The circular law for random matrices

Friedrich Götze, Alexander Tikhomirov

Introduction

Let Xjk,1≤j,k<∞X_{jk},1\leq j,k<\infty, be complex random variables with EXjk=0\mathbf{E}X_{jk}=0 and E∣Xjk∣2=1\mathbf{E}|X_{jk}|^{2}=1. For a fixed n≥1n\geq 1, denote by λ1,…,λn\lambda_{1},\ldots,\lambda_{n} the eigenvalues of the n×nn\times n matrix

and define its empirical spectral distribution function by

The main result of our paper is the following:

Let φ(x)\varphi(x) denote the function (ln⁡(1+∣x∣))19+η(\ln(1+|x|))^{19+\eta}, η>0\eta>0, arbitrary, small and fixed. Let Xjk,j,k∈NX_{jk},j,k\in\mathbf{N}, denote independent complex random variables with

Then EGn(x,y)\mathbf{E}G_{n}(x,y) converges weakly to the distribution function G(x,y)G(x,y) as n→∞n\to\infty.

We shall prove the same result for the following class of sparse matrices. Let εjk\varepsilon_{jk}, j,k=1,…,nj,k=1,\ldots,n, denote a triangular array of Bernoulli random variables (taking values 0,10,1 only) which are independent in aggregate and independent of (Xjk)j,k=1n(X_{jk})_{j,k=1}^{n} with common success probability pn:=Pr⁡{εjk=1}p_{n}:=\Pr\{\varepsilon_{jk}=1\} depending on nn. Consider the sequence of matrices X(ε)=1npn(εjkXjk)j,k=1n\mathbf{X}^{(\varepsilon)}=\frac{1}{\sqrt{np_{n}}}(\varepsilon_{jk}X_{jk})_{j,k=1}^{n}. Let λ1(ε),…,λn(ε)\lambda_{1}^{(\varepsilon)},\ldots,\lambda_{n}^{(\varepsilon)} denote the (complex) eigenvalues of the matrix X(ε)\mathbf{X}^{(\varepsilon)} and denote by Gn(ε)(x,y)G_{n}^{(\varepsilon)}(x,y) the empirical spectral distribution function of the matrix X(ε)\mathbf{X}^{(\varepsilon)}, that is,

For η>0\eta>0 define φ(x)=(ln⁡(1+∣x∣))19+η\varphi(x)=(\ln(1+|x|))^{19+\eta}. Let Xjk,j,k∈NX_{jk},j,k\in\mathbf{N}, denote independent complex random variables with

Assume that there is a θ∈(0,1]\theta\in(0,1] such that pn−1=O(n1−θ)p_{n}^{-1}=\mathcal{O}(n^{1-\theta}) as n→∞n\to\infty. Then EGn(ε)(x,y)\mathbf{E}G_{n}^{(\varepsilon)}(x,y) converges weakly to the distribution function G(x,y)G(x,y) as n→∞n\to\infty.

The crucial problem of the proofs of Theorems 1.1 and 1.2 is to bound the smallest singular values sn(z)s_{n}(z), respectively, sn(ε)(z)s_{n}^{(\varepsilon)}(z) of the shifted matrices X−zI\mathbf{X}-z\mathbf{I}, respectively, X(ε)−zI\mathbf{X}^{(\varepsilon)}-z\mathbf{I}. (See also Zei04 , page 1561.) These bounds are based on the results obtained by Rudelson and Vershynin in RV . In a previous version of this paper GT07 we have used the corresponding results of Rudelson rud06 proving the circular law in the case of i.i.d. sub-Gaussian random variables. In fact, the results in GT07 actually imply the circular law for i.i.d. random variables with sup⁡j,kE∣Xjk∣4≤ϰ4<∞\sup_{j,k}\mathbf{E}|X_{jk}|^{4}\leq\varkappa_{4}<\infty in view of the fact (explicitly stated by Rudelson in rud06 ) that in his results the sub-Gaussian condition is needed for the proof of Pr⁡{∥X∥>K}≤Cexp⁡{−cn}\Pr\{\|\mathbf{X}\|>K\}\leq C\exp\{-cn\} only. Restricting oneself to the set Ωn(z)={sn(z)≤cn−3;∥X∥≤K}\Omega_{n}(z)=\{s_{n}(z)\leq cn^{-3};\|\mathbf{X}\|\leq K\} for the investigation of the smallest singular values, the inequality Pr⁡{Ωn(z)c}≤cn−1/2\Pr\{\Omega_{n}(z)^{c}\}\leq cn^{-1/2} follows from the results of Rudelson rud06 without the assumption of sub-Gaussian tails for the matrix X\mathbf{X}. A similar result has been proved by Pan and Zhou in PanZhou2007 based on results of Rudelson and Vershynin RV and Bai and Silverstein Bainn .

The strong circular law assuming moment condition of order larger than 22 only and comparable sparsity assumptions was proved independently by Tao and Vu in TaoVu2007 based on their results in TaoVu2007a in connection with the multivariate Littlewood Offord problem.

The approach in this paper though is based on the fruitful idea of Rudelson and Vershynin to characterize the vectors leading to small singular values of matrices with independent entries via “compressible” and “incompressible” vectors (see RV , Section 3.2, page 15). For the approximation of the distribution of singular values of X−zI\mathbf{X}-z\mathbf{I} we use a scheme different from the approach used in Bai Bai1997 .

The investigation of the convergence the spectral distribution functions of real or complex (nonsymmetric and non-Hermitian) random matrices with independent entries has a long history. Ginibre’s ginibre , in 1965, studied the real, complex and quaternion matrices with i.i.d. Gaussian entries. He derived the joint density for the distribution of eigenvalues of matrix. Applying Ginibre’s formula, Mehta Me , in 1967, determined the density of the expected spectral distribution function of random matrices with Gaussian entries with independent real and imaginary parts and deduced the circle law. Pastur suggested in 1973 the circular law for the general case (see Pastur1973 , page 64). Using the Ginibre results, Edelman edelman , in 1997, proved the circular law for the matrices with i.i.d. Gaussian real entries. Rider proved in Rider2003 and Rider2006 results about the spectral radius and about linear statistics of eigenvalues of non-Hermitian matrices with Gaussian entries.

Girko Girko1984a , in 1984, investigated the circular law for general matrices with independent entries assuming that the distribution of the entries has densities. As pointed out by Bai Bai1997 , Girko’s proof had serious gaps. Bai in Bai1997 gave a proof of the circular law for random matrices with independent entries assuming that the entries had bounded densities and finite sixth moments. His result does not cover the case of the Wigner ensemble and in particular ensembles of matrices with Rademacher entries. These ensembles are of some interest in various applications (see, e.g., Timm2004 ). Girko’s Girko1984a approach using families of spectra of Hermitian matrices for a characterization of the circular law based on the so-called V-transform was fruitful for all later work. See, for example, Girko’s Lemma 1 in Bai1997 . In fact, Girko Girko1984a was the first who used the logarithmic potential to prove the circular law. We shall outline his approach using logarithmic potential theory. Let ξ\xi denote a random variable uniformly distributed over the unit disc and independent of the matrix X\mathbf{X}. For any r>0r>0, consider the matrix

where I\mathbf{I} denotes the identity matrix of order nn. Let μn(r)\mu_{n}^{(r)} (resp., μn\mu_{n}) be empirical spectral measure of matrix X(r)\mathbf{X}(r) (resp., X\mathbf{X}) defined on the complex plane as empirical measure of the set of eigenvalues of matrix. We define a logarithmic potential of the expected spectral measure Eμn(r)(ds,dt)\mathbf{E}\mu_{n}^{(r)}(ds,dt) as

where λ1,…,λn\lambda_{1},\ldots,\lambda_{n} are the eigenvalues of the matrix X\mathbf{X}. Note that the expected spectral measure Eμn(r)\mathbf{E}\mu_{n}^{(r)} is the convolution of the measure Eμn\mathbf{E}\mu_{n} and the uniform distribution on the disc of radius rr (see Lemma .4 in the Appendix for details).

Assume that the sequence Eμn(r)\mathbf{E}\mu_{n}^{(r)} converges weakly to a measure μ\mu as n→∞n\to\infty and r→0r\to 0. Then

Let JJ be a random variable which is uniformly distributed on the set {1,…,n}\{1,\ldots,n\} and independent of the matrix X\mathbf{X}. We may represent the measure Eμn(r)\mathbf{E}\mu_{n}^{(r)} as the distribution of a random variable λJ+rξ\lambda_{J}+r\xi where λJ\lambda_{J} and ξ\xi are independent. Computing the characteristic function of this measure and passing first to the limit with respect to n→∞n\to\infty and then with respect to r→0r\to 0 (see also Lemma .5 in the Appendix), we conclude the result.

Now we may fix r>0r>0 and consider the measures Eμn(r)\mathbf{E}\mu_{n}^{(r)}. They have bounded densities. Assume that the measures Eμn\mathbf{E}\mu_{n} have supports in a fixed compact set and that Eμn\mathbf{E}\mu_{n} converges weakly to a measure μ\mu. Applying Theorem 6.9 (Lower envelope theorem) from saff , page 73 (see also Section 3.8 in the Appendix), we obtain that under these assumptions

Applying Theorem 1.2 in saff , page 84, we get

Let s1(X)≥⋯≥sn(X)s_{1}(\mathbf{X})\geq\cdots\geq s_{n}(\mathbf{X}) denote the singular values of the matrix X\mathbf{X}.

Since E1nTr⁡XX∗=1\mathbf{E}\frac{1}{n}\operatorname{Tr}\mathbf{X}\mathbf{X}^{*}=1 the sequence of measures Eμn\mathbf{E}\mu_{n} is weakly relatively compact. These results imply that for any η>0\eta>0 we may restrict the measures Eμn\mathbf{E}\mu_{n} to some compact set KηK_{\eta} such that sup⁡nEμn(Kη(c))<η\sup_{n}\mathbf{E}\mu_{n}(K_{\eta}^{(c)})<\eta. Moreover, Lemma .2 implies the existence of a compact KK such thatlim⁡n→∞sup⁡nEμn(K(c))=0\lim_{n\to\infty}\sup_{n}\mathbf{E}\mu_{n}(K^{(c)})=0. If we take some subsequence of the sequence of restricted measures Eμn\mathbf{E}\mu_{n} which converges to some measure μ\mu, thenlim inf⁡n→∞Uμn(r)(z)=Uμ(r)(z)\liminf_{n\to\infty}U_{\mu_{n}}^{(r)}(z)=U_{\mu}^{(r)}(z), r>0r>0, and lim⁡r→0Uμ(r)(z)=Uμ(z)\lim_{r\to 0}U_{\mu}^{(r)}(z)=U_{\mu}(z). If we prove that lim inf⁡n→∞Uμn(r)(z)\liminf_{n\to\infty}U_{\mu_{n}}^{(r)}(z) exists and Uμ(z)U_{\mu}(z) is equal to the logarithmic potential corresponding the uniform distribution on the unit disc [see Section 3, equality (68)], then the sequence of measures Eμn\mathbf{E}\mu_{n} weakly converges to the uniform distribution on the unit disc. Moreover, it is enough to prove that for some sequence r=r(n)→0r=r(n)\to 0, lim⁡n→∞Uμn(r)(z)=Uμ(z)\lim_{n\to\infty}U_{\mu_{n}}^{(r)}(z)=U_{\mu}(z).

Furthermore, let s1(ε)(z,r)≥⋯≥sn(ε)(z,r)s_{1}^{(\varepsilon)}(z,r)\geq\cdots\geq s_{n}^{(\varepsilon)}(z,r) denote the singular values of matrix X(ε)(z,r)=X(ε)(r)−zI\mathbf{X}^{(\varepsilon)}(z,r)=\mathbf{X}^{(\varepsilon)}(r)-z\mathbf{I}. We shall investigate the logarithmic potential Uμn(r)(z)U_{\mu_{n}}^{(r)}(z). Using elementary properties of singular values (see, e.g., Goh69 , Lemma 3.3, page 35), we may represent the function Uμn(r)(z)U_{\mu_{n}}^{(r)}(z) as follows:

where νn(ε)(⋅,z,r)\nu_{n}^{(\varepsilon)}(\cdot,z,r) denotes the expected spectral measure of the matrixHn(ε)(z,r)=(X(ε)(r)−zI)(X(ε)(r)−zI)∗\mathbf{H}_{n}^{(\varepsilon)}(z,r)=(\mathbf{X}^{(\varepsilon)}(r)-z\mathbf{I})(\mathbf{X}^{(\varepsilon)}(r)-z\mathbf{I})^{*}, which is the expectation of the counting measure of the set of eigenvalues of the matrix Hn(ε)(z,r)\mathbf{H}_{n}^{(\varepsilon)}(z,r).

In Section 2 we investigate convergence of the measure νn(ε)(⋅,z):=\breakν(ε)(⋅,z,0)\nu_{n}^{(\varepsilon)}(\cdot,z):=\break\nu^{(\varepsilon)}(\cdot,z,0). In Section 3 we study the properties of the limit measures ν(⋅,z)\nu(\cdot,z). But the crucial problem for the proof of the circular law is the so-called “regularization of the potential.” We solve this problem using bounds for the minimal singular values of the matrices X(ε)(z):=X(ε)−zI\mathbf{X}^{(\varepsilon)}(z):=\mathbf{X}^{(\varepsilon)}-z\mathbf{I} based on techniques developed in Rudelson rud06 and Rudelson and Vershynin RV . The bounds of minimal singular values of matrices X(ε)\mathbf{X}^{(\varepsilon)} are given in Section 4 and in the Appendix, Theorem 1.2. In Section 5 we give the proof of the main theorem. In the Appendix we combine precise statements of relevant results from potential theory and some auxiliary inequalities for the resolvent matrices.

In the what follows we shall denote by CC and cc or α,β,δ,ρ,η\alpha,\beta,\delta,\rho,\eta (without indices) some general absolute constant which may be changed from line to line. To specify a constant we shall use subindices. By IAI_{A} we shall denote the indicator of an event AA. For any matrix G\mathbf{G} we denote the Frobenius norm by ∥G∥2\|\mathbf{G}\|_{2}, and we denote by ∥G∥\|\mathbf{G}\| the operator norm.

Denote by Fn(ε)(x,z)F_{n}^{(\varepsilon)}(x,z) the distribution function of the measure νn(ε)(⋅,z)\nu_{n}^{(\varepsilon)}(\cdot,z), that is,

where s1(ε)(z)≥⋯≥sn(ε)(z)≥0s_{1}^{(\varepsilon)}(z)\geq\cdots\geq s_{n}^{(\varepsilon)}(z)\geq 0 denote the singular values of the matrix X(ε)(z)=X(ε)−zI\mathbf{X}^{(\varepsilon)}(z)=\mathbf{X}^{(\varepsilon)}-z\mathbf{I}. For a positive random variable ξ\xi and a Rademacher random variable (r.v.) κ\kappa consider the transformed r.v. ξ~=κξ\widetilde{\xi}=\kappa\sqrt{\xi}. If ζ\zeta has distribution function Fn(ε)(x,z){F}_{n}^{(\varepsilon)}(x,z), the variable ζ~\widetilde{\zeta} has distribution function F~n(ε)(x,z)\widetilde{F}_{n}^{(\varepsilon)}(x,z), given by

for all real xx. Note that this induces a one-to-one corresponds between the respective measures νn(ε)(⋅,z)\nu_{n}^{(\varepsilon)}(\cdot,z) and ν~n(ε)(⋅,z){\widetilde{\nu}}_{n}^{(\varepsilon)}(\cdot,z). The limit distribution function of Fn(ε)(x,z){F}_{n}^{(\varepsilon)}(x,z) as n→∞n\to\infty, is denoted by F(⋅,z)F(\cdot,z). The corresponding symmetrization F~(x,z)\widetilde{F}(x,z) is the limit of F~n(ε)(x,z)\widetilde{F}_{n}^{(\varepsilon)}(x,z) as n→∞n\to\infty. We have

Denote by sn(ε)(α,z)s_{n}^{(\varepsilon)}(\alpha,z) [resp., s(α,z)s(\alpha,z)] and Sn(ε)(x,z)S_{n}^{(\varepsilon)}(x,z) [resp., S(x,z)S(x,z)] the Stieltjes transforms of the measures νn(ε)(⋅,z)\nu_{n}^{(\varepsilon)}(\cdot,z) [resp., ν(⋅,z)\nu(\cdot,z)] and ν~n(ε)(⋅,z)\widetilde{\nu}_{n}^{(\varepsilon)}(\cdot,z) [resp., ν~(⋅,z)\widetilde{\nu}(\cdot,z)] correspondingly. Then we have

As shown in Bai Bai1997 , the measure ν(⋅,z)\nu(\cdot,z) has a density p(x,z)p(x,z) with bounded support. More precisely, p(x,z)≤Cmax⁡{1,1x}p(x,z)\leq C\max\{1,\frac{1}{\sqrt{x}}\}. Thus the measure ν~(⋅,z)\widetilde{\nu}(\cdot,z) has bounded support and bounded density p~(x,z)=∣x∣p(x2,z)\widetilde{p}(x,z)=|x|p(x^{2},z).

Let EXjk=0\mathbf{E}X_{jk}=0, E∣Xjk∣2=1\mathbf{E}|X_{jk}|^{2}=1. Assume for some function φ(x)>0\varphi(x)>0 such that φ(x)→∞\varphi(x)\to\infty as x→∞x\to\infty and such that the function x/φ(x)x/\varphi(x) is nondecreasing we have

Let EXjk=0\mathbf{E}X_{jk}=0, E∣Xjk∣2=1\mathbf{E}|X_{jk}|^{2}=1, and

To bound the distance between the distribution functionsF~n(ε)(x,z)\widetilde{F}_{n}^{(\varepsilon)}(x,z) and F~(x,z)\widetilde{F}(x,z) we investigate the distance between their the Stieltjes transforms. Introduce the Hermitian 2n×2n2n\times 2n matrix

where On\mathbf{O}_{n} denotes n×nn\times n matrix with zero entries. Using the inverse of the partial matrix (see, e.g., HoJohn91 , Chapter 08, page 18) it follows that, for α=u+iv\alpha=u+iv, v>0v>0,

where X(ε)(z)=X(ε)−zI\mathbf{X}^{(\varepsilon)}(z)=\mathbf{X}^{(\varepsilon)}-z\mathbf{I} and I2n\mathbf{I}_{2n} denotes the unit matrix of order 2n2n. By definition of Sn(ε)(α,z)S_{n}^{(\varepsilon)}(\alpha,z), we have

Set R(α,z):=(Rj,k(α,z))j,k=12n=(W−αI2n)−1\mathbf{R}(\alpha,z):=(R_{j,k}(\alpha,z))_{j,k=1}^{2n}=(\mathbf{W}-\alpha\mathbf{I}_{2n})^{-1}. It is easy to check that

With this notation we rewrite equality (2) as follows:

In what follows we shall use a simple resolvent equality. For two matrices U\mathbf{U} and V\mathbf{V} let RU=(U−αI)−1\mathbf{R}_{U}=(\mathbf{U}-\alpha\mathbf{I})^{-1}, RU+V=(U+V−αI)−1\mathbf{R}_{U+V}=(\mathbf{U}+\mathbf{V}-\alpha\mathbf{I})^{-1}, then

Using this representation and the resolvent equality, we get

Here, and in what follows, we omit the arguments α\alpha and zz in the notation of resolvent matrices. For any vector a\mathbf{a}, let aT\mathbf{a}^{T} denote the transposed vector a\mathbf{a}. Applying the resolvent equality again, we obtain

Applying this notation to equality (2) and taking into account that XjkX_{jk} and R(jk)\mathbf{R}^{(jk)} are independent, we get

From (2) it follows immediately that for any p,q=1,…,2np,q=1,\ldots,2n, j,k=1,…,nj,k=1,\ldots,n,

Since ∑m,l=1n∣Rm,l∣2≤n/v2{\sum_{m,l=1}^{n}}|R_{m,l}|^{2}\leq n/v^{2} and ∑m,l=1n∣Rm,l(jk)∣2≤n/v2{\sum_{m,l=1}^{n}}|R^{(jk)}_{m,l}|^{2}\leq n/v^{2}, equality (2) implies

By definition (2) of T(j,k)\mathbf{T}^{(j,k)}, applying standard resolvent properties, we obtain the following bounds, for any z=u+iv,v>0z=u+iv,v>0,

For the proof of this inequality see Lemma .3 in the Appendix. Using the last inequalities we obtain, that for v>0v>0

Since 1n∑j=1nRjj=1n∑k=1nRk+n,k+n=12nTr⁡R(α,z)\frac{1}{n}\sum_{j=1}^{n}R_{jj}=\frac{1}{n}\sum_{k=1}^{n}R_{k+n,k+n}=\frac{1}{2n}\operatorname{Tr}\mathbf{R}(\alpha,z), we obtain

Note that for any Hermitian random matrix W\mathbf{W} with independent entries on and above the diagonal we have

The proof of this inequality is easy and due to a martingale-type expansion already used by Girko. Inequalities (24) and (25) together imply that for v>0v>0

Denote by r(α,z)r(\alpha,z) some generic function with ∣r(α,z)∣≤1|r(\alpha,z)|\leq 1 which may vary from line to line. We may now rewrite equality (2) as follows:

We now investigate the functions T(α,z)=1nETr⁡BT(\alpha,z)=\frac{1}{n}\mathbf{E}\operatorname{Tr}\mathbf{B} and V(α,z)=1nETr⁡DV(\alpha,z)=\frac{1}{n}\mathbf{E}\operatorname{Tr}\mathbf{D}. Since the arguments for both functions are similar we provide it for the first one only. By definition of the matrix B\mathbf{B}, we have

Using the resolvent equality (2) and Lemma .3, we get, for v>c×\breakφ(npn)/nv>c\times\break\varphi(\sqrt{np_{n}})/n

Inequalities (28) and (29) together imply, for v>cφ(npn)/nv>c\varphi(\sqrt{np_{n}})/n,

where δ~n(α,z)=θCϰr(α,z)φ(npn)v3\widetilde{\delta}_{n}(\alpha,z)=\theta\frac{C\varkappa r(\alpha,z)}{\varphi(\sqrt{np_{n}})v^{3}}.

Furthermore, we prove the following simple lemma.

Let α=u+iv\alpha=u+iv, v>0v>0. Let S(α,z)S(\alpha,z) satisfy the equation

and Im⁡{S(α,z)}>0\operatorname{Im}\{S(\alpha,z)\}>0. Then the inequality

For α=u+iv\alpha=u+iv with v>0v>0, the Stieltjes transform S(α,z)S(\alpha,z) satisfies the following equation:

Comparing the imaginary parts of both sides of this equation, we get

Since v>0v>0 and Im⁡{α+S(α,z)}>0\operatorname{Im}\{\alpha+S(\alpha,z)\}>0, it follows that

Equality (40) and the last remark together imply

To compare the functions S(α,z)S(\alpha,z) and Sn(α,z)S_{n}(\alpha,z) we prove:

Repeating the arguments of Lemma 2.2 completes the proof.

The next lemma provides a bound for the distance between the Stieltjes transforms S(α,z)S(\alpha,z) and Sn(ε)(α,z)S_{n}^{(\varepsilon)}(\alpha,z).

Note that S(α,z)S(\alpha,z) and Sn(ε)(α,z)S_{n}^{(\varepsilon)}(\alpha,z) satisfy the equations

respectively. These equations together imply

Applying inequality ∣ab∣≤12(a2+b2)|ab|\leq\frac{1}{2}{(a^{2}+b^{2})}, we get

The last inequality and Lemmas 2.2 and 2.3 together imply

To bound the distance between the distribution function Fn(x,z)F_{n}(x,z) and the distribution function F(x,z)F(x,z) corresponding the Stieltjes transforms Sn(α,z)S_{n}(\alpha,z) and S(α,z)S(\alpha,z) we use Corollary 2.3 from GT03 . In the next lemma we give an integral bound for the distance between the Stieltjes transforms S(α,z)S(\alpha,z) and Sn(ε)(α,z)S_{n}^{(\varepsilon)}(\alpha,z).

For v≥v0(n)=c(φ(npn))−1/6v\geq v_{0}(n)=c(\varphi(\sqrt{np_{n}}))^{-1/6} the inequality

It follows from here that ∣δ^n(α,z)∣≤Cv5φ(npn)|\widehat{\delta}_{n}(\alpha,z)|\leq\frac{C}{v^{5}\varphi(\sqrt{np_{n}})} and

for v≥c(φ(npn))−1/6v\geq c(\varphi(\sqrt{np_{n}}))^{-1/6}. Lemma 2.4 implies that it is enough to prove the inequality

where γn=Cv6φ(npn)\gamma_{n}=\frac{C}{v^{6}\varphi(\sqrt{np_{n}})}. By definition of δ^(α,z)\widehat{\delta}(\alpha,z), we have

Furthermore, representation (35) implies that

It follows from relation (32) that for v>c(φ(npn))−1/6v>c(\varphi(\sqrt{np_{n}}))^{-1/6},

The last two inequalities together imply that for sufficiently large nn and v>c(φ(npn))−1/6v>c(\varphi(\sqrt{np_{n}}))^{-1/6},

Inequalities (47), (45) and the definition of δ^n(α,z)\widehat{\delta}_{n}(\alpha,z) together imply

If we choose vv such that Cϰv4φ(npn)<12\frac{C\varkappa}{v^{4}\varphi(\sqrt{np_{n}})}<\frac{1}{2} we obtain

In Section 3 we show that the measure ν~(⋅,z)\widetilde{\nu}(\cdot,z) has bounded support and bounded density for any zz. To bound the distance between the distribution functions F~n(ε)(x,z)\widetilde{F}_{n}^{(\varepsilon)}(x,z) and F~(x,z)\widetilde{F}(x,z) we may apply Corollary 3.2 from GT03 (see also Lemma .6 in the Appendix). We take V=1V=1 and v0=C(φ(npn))−1/6v_{0}=C(\varphi(\sqrt{np_{n}}))^{-1/6}. Then Lemmas 2.2 and 2.3 together imply

Properties of the measure ν~​(⋅,z)~𝜈⋅𝑧\widetilde{\nu}(\cdot,z)

In this section we investigate the properties of the measure ν~(⋅,z)\widetilde{\nu}(\cdot,z). At first note that there exists a solution S(α,z)S(\alpha,z) of the equation

Set y=S(x,z)+xy=S(x,z)+x and consider equation (41) on the real line

It is straightforward to check that 3(1−∣z∣2)≤∣x1∣\sqrt{3(1-|z|^{2})}\leq|x_{1}| and x22<0x_{2}^{2}<0 for ∣z∣<1|z|<1 and x22=0x_{2}^{2}=0 for ∣z∣=1|z|=1, and x22>0x_{2}^{2}>0 for ∣z∣>1|z|>1.

In the case ∣z∣≤1|z|\leq 1 equation (57) has one real root for ∣x∣≤∣x1∣|x|\leq|x_{1}| and three real roots for ∣x∣>∣x1∣|x|>|x_{1}|. In the case ∣z∣>1|z|>1 equation (57) has one real root for ∣x2∣≤x≤∣x1∣|x_{2}|\leq x\leq|x_{1}| and three real roots for ∣x∣≤∣x2∣|x|\leq|x_{2}| or for ∣x∣≥∣x1∣|x|\geq|x_{1}|.

This implies that, for ∣z∣≤1|z|\leq 1 and for

equation (57) has one real root. Furthermore, direct calculations show that

Solving the equation L(y1)L(y2)=0L(y_{1})L(y_{2})=0 with respect to xx, we get for ∣z∣≤1|z|\leq 1 and 3(1−∣z∣2)≤∣x∣≤∣x1∣\sqrt{3(1-|z|^{2})}\leq|x|\leq|x_{1}|

and for ∣z∣≤1|z|\leq 1 and ∣x∣>20+8∣z∣28+(1+8∣z∣2)3/2−18∣z∣2|x|>\sqrt{\frac{20+8|z|^{2}}{8}+\frac{(1+8|z|^{2})^{3/2}-1}{8|z|^{2}}}

These relations imply that for ∣z∣≤1|z|\leq 1 the function L(y)L(y) has three real roots for ∣x∣≥∣x1∣|x|\geq|x_{1}| and one real root for ∣x∣<∣x1∣|x|<|x_{1}|.

Consider the case ∣z∣>1|z|>1 now. In this case y1,2y_{1,2} are real for all xx and x22>0x_{2}^{2}>0. Note that

for ∣x∣≤∣x2∣|x|\leq|x_{2}| and for ∣x∣≥∣x1∣|x|\geq|x_{1}| and

for ∣x2∣<x<∣x1∣|x_{2}|<x<|x_{1}|. These implies that for ∣z∣>1|z|>1 and for ∣x2∣<x<∣x1∣|x_{2}|<x<|x_{1}| the function L(y)L(y) has one real root and for ∣x∣≤∣x2∣|x|\leq|x_{2}| or for ∣x∣≥∣x1∣|x|\geq|x_{1}| the function L(y)L(y) has three real roots. The lemma is proved.

From Lemma 3.1 it follows that the measure ν~(x,z)\widetilde{\nu}(x,z) has a density p(x,z)=lim⁡v→0Im⁡S(α,z)p(x,z)=\lim_{v\to 0}\operatorname{Im}{S(\alpha,z)} and:

for ∣z∣≤1|z|\leq 1, if ∣x∣≥x1|x|\geq x_{1}, then p(x,z)=0p(x,z)=0;

for ∣z∣≥1|z|\geq 1, if ∣x∣≥x1|x|\geq x_{1} or ∣x∣≤x2|x|\leq x_{2}, then p(x,z)=0p(x,z)=0;

It is well known that for z=s+itz=s+it the logarithmic potential of uniform distribution on the unit disc is

According to Lemma 4.4 in Bai Bai1997 , we have, for z=s+itz=s+it,

According to Remark 3.1, we have, for ∣z∣≥1|z|\geq 1,

Comparing equalities (63) and (61) and using relation (65), we obtain

The smallest singular value

Let X(ε)=1npn(εjkXjk)j,k=1n\mathbf{X}^{(\varepsilon)}=\frac{1}{\sqrt{np_{n}}}(\varepsilon_{jk}X_{jk})_{j,k=1}^{n} be an n×nn\times n matrix with independent entries εjkXjk\varepsilon_{jk}X_{jk}, j,k=1,…,nj,k=1,\ldots,n. Assume that EXjk=0\mathbf{E}X_{jk}=0 and EXjk2=1\mathbf{E}X_{jk}^{2}=1 and let εjk\varepsilon_{jk} denote Bernoulli random variables with pn=Pr⁡{εjk=1}p_{n}=\Pr\{\varepsilon_{jk}=1\}, j,k=1,…,nj,k=1,\ldots,n. Denote by s1(ε)(z)≥⋯≥sn(ε)(z)s_{1}^{(\varepsilon)}(z)\geq\cdots\geq s_{n}^{(\varepsilon)}(z) the singular values of the matrix X(ε)(z):=X(ε)−zI\mathbf{X}^{(\varepsilon)}(z):=\mathbf{X}^{(\varepsilon)}-z\mathbf{I}. In this section we prove a bound for the minimal singular value of the matrices X(ε)(z)\mathbf{X}^{(\varepsilon)}(z). We prove the following result.

Let Xjk,j,k∈NX_{jk},j,k\in\mathbf{N}, be independent random complex variables with EXjk=0\mathbf{E}X_{jk}=0 and E∣Xjk∣2=1\mathbf{E}|X_{jk}|^{2}=1, which are uniformly integrable, that is,

Let XjkX_{jk} be i.i.d. random variables with EXjk=0\mathbf{E}X_{jk}=0 and E∣Xjk∣2=1\mathbf{E}|X_{jk}|^{2}=1. Then condition (69) holds.

Consider the event AA that there exists at least one row with zero entries only. Its probability is given by

Simple calculations show that if npn≤ln⁡nnp_{n}\leq\ln n for all n≥1n\geq 1, then

Hence in the case npn≤ln⁡nnp_{n}\leq\ln n and npn→∞np_{n}\to\infty we have no invertibility with positive probability.

The proof of Theorem 4.1 uses ideas of Rudelson and Vershynin RV , to classify with high probability vectors x\mathbf{x} in the (n−1)(n-1)-dimensional unit sphere Sn−1\mathcal{S}^{n-1} such that ∥X(ε)(z)x∥2\|\mathbf{X}^{(\varepsilon)}(z)\mathbf{x}\|_{2} is extremely small into two classes, called compressible and incompressible vectors.

We develop our approach for shifted sparse and normalized matrices X(ε)(z)\mathbf{X}^{(\varepsilon)}(z). The generalization to the case of complex sparse and shifted matrices X(ε)(z)\mathbf{X}^{(\varepsilon)}(z) is straightforward. For details see, for example, the paper of Götze and Tikhomirov GT07 and the proof of the Lemma 4.1 below.

We may relax the condition pn−1=O(n1−θ)p_{n}^{-1}=\mathcal{O}(n^{1-\theta}) to pn−1=o(n/\breakln⁡2n)p_{n}^{-1}=o(n/\break\ln^{2}{n}). The quantity BB in Theorem 4.1 should be of order ln⁡n\ln n in this case. See Remark 4.9 for details.

Let x=(x1,…,xn)∈Sn−1\mathbf{x}=(x_{1},\ldots,x_{n})\in\mathcal{S}^{n-1} be a fixed unit vector and X(ε)(z)\mathbf{X}^{(\varepsilon)}(z) be a matrix as in Theorem 4.1. Then there exist some positive absolute constants γ0\gamma_{0} and c0c_{0} such that for any 0<τ≤γ00<\tau\leq\gamma_{0}

Recall that EXij=0\mathbf{E}X_{ij}=0 and E∣Xij∣2=1\mathbf{E}|X_{ij}|^{2}=1. Assume first that XijX_{ij} are real independent r.v. with mean zero, and variance at least 11. Let Xij(ε)=XijεijX^{(\varepsilon)}_{ij}=X_{ij}\varepsilon_{ij} with independent Bernoulli variables which are independent of XijX_{ij} in aggregate and let z=0z=0. Assume also that x\mathbf{x} is a real vector. Then

Using e−t2/2=Eexp⁡{itξ}e^{-t^{2}/2}=\mathbf{E}\exp\{it\xi\}, where ξ\xi is a standard Gaussian random variable, we obtain

where ξj\xi_{j}, j=1,…,nj=1,\ldots,n, denote i.i.d. standard Gaussian r.v.s and EZ\mathbf{E}_{Z} denotes expectation with respect to ZZ conditional on all other r.v.s. For every α,x∈\alpha,x\in and ρ∈(0,1)\rho\in(0,1) the following inequality holds:

(see Bickel , inequality (3.7)). Take α=Pr⁡{∣ξj∣≤C1}\alpha=\Pr\{|\xi_{j}|\leq C_{1}\} for some absolute positive constant C1C_{1} which will be chosen later. Then it follows from (4) that

where fjk(u)=Eexp⁡{iuXjk}f_{jk}(u)=\mathbf{E}\exp\{iuX_{jk}\}. Assuming (69), choose a constant M>0M>0 such that

Since 1−cos⁡x≥11/24x21-\cos x\geq 11/24x^{2} for ∣x∣≤1|x|\leq 1, conditioning on the event ∣ξj∣≤C1|\xi_{j}|\leq C_{1}, we get for 0<t≤1/(MC1)0<t\leq 1/(MC_{1})

It follows from (4) for 0<t<1/(MC1)0<t<1/{(MC_{1})} and for some constant c>0c>0

This implies that conditionally on ∣ξj∣≤C1|\xi_{j}|\leq C_{1} and for 0<t≤1/(MC1)0<t\leq 1/(MC_{1})

Let Φ0(x):=2Φ(x)−1\Phi_{0}(x):=2\Phi(x)-1, x>0x>0, where Φ(x)\Phi(x) denotes the standard Gaussian distribution function. It is straightforward to show that

We may choose C1C_{1} large enough such that following inequalities hold:

for all ∣t∣≤1/(MC1)|t|\leq 1/(MC_{1}). Inequalities (4), (77), (4), (86) together imply that for any β∈(0,1)\beta\in(0,1)

Without loss of generality we may take C1C_{1} sufficiently large, such that α≥4/5\alpha\geq 4/5 and choose β=2/5\beta=2/5. Then we obtain

For τ<c60\tau<\frac{\sqrt{c}}{\sqrt{60}} we conclude from here that for ∣t∣≤1/(MC1)|t|\leq 1/(MC_{1})

Inequality (89) implies that inequality (73) holds with some positive constant c0>0c_{0}>0. This completes the proof in the real case.

Consider now the general case. Let Xjk=ξjk+iηjkX_{jk}=\xi_{jk}+i\eta_{jk} with i=−1i=\sqrt{-1} with E∣Xjk∣2=1\mathbf{E}|X_{jk}|^{2}=1 and xk=uk+ivkx_{k}=u_{k}+iv_{k} and z=u+ivz=u+iv. In this notation we have

For any j=1,…,nj=1,\ldots,n we introduce the set AjA_{j} as follows:

It is straightforward to check that for any k∉Ajk\notin A_{j}

According to inequality (91), for any j=1,…,nj=1,\ldots,n, there exists a set BjB_{j} such that

Introduce the following random variables for any j,k=1,…,nj,k=1,\ldots,n

Inequalities (95) and (96) together imply that one of the following two inequalities

holds. If (99) holds we shall bound the first term on the right-hand side of (4). In the other case we shall bound the second term. In what follows we may repeat the arguments leading to inequalities (4)–(84). Thus the lemma is proved.

For any qn∈(0,1)q_{n}\in(0,1) and K>0K>0 to be chosen later we define Kn:=KnpnK_{n}:=Kn\sqrt{p_{n}}, q^n:=qn/(ln⁡(2/pn)ln⁡Kn)\widehat{q}_{n}:=q_{n}/(\ln(2/p_{n})\ln K_{n}) and p^n:=pn/(ln⁡(2/pn)ln⁡Kn)\widehat{p}_{n}:=p_{n}/(\ln(2/p_{n})\ln K_{n}). Without loss of generality we shall assume that

Assume there exist an absolute constant c>0c>0 and values γn,qn∈(0,1)\gamma_{n},q_{n}\in(0,1) such that for any x∈C⊂S(n−1)\mathbf{x}\in\mathcal{C}\subset\mathcal{S}^{(n-1)}

holds. Then there exists a constant δ0>0\delta_{0}>0 depending on KK and cc only such that, for k<δ0nq^nk<\delta_{0}n\widehat{q}_{n},

Let η>0\eta>0 to be chosen later. There exists an η\eta-net N\mathcal{N} in Sk−1∩C\mathcal{S}^{k-1}\cap\mathcal{C} of cardinality ∣N∣≤(3η)2k|\mathcal{N}|\leq(\frac{3}{\eta})^{2k} (see, e.g., Lemma 3.4 in rud06 ). By condition (102), we have for τ≤γn\tau\leq\gamma_{n}

Let VV be the event that ∥X(ε)(z)∥≤Kn\|\mathbf{X}^{(\varepsilon)}(z)\|\leq K_{n} and ∥X(ε)(z)y∥2≤12τ\|\mathbf{X}^{(\varepsilon)}(z)\mathbf{y}\|_{2}\leq\frac{1}{2}\tau for some point y∈S(k−1)∩C\mathbf{y}\in\mathcal{S}^{(k-1)}\cap\mathcal{C}. Assume that VV occurs and choose a point x∈N\mathbf{x}\in\mathcal{N} such that ∥y−x∥2≤η\|\mathbf{y}-\mathbf{x}\|_{2}\leq\eta. Then

Choosing δ0=c80\delta_{0}=\frac{c}{80} and τ=γn\tau=\gamma_{n}, we complete the proof.

Following Rudelson and Vershynin RV , we shall partition the unit sphere S(n−1)\mathcal{S}^{(n-1)} into the two sets of so-called compressible and incompressible vectors, and we will show the invertibility of X\mathbf{X} on each set separately.

The sets of sparse, compressible and incompressible vectors depending on δ\delta and ρ\rho will be denoted by

Let X(ε)(z)\mathbf{X}^{(\varepsilon)}(z) be a random matrix as in Theorem 1.2, and let Kn=KnpnK_{n}=Kn\sqrt{p_{n}} with a constant K≥1K\geq 1. Assume there exist an absolute constant c>0c>0 and values γn,qn∈(0,1)\gamma_{n},q_{n}\in(0,1) such that for any x∈C⊂S(n−1)\mathbf{x}\in\mathcal{C}\subset\mathcal{S}^{(n-1)}

holds. Then there exist δ1,c1\delta_{1},c_{1} that depend on KK and cc only, such that

At first we estimate the invertibility for sparse vectors. Let k=[δ1nq^n]k=[\delta_{1}n\widehat{q}_{n}] with some positive constant δ1\delta_{1} which will be chosen later. According to Proposition 4.6 for any δ1≤δ0\delta_{1}\leq\delta_{0} and for any τ≤γn/2\tau\leq\gamma_{n}/2, we have the following inequality:

Using Stirling’s formula, we get for some absolute positive constant CC

We may choose δ1\delta_{1} small enough that

Choose ρ:=γ:=γn/4\rho:=\gamma:=\gamma_{n}/4. Let VV be the event that ∥X(ε)(z)∥≤Kn\|\mathbf{X}^{(\varepsilon)}(z)\|\leq K_{n} and ∥X(ε)(z)y∥2≤γ1\|\mathbf{X}^{(\varepsilon)}(z)\mathbf{y}\|_{2}\leq\gamma_{1} for some point y∈Comp(δ1p^n,ρKn−1)\mathbf{y}\in\mathit{Comp}(\delta_{1}\widehat{p}_{n},\rho K_{n}^{-1}). Assume that VV occurs and choose a point x∈Sparse(δ1p^n)\mathbf{x}\in\mathit{Sparse}(\delta_{1}\widehat{p}_{n}) such that ∥y−x∥2≤ρKn−1\|\mathbf{y}-\mathbf{x}\|_{2}\leq\rho K_{n}^{-1}. Then

Let δ,ρ∈(0,1)\delta,\rho\in(0,1). Let x∈Incomp(δ,ρ)\mathbf{x}\in\mathit{Incomp}(\delta,\rho). Then there exists a set σ(x)⊂{1,…,n}\sigma(\mathbf{x})\subset\{1,\ldots,n\} of cardinality ∣σ(x)∣≥12nδ|\sigma(\mathbf{x})|\geq\frac{1}{2}n\delta such that

which we shall call “spread set of xx” henceforth.

See proof of Lemma 3.4 RV , page 16. For the reader’s convenience we repeat this proof here. Consider the subsets of {1,…,n}\{1,\ldots,n\} defined by

If x∈Incomp(δp^n,ρ)\mathbf{x}\in\mathit{Incomp}(\delta\widehat{p}_{n},\rho) then there exists a set σ(x)\sigma(\mathbf{x}) with cardinality ∣σ(x)∣≥12nδp^n|\sigma(\mathbf{x})|\geq\frac{1}{2}n\delta\widehat{p}_{n} such that

We shall now bound this concentration function and prove a tensorization lemma for incompressible vectors.

Let δn\delta_{n} and ρn\rho_{n} be some functions of nn such that ρn,δn∈(0,1)\rho_{n},\delta_{n}\in(0,1). Let η0\eta_{0} and r0r_{0} as in Lemma .7. Let x∈Incomp(δn,ρn)\mathbf{x}\in\mathit{Incomp}(\delta_{n},\rho_{n}). Then there exists positive constants r1r_{1} and r2r_{2} depending on r0r_{0} such that for any 0<η≤η00<\eta\leq\eta_{0} we have

Introduce σ(x):={k∈{1,…,n}\dvtxρn/2n≤∣xk∣≤1/m/2}\sigma(\mathbf{x}):=\{k\in\{1,\ldots,n\}\dvtx\rho_{n}/{\sqrt{2n}}\leq|x_{k}|\leq 1/\sqrt{m/2}\}. Since x∈\breakIncomp(δn,ρn)\mathbf{x}\in\break\mathit{Incomp}(\delta_{n},\rho_{n}) the cardinality of σ(x)\sigma(\mathbf{x}) is at least m/2m/2. Using that the concentration function of sum of independent random variables is less then concentration function of its summands, we obtain

According to Lemma .7 in the Appendix for any η≤η0\eta\leq\eta_{0}, we have Q(η)≤r0<1Q(\eta)\leq r_{0}<1. Assume that mpn≥1/3mp_{n}\geq 1/3. Then we have

If mpn≤1/3mp_{n}\leq 1/3 then (1−pn)m≤1−mpn/3(1-p_{n})^{m}\leq 1-mp_{n}/3 and

Let ζ1,…,ζn\zeta_{1},\ldots,\zeta_{n} be independent nonnegative random variables. Assume that

for some positive qn∈(0,1)q_{n}\in(0,1) and λn>0\lambda_{n}>0. Then there exists positive absolute constants K1K_{1} and K2K_{2} such that

We repeat the proof of Lemma 4.4 in LPRT . Let t=K1qnλnt=K_{1}\sqrt{q_{n}}\lambda_{n}. For any τ>0\tau>0 we have

Choosing τ:=qn/4\tau:=q_{n}/4 and K12:=14ln⁡2K_{1}^{2}:=\frac{1}{4\ln 2}, we get

Recall that we assume pn−1=O(n1−θ),1≥θ>0p_{n}^{-1}=O(n^{1-\theta}),1\geq\theta>0. For this fixed θ\theta consider L:=[1θ]L:=[\frac{1}{\theta}]. Hence by definition pn,l:=(np^n)lpn→0,n→∞p_{n,l}:=(n\widehat{p}_{n})^{l}p_{n}\to 0,n\to\infty for l=1,…,L−1l=1,\ldots,L-1 and lim sup⁡n→∞(npn)Lpn>0\limsup_{n\to\infty}(np_{n})^{L}p_{n}>0. We put pn,L:=1p_{n,L}:=1.

We shall assume that nn is large enough such that (npn)Lpn≥q1>0(np_{n})^{L}p_{n}\geq q_{1}>0 for some constant q1>0q_{1}>0. Starting with a decomposition of C0:=S(n−1)\mathcal{C}_{0}:=\mathcal{S}^{(n-1)} into compressible vectors x\mathbf{x} in C^1:=C0∩Comp(δ1pn,1,ρn,1)\widehat{\mathcal{C}}_{1}:=\mathcal{C}_{0}\cap\mathit{Comp}(\delta_{1}p_{n,1},\rho_{n,1}), where pn,1=p^np_{n,1}=\widehat{p}_{n}, ρn,1=γ0/(4Kn)\rho_{n,1}=\gamma_{0}/(4K_{n}), and the constants γ0\gamma_{0} and δ1\delta_{1} are chosen as in Lemmas 4.1 and 4.2, respectively. Then Lemma 4.1 implies inequality (108) with qnq_{n} replaced by pnp_{n} and γn\gamma_{n} replaced by γ0\gamma_{0}. Hence, using Lemma 4.2, one obtains the claim for the subset of vectors C^1\widehat{\mathcal{C}}_{1}. The remaining vectors x\mathbf{x} in C0\mathcal{C}_{0} lie in C1:=Incomp(δ1pn,1,ρn,1)\mathcal{C}_{1}:=\mathit{Incomp}(\delta_{1}p_{n,1},\rho_{n,1}). According to Lemmas 4.4, 4.5 inequality (108) holds again for these vectors but with new parameters qn=npnδ1pn,1q_{n}=np_{n}\delta_{1}p_{n,1} and γn=cρn,1δ1pn,1\gamma_{n}=c\rho_{n,1}\sqrt{\delta_{1}p_{n,1}}. Thus we may again subdivide the vectors in C1\mathcal{C}_{1} into the vectors within distance ρn,2\rho_{n,2} from these sparse ones, that is, C^2:=C1∩Comp(δ2pn,2,ρn,2)\widehat{\mathcal{C}}_{2}:={\mathcal{C}}_{1}\cap\mathit{Comp}(\delta_{2}p_{n,2},\rho_{n,2}) and the remaining ones, that is, C2:=C1∩Incomp(δ2pn,2,ρn,2){\mathcal{C}}_{2}:={\mathcal{C}}_{1}\cap\mathit{Incomp}(\delta_{2}p_{n,2},\rho_{n,2}). Iterating this procedure LL times we arrive at the incompressible set CL{\mathcal{C}}_{L} of vectors x\mathbf{x} where Lemmas 4.4, 4.5 and Proposition 4.6 yield the required bound of order exp⁡{−δn}\exp\{-\delta n\}, for a sufficiently small absolute constant δ>0\delta>0.

Summarizing, we will determine iteratively constants δl,ρn,l\delta_{l},\rho_{n,l}, for l=1,…,Ll=1,\ldots,L and the following sets of vectors:

The main bounds to carry out this procedure are given in the following Lemmas 4.6 and 4.7.

Let δn,ρn∈(0,1)\delta_{n},\rho_{n}\in(0,1) and let x∈Incomp(δn,ρn)\mathbf{x}\in\mathit{Incomp}(\delta_{n},\rho_{n}) and X(ε)(z)\mathbf{X}^{(\varepsilon)}(z) be a matrix as in Theorem 4.1. Then there exist some positive constants c1c_{1} and c2c_{2} depending on KK, r0r_{0}, η0\eta_{0} such that for any 0<τ≤γn0<\tau\leq\gamma_{n}

where a∧ba\wedge b denotes the minimum of aa and bb.

Assume at first that nδnpn≤1/3n\delta_{n}p_{n}\leq 1/3. According to Lemma 4.4, we have, for any j=1,…,nj=1,\ldots,n,

Applying Lemma 4.5 with qn=r1δnnpnq_{n}=r_{1}\delta_{n}np_{n}, we get

Consider now the case nδnpn≥1/3n\delta_{n}p_{n}\geq 1/3. According to Lemma 4.4, we have

Applying Lemma 4.5 with qn=r1δnnpnq_{n}=r_{1}\delta_{n}np_{n}, we get

For l=2,…,Ll=2,\ldots,L assume that δi,ρn,i\delta_{i},\rho_{n,i} have been already determined for i=1,…,l−1i=1,\ldots,l-1. Then there exist absolute constants c^l>0\widehat{c}_{l}>0 and c‾l>0\overline{c}_{l}>0 and δl>0\delta_{l}>0 such that

where C^l:=Cl−1∩Comp(δlpn,l,ρn,l)\widehat{\mathcal{C}}_{l}:=\mathcal{C}_{l-1}\cap\mathit{Comp}(\delta_{l}p_{n,l},\rho_{n,l}).

There exists some absolute constant c>0c>0 that

Note that pn,l−1=O(n1−lθ)p_{n,l}^{-1}=\mathcal{O}(n^{1-l\theta}). This implies that

According to Lemmas 4.1 and 4.2, we have ρn1−1=O(n(3−θ)/2)\rho_{n1}^{-1}=\mathcal{O}(n^{({3-\theta})/2}). After simple calculations we get

Proof of Lemma 4.7 To prove of this lemma we may use arguments similar to those in the proofs of Lemmas 2.6 and 3.3 in RV . From x∈Cl\mathbf{x}\in\mathcal{C}_{l} it follows that x∈Incomp(δl−1pn,l−1,ρn,l−1)\mathbf{x}\in\mathit{Incomp}(\delta_{l-1}p_{n,l-1},\rho_{n,l-1}). Applying Lemma 4.6 with δn=pn,l−1\delta_{n}=p_{n,l-1} and ρn=ρn,l−1\rho_{n}=\rho_{n,l-1}, we get

Inequality (4) and Lemma 4.2 together imply

with δl\delta_{l} defined in Lemma 4.2 and

The next lemma gives an estimate of small ball probabilities adapted to our case.

Let x∈Incomp(δ,ρn,L)\mathbf{x}\in\mathit{Incomp}(\delta,\rho_{n,L}). Let X1,…,XnX_{1},\ldots,X_{n} be random variables with zero mean and variance at least 1. Assume that the following condition holds:

Then there exist some constants C>0C>0 depending on δ\delta such that for every ε>0\varepsilon>0

Put L1:=[−log⁡2(ρn,L2δ)]L_{1}:=[-\log_{2}(\rho_{n,L}\sqrt{2\delta})]. Note that

According to Remark 4.9, we have ρn,L≥cn−L/2\rho_{n,L}\geq cn^{-L/2}. This implies L1≤Cln⁡nL_{1}\leq C\ln n. Let σ(x)\sigma(\mathbf{x}) denote the spread set of the vector x\mathbf{x}, that is,

We divide the spread interval of the vector x\mathbf{x} into L1+2L_{1}+2 intervals Δl\Delta_{l}, l=0,…,L1+1l=0,\ldots,L_{1}+1 by

Note that there exists an l0∈{0,…,L1+1}l_{0}\in\{0,\ldots,L_{1}+1\} such that

Let y=PΔl0x\mathbf{y}=P_{\Delta_{l_{0}}}\mathbf{x}. Put al:=min⁡k∈Δl∣xk∣a_{l}:={\min_{k\in\Delta_{l}}}|x_{k}| and bl:=max⁡k∈Δl∣xk∣b_{l}:=\max_{k\in\Delta_{l}}|x_{k}|. Choose a constant MM such that L(M)≤1/2L(M)\leq 1/2. By the properties of concentration functions, we have

By definition of Δl0\Delta_{l_{0}}, we have

and introduce for a random variable ξ\xi, ξ~:=ξ−ξ^\widetilde{\xi}:=\xi-\widehat{\xi} where ξ^\widehat{\xi} denotes an independent copy of ξ\xi. Put ξk:=xkεkXk\xi_{k}:=x_{k}\varepsilon_{k}X_{k}. We use the following inequality for a concentration function of a sum of independent random variables:

with λk≤Mbl0\lambda_{k}\leq Mb_{l_{0}}. See Petrov Petrov75 , page 43, Theorem 3. Put λk=M∣xk∣\lambda_{k}=M|x_{k}|. It is straightforward to check that

Combining this inequality with (164) and (160) we obtain

Let X1,X2,…,Xn\mathbf{X}_{1},\mathbf{X}_{2},\ldots,\mathbf{X}_{n} denote the columns of npnX(ε)(z)\sqrt{np_{n}}\mathbf{X}^{(\varepsilon)}(z), and let Hk\mathcal{H}_{k} denotes the span of all column vectors except the kkth. Then for every δ,ρ∈(0,1)\delta,\rho\in(0,1) and every η>0\eta>0 one has

For the upper bound of the r.h.s. of (4) (see RV , proof of Lemma 3.5). For the reader’s convenience we repeat this proof. Introduce the matrix G:=npnX(ε)(z)\mathbf{G}:=\sqrt{np_{n}}\mathbf{X}^{(\varepsilon)}(z). Recall that X1,…,Xn\mathbf{X}_{1},\ldots,\mathbf{X}_{n} denote the column vector of the matrix G\mathbf{G} and Hk\mathcal{H}_{k} denotes the span of all column vectors except the kkth. Writing Gx=∑k=1nxkXk\mathbf{G}\mathbf{x}=\sum_{k=1}^{n}x_{k}\mathbf{X}_{k}, we have

Denote by UU the event that the set σ1:={k\dvtxdist⁡(Xk,Hk)≥ηρn,L/n}\sigma_{1}:=\{k\dvtx\operatorname{dist}(\mathbf{X}_{k},H_{k})\geq\eta\rho_{n,L}/\sqrt{n}\} contains more than (1−δL)n(1-\delta_{L})n elements. Then by Chebyshev’s inequality

On the other hand, for every incompressible vector x\mathbf{x}, the set σ2(x):={k\dvtx∣xk∣≥ρn,L/n}\sigma_{2}(\mathbf{x}):=\{k\dvtx|x_{k}|\geq\rho_{n,L}/\sqrt{n}\} contains at least nδLn\delta_{L} elements. (Otherwise, since∥Pσ2(x)cx∥2≤ρn,L\|P_{\sigma_{2}(\mathbf{x})^{c}}\mathbf{x}\|_{2}\leq\rho_{n,L}, we have ∥x−y∥2≤ρn,L\|\mathbf{x}-\mathbf{y}\|_{2}\leq\rho_{n,L} for the sparse vector y:=Pσ2(x)x\mathbf{y}:=P_{\sigma_{2}(\mathbf{x})}\mathbf{x}, which would contradict the incompressibility of x\mathbf{x}.)

Assume that the event UU occurs. Fix any incompressible vector x\mathbf{x}. Then ∣σ1∣+∣σ2(x)∣>(1−δL)n+nδL>n|\sigma_{1}|+|\sigma_{2}(\mathbf{x})|>(1-\delta_{L})n+n\delta_{L}>n, so the sets σ1\sigma_{1} and σ2(x)\sigma_{2}(\mathbf{x}) have nonempty intersection. Let k∈σ1∩σ2(x)k\in\sigma_{1}\cap\sigma_{2}(\mathbf{x}). Then by (169) and by definitions of the sets σ1\sigma_{1} and σ2(x)\sigma_{2}(\mathbf{x}), we have

We now reformulate Lemma 3.6 from RV . Let Xn∗\mathbf{X}_{n}^{*} be any unit vector orthogonal to X1,…,Xn−1\mathbf{X}_{1},\ldots,\mathbf{X}_{n-1}. Consider the subspace Hn=span⁡(X1,…,Xn−1)\mathcal{H}_{n}=\operatorname{span}(\mathbf{X}_{1},\ldots,\mathbf{X}_{n-1}).

Let δl,ρl,cl\delta_{l},\rho_{l},c_{l}, l=1,…,L−1l=1,\ldots,L-1, be as in Lemma 4.2 and δL\delta_{L}, ρL,c‾L\rho_{L},\overline{c}_{L} as in Lemma 4.7. Then there exists an absolute constant c^L>0\widehat{c}_{L}>0 such that

The event {X∗∉CL\mboxand∥X(ε)(z)∥≤Kn}\{\mathbf{X}^{*}\notin{\mathcal{C}}_{L}\mbox{ and }\|\mathbf{X}^{(\varepsilon)}(z)\|\leq K_{n}\} implies that the event

occurs for any positive cc. This implies, for c>0c>0,

Now choose c:=min⁡{γn,l,l=1,…,L−1}c:=\min\{\gamma_{n,l},l=1,\ldots,L-1\}. Applying Lemma 4.7 proves the claim.

Let X(ε)(z)\mathbf{X}^{(\varepsilon)}(z) be a random matrix as in Theorem 1.2. Let X1,…,Xn\mathbf{X}_{1},\ldots,\mathbf{X}_{n} denote column vectors of the matrix npnX(ε)(z)\sqrt{np_{n}}\mathbf{X}^{(\varepsilon)}(z), and consider the subspace Hn=span⁡(X1,…,Xn−1)\mathcal{H}_{n}=\operatorname{span}(\mathbf{X}_{1},\ldots,\mathbf{X}_{n-1}). Let Kn=KnpnK_{n}=Kn\sqrt{p_{n}}. Then we have

We repeat Rudelson and Vershynin’s proof of Lemma 3.8 in RV . Let X∗\mathbf{X}^{*} be any unit vector orthogonal to X1,X2,…,Xn−1\mathbf{X}_{1},\mathbf{X}_{2},\ldots,\mathbf{X}_{n-1}. We can choose X∗\mathbf{X}^{*} so that it is a random vector that depends on X1,X2,…,Xn−1\mathbf{X}_{1},\mathbf{X}_{2},\ldots,\mathbf{X}_{n-1} only and is independent of Xn\mathbf{X}_{n}. We have

We denote the probability with respect to Xn\mathbf{X}_{n} by Pr⁡n{\Pr}_{n} and the expectation with respect to X1,…,Xn−1\mathbf{X}_{1},\ldots,\mathbf{X}_{n-1} by E1,…,n−1\mathbf{E}_{1,\ldots,n-1}. Then

According to Lemma 4.10, the second term in the right-hand side of the last inequality is less then exp⁡{−c^Ln}\exp\{-\widehat{c}_{L}n\}. Since the vectors X∗=(a1,…,an)∈S(n−1)\mathbf{X}^{*}=(a_{1},\ldots,a_{n})\in\mathcal{S}^{(n-1)} and Xn=(ε1ξ1,…,εnξn)\mathbf{X}_{n}=(\varepsilon_{1}\xi_{1},\ldots,\varepsilon_{n}\xi_{n}) are independent, we may use small ball probability estimates. We have

By Lemma 4.8, we have for some absolute constant C>0C>0

Let X(ε)(z)\mathbf{X}^{(\varepsilon)}(z) be a random matrix as in Theorem 4.1. Let δL,ρn,L∈(0,1)\delta_{L},\rho_{n,L}\in(0,1). Let X1,…,Xn\mathbf{X}_{1},\ldots,\mathbf{X}_{n} denote column vectors of matrix npnX(ε)(z)\sqrt{np_{n}}\mathbf{X}^{(\varepsilon)}(z). Let Kn=KnpnK_{n}=Kn\sqrt{p_{n}} with K≥1K\geq 1. Then we have

Applying Lemma 4.9 with η=pn\eta=\sqrt{p_{n}}, we get

Thus the lemma is proved. {pf*}Proof of Theorem 4.1 By definition of the minimal singular value, we have

Furthermore, using the decomposition of the sphere S(n−1)=⋃l=1L−1C^l∪CL\mathcal{S}^{(n-1)}=\bigcup_{l=1}^{L-1}\widehat{\mathcal{C}}_{l}\cup\mathcal{C}_{L} into compressible and incompressible vectors, we get

The last two inequalities together imply the result.

To relax the condition pn−1=O(n1−θ)p_{n}^{-1}=\mathcal{O}(n^{1-\theta}) of Theorem 4.1 to pn−1=o(n/ln⁡2n)p_{n}^{-1}=o(n/\ln^{2}{n}) we should put L=ln⁡nL=\ln{n}. Then the value L1L_{1} in Lemma 4.8 is at most C(ln⁡n)2C(\ln{n})^{2}, and hence we get the bound Cln⁡n/npnC\ln{n}/\sqrt{np_{n}} in (153). This yields the bound Cln⁡n/npn+exp⁡{−c^Ln}C\ln{n}/\sqrt{np_{n}}+\exp\{-\widehat{c}_{L}n\} in (4). Thus Theorem 4.1 holds with BB chosen to be of order Cln⁡nC\ln{n}.

Proof of the main theorem

According to Theorem 4.1 with ε=c\varepsilon=c, we have

According to Lemma .2 with q=18q=18, we have

Let r=r(n)r=r(n) be such that r(n)→0r(n)\to 0 as n→∞n\to\infty. A more specific choice will be made later. Consider the potential Uμn(r)U_{\mu_{n}}^{(r)}. We have

where IAI_{A} denotes an indicator function of an event AA and Ωn(z)c{\Omega_{n}(z)}^{c} denotes the complement of Ωn(z)\Omega_{n}(z).

Assuming the conditions of Theorem 4.1, for rr such that

Applying Cauchy’s inequality, we get, for any τ>0\tau>0,

Furthermore, since ξ\xi is uniformly distributed in the unit disc and independent of λj\lambda_{j}, we may write

Since for any b>0b>0, the function −ublog⁡u-u^{b}\log u is not decreasing on the interval [0,exp⁡{−1b}][0,\exp\{-\frac{1}{b}\}], we have for 0<u≤ε<exp⁡{−1b}0<u\leq\varepsilon<\exp\{-\frac{1}{b}\},

Using this inequality, we obtain, for b(1+τ)<2b(1+\tau)<2,

If we choose ε=r\varepsilon=r, then we get

The following bound holds for 1n∑j=1nEJ3(j)\frac{1}{n}\sum_{j=1}^{n}\mathbf{E}J_{3}^{(j)}. Note that ∣log⁡x∣1+τ≤ε2×\break∣log⁡ε∣1+τx2|{\log x}|^{1+\tau}\leq\varepsilon^{2}\times\break|{\log\varepsilon}|^{1+\tau}x^{2} for x≥1εx\geq\frac{1}{\varepsilon} and sufficiently small ε\varepsilon. Using this inequality, we obtain

Furthermore, inequalities (188), (190), (5) and (196) together imply

We choose τ=18\tau=18 and rewrite the last inequality as follows:

If we choose r=1npnr=\frac{1}{\sqrt{np_{n}}} we obtain log⁡(1/r)((φ(npn))−1/19→0\log(1/r)((\varphi(\sqrt{np_{n}}))^{-{1}/{19}}\to 0, then (189) holds and the lemma is proved.

We shall investigate U‾μn(r)\overline{U}{}^{(r)}_{\mu_{n}} now. We may write

where F‾n(ε)(⋅,z,r)\overline{F}{}^{(\varepsilon)}_{n}(\cdot,z,r) is the distribution function corresponding to the restriction of the measure νn(ε)(⋅,z,r)\nu_{n}^{(\varepsilon)}(\cdot,z,r) to the set Ωn(z)\Omega_{n}(z). Introduce the notation

Note that, for any r>0r>0, ∣sj(ε)(z)−sj(ε)(z,r)∣≤r|s_{j}^{(\varepsilon)}(z)-s_{j}^{(\varepsilon)}(z,r)|\leq r. This implies that

Since the distribution function F(x,z)F(x,z) has a density p(x,z)p(x,z) which is bounded (see Remark 3.1) we obtain

Choose r=1npnr=\frac{1}{\sqrt{np_{n}}}. Inequalities (203) and (53) together imply

From inequalities (204) and (200) it follows that

Furthermore, let μ‾n(r){\overline{\mu}}_{n}^{(r)} and μ^n(r){\widehat{\mu}}_{n}^{(r)} be probability measures supported on the compact set KK and K(c)K^{(c)}, respectively, such that

Introduce the logarithmic potential of the measure μ‾n(r){\overline{\mu}}_{n}^{(r)},

Similar to the proof of Lemma 5.1 we show that

in the weak topology. Inequality (205) and relations (206) and (206) together imply that

in the weak topology. Finally, by Lemma 1.1 we get

in the weak topology. Thus Theorem 1.2 is proved.

Appendix

In this appendix we collect some technical results.

Recall that ∣λ1(ε)∣≥⋯≥∣λn(ε)∣|\lambda_{1}^{(\varepsilon)}|\geq\cdots\geq|\lambda_{n}^{(\varepsilon)}| denote the eigenvalues of the matrix X(ε)\mathbf{X}^{(\varepsilon)} ordered via decreasing absolute values, and let s1(ε)≥⋯≥sn(ε)s_{1}^{(\varepsilon)}\geq\cdots\geq s_{n}^{(\varepsilon)} denote the singular values of the matrix X(ε)\mathbf{X}^{(\varepsilon)}.

Under condition of Theorem 1.1 for sufficiently large K≥1K\geq 1 we have

Assume that max⁡j,kE∣Xjk∣2φ(Xjk)≤C\max_{j,k}\mathbf{E}|X_{jk}|^{2}\varphi(X_{jk})\leq C with φ(x):=(ln⁡(1+∣x∣))q\varphi(x):=(\ln(1+|x|))^{q}, q≥7q\geq 7, and Δn:=sup⁡x∣Fn(ε)(x,z)−F(x,z)∣\Delta_{n}:=\sup_{x}|F_{n}^{(\varepsilon)}(x,z)-F(x,z)|. Then there exists some absolute positive constant RR such that

where k1:=[Δn(q+6)/(2q)nln⁡n]k_{1}:=[\Delta_{n}^{{(q+6)}/{(2q)}}n\ln{n}].

Let us introduce k0:=[Δn(q+6)/(2q)n]k_{0}:=[\Delta_{n}^{{(q+6)}/{(2q)}}n]. Using Chebyshev’s inequality we obtain, for sufficiently large R>0R>0,

Furthermore, for any value R1≥1R_{1}\geq 1, splitting into the events sk0(ε)>Rs_{k_{0}}^{(\varepsilon)}>R and sk0(ε)≤Rs_{k_{0}}^{(\varepsilon)}\leq R, we get

Now choose R1:=R2R_{1}:=R^{2}. Thus, since k1/k0∼ln⁡nk_{1}/k_{0}\sim\ln n,

Taking into account Lemma .1 and inequality (53) we obtain

for some positive constant C>0C>0, thus proving the lemma.

Let ϰ=max⁡j,kE∣Xjk∣2φ(Xjk)\varkappa=\max_{j,k}\mathbf{E}|X_{jk}|^{2}\varphi(X_{jk}). The following inequality holds:

Since the function ∣x∣/φ(x)|x|/\varphi(x) not decreasing, it follows from inequality (2) that

Let μn\mu_{n} be the empirical spectral measure of the matrix X\mathbf{X} and νr\nu_{r} be the uniform distribution on the disc of radius rr. Let μn(r)\mu_{n}^{(r)} be the empirical spectral measure of the matrix X(r)=X−rξI\mathbf{X}(r)=\mathbf{X}-r\xi\mathbf{I}, where ξ\xi is a random variable which is uniformly distributed on the unit disc. Then the measure Eμn(r)\mathbf{E}\mu_{n}^{(r)} is the convolution of the measures Eμn\mathbf{E}\mu_{n} and νr\nu_{r}, that is,

Let JJ be a random variable which is uniformly distributed on the set {1,…,n}\{1,\ldots,n\}. Let λ1,…,λn\lambda_{1},\ldots,\lambda_{n} be the eigenvalues of the matrix X\mathbf{X}. Then λ1+rξ,…,λn+rξ\lambda_{1}+r\xi,\ldots,\lambda_{n}+r\xi are eigenvalues of the matrix X(r)\mathbf{X}(r). Let δx\delta_{x} be denote the Dirac measure. Then

Denote by μnj\mu_{nj} the distribution of λj\lambda_{j}. Then

Denote by h(t,v)h(t,v) the characteristic function of the joint distribution of the real and imaginary parts of ξ\xi,

If for any t,vt,v there exists lim⁡n→∞fn(t,v)\lim_{n\to\infty}f_{n}(t,v), then

The first equality follows immediately from the independence of the random variable ξ\xi and the matrix X\mathbf{X}. Since lim⁡r→0h(rt,rv)=h(0,0)=1\lim_{r\to 0}h(rt,rv)=h(0,0)=1 the first equality implies the second one.

Let FF and GG be distribution functions with Stieltjes transforms SF(z)S_{F}(z) and SG(z)S_{G}(z), respectively. Assume that ∫−∞∞∣F(x)−G(x)∣ dx<∞\int_{-\infty}^{\infty}|F(x)-G(x)|\,dx<\infty. Let G(x)G(x) have a bounded support JJ and density bounded by some constant KK. Let V>v0>0V>v_{0}>0 and aa be positive numbers such that

Then there exist some constants C1,C2,C3C_{1},C_{2},C_{3} depending on JJ and KK only such that

Let XjkX_{jk}, 1≤j,k≤n1\leq j,k\leq n, be independent complex random variables with EXj,k=0\mathbf{E}X_{j,k}=0 and E∣Xj,k∣2=1\mathbf{E}|X_{j,k}|^{2}=1. Assume furthermore that

Then we have, for some positive r0r_{0} and η0\eta_{0},

First we note, that there exists a positive number MM such that

Let η0{\eta_{0}} be a small positive number. For ∣u∣>M+η0|u|>M+{\eta_{0}} we have

Consider now ∣u∣≤M+η0|u|\leq M+{\eta_{0}}. Then

Combining inequalities (The largest singular value) and (The largest singular value) we obtain the claim.

Acknowledgments

The authors would like to thank Terence Tao for drawing their attention to a gap in a previous version of the paper and Dmitry Timushev for a careful reading of this manuscript.

References