Page 141 - Textos de Matemática Vol. 39

MATRIX PENCILS 131 Clearly, Z1, . . . , Zt are solutions of HX = 0. By the definition of fundamental
series of solutions,
k1 =degX1 ≤degZ1 =degY1 =c1.
Let i ∈ {1,...,t − 1}. As Z1,...,Zi+1 are linearly independent, there exists j ∈ {1,...,i + 1} such that Zj is not a linear combination (as a vector in the F(x)-space F(x)h×1) of X1,...,Xi. Thus X1,...,Xi,Zj are linearly indepen- dent. By the definition of fundamental series of solutions,
ki+1 = degXi+1 ≤ degZj = degYj ≤ degYi+1 = ci+1,
and the conclusion follows.
Lemma 2.3. Let H = [ G ∗ ] ∈ F[x]m×h be a matrix pencil, where G ∈ F[x]m×g. Let k1′ ≤ ··· ≤ ks′′ and c′1 ≤ ··· ≤ c′t′ be the nonzero col- umn minimal indices of H and G, respectively. Then t′ ≤ s′ and ki′ ≤ c′i, i ∈ {1,...,t′}.
Proof. Let
M= Y R ∈F[x]h×h, 0W
where Y ∈ Fg×g, be a nonsingular matrix such that
HM = 0m,g−rankG G1 G2 0 ,
where G1 ∈ F[x]m×rankG and the columns of [ G1 G2 ] are linearly indepen- dent as vectors in the F-space F[x]m×1. Then k1′ ≤ ··· ≤ ks′′ and c′1 ≤ ··· ≤ c′t′ are the column minimal indices of [ G1 G2 ] and G1, respectively. The con- clusion follows from the previous lemma.
Proof of Theorem 2.1. Suppose that there exists a matrix pencil strictly equivalent to Ax + B containing a zero subpencil of size p × q. Then Ax + B is strictly equivalent to a pencil of the form
H=G ∗, where G=0p,q . E
Let j be the number of nonzero column minimal indices of E. Clearly E and G have the same column minimal indices and the sum of these indices does not exceed m − p. It follows, from Lemma 2.3, that j ≤ jc. Then
dc(Ax + B) ≤ dc(E) + n − q = rankE + j + n − q ≤ m + n − p − q + jc. Using (2.3), we deduce (2.1). The proof of (2.2) is analogous.
Conversely, suppose that (2.1) and (2.2) are satisfied. The proof is by induction on m + n. When m + n = 2, the proof is trivial. Suppose that m + n ≥ 3.