Page 56 - Textos de Matemática Vol. 39

46 RAFAEL CANTO´ AND CHARLES R. JOHNSON
Then P [{1, 2, 3, 4, 5}] is a triangle, and, as an exhaustive research shows, there is no 6-triangle, this means MT(P ) = 5. However, we may use Lemma 1.6 to see that mr(P ) = 6. Of course, mr(P ) ̸= 7 because of Lemma 1.5. Via diagonal scaling, any A11 ∈ P [{1, 2, 3, 4, 5}] may be put in the form
⎡⎢ 1 0 0 0 0 ⎤⎥
⎢−1 1 000⎥ A 1 1 = ⎢⎣ z − 1 1 0 0 ⎥⎦ ,
0 0 −1 1 0 0 −1 0 0 1
⎡⎢ 1 0 0 0 0 ⎤⎥
Letting A21 ∈ P [{6, 7}, {1, 2, 3, 4, 5}] be
A21=0x1 x2 x3 0
for which
we have that a general matrix in
⎢ 1 1 0 0 0 ⎥ A−1= 1−z 1 1 0 0 .
11 ⎢⎣1−z1110⎥⎦ 1 1001
0 0 0 0 x4
and A12 ∈ P[{1,2,3,4,5},{6,7}] be
⎡⎢y1 0⎤⎥
⎢y2 0⎥ A12=⎢0 y3⎥,
⎣0 0⎦ 0 y4
P[{6,7},{1,2,3,4,5}]P[{1,2,3,4,5}]−1P[{1,2,3,4,5},{6,7}] is of the form
A/A11 = (y1 +y2)(x1 +x2 +x3)−zy1(x2 +x3) y3(x2 +x3) , (y1 + y2)x4 y4x4
There is no difficulty with the 2,2 position. However, the 2,1 position implies that y1 + y2 = 0, and the 1,2 position that x2 + x3 = 0. But then the 1,1 entry of A/A11 is 0, contrary to the ∗ in the 1,1 position of P[{6,7}]. According to Lemma 1.6, mr(P ) > MT(P ) = 5, and the only possibility is that mr(P ) = 6, by Lemma 1.5.
in which each variable is independent and nonzero. Can A take on the form
P[{6,7}]= ∗ 0 ? 0∗
11