Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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Trace Holder Inequality

Statement

For finite-dimensional complex inner-product spaces H,K and compatible linear maps A,B:H→K (Linear map between vector spaces over the same field), the composite B∗A is an endomorphism of H and

∣tr⁡H(B∗A)∣≤∥A∥1 ∥B∥.

Facts & Assumptions

Given: The finite-dimensional spaces and maps in the Statement.

[F1]

Trace-norm duality says ∥A∥1=max⁡∥C∥≤1∣tr⁡H(C∗A)∣ (Trace Norm Duality And Norm Properties).

[F2]

The operator norm is 0 when H=0 and otherwise is defined as the maximum of ∥Bv∥ over unit vectors (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

[F3]
[F4]

The trace is computed by diagonal sums in any ordered basis and vanishes on the zero space (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[F5]

The root inner product is linear in its first argument, and the induced length satisfies ∥v∥2=⟨v,v⟩ (Real and complex inner-product spaces and their induced length).

[F7]

An SVD expresses B using its positive singular values and orthonormal domain and codomain vectors (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[F8]

For an orthonormal basis, the sum of the squared coordinate moduli equals the squared vector length, including the empty-basis case (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis).

[F9]

In an ordered basis, the j-th matrix column of an endomorphism is the coordinate column of its image on the j-th basis vector (Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases).

Proof

1.1givenF2F3F4F7F8algebra

Put β=∥B∥. If H=0, [F2] defines β=0. If H≠0, choose an SVD of B from [F7], with positive singular values t1≥⋯≥tq>0 and orthonormal domain vectors (uj). If q>0, [F8] gives ∥Bv∥2=∑j≤qtj2∣⟨v,uj⟩∣2≤t12 for every unit vector v, with equality at u1. If q=0, the SVD gives B=0, and every unit vector attains 0. Thus the maximum in [F2] exists in all cases. When β=0, every unit vector has Bv=0; scaling and linearity give B=0 (also when H=0). By [F3], B∗=0, and then B∗A=0, whose trace is 0 by [F4]; both sides of the inequality are 0.

2.1F1F2F5step 1.1algebra

Suppose β>0 and put C=B/β. For every unit vector v, ∥Cv∥2=⟨β−1Bv,β−1Bv⟩=β−2∥Bv∥2, so nonnegativity gives ∥Cv∥=∥Bv∥/β≤1 by [F2]. Applying the argument in step 1.1 with C in place of B shows its defining maximum exists and is at most 1, hence ∥C∥≤1. Trace-norm duality [F1] gives ∣tr⁡H(C∗A)∣≤∥A∥1.

3.1F3F4F6F9step 2.1algebra∎

Since β is a positive real, [F3] gives C∗=β−1B∗. Fix an ordered basis of H. The coordinate column of β−1(B∗A)ej is β−1 times the coordinate column of (B∗A)ej, by uniqueness of coordinates and [F9]; hence their diagonal entries have the same scaling. Summing them by [F4] gives tr⁡H(C∗A)=β−1tr⁡H(B∗A). Thus step 2.1 and modulus homogeneity [F6] yield β−1∣tr⁡H(B∗A)∣≤∥A∥1. Multiplying by β proves the inequality when β>0; step 1.1 proves the β=0 case.

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