Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The normalized Hermitian form on three points

Example

For X={1,2,3}, identify functions with their value triples and put f,h=13j=13f(j)h(j). Its Gram matrix in the point-indicator basis is I3/3. For f=(1,i,0) and h=(i,1,1), one has f,h=0 and f,f=2/3. Multiplying each point indicator by 3 gives an orthonormal basis.

Facts & Assumptions

Given: X={1,2,3}, point indicators e1,e2,e3, and the displayed normalized form and vectors f,h.

[F1]

The normalized finite function-space form is an inner product linear in its first variable (The normalized Hermitian form on a finite function space).

Verification

technique · direct
1.1

The set X is nonempty of cardinality three, so F1 applies with normalization 1/3. The point indicators are a basis: v=v(1)e1+v(2)e2+v(3)e3 for every function v, and evaluating a zero combination at each point makes all coefficients zero.

F1givenalgebra
2.1

For j=k, only the value at point j contributes to ej,ek, giving 1/3. For jk, the two indicators never both have nonzero value at the same point, so the pairing is zero. Thus ej,ek=δjk/3, which is the matrix I3/3.

step 1.1algebra
2.2

Since i=i, f,h=(1(i)+i(1)+0(1))/3=0. Also f,f=(11+i(i)+0)/3=(1+1)/3=2/3. In particular this nonzero vector has strictly positive diagonal value.

step 1.1givenalgebra
3.1

Put uj=3ej, where 3 is the positive real square root. Direct substitution gives uj,uk=3ej,ek=δjk. Scaling each basis vector by the nonzero scalar 3 preserves spanning and independence (divide the coefficients by 3), so (u1,u2,u3) is an orthonormal basis.

step 1.1step 2.1algebra

Sources

Axler, 6.3(b), p. 184, gives positive weighted inner products. The equal weights 1/3 and the displayed vectors are the local example.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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