Page 52 - Textos de Matemática Vol. 39

42 RAFAEL CANTO´ AND CHARLES R. JOHNSON
Only case 1.b remains. In this case, again by minimality, Q is square and
appears as
Q=⎢ . . . . ⎥, ⎢⎣ ∗ 0 · · · ∗ ∗ ⎥⎦
0∗···∗∗
in which event mr(Q) = 2, a contradiction, completing the proof. That mr(Q) = 2 follows from [2, Corollary 3.3] or may be seen by construction of a particular example.
Two more lemmas will complete our primary needs. The first is verified in [2, Theorem 4.2].
Lemma 1.5. If m ≤ n, then t(m,n,m) = m.
For the second, we say that one pattern Q is “allowed” by an algebraic expression in others, if it is possible to attain a matrix of the pattern Q from matrices of the other patterns substituted into the algebraic expression. Re- call that if A is an m-by-n matrix, the Schur complement of the invertible submatrix A[α,β], | α |=| β |, in A is
A/A[α, β] = A[αc, βc] − A[αc, α]A[α, β]−1A[β, βc],
in which α is a subset of the row indices and β of the column indices of A. It is well known [1, Section 2] that rank A = rank A[α, β] + rank A/A[α, β] =| α | + rank A/A[α, β]. Using the fact that any matrix of the pattern of a k-triangle is invertible, we then have
Lemma 1.6. Let P be an m-by-n pattern and T = P [α, β] a maximum triangle of P (| α |= MT(P)). Then, mr(P) = MT(P) if and only if P[αc,βc] is allowed by P[αc,α]T−1P[β,βc].
We note that it now follows from Lemma 1.5 that
Corollary 1.7. If P is an m-by-n pattern with MT(P) = n − 1, then
mr(P ) = MT(P ) = n − 1.
Thus, for an m-by-n pattern P, m ≤ n, to have mr(P) = m−1, but mr(P) ̸= MT(P), we must have MT(P) ≤ m − 2. In examples of this sort that follow, equality occurs. This raises the interesting question whether for P , m-by-n, mr(P ) = m − 1 implies MT(P ) = m − 1 or m − 2.
Lemma 1.4 together with either the simple observations preceding it or the earlier lemmas verify the first few rows of our table, but, in order to show that t(m,n,r) = r fails for some cells, we need specific examples that then imply
⎡⎢ ∗ ∗ · · · ∗ 0 ⎤⎥ ⎢∗∗···0∗⎥