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Clearing dyadic row denominators preserves determinant sign
Statement
For every finite matrix over dyadic rationals with , 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 matrix with dyadic entries, and the Leibniz definition of determinant from Dyadic orientation and lifted sphere determinants.
Dyadic floating-point format and directed rounding: an exact dyadic rational is for integers .
Dyadic orientation and lifted sphere determinants: the determinant is the finite signed sum over permutations, with one entry from each row and column.
Proof
For each row , write its entries as using [F1], and choose . The row is finite, so this integer exists; then every is an integer.
In every Leibniz term of , the product contains exactly one entry from each row, so it equals times the corresponding term of . Summing the finite terms using [F2] gives .
Every factor is positive. Therefore and have the same sign and vanish simultaneously.
Depends on
Used by
Dependency tree · two levels
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Sources
- Jonathan R. Shewchuk, Adaptive Precision Floating-Point Arithmetic and Fast Robust Geometric Predicates, §4.2 (standard reference, not scraped)
- CGAL Developer Manual, Robustness Issues, §11.1 (standard reference, not scraped)