ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

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The normalized Hermitian form on three points

Example

For $X = {1, 2, 3}$ , identify functions with their value triples and put $⟨ f, h ⟩ = \frac{1}{3} \sum_{j = 1}^{3} f (j) \overline{h (j)}$ . Its Gram matrix in the point-indicator basis is $I_{3} / 3$ . 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 $\sqrt{3}$ gives an orthonormal basis.

Facts & Assumptions

Given: $X = {1, 2, 3}$ , point indicators $e_{1}, e_{2}, e_{3}$ , 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

1.1F1givenalgebra

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

2.1step 1.1algebra

For $j = k$ , only the value at point $j$ contributes to $⟨ e_{j}, e_{k} ⟩$ , giving $1 / 3$ . For $j \neq k$ , the two indicators never both have nonzero value at the same point, so the pairing is zero. Thus $⟨ e_{j}, e_{k} ⟩ = δ_{j k} / 3$ , which is the matrix $I_{3} / 3$ .

2.2step 1.1givenalgebra

Since $\overline{i} = - i$ , $⟨ f, h ⟩ = (1 (- i) + i (1) + 0 (1)) / 3 = 0$ . Also $⟨ f, f ⟩ = (1 \cdot 1 + i (- i) + 0) / 3 = (1 + 1) / 3 = 2 / 3$ . In particular this nonzero vector has strictly positive diagonal value.

3.1step 1.1step 2.1algebra∎

Put $u_{j} = \sqrt{3} e_{j}$ , where $\sqrt{3}$ is the positive real square root. Direct substitution gives $⟨ u_{j}, u_{k} ⟩ = 3 ⟨ e_{j}, e_{k} ⟩ = δ_{j k}$ . Scaling each basis vector by the nonzero scalar $\sqrt{3}$ preserves spanning and independence (divide the coefficients by $\sqrt{3}$ ), so $(u_{1}, u_{2}, u_{3})$ is an orthonormal basis.

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

The normalized Hermitian form on a finite function space

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

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)