Projection-matrix AM-GM refutation
Abstract
A rank-one projection family gives a negative quadratic form for the matrix AM-GM difference.
Definition 1.1 (The six permutations of three points).
Formalization. D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.sixPerms (✓ std3).
Source. Repository-derived.
Commentary.
The identity, the three transpositions, and the two products of adjacent transpositions list all permutations of Fin 3.
Theorem 1.2 (The list contains every permutation).
Proof. Machine-checked in Lean as D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.sixPerms_univ (✓ std3). ∎
Source. Repository-derived.
Commentary.
The finite set sixPerms equals the finite set of all permutations of Fin 3.
Definition 1.3 (Arithmetic mean of projections).
Formalization. D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.meanProj (✓ std3).
Citation. R. Sun; Z.-Q. Luo; Y. Ye (2020). On the efficiency of random permutation for ADMM and coordinate descent. DOI: 10.1287/moor.2019.0990. URL: https://arxiv.org/abs/1503.06387v3.
Commentary.
meanProj scales the finite sum of a projection family by the inverse of its cardinality.
Definition 1.4 (Symmetrized projection product).
Formalization. D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.symProd (✓ std3).
Citation. R. Sun; Z.-Q. Luo; Y. Ye (2020). On the efficiency of random permutation for ADMM and coordinate descent. DOI: 10.1287/moor.2019.0990. URL: https://arxiv.org/abs/1503.06387v3.
Commentary.
symProd scales the sum over all permutation words by the inverse factorial; List.reverse makes the first permutation entry the rightmost factor.
Definition 1.5 (Matrix AM-GM positive-semidefinite claim).
Formalization. D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.claim (✓ std3).
Citation. R. Sun; Z.-Q. Luo; Y. Ye (2020). On the efficiency of random permutation for ADMM and coordinate descent. DOI: 10.1287/moor.2019.0990. URL: https://arxiv.org/abs/1503.06387v3.
Commentary.
Sun, Luo, and Ye state in Section 3.3: “We observe that we only need to prove a matrix AM-GM inequality for projection matrices. We conjecture that the following matrix AM-GM inequality holds. Conjecture (matrix AM-GM inequality for projection matrices). Suppose Pᵢ ∈ ℝᴺˣᴺ, i=1,…,n are projection matrices, then (1/n!) Σ_{σ=(σ₁,…,σₙ)∈Γ} P_{σₙ} P_{σₙ₋₁} ⋯ P_{σ₁} ⪯ ((1/n) Σᵢ Pᵢ)ⁿ.” Here Γ is the set of all permutations of {1,…,n}, and X ⪯ Y means that Y−X is positive semidefinite. The displayed formula is its real finite-dimensional encoding.
Theorem 1.6 (A negative quadratic form at dimension three).
Proof. Machine-checked in Lean as D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.result (✓ std3). ∎
Resolves. Problems/sun-luo-ye-2020-projection-matrix-amgm (refuted) by D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.result.
Source. Repository-derived.
Acknowledgement. R. Sun; Z.-Q. Luo; Y. Ye (2020). On the efficiency of random permutation for ADMM and coordinate descent. DOI: 10.1287/moor.2019.0990. URL: https://arxiv.org/abs/1503.06387v3.
Commentary.
The explicit matrices on Fin 3 satisfy the projection hypotheses, but z=(1,1,1) gives zᵀ((meanProj P)^3−symProd P)z=−8/243. Positive semidefiniteness would force this quadratic form to be nonnegative, so the universal statement is false.
The scalar200 basis counterexample
The dimension-three kernel-checked result above is the existing public refutation. The following complete scalar comparison is an existing-owner exposition of a separate private native-verified candidate; it adds no public declaration, freeze, deposit or second open-problem-resolution binding.
Fix n=200, K=1000000000 and s=sqrt(K^2+1). In the standard orthonormal basis e0,…,e199, take a0=e1, ai=(K e1+e(i+1))/s for 1<=i<=198, and a199=(e0-e1)/sqrt(2). Take z=e0+200e1. The submitted family is this finite real unit-column basis; the dependent auxiliary family below is only a comparison.
Every column has squared length one. The spanning identities e1=a0, e(i+1)=s ai-K a0 and e0=sqrt(2)a199+a0 recover every standard vector at the fixed positive finite K. Finite-dimensional spanning with 200 columns gives linear independence. Thus the witness is an actual basis in R^200, and each complement I-ai ai^T is a self-adjoint idempotent contraction.
For the auxiliary family, retain the last projection Q and replace the first 199 projections by P=I-e1 e1^T. The unique Q position occurs 199! times in the full labelled permutation sum. Therefore the complete 200! reversed-permutation average is (1/200) sum from k=0 to 199 of P^(199-k) Q P^k, equivalently (PQ+QP+198PQP)/200; no permutation terms are discarded or enumerated. On span(e0,e1), the mean is B=[[399,1],[1,1]]/400 and the product average is [[1/2,1/400],[1/400,0]], whose quadratic form at (1,200) is exactly 3/2.
The two-dimensional Euclidean operator norm of B is at most 401/402, by an exact positive quadratic-form estimate. The first three binomial terms give (402/401)^196 >= 258507/160801 > 8/5, hence (401/402)^196 < 5/8. Direct multiplication gives B^2(1,200)=(119601/80000,1/200), with squared length below 9/4. Symmetry and B^100=B^98 B^2 imply (1,200)^T B^200(1,200) < 45/32, so the auxiliary difference has quadratic form below -3/32.
Exactly 198 factors change. Each changed projection difference is supported on span(e1,e(i+1)); its matrix there is [[1,-K],[-K,-1]]/(K^2+1), whose square is the plane identity divided by K^2+1. Hence its norm is at most 1/s. Product telescoping for contractions bounds every original permutation word against its auxiliary word by 198/s, and the full uniform average has the same bound. The mean difference is at most 198/(200s); telescoping its 200th power gives 198/s. The two original AM-GM differences therefore differ in norm by at most 396/s < 396/K.
The squared length of z is 40001. Cauchy-Schwarz gives the strict original certificate z^T(M^200-G)z < -3/32 + 15840396/1000000000 < -1/16 < 0. This contradicts positive semidefiniteness and refutes the literal all-order scalar basis assertion at the even dimension n=200. The argument preserves the full-space estimates and the complete factorially weighted average.
For the source correspondence with a Hessian square-root convention, put C=A^T A and U=A C^(-1/2). Then U is orthogonal and A=U C^(1/2); all complement projections, means, powers and individual permutation words are carried by the same orthogonal conjugation, and U^T z has the identical negative form. This bridge does not assign a sign to the separate expected-inverse reformulation.
The private source snapshot is native-verified on Mac Studio under Lean 4.35 with no sorryAx in its scoped report and with the literal result type Not claimScalar; the source SHA-256 is 1067ac81d5cb1acca53faf795d906c6d832b55ab7c16c3890f44a6c143ca15ec. This paragraph records the proof and its evidence boundary without claiming that the unpublished candidate is a new admitted public declaration. The smallest failing dimension, the optimal random-permutation coordinate-descent rate, and the separate second-moment comparison remain open.
References
- Truth anchor:
D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.claim - Truth anchor:
D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.meanProj - Truth anchor:
D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.result - Truth anchor:
D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.sixPerms - Truth anchor:
D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.sixPerms_univ - Truth anchor:
D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.symProd