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Clearing dyadic row denominators preserves determinant sign

Statement

For every finite s×s matrix over dyadic rationals with s≥1, multiplying each row by a suitable positive power of two gives an integer matrix whose determinant has the same sign and is zero exactly when the original determinant is zero.

Facts & Assumptions

Given: A finite s×s matrix M with dyadic entries, and the Leibniz definition of determinant from Dyadic orientation and lifted sphere determinants.

[F1]

Dyadic floating-point format and directed rounding: an exact dyadic rational is m2e for integers m,e.

[F2]

Dyadic orientation and lifted sphere determinants: the determinant is the finite signed sum over permutations, with one entry from each row and column.

Proof

1.1givenF1choosealgebra

For each row i, write its entries as Mij=mij2eij using [F1], and choose ki=max⁡(0,−min⁡jeij). The row is finite, so this integer exists; then every Nij=2kiMij is an integer.

2.1givenstep 1.1F2algebra

In every Leibniz term of det⁡N, the product contains exactly one entry from each row, so it equals (∏i2ki) times the corresponding term of det⁡M. Summing the finite terms using [F2] gives det⁡N=(∏i2ki)det⁡M.

3.1step 2.1algebra∎

Every factor 2ki is positive. Therefore det⁡N and det⁡M have the same sign and vanish simultaneously.

Depends on

Used by

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