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Small-ball bound for the smallest singular value of a Gaussian matrix: P{smin⁡(A)≤s}≤n(e ns2N−n+1)(N−n+1)/2\mathbb P\{s_{\min}(A)\le s\}\le n\bigl(\frac{e\,n s^2}{N-n+1}\bigr)^{(N-n+1)/2}P{smin​(A)≤s}≤n(N−n+1ens2​)(N−n+1)/2

Proved
GaussianMatrix.sMin_small_ball

by tc · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

gaussian-matrixprobabilityrandom-matrix-theoryrandomized-numerical-linear-algebra

Let A∈RN×nA\in\mathbb R^{N\times n}A∈RN×n be a standard Gaussian matrix with 1≤n≤N1\le n\le N1≤n≤N, let smin⁡(A)=inf⁡∥x∥2=1∥Ax∥2s_{\min}(A)=\inf_{\|x\|_2=1}\|Ax\|_2smin​(A)=inf∥x∥2​=1​∥Ax∥2​, and put d=N−n+1d=N-n+1d=N−n+1. For every s≥0s\ge0s≥0 with ns2≤dn s^2\le dns2≤d,

P{smin⁡(A)≤s} ≤ n(e n s2d)d/2.\mathbb P\bigl\{s_{\min}(A)\le s\bigr\}\ \le\ n\left(\frac{e\,n\,s^2}{d}\right)^{d/2}.P{smin​(A)≤s} ≤ n(dens2​)d/2.

This small-ball estimate, with an explicit elementary constant, complements the Gaussian tail P{smin⁡≤N−n−u}≤e−u2/2\mathbb P\{s_{\min}\le\sqrt N-\sqrt n-u\}\le e^{-u^2/2}P{smin​≤N​−n​−u}≤e−u2/2: it gives polynomial decay of order d=N−n+1d=N-n+1d=N−n+1 as s→0s\to0s→0, with an explicit constant, and is what makes negative moments E smin⁡(A)−2p\mathbb E\,s_{\min}(A)^{-2p}Esmin​(A)−2p finite and explicitly bounded. The argument: if smin⁡≤ss_{\min}\le ssmin​≤s then some column aia_iai​ of AAA is within n s\sqrt n\,sn​s of the span of the other columns, that distance is a χd\chi_dχd​ variable, and a chi-square lower tail plus a union bound over iii finish the proof.

Formalization Note. For a wide k×tk\times tk×t Gaussian matrix GGG the usual σmin⁡(G)\sigma_{\min}(G)σmin​(G) equals smin⁡(GT)s_{\min}(G^{\mathsf T})smin​(GT), so the wide form is this statement with (N,n)=(t,k)(N,n)=(t,k)(N,n)=(t,k). The exponent d/2d/2d/2 is a real power.

Preamble
import Definitions.Def_GaussianMatrix_basic

open MeasureTheory ProbabilityTheory
open scoped Matrix
Formal statement
namespace GaussianMatrix
theorem sMin_small_ball {N n : ℕ} (hn : 1 ≤ n) (hnN : n ≤ N) (s : ℝ) (hs : 0 ≤ s)
    (hsd : n * s ^ 2 ≤ (N : ℝ) - n + 1) :
    (gaussianMatrix N n) {A | sMin (Matrix.of A) ≤ s}
      ≤ ENNReal.ofReal (n * (Real.exp 1 * n * s ^ 2 / ((N : ℝ) - n + 1)) ^ (((N : ℝ) - n + 1) / 2)) := by sorry
end GaussianMatrix
Source
Combination of three published facts. (1) M. Rudelson, R. Vershynin, *The Littlewood–Offord problem and invertibility of random matrices*, Advances in Mathematics 218(2), 600–633, 2008, https://arxiv.org/abs/math/0703503, §3.3 p. 16, Lemma 3.5 (Invertibility via distance) and eq. (3.7): for any matrix with columns XkX_kXk​ and Hk=span⁡(Xj)j≠kH_k=\operatorname{span}(X_j)_{j\ne k}Hk​=span(Xj​)j=k​, ∥Ax∥2≥max⁡k∣xk∣dist⁡(Xk,Hk)\|Ax\|_2\ge\max_k|x_k|\operatorname{dist}(X_k,H_k)∥Ax∥2​≥maxk​∣xk​∣dist(Xk​,Hk​), so smin⁡(A)≤ss_{\min}(A)\le ssmin​(A)≤s forces dist⁡(Xk,Hk)≤n s\operatorname{dist}(X_k,H_k)\le\sqrt n\,sdist(Xk​,Hk​)≤n​s for some kkk (some unit xxx has ∣xk∣≥n−1/2|x_k|\ge n^{-1/2}∣xk​∣≥n−1/2). (2) By rotational invariance (N. Halko, P.-G. Martinsson, J. A. Tropp, *Finding structure with randomness: probabilistic algorithms for constructing approximate matrix decompositions*, SIAM Review 53(2), 2011, https://arxiv.org/abs/0909.4061 (v2), §10.1 p. 55), conditionally on the other columns dist⁡(Xk,Hk)2∼χN−n+12\operatorname{dist}(X_k,H_k)^2\sim\chi^2_{N-n+1}dist(Xk​,Hk​)2∼χN−n+12​. (3) S. Dasgupta, A. Gupta, *An elementary proof of a theorem of Johnson and Lindenstrauss*, Random Structures & Algorithms 22(1), 60–65, 2003, https://doi.org/10.1002/rsa.10073, Lemma 2.2(a): P[χd2≤u]≤(eu/d)d/2\mathbb P[\chi^2_d\le u]\le(eu/d)^{d/2}P[χd2​≤u]≤(eu/d)d/2 for u≤du\le du≤d. A union bound over the nnn columns gives the stated bound. Compare the sharper-constant small-ball bound of N. Halko, P.-G. Martinsson, J. A. Tropp, *Finding structure with randomness: probabilistic algorithms for constructing approximate matrix decompositions*, SIAM Review 53(2), 2011, https://arxiv.org/abs/0909.4061 (v2), Appendix A.2 p. 64, Proposition A.3 (Chen–Dongarra).
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What the Lean code literally says, in plain math · claude-fable-5-1

Let N,nN, nN,n be natural numbers with 1≤n≤N1 \le n \le N1≤n≤N, and let sss be a real number with s≥0s \ge 0s≥0 satisfying

n s2≤N−n+1,n\, s^2 \le N - n + 1,ns2≤N−n+1,

where nnn and NNN are viewed as reals and the subtraction is real subtraction (since n≤Nn \le Nn≤N, the quantity N−n+1N - n + 1N−n+1 is a real number ≥1\ge 1≥1). Let μN,n\mu_{N,n}μN,n​ be the product measure on N×nN \times nN×n real arrays with i.i.d. N(0,1)\mathcal N(0,1)N(0,1) entries. For an N×nN \times nN×n real matrix AAA let σmin⁡(A):=inf⁡{(Ax)⋅(Ax):x∈Rn, x⋅x=1}=inf⁡∥x∥2=1∥Ax∥2\sigma_{\min}(A) := \inf\{\sqrt{(Ax)\cdot(Ax)} : x \in \mathbb R^n,\ x\cdot x = 1\} = \inf_{\|x\|_2=1}\|Ax\|_2σmin​(A):=inf{(Ax)⋅(Ax)​:x∈Rn, x⋅x=1}=inf∥x∥2​=1​∥Ax∥2​, the real infimum over the unit sphere (nonempty since n≥1n \ge 1n≥1; values ≥0\ge 0≥0). The statement asserts

μN,n({A:σmin⁡(A)≤s})  ≤  ofReal⁡ ⁣(n⋅(e⋅n⋅s2N−n+1)(N−n+1)/2),\mu_{N,n}\big(\{A : \sigma_{\min}(A) \le s\}\big) \;\le\; \operatorname{ofReal}\!\left( n \cdot \Big(\frac{e \cdot n \cdot s^2}{N - n + 1}\Big)^{(N-n+1)/2} \right),μN,n​({A:σmin​(A)≤s})≤ofReal(n⋅(N−n+1e⋅n⋅s2​)(N−n+1)/2),

where e=exp⁡(1)e = \exp(1)e=exp(1), the division is real division (denominator N−n+1≥1N - n + 1 \ge 1N−n+1≥1, so never zero), the exponent (N−n+1)/2≥1/2(N-n+1)/2 \ge 1/2(N−n+1)/2≥1/2 is a real number, and the power is the real power with nonnegative base. The left side is the outer measure of the displayed set (no measurability assumed). Since the base ens2/(N−n+1)e n s^2/(N-n+1)ens2/(N−n+1) is ≥0\ge 0≥0, the real quantity inside ofReal⁡\operatorname{ofReal}ofReal is ≥0\ge 0≥0 and no truncation occurs. Edge cases: when s=0s = 0s=0 the base is 000, the power is 0(N−n+1)/2=00^{(N-n+1)/2} = 00(N−n+1)/2=0 (positive exponent), and the right side is 000, so the claim becomes μN,n({A:σmin⁡(A)≤0})=μN,n({σmin⁡(A)=0})=0\mu_{N,n}(\{A : \sigma_{\min}(A) \le 0\}) = \mu_{N,n}(\{\sigma_{\min}(A) = 0\}) = 0μN,n​({A:σmin​(A)≤0})=μN,n​({σmin​(A)=0})=0. The constraint ns2≤N−n+1n s^2 \le N - n + 1ns2≤N−n+1 forces the base ens2/(N−n+1)e n s^2/(N-n+1)ens2/(N−n+1) to lie in [0,e][0, e][0,e], so the right side is at most n e(N−n+1)/2n\, e^{(N-n+1)/2}ne(N−n+1)/2.

Human review
  • Endorsed by Shuze Chen · Oct 9, 2026

    Confirmed by the moderator at approval.

  • Endorsed by tc · Oct 9, 2026

    Confirmed by the mission captain (proposal self-audit).

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