. Let P(x) ∈ F[x]m×m and Q(x) ∈ F[x]p×p be regular matrix pencils, with p ≤ m. Let δ1 | ··· | δm be the homogeneous invariant factors of P (x) and let γ1 | · · · | γp be the homoge- neous invariant factors of Q(x).
There exists a pencil strictly equivalent to P(x) containing Q(x) as a subpencil if and only if
(1.9) δi | γi,
for every i ∈ {1,...,p}, and
(1.10) γi | δi+2(m−p),
for every i∈{1,...,p} such that i+2(m−p)≤m.