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

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Finite-dimensional linear isometries satisfy the adjoint identity

Statement

Let $H, K$ be finite-dimensional (have finite bases by Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis) complex inner-product spaces and let $V : H \to K$ be a linear isometry. Then $V^{†} V = I_{H}$ . If $V$ is also surjective, then $V V^{†} = I_{K}$ .

Facts & Assumptions

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

[F1]

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

[F2]

Jordan–von Neumann: a norm is induced by an inner product exactly when it satisfies the parallelogram law: for the first-argument-linear complex pairing, $⟨ x, y ⟩ = \frac{1}{4} (∥ x + y ∥^{2} - ∥ x - y ∥^{2} + i ∥ x + i y ∥^{2} - i ∥ x - i y ∥^{2})$ .

[F3]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ : the adjoint satisfies $⟨ V x, y ⟩_{K} = ⟨ x, V^{†} y ⟩_{H}$ .

[F4]

Every linear map between finite-dimensional inner product spaces has a unique adjoint: every linear map between these finite-dimensional inner-product spaces has a unique adjoint.

[F5]

Inner products separate vectors, and the induced norm is homogeneous: $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ : if $⟨ u, v ⟩ = 0$ for every $v$ , then $u = 0$ .

Proof

1.1givenF1F2algebra

For all $x, y \in H$ , [F2] expresses each inner product using the four displayed norm squares. Applying [F1] to $x + y$ , $x - y$ , $x + i y$ , and $x - i y$ therefore gives $⟨ V x, V y ⟩_{K} = ⟨ x, y ⟩_{H}$ .

2.1givenF3F4F5step 1.1algebra

The adjoint exists by [F4]. By [F3] and step 1.1, $⟨ x, (V^{†} V - I_{H}) y ⟩_{H} = 0$ for every $x, y \in H$ . For each fixed $y$ , conjugate symmetry and [F5] force $(V^{†} V - I_{H}) y = 0$ , hence $V^{†} V = I_{H}$ .

3.1givenstep 2.1choosealgebra∎

If $V$ is surjective, choose for arbitrary $z \in K$ a vector $x \in H$ with $V x = z$ . Step 2.1 gives $V V^{†} z = V V^{†} V x = V x = z$ , so $V V^{†} = I_{K}$ . The proof includes zero-dimensional spaces and uses only pointwise preimages, not a choice axiom.

Depends on

Used by

Clean-ancilla zero-branch postselection Definition
Isometric Stinespring representation Definition
Unitary conjugation map Definition
Complete positivity implies Choi positivity Lemma
Robust one-step OAA including ancilla leakage Lemma
Unitary conjugation is invariant under a global phase Proposition
An isometry followed by partial trace is a channel Theorem
Clean-ancilla postselection has normalized branch weights Theorem
Every finite-dimensional channel has a finite isometric dilation Theorem
Oblivious-amplification block identity Theorem

Dependency tree · two levels

34 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

John Watrous, The Theory of Quantum Information (2018), Theorem 2.22 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 3, §3.3.2 (standard reference, not scraped)