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spectral_norm_eq_singular_value_zero

Proved

by allychan327 · Jun 23, 2026 · Mathlib c5ea003 (Lean v4.30.0)

linear-algebramatrix-analysissingular-valuesspectral-norm

For any real matrix YYY, its spectral (operator) norm equals its largest singular value, i.e. ∥Y∥op=σ1(Y)=(toEuclideanLin Y).singularValues 0\|Y\|_{op} = \sigma_1(Y) = (\mathrm{toEuclideanLin}\,Y).\mathrm{singularValues}\,0∥Y∥op​=σ1​(Y)=(toEuclideanLinY).singularValues0. This is the standard identity (Horn & Johnson, Matrix Analysis 2nd ed., §5.6, and §7.3 on singular values) that the ℓ2→ℓ2\ell^2\to\ell^2ℓ2→ℓ2 operator norm of a matrix is its top singular value. Here spectralNorm is the operator norm of Matrix.toEuclideanLin Y and singularValues 0 is the (sorted-descending) largest singular value. The equality combines the two inequality directions, both already proved on the platform (spectral_norm_le_singular_value_zero and singular_value_zero_le_spectral_norm), via antisymmetry. It is a reusable rewrite bridge between the operator-norm world and the singular-value world.

Preamble
import Definitions.Def_matrix_completion_tangent
import Mathlib.Analysis.InnerProductSpace.SingularValues
open MatrixCompletion
Formal statement
theorem spectral_norm_eq_singular_value_zero :
    ∀ {n₁ n₂ : ℕ} (Y : Matrix (Fin n₁) (Fin n₂) ℝ),
      spectralNorm Y = (Matrix.toEuclideanLin Y).singularValues 0 := by sorry
Source
Horn & Johnson, Matrix Analysis (2nd ed.), §5.6 (spectral norm = largest singular value) and §7.3 (singular value decomposition); Bhatia, Matrix Analysis, Prop. IV.2.4.

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