LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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.

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Projection onto a finite-dimensional subspace of an inner-product space

Statement

Let $H$ be a real inner-product space and $V \subseteq H$ a finite-dimensional subspace. For each $f \in H$ there is a unique $p \in V$ such that $f - p$ is orthogonal to $V$ , meaning $⟨ f - p, v ⟩ = 0$ for every $v \in V$ . For every $v \in V$ , $∥ f - v ∥^{2} = ∥ f - p ∥^{2} + ∥ p - v ∥^{2} .$ In particular, $∥ f ∥^{2} = ∥ p ∥^{2} + ∥ f - p ∥^{2}$ .

Statement references

Every finite-dimensional inner-product space has an orthonormal basis, with the empty basis in dimension zero.

Treil, Linear Algebra Done Wrong, Chapter 5 §5.3.2

Facts & Assumptions

Given: A real inner-product space $H$ , a finite-dimensional subspace $V \subseteq H$ , and $f \in H$ .

[F1]

Every finite-dimensional real inner-product space has an orthonormal basis; the basis is empty when the dimension is zero. Every finite-dimensional real or complex inner product space has an orthonormal basis

Proof

Given: The hypotheses in the Statement.

1.1F1givenchooseconstructalgebra

Choose an orthonormal basis $e_{1}, \dots, e_{r}$ of $V$ by [F1], allowing $r = 0$ , and set $p = \sum_{j = 1}^{r} ⟨ f, e_{j} ⟩ e_{j}$ . For each $ℓ$ , orthonormality gives $⟨ f - p, e_{ℓ} ⟩ = ⟨ f, e_{ℓ} ⟩ - \sum_{j = 1}^{r} ⟨ f, e_{j} ⟩ ⟨ e_{j}, e_{ℓ} ⟩ = 0$ ; since the basis spans $V$ , $f - p$ is orthogonal to $V$ .

2.1givenstep 1.1algebra∎

For any $v \in V$ , $p - v \in V$ and $f - p$ is orthogonal to $V$ . Expanding the squared norm of $f - v = (f - p) + (p - v)$ therefore gives $∥ f - v ∥^{2} = ∥ f - p ∥^{2} + ∥ p - v ∥^{2}$ . If $q \in V$ also has $f - q$ orthogonal to $V$ , then $p - q = (f - q) - (f - p)$ lies in $V^{⊥}$ and in $V$ , hence $∥ p - q ∥^{2} = 0$ and $p = q$ . Taking $v = 0$ gives the stated norm identity.

Depends on

Every finite-dimensional real or complex inner product space has an orthonormal basis

Used by

Dependency tree · two levels

3 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

Sergei Treil, Linear Algebra Done Wrong, Chapter 5 §5.3.2 (standard reference, not scraped)