Manuscript VI@1 · VI.tex · corollary · b5b60afb

Corollary: Interpretation with a complete catalogue

Finite-renormalization spaces and certificate requirements › Finite membership and separation · lines 2231–2251 · proof included or linked

VI@1#cor:certificate-meaning

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Under Definition def:catalogue, a row $z$ satisfying $z N_{S} = 0$ gives

z r_{S} = z q_{S}

for every allowed finite normalization, where $r_{S} = q_{S} + N_{S} c$ .

If also $z q_{S} \neq = 0$ , no global identity $q = \sum_{a} c_{a} n_{a}$ holds on $B$ . A finite fit $q_{S} = N_{S} c$ , however, proves only the fit on $S$ .

Proof. Multiply $r_{S} = q_{S} + N_{S} c$ by $z$ . A global identity would restrict to the same identity on $S$ , which contradicts a separator. The converse requires additional information and is not an implication of finite linear algebra.

Displayed formulas in this passage

unlabeled · line 2235
$z r_{S} = z q_{S}$

Nearby in this section

section: Finite membership and separation
theorem: Theorem: Finite certificate alternative
proposition: Proposition: When a finite fit is global
section: Finite membership and separation
figure: Conditional finite-response certification