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nuclear_norm_dual_achiever_contraction

Proved

by Harry_Xu · Jun 21, 2026 · Mathlib 0df444a (Lean v4.33.1)

Nuclear-norm dual achiever (Candès–Recht 2009, arXiv:0805.4471, Lemma 3.2, p.15). For every real matrix NNN there is a contraction ZZZ in the operator (spectral) norm, ∥Z∥≤1\lVert Z\rVert\le 1∥Z∥≤1, that realizes the nuclear norm of NNN as a Frobenius inner product: ⟨Z,N⟩=∥N∥∗\langle Z, N\rangle = \lVert N\rVert_*⟨Z,N⟩=∥N∥∗​. Concretely, if N=∑ℓσ~ℓu~ℓv~ℓ⊤N=\sum_\ell\tilde\sigma_\ell\tilde u_\ell\tilde v_\ell^\topN=∑ℓ​σ~ℓ​u~ℓ​v~ℓ⊤​ is an SVD of NNN, then Z=sign⁡(N)=∑ℓu~ℓv~ℓ⊤Z=\operatorname{sign}(N)=\sum_\ell\tilde u_\ell\tilde v_\ell^\topZ=sign(N)=∑ℓ​u~ℓ​v~ℓ⊤​ is a partial isometry with ∥Z∥=1\lVert Z\rVert=1∥Z∥=1 (or 000 when N=0N=0N=0) and ⟨Z,N⟩=∑ℓσ~ℓ=∥N∥∗\langle Z,N\rangle=\sum_\ell\tilde\sigma_\ell=\lVert N\rVert_*⟨Z,N⟩=∑ℓ​σ~ℓ​=∥N∥∗​. This is the dual-norm duality ∥N∥∗=max⁡∥Z∥≤1⟨Z,N⟩\lVert N\rVert_*=\max_{\lVert Z\rVert\le 1}\langle Z,N\rangle∥N∥∗​=max∥Z∥≤1​⟨Z,N⟩, achieved at the sign matrix; it is the achiever half of trace duality (the equality case complementing the von Neumann inequality ⟨Z,N⟩≤∥Z∥ ∥N∥∗\langle Z,N\rangle\le\lVert Z\rVert\,\lVert N\rVert_*⟨Z,N⟩≤∥Z∥∥N∥∗​).

Preamble
import Definitions.Def_matrix_completion_tangent
open MatrixCompletion
Formal statement
theorem nuclear_norm_dual_achiever_contraction {n₁ n₂ : ℕ} (N : Matrix (Fin n₁) (Fin n₂) ℝ) : ∃ Z : Matrix (Fin n₁) (Fin n₂) ℝ, spectralNorm Z ≤ 1 ∧ matrixInner Z N = nuclearNorm N := by sorry
Source
Candès–Recht 2009, arXiv:0805.4471, Lemma 3.2 (p.15)

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