Page 139 - Textos de Matemática Vol. 39

Theorem 1.1 that P−1P is similar to a matrix of the form
MATRIX PENCILS 129
Proof. The necessity is a particular case of Theorem 1.3. Conversely, suppose that (1.9) and (1.10) are satisfied.
Firstly, suppose that P (x) = P1x + P2 and Q(x) = Q1x + Q2, where
P1, P2 ∈ F m×m, Q1, Q2 ∈ F p×p, do not have infinite elementary divisors. Then
P1 and Q1 are nonsingular. Let β1 | · · · | βm be the invariant factors of P (x)
and let α1 | · · · | αp be the invariant factors of Q(x). From (1.9) and (1.10), we
deduce that (1.1) and (1.2) are satisfied. Note that P(x) is strictly equivalent
to Imx+P−1P2 and Q(x) is strictly equivalent to Ipx+Q−1Q2. It follows from 11
12
Q−1Q ∗
12. ∗∗
Then P(x) is strictly equivalent to
Q−1Q ∗
Imx+ 1 2 . ∗∗
Clearly, this last pencil is strictly equivalent to a pencil containing Q(x) as a subpencil.
Now consider the general case. As F is infinite, there exists a nonsingular matrix S ∈ F2×2 such that the coefficient of x of the pencil PS(P(x)) is non- singular and the coefficient of x of the pencil PS (Q(x)) is nonsingular. That is, PS(P(x)) and PS(Q(x)) do not have infinite elementary divisors. From (1.9) and (1.10), it follows that
ΠS(δi) | ΠS(γi), for every i ∈ {1,...,p}, and
ΠS(γi) | ΠS(δi+2(m−p)),
for every i ∈ {1,...,p} such that i+2(m−p) ≤ m. According to the previous case, PS(P(x)) is strictly equivalent to a pencil R(x) containing PS(Q(x)) as a subpencil. Then P(x) = PS−1(PS(P(x))) is strictly equivalent to PS−1(R(x)) and PS−1 (R(x)) contains Q(x) = PS−1 (PS (Q(x))) as a subpencil.
Theorem 1.4 is valid over arbitrary fields, but the proof, in this case, follows from more general results, in [4, 5] or [11, 12], with much longer proofs. Theorems 1.3 and 1.4 illustrate how Theorems 1.1 and 1.2 are an inspira-
tion for the study of completions of matrix pencils.
The general problem of finding the possible strict equivalence classes of
a pencil with a prescribed arbitrary subpencil was described as a challenge problem in Linear Algebra and Its Applications [8]. In [8], many references of partial and related results are given. Of the more recent results, [4, 5] should be mentioned. The main theorem in these references solves the challenge problem