LemmaStatement: AI-generatedProof: AI-generated

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.

Reference-uniform trace-norm sandwich estimates

Statement

On finite complex matrix spaces, use spectral operator norm $∥ A ∥$ and trace norm $∥ X ∥_{1}$ . For compatible matrices, $∥ A X B ∥_{1} \leq ∥ A ∥ ∥ B ∥ ∥ X ∥_{1}$ . Consequently $∥ A X A^{†} - B X B^{†} ∥_{1} \leq (∥ A ∥ + ∥ B ∥) ∥ A - B ∥ ∥ X ∥_{1}$ . Both estimates hold after adjoining an identity on any finite reference space with the same constants. An isometric encoding preserves trace norm, and unitary conjugation does also.

Facts & Assumptions

[F1]

Finite matrices admit a singular-value decomposition with orthonormal domain and codomain bases. Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition

[F2]

A rank-one outer product has trace norm equal to the product of its vector norms. Trace Norm Rank One Formula

[F3]

Trace norm is homogeneous and obeys the triangle inequality. Trace Norm Duality And Norm Properties

[F4]

Trace norm is the sum of the domain singular values, including zeros; the sum is zero on a zero domain. Trace Norm

[F5]

The operator norm is zero on a zero domain and otherwise is the maximum over unit vectors. The induced vector norm is homogeneous. The operator norm is zero on the zero domain and otherwise is max_‖ v‖=1‖ Tv‖ Inner products separate vectors, and the induced norm is homogeneous: ‖λ v‖=|λ|‖ v‖

[F6]

The pairing is linear in the first argument, conjugate-symmetric and positive definite, with $∥ z ∥^{2} = ⟨ z, z ⟩$ . Finite-dimensional adjoints exist uniquely and satisfy $⟨ T z, w ⟩ = ⟨ z, T^{†} w ⟩$ ; adjoints preserve sums, reverse composition and satisfy $(T^{†})^{†} = T$ . Cauchy–Schwarz bounds the absolute pairing by the product of the vector norms. Real and complex inner-product spaces and their induced length The adjoint T^*:W→ V is characterised by ⟨ Tv,w⟩_W=⟨ v,T^*w⟩_V Every linear map between finite-dimensional inner product spaces has a unique adjoint Adjoints satisfy (S+T)^=S^+T^, (λ T)^=λ̄ T^, (ST)^=T^S^, and T^**=T Cauchy–Schwarz: |⟨ x,y⟩|≤‖x‖ ‖y‖, with equality exactly for dependent pairs

[F7]

Finite tensor spaces carry the product inner product and product orthonormal basis, allowing zero-dimensional factors. Tensor maps act factorwise on elementary tensors and preserve composition, identities and adjoints. Ordered tensor registers, their inner product, and wire permutations Tensor products of finitely many linear maps Tensor maps preserve composition, identities, and adjoints

[F8]

A linear isometry preserves every vector norm and, over the complex field, every inner product. A unitary operator is an invertible linear isometry. Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces Finite complex linear isometries preserve inner products

[F9]

The outer product $∣ u ⟩ ⟨ v ∣$ denotes the map $z \mapsto ⟨ z, v ⟩ u$ in the first-argument-linear convention. Dirac Notation Conversion

Proof

Given: The finite compatible matrices and any finite reference. Goal: Establish all stated trace-norm controls.

1.1F5F6algebra

For a finite map $T$ , scaling a nonzero vector to a unit vector in [F5] gives $∥ T z ∥ \leq ∥ T ∥ ∥ z ∥$ ; for $z = 0$ the inequality is immediate. To bound its adjoint, put $q = T^{†} w$ . If $q \neq 0$ , [F6] gives $∥ q ∥^{2} = ∣ ⟨ T q, w ⟩ ∣ \leq ∥ T ∥ ∥ q ∥ ∥ w ∥$ , hence $∥ T^{†} w ∥ \leq ∥ T ∥ ∥ w ∥$ ; the same inequality holds if $q = 0$ . Taking the maximum over unit $w$ , or using the zero-domain convention when there are none, yields $∥ T^{†} ∥ \leq ∥ T ∥$ . Applying this to $T^{†}$ and using its double-adjoint identity gives $∥ T^{†} ∥ = ∥ T ∥$ .

2.1F1F2F3F4F6F9step 1.1algebra

Write an SVD $X = \sum_{j} s_{j} ∣ u_{j} ⟩ ⟨ v_{j} ∣$ , with orthonormal vectors and $\sum_{j} s_{j} = ∥ X ∥_{1}$ by [F1] and [F4]. The outer-product and adjoint rules [F6] and [F9] give $A X B = \sum_{j} s_{j} ∣ A u_{j} ⟩ ⟨ B^{†} v_{j} ∣$ . Thus [F2], [F3] and step 1.1 imply $∥ A X B ∥_{1} \leq \sum_{j} s_{j} ∥ A u_{j} ∥ ∥ B^{†} v_{j} ∥ \leq ∥ A ∥ ∥ B ∥ ∥ X ∥_{1}$ . The empty SVD sum gives the same result for zero rank or a zero domain. The identity $A X A^{†} - B X B^{†} = (A - B) X A^{†} + B X (A - B)^{†}$ , the triangle inequality and the just-proved sandwich estimate yield $∥ A X A^{†} - B X B^{†} ∥_{1} \leq (∥ A ∥ + ∥ B ∥) ∥ A - B ∥ ∥ X ∥_{1}$ .

3.1F1F5F7step 1.1step 2.1choosealgebra

Fix any finite reference space $R$ and choose an orthonormal basis $(e_{j})$ , available by applying [F1] to its identity map. In the product basis [F7], every vector has the form $z = \sum_{j} z_{j} \otimes e_{j}$ . The product inner product and tensor-map rule give $∥ (T \otimes I_{R}) z ∥^{2} = \sum_{j} ∥ T z_{j} ∥^{2} \leq ∥ T ∥^{2} \sum_{j} ∥ z_{j} ∥^{2} = ∥ T ∥^{2} ∥ z ∥^{2}$ by step 1.1. Consequently $∥ T \otimes I_{R} ∥ \leq ∥ T ∥$ , also when $R = 0$ or the domain of $T$ is zero, since then the product domain is zero and its operator norm is zero. Expanding instead as $z = \sum_{j} e_{j} \otimes z_{j}$ proves the identical bound for $I_{R} \otimes T$ . Apply step 2.1 to the amplified matrices. The composition, identity and adjoint rules in [F7], together with the tensor-map rule for differences, identify those sandwiches and congruences with the desired reference extensions. The upper bounds for $A$ , $B$ and $A - B$ retain exactly the original constants.

4.1F5F6F7F8step 1.1step 2.1step 3.1algebra∎

Let $V : H \to K$ be a linear isometry and $X : H \to H$ . By [F8] and the adjoint rule [F6], $⟨ x, V^{†} V y ⟩ = ⟨ V x, V y ⟩ = ⟨ x, y ⟩$ . For each $y$ , take $x = (V^{†} V - I_{H}) y$ in the resulting zero pairing; positive definiteness gives $V^{†} V = I_{H}$ . Norm preservation and [F5] give $∥ V ∥ \leq 1$ , including $H = 0$ ; step 1.1 gives $∥ V^{†} ∥ \leq 1$ . Step 2.1 now gives $∥ V X V^{†} ∥_{1} \leq ∥ X ∥_{1}$ and, in the reverse direction, $∥ X ∥_{1} = ∥ V^{†} (V X V^{†}) V ∥_{1} \leq ∥ V X V^{†} ∥_{1}$ . Hence encoding preserves trace norm. The tensor expansion in step 3.1 and vector norm preservation show that $V \otimes I_{R}$ and $I_{R} \otimes V$ are isometries even for $R = 0$ , so this preservation also holds with references. A unitary operator is an isometry by [F8], giving unitary-conjugation invariance as well.

Depends on

Used by · 3

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.