16
+-•••-?
- + ∙∙∙ - ?
- - ∙∙∙ + ?
о о ••• о -
The submatrix of A, which is obtained by deleting the last column and the last row,
is a productive Leontief matrix. Thus there exists a positive integral combination
of columns one through n — 1 for which the last element is zero, the и-th element
is strictly negative and the other elements are strictly positive. We can therefore
transform A to the matrix B = AV by subtracting a large positive integral multiple
of this combination from the last column.
Next we shall give a procedure to transform the matrix B = (ðʊ) to the standard
form. The procedure is described as follows:
Step (a) If there are indices i and j (г ψ j^) for some 1 < i,j ≤ n with Ьц ≤ ∣‰7∙∣, we
can find a positive integer c and an integer d ∈ { 0,1,..., ⅛ — 1 } such that
∖bij I = сЬц + d, where c is the lower integer part of ()'ɪ, and then add c multiple
of column i to column j.
Step (b) Repeat Step (a) until there are no indices i and j (г ψ j) for 1 ≤ i,j' ≤ n with
Ьц N I b{j i.
It is obvious that the above operation is a unimodular transformation. We still
have to demonstrate that the procedure is feasible. Recall that the origin of Rn is
in the interior of the convex hull of the vectors α1,...,αra+1. It implies that there are
n ⅛ 1 strictly positive convex combination coefficients ʌɪ,..., Λra+ι such that
∏+1
∑aλ = o. (4.1)
г = 1