Page 142 - Textos de Matemática Vol. 39

132 ALICIA ROCA AND FERNANDO C. SILVA
Case 1. Suppose that p = m. Then jc = 0 and rank(Ax+B)+u ≤ n−q. In this case, the Kronecker canonical form of Ax + B contains a zero submatrix of size p×q.
The case q = n is analogous. From now on, assume that p < m and q < n.
Case 2. Suppose that Ax+B is regular. Then p+q ≤ m = n. In this case, the Kronecker canonical form of Ax + B contains a zero submatrix of size p×q.
Case 3. Suppose that Ax + B has a column minimal index.
Subcase 3.1. Suppose that Ax + B has a column minimal index equal to zero. In this case, the Kronecker canonical form of Ax + B is permutation strictly equivalent to a pencil of the form
0m,1 G .
If q = 1, the proof is complete. If q ≥ 2, then, according to the induction assumption, G is strictly equivalent to a pencil containing a zero subpencil of size p × (q − 1). It is easy to complete the proof.
Subcase 3.2. Suppose that jc > 0. Let k1 ≤ ··· ≤ ku be the nonzero column minimal indices of Ax + B. Let
⎡x10⎤
K = ⎢⎣ ... ... ⎥⎦ ∈ F[x]k1×(k1+1).
0x1
In this case, the Kronecker canonical form of Ax + B is permutation strictly
equivalent to a pencil of the form 0G.
K∗
If q ≤ k1 +1, the proof is complete. If q > k1 +1, then, according to the induction assumption, G is strictly equivalent to a pencil containing a zero subpencil of size p × (q − k1 + 1). It is easy to complete the proof.
Subcase3.3.Supposethatjc =0andu>0.Letk1 ≤···≤ku bethe nonzero column minimal indices of Ax+B. In this case, the Kronecker canonical form of Ax + B is permutation strictly equivalent to a pencil of the form
0G, x10···0
where the nonzero column minimal indices of G are
• k1 − 1, k2, . . . , ku, if k1 > 1, • k2,...,ku,ifk1 =1.