PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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.

Pythagoras, the parallelogram identity, and the real and complex polarisation identities

Statement

For vectors $u, v$ in an inner product space:

if $⟨ u, v ⟩ = 0$ , then $∥ u + v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2}$ ;
$∥ u + v ∥^{2} + ∥ u - v ∥^{2} = 2 ∥ u ∥^{2} + 2 ∥ v ∥^{2}$ ;
over $R$ , $⟨ u, v ⟩ = \frac{1}{4} (∥ u + v ∥^{2} - ∥ u - v ∥^{2})$ ;
over $C$ with the linear-first convention, $⟨ u, v ⟩ = \frac{1}{4} (∥ u + v ∥^{2} - ∥ u - v ∥^{2} + i ∥ u + i v ∥^{2} - i ∥ u - i v ∥^{2}) .$

Facts & Assumptions

Given: Vectors $u, v$ in a real or complex inner product space.

[L1]

The inner product is linear first, conjugate-linear second, and conjugate symmetric (Real and complex inner product spaces, with the inner product linear in the first argument).

[L2]

Squared norm is $∥ w ∥^{2} = ⟨ w, w ⟩$ (The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product).

Proof

technique · direct

1.1L1L2algebra

Expanding by [L1] and [L2] gives $∥ u + v ∥^{2} = ∥ u ∥^{2} + ⟨ u, v ⟩ + \overline{⟨ u, v ⟩} + ∥ v ∥^{2}$ . Orthogonality removes the middle terms and proves Pythagoras.

2.1step 1.1L1L2algebra

Expanding $∥ u - v ∥^{2}$ changes the signs of both middle terms. Adding this expansion to step 1.1 proves the parallelogram identity; subtracting gives $4 Re ⟨ u, v ⟩$ .

3.1step 2.1L1L2algebra∎

Over $R$ , the real part is the scalar itself, giving claim 3. Over $C$ , the same expansion with $i v$ gives $∥ u + i v ∥^{2} - ∥ u - i v ∥^{2} = 4 Im ⟨ u, v ⟩$ under the linear-first convention. Combining real and imaginary parts gives claim 4.

Depends on

Used by

Dependency tree · two levels

5 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, 4th ed., §6A (standard reference, not scraped)
Sergei Treil, Linear Algebra Done Wrong, Ch. 5, §5.1 (standard reference, not scraped)