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A nonzero 2-by-2 or 3-by-3 dyadic determinant gives a unique real solution

Statement

Let $d \in {2, 3}$ , let $A = (A_{i j})_{1 \leq i, j \leq d}$ have entries in $D$ , use the determinant defined by Dyadic orientation and lifted sphere determinants, and assume $\det A \neq 0$ . For every $b = (b_{i})_{1 \leq i \leq d} \in R^{d}$ , there is a unique $x = (x_{j})_{1 \leq j \leq d} \in R^{d}$ satisfying $(A x)_{i} = \sum_{j = 1}^{d} A_{i j} x_{j} = b_{i}$ for all $i$ . Matrix products use $(A B)_{i j} = \sum_{k = 1}^{d} A_{i k} B_{k j}$ , and $I_{d}$ is the matrix with ones on the diagonal and zeros elsewhere. The transpose is defined by $(A^{T})_{i j} = A_{j i}$ .

Facts & Assumptions

Given: $d \in {2, 3}$ , a dyadic matrix $A$ of size $d$ with nonzero Leibniz determinant, and a real vector $b$ of size $d$ .

[F1]

Dyadic orientation and lifted sphere determinants: determinants are the finite signed Leibniz sums, with permutation sign $(- 1)^{inv (π)}$ .

[F2]

Dedekind-complete ordered real field: $R$ is an ordered field; every nonzero real scalar has an inverse, and field operations preserve the order.

Proof

1.1givenF1F2algebra

If $d = 2$ , write $A = (\begin{array}{cc} a & b_{0} \\ c & d_{0} \end{array})$ and $Δ = a d_{0} - b_{0} c = \det A$ by [F1]. For $B = (\begin{array}{cc} d_{0} & - b_{0} \\ - c & a \end{array})$ , entrywise multiplication gives $A B = (\begin{array}{cc} a d_{0} - b_{0} c & - a b_{0} + a b_{0} \\ c d_{0} - d_{0} c & - c b_{0} + d_{0} a \end{array}) = Δ I_{2}$ and $B A = (\begin{array}{cc} d_{0} a - b_{0} c & d_{0} b_{0} - b_{0} d_{0} \\ - c a + a c & - c b_{0} + a d_{0} \end{array}) = Δ I_{2}$ . Since $Δ \neq 0$ , [F2] gives $Δ^{- 1}$ , and $Δ^{- 1} B$ is a two-sided inverse of $A$ .

1.2givenF1F2algebra

If $d = 3$ , write $A = (\begin{array}{ccc} a & b_{0} & c \\ d_{0} & e & f \\ g & h & i \end{array})$ and $Δ = a e i + b_{0} f g + c d_{0} h - c e g - b_{0} d_{0} i - a f h = \det A$ by [F1]. Set $B = (\begin{array}{ccc} e i - f h & c h - b_{0} i & b_{0} f - c e \\ f g - d_{0} i & a i - c g & c d_{0} - a f \\ d_{0} h - e g & b_{0} g - a h & a e - b_{0} d_{0} \end{array})$ . Direct multiplication gives diagonal entries of $A B$ equal respectively to $a e i - a f h + b_{0} f g - b_{0} d_{0} i + c d_{0} h - c e g$ , $c d_{0} h - b_{0} d_{0} i + a e i - c e g + b_{0} f g - a f h$ , and $b_{0} f g - c e g + c d_{0} h - a f h + a e i - b_{0} d_{0} i$ , each $Δ$ . Its off-diagonal entries, in row order, are $a c h - a b_{0} i + a b_{0} i - b_{0} c g + b_{0} c g - a c h = 0$ , $a b_{0} f - a c e + c b_{0} d_{0} - a b_{0} f + a c e - c b_{0} d_{0} = 0$ , $d_{0} e i - d_{0} f h + e f g - d_{0} e i + d_{0} f h - e f g = 0$ , $d_{0} b_{0} f - d_{0} c e + e c d_{0} - e a f + f a e - f b_{0} d_{0} = 0$ , $g e i - g f h + h f g - h d_{0} i + i d_{0} h - i e g = 0$ , and $g c h - g b_{0} i + h a i - h c g + i b_{0} g - i a h = 0$ . Thus the displayed cofactor formula, denoted $B (C)$ when applied to any $3 \times 3$ matrix $C$ , satisfies $C B (C) = \det (C) I_{3}$ by the same entrywise calculation. Substitution of $A^{T}$ into that formula gives $B (A^{T}) = (\begin{array}{ccc} e i - h f & g f - d_{0} i & d_{0} h - g e \\ h c - b_{0} i & a i - g c & g b_{0} - a h \\ b_{0} f - e c & d_{0} c - a f & a e - d_{0} b_{0} \end{array}) = B^{T},$ where the equality uses commutativity of the real field. In the Leibniz sum, reindexing $π$ by $σ = π^{- 1}$ gives $\det (A^{T}) = \det A$ : the product $\prod_{i} A_{π (i), i}$ becomes $\prod_{i} A_{i, σ (i)}$ , and $sgn (π) = sgn (σ)$ because the inversion map $(i, j) \mapsto (π (j), π (i))$ bijects inversions of $π$ with inversions of $π^{- 1}$ . Therefore applying the displayed cofactor identity to $A^{T}$ yields $A^{T} B^{T} = \det (A^{T}) I_{3} = Δ I_{3}$ . Transposing this equality reverses the factors by the finite matrix-product formula, so $B A = Δ I_{3}$ . Since $Δ \neq 0$ , $Δ^{- 1} B$ is a two-sided inverse of $A$ by [F2].

2.1givenstep 1.1step 1.2algebra∎

In either dimension, let $C$ be the two-sided inverse obtained in [step 1.1] or [step 1.2]. Then $x = C b$ satisfies $A x = (A C) b = I_{d} b = b$ . If $A x = A y$ , multiplying by $C$ gives $x = (C A) x = (C A) y = y$ . Hence the solution exists and is unique.

Depends on

Used by

A nondegenerate dyadic simplex determines a unique sphere Lemma

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jonathan R. Shewchuk, Adaptive Precision Floating-Point Arithmetic and Fast Robust Geometric Predicates, §4.2 (standard reference, not scraped)
Mark de Berg et al., Computational Geometry: Algorithms and Applications, 2nd ed., §1.2 (standard reference, not scraped)