LemmaStatement: AI-adaptedProof: AI-generatedPipeline-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.

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

Finite complex linear isometries preserve inner products

Statement

Let $V : H \to K$ be a linear isometry between finite-dimensional complex inner-product spaces whose pairings are linear in the first argument. Then $⟨ V x, V y ⟩_{K} = ⟨ x, y ⟩_{H}$ for all $x, y \in H$ .

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces $H, K$ with first-argument-linear pairings and a linear isometry $V : H \to K$ .

[F1]

Real and complex inner-product spaces and their induced length: The complex pairing is linear in its first argument and conjugate-linear in its second, with conjugate symmetry.

[F2]

Real and complex inner-product spaces and their induced length: The induced norm satisfies $∥ x ∥^{2} = ⟨ x, x ⟩$ .

[F3]

Real and imaginary parts, complex conjugation, and modulus: Every complex number has unique real and imaginary parts, and conjugation sends $a + b i$ to $a - b i$ .

[L1]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces: A linear isometry is linear and preserves the induced norm of every vector.

Proof

1.1F1F2F3givenalgebra

Put $z = ⟨ x, y ⟩$ . Expanding with [F1] and [F2] gives $∥ x \pm y ∥^{2} = ∥ x ∥^{2} + ∥ y ∥^{2} \pm z \pm \overline{z}$ , so $∥ x + y ∥^{2} - ∥ x - y ∥^{2} = 2 (z + \overline{z}) = 4 Re z$ by [F3]. Also $⟨ x, i y ⟩ = - i z$ , $⟨ i y, x ⟩ = i \overline{z}$ , and $∥ i y ∥^{2} = ∥ y ∥^{2}$ ; hence $∥ x \pm i y ∥^{2} = ∥ x ∥^{2} + ∥ y ∥^{2} \pm i (\overline{z} - z)$ and $∥ x + i y ∥^{2} - ∥ x - i y ∥^{2} = 2 i (\overline{z} - z) = 4 Im z$ by [F3].

2.1F1F2F3L1givenstep 1.1algebra∎

Linearity and norm preservation [L1] give $V (x \pm y) = V x \pm V y$ and $V (x \pm i y) = V x \pm i V y$ and equate the four squared norms for $(V x, V y)$ with those for $(x, y)$ . Applying step 1.1 in each space shows that $⟨ V x, V y ⟩$ and $⟨ x, y ⟩$ have the same real and imaginary parts; their equality follows from [F3].

Remarks

The proof is a direct finite algebra expansion under the root convention and uses no choice principle.

Depends on

Used by

A preserved unknown state cannot program its own reflection Counterexample
An isometric quantum encoding gives an orthogonal code projector Theorem

Dependency tree · two levels

9 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., §7D, Theorem 7.49 (standard reference, not scraped)