Manuscript II@2 · II.tex · theorem · 50d64def

Theorem: Complete-catalogue finite fit

Complete matrix encoding and the finite certificate › Rank factorization with an original minor · lines 1596–1634 · proof included or linked

II@2#thm:certificate, II@2#eq:certificate, II@2#eq:left-kernel

Back to manuscript

Let $α_{0} \in Q^{15120}$ have exactly

(α_{0})_{ℓ_{a}} = (α_{0})_{ℓ_{b}} = 2/3

as its nonzero coordinates. Then

N_{S} α_{0} = m_{S}, rank [N_{S} ∣ m_{S}] = rank N_{S} = 2.

The complete left kernel is

{z \in R^{1 \times 3} : z N_{S} = 0} = span_{R} {(1, 0, 0)} .

No separator exists on $S$ against the declared contact class.

Proof. The original pivots give

N_{S} α_{0} = \frac{2}{3} 080 + \frac{2}{3} 008 = 0 16/3 16/3 = m_{S} .

Appending a fitted column does not increase rank. If $z = (z_{1}, z_{2}, z_{3})$ annihilates the matrix, the pivots force $8 z_{2} = 8 z_{3} = 0$ . Every $(z_{1}, 0, 0)$ annihilates its zero first row. The target also has zero first entry, so every left annihilator annihilates the target.

Displayed formulas in this passage

II@2#eq:certificate · line 1604
$N_{S} α_{0} = m_{S}, rank [N_{S} ∣ m_{S}] = rank N_{S} = 2.$
II@2#eq:left-kernel · line 1612
${z \in R^{1 \times 3} : z N_{S} = 0} = span_{R} {(1, 0, 0)} .$
unlabeled · line 1599
$(α_{0})_{ℓ_{a}} = (α_{0})_{ℓ_{b}} = 2/3$
unlabeled · line 1622
$N_{S} α_{0} = \frac{2}{3} 080 + \frac{2}{3} 008 = 0 16/3 16/3 = m_{S} .$

Theorem: Complete-catalogue finite fit

Displayed formulas in this passage

Nearby in this section