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15

Theorem 4.2 Transformation Theorem

For any given п-dimensional simplex

P = {x E RnAx <b}

there exists a unimodular matrix U such that A = AU has the standard form.

Proof: We shall first prove by induction that there is a unimodular matrix V such
that

AV =

where ” ⅛ ” stands for a positive entry, and ” — ” for a zero or negative entry. The
above result is basically equivalent to an example given by White [16] on page 51.
For completeness, we shall give a proof of it. Let us consider the last row of
A.
By applying reversals of sign, we may assume that each entry of the last row is
less than or equal to zero. Take any two negative entries α(ra+1j⅛ and α(ra+1y such
that θ(ra+1j⅛ ≥ θ(ra+1y. Replace column
I by column I minus column k. Repeat this
simple transformation. We can reduce all entries of the last row but one to zero.
By an interchange we may assume that this element is the last entry of the last row.
So after a finite number of steps we end with the last row having the sign pattern
(0,0, ∙ ∙ ∙ ,0, —Notice that the submatrix obtained by deleting the last row and
the last column of
A must also have the property that the zero row vector in Rn~1
is in the interior of the convex hull of all rows of this submatrix. Then by induction
on n, there exists a unimodular matrix which transforms
A into the form:



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