The circular law for random matrices
Friedrich Götze, Alexander Tikhomirov
Introduction
Let , be complex random variables with and . For a fixed , denote by the eigenvalues of the matrix
and define its empirical spectral distribution function by
The main result of our paper is the following:
Let denote the function , , arbitrary, small and fixed. Let , denote independent complex random variables with
Then converges weakly to the distribution function as .
We shall prove the same result for the following class of sparse matrices. Let , , denote a triangular array of Bernoulli random variables (taking values only) which are independent in aggregate and independent of with common success probability depending on . Consider the sequence of matrices . Let denote the (complex) eigenvalues of the matrix and denote by the empirical spectral distribution function of the matrix , that is,
For define . Let , denote independent complex random variables with
Assume that there is a such that as . Then converges weakly to the distribution function as .
The crucial problem of the proofs of Theorems 1.1 and 1.2 is to bound the smallest singular values , respectively, of the shifted matrices , respectively, . (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 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 only. Restricting oneself to the set for the investigation of the smallest singular values, the inequality follows from the results of Rudelson rud06 without the assumption of sub-Gaussian tails for the matrix . 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 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 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 denote a random variable uniformly distributed over the unit disc and independent of the matrix . For any , consider the matrix
where denotes the identity matrix of order . Let (resp., ) be empirical spectral measure of matrix (resp., ) 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 as
where are the eigenvalues of the matrix . Note that the expected spectral measure is the convolution of the measure and the uniform distribution on the disc of radius (see Lemma .4 in the Appendix for details).
Assume that the sequence converges weakly to a measure as and . Then
Let be a random variable which is uniformly distributed on the set and independent of the matrix . We may represent the measure as the distribution of a random variable where and are independent. Computing the characteristic function of this measure and passing first to the limit with respect to and then with respect to (see also Lemma .5 in the Appendix), we conclude the result.
Now we may fix and consider the measures . They have bounded densities. Assume that the measures have supports in a fixed compact set and that converges weakly to a measure . 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 denote the singular values of the matrix .
Since the sequence of measures is weakly relatively compact. These results imply that for any we may restrict the measures to some compact set such that . Moreover, Lemma .2 implies the existence of a compact such that. If we take some subsequence of the sequence of restricted measures which converges to some measure , then, , and . If we prove that exists and is equal to the logarithmic potential corresponding the uniform distribution on the unit disc [see Section 3, equality (68)], then the sequence of measures weakly converges to the uniform distribution on the unit disc. Moreover, it is enough to prove that for some sequence , .
Furthermore, let denote the singular values of matrix . We shall investigate the logarithmic potential . Using elementary properties of singular values (see, e.g., Goh69 , Lemma 3.3, page 35), we may represent the function as follows:
where denotes the expected spectral measure of the matrix, which is the expectation of the counting measure of the set of eigenvalues of the matrix .
In Section 2 we investigate convergence of the measure . In Section 3 we study the properties of the limit measures . 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 based on techniques developed in Rudelson rud06 and Rudelson and Vershynin RV . The bounds of minimal singular values of matrices 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 and or (without indices) some general absolute constant which may be changed from line to line. To specify a constant we shall use subindices. By we shall denote the indicator of an event . For any matrix we denote the Frobenius norm by , and we denote by the operator norm.
Denote by the distribution function of the measure , that is,
where denote the singular values of the matrix . For a positive random variable and a Rademacher random variable (r.v.) consider the transformed r.v. . If has distribution function , the variable has distribution function , given by
for all real . Note that this induces a one-to-one corresponds between the respective measures and . The limit distribution function of as , is denoted by . The corresponding symmetrization is the limit of as . We have
Denote by [resp., ] and [resp., ] the Stieltjes transforms of the measures [resp., ] and [resp., ] correspondingly. Then we have
As shown in Bai Bai1997 , the measure has a density with bounded support. More precisely, . Thus the measure has bounded support and bounded density .
Let , . Assume for some function such that as and such that the function is nondecreasing we have
Let , , and
To bound the distance between the distribution functions and we investigate the distance between their the Stieltjes transforms. Introduce the Hermitian matrix
where denotes matrix with zero entries. Using the inverse of the partial matrix (see, e.g., HoJohn91 , Chapter 08, page 18) it follows that, for , ,
where and denotes the unit matrix of order . By definition of , we have
Set . 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 and let , , then
Using this representation and the resolvent equality, we get
Here, and in what follows, we omit the arguments and in the notation of resolvent matrices. For any vector , let denote the transposed vector . Applying the resolvent equality again, we obtain
Applying this notation to equality (2) and taking into account that and are independent, we get
From (2) it follows immediately that for any , ,
Since and , equality (2) implies
By definition (2) of , applying standard resolvent properties, we obtain the following bounds, for any ,
For the proof of this inequality see Lemma .3 in the Appendix. Using the last inequalities we obtain, that for
Since , we obtain
Note that for any Hermitian random matrix 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
Denote by some generic function with which may vary from line to line. We may now rewrite equality (2) as follows:
We now investigate the functions and . Since the arguments for both functions are similar we provide it for the first one only. By definition of the matrix , we have
Using the resolvent equality (2) and Lemma .3, we get, for
Inequalities (28) and (29) together imply, for ,
where .
Furthermore, we prove the following simple lemma.
Let , . Let satisfy the equation
and . Then the inequality
For with , the Stieltjes transform satisfies the following equation:
Comparing the imaginary parts of both sides of this equation, we get
Since and , it follows that
Equality (40) and the last remark together imply
To compare the functions and 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 and .
Note that and satisfy the equations
respectively. These equations together imply
Applying inequality , we get
The last inequality and Lemmas 2.2 and 2.3 together imply
To bound the distance between the distribution function and the distribution function corresponding the Stieltjes transforms and we use Corollary 2.3 from GT03 . In the next lemma we give an integral bound for the distance between the Stieltjes transforms and .
For the inequality
It follows from here that and
for . Lemma 2.4 implies that it is enough to prove the inequality
where . By definition of , we have
Furthermore, representation (35) implies that
It follows from relation (32) that for ,
The last two inequalities together imply that for sufficiently large and ,
Inequalities (47), (45) and the definition of together imply
If we choose such that we obtain
In Section 3 we show that the measure has bounded support and bounded density for any . To bound the distance between the distribution functions and we may apply Corollary 3.2 from GT03 (see also Lemma .6 in the Appendix). We take and . 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 . At first note that there exists a solution of the equation
Set and consider equation (41) on the real line
It is straightforward to check that and for and for , and for .
In the case equation (57) has one real root for and three real roots for . In the case equation (57) has one real root for and three real roots for or for .
This implies that, for and for
equation (57) has one real root. Furthermore, direct calculations show that
Solving the equation with respect to , we get for and
and for and
These relations imply that for the function has three real roots for and one real root for .
Consider the case now. In this case are real for all and . Note that
for and for and
for . These implies that for and for the function has one real root and for or for the function has three real roots. The lemma is proved.
From Lemma 3.1 it follows that the measure has a density and:
for , if , then ;
for , if or , then ;
It is well known that for the logarithmic potential of uniform distribution on the unit disc is
According to Lemma 4.4 in Bai Bai1997 , we have, for ,
According to Remark 3.1, we have, for ,
Comparing equalities (63) and (61) and using relation (65), we obtain
The smallest singular value
Let be an matrix with independent entries , . Assume that and and let denote Bernoulli random variables with , . Denote by the singular values of the matrix . In this section we prove a bound for the minimal singular value of the matrices . We prove the following result.
Let , be independent random complex variables with and , which are uniformly integrable, that is,
Let be i.i.d. random variables with and . Then condition (69) holds.
Consider the event that there exists at least one row with zero entries only. Its probability is given by
Simple calculations show that if for all , then
Hence in the case and 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 in the -dimensional unit sphere such that is extremely small into two classes, called compressible and incompressible vectors.
We develop our approach for shifted sparse and normalized matrices . The generalization to the case of complex sparse and shifted matrices 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 to . The quantity in Theorem 4.1 should be of order in this case. See Remark 4.9 for details.
Let be a fixed unit vector and be a matrix as in Theorem 4.1. Then there exist some positive absolute constants and such that for any
Recall that and . Assume first that are real independent r.v. with mean zero, and variance at least . Let with independent Bernoulli variables which are independent of in aggregate and let . Assume also that is a real vector. Then
Using , where is a standard Gaussian random variable, we obtain
where , , denote i.i.d. standard Gaussian r.v.s and denotes expectation with respect to conditional on all other r.v.s. For every and the following inequality holds:
(see Bickel , inequality (3.7)). Take for some absolute positive constant which will be chosen later. Then it follows from (4) that
where . Assuming (69), choose a constant such that
Since for , conditioning on the event , we get for
It follows from (4) for and for some constant
This implies that conditionally on and for
Let , , where denotes the standard Gaussian distribution function. It is straightforward to show that
We may choose large enough such that following inequalities hold:
for all . Inequalities (4), (77), (4), (86) together imply that for any
Without loss of generality we may take sufficiently large, such that and choose . Then we obtain
For we conclude from here that for
Inequality (89) implies that inequality (73) holds with some positive constant . This completes the proof in the real case.
Consider now the general case. Let with with and and . In this notation we have
For any we introduce the set as follows:
It is straightforward to check that for any
According to inequality (91), for any , there exists a set such that
Introduce the following random variables for any
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 and to be chosen later we define , and . Without loss of generality we shall assume that
Assume there exist an absolute constant and values such that for any
holds. Then there exists a constant depending on and only such that, for ,
Let to be chosen later. There exists an -net in of cardinality (see, e.g., Lemma 3.4 in rud06 ). By condition (102), we have for
Let be the event that and for some point . Assume that occurs and choose a point such that . Then
Choosing and , we complete the proof.
Following Rudelson and Vershynin RV , we shall partition the unit sphere into the two sets of so-called compressible and incompressible vectors, and we will show the invertibility of on each set separately.
The sets of sparse, compressible and incompressible vectors depending on and will be denoted by
Let be a random matrix as in Theorem 1.2, and let with a constant . Assume there exist an absolute constant and values such that for any
holds. Then there exist that depend on and only, such that
At first we estimate the invertibility for sparse vectors. Let with some positive constant which will be chosen later. According to Proposition 4.6 for any and for any , we have the following inequality:
Using Stirling’s formula, we get for some absolute positive constant
We may choose small enough that
Choose . Let be the event that and for some point . Assume that occurs and choose a point such that . Then
Let . Let . Then there exists a set of cardinality such that
which we shall call “spread set of ” henceforth.
See proof of Lemma 3.4 RV , page 16. For the reader’s convenience we repeat this proof here. Consider the subsets of defined by
If then there exists a set with cardinality such that
We shall now bound this concentration function and prove a tensorization lemma for incompressible vectors.
Let and be some functions of such that . Let and as in Lemma .7. Let . Then there exists positive constants and depending on such that for any we have
Introduce . Since the cardinality of is at least . 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 , we have . Assume that . Then we have
If then and
Let be independent nonnegative random variables. Assume that
for some positive and . Then there exists positive absolute constants and such that
We repeat the proof of Lemma 4.4 in LPRT . Let . For any we have
Choosing and , we get
Recall that we assume . For this fixed consider . Hence by definition for and . We put .
We shall assume that is large enough such that for some constant . Starting with a decomposition of into compressible vectors in , where , , and the constants and are chosen as in Lemmas 4.1 and 4.2, respectively. Then Lemma 4.1 implies inequality (108) with replaced by and replaced by . Hence, using Lemma 4.2, one obtains the claim for the subset of vectors . The remaining vectors in lie in . According to Lemmas 4.4, 4.5 inequality (108) holds again for these vectors but with new parameters and . Thus we may again subdivide the vectors in into the vectors within distance from these sparse ones, that is, and the remaining ones, that is, . Iterating this procedure times we arrive at the incompressible set of vectors where Lemmas 4.4, 4.5 and Proposition 4.6 yield the required bound of order , for a sufficiently small absolute constant .
Summarizing, we will determine iteratively constants , for 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 and let and be a matrix as in Theorem 4.1. Then there exist some positive constants and depending on , , such that for any
where denotes the minimum of and .
Assume at first that . According to Lemma 4.4, we have, for any ,
Applying Lemma 4.5 with , we get
Consider now the case . According to Lemma 4.4, we have
Applying Lemma 4.5 with , we get
For assume that have been already determined for . Then there exist absolute constants and and such that
where .
There exists some absolute constant that
Note that . This implies that
According to Lemmas 4.1 and 4.2, we have . 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 it follows that . Applying Lemma 4.6 with and , we get
Inequality (4) and Lemma 4.2 together imply
with defined in Lemma 4.2 and
The next lemma gives an estimate of small ball probabilities adapted to our case.
Let . Let be random variables with zero mean and variance at least 1. Assume that the following condition holds:
Then there exist some constants depending on such that for every
Put . Note that
According to Remark 4.9, we have . This implies . Let denote the spread set of the vector , that is,
We divide the spread interval of the vector into intervals , by
Note that there exists an such that
Let . Put and . Choose a constant such that . By the properties of concentration functions, we have
By definition of , we have
and introduce for a random variable , where denotes an independent copy of . Put . We use the following inequality for a concentration function of a sum of independent random variables:
with . See Petrov Petrov75 , page 43, Theorem 3. Put . It is straightforward to check that
Combining this inequality with (164) and (160) we obtain
Let denote the columns of , and let denotes the span of all column vectors except the th. Then for every and every 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 . Recall that denote the column vector of the matrix and denotes the span of all column vectors except the th. Writing , we have
Denote by the event that the set contains more than elements. Then by Chebyshev’s inequality
On the other hand, for every incompressible vector , the set contains at least elements. (Otherwise, since, we have for the sparse vector , which would contradict the incompressibility of .)
Assume that the event occurs. Fix any incompressible vector . Then , so the sets and have nonempty intersection. Let . Then by (169) and by definitions of the sets and , we have
We now reformulate Lemma 3.6 from RV . Let be any unit vector orthogonal to . Consider the subspace .
Let , , be as in Lemma 4.2 and , as in Lemma 4.7. Then there exists an absolute constant such that
The event implies that the event
occurs for any positive . This implies, for ,
Now choose . Applying Lemma 4.7 proves the claim.
Let be a random matrix as in Theorem 1.2. Let denote column vectors of the matrix , and consider the subspace . Let . Then we have
We repeat Rudelson and Vershynin’s proof of Lemma 3.8 in RV . Let be any unit vector orthogonal to . We can choose so that it is a random vector that depends on only and is independent of . We have
We denote the probability with respect to by and the expectation with respect to by . Then
According to Lemma 4.10, the second term in the right-hand side of the last inequality is less then . Since the vectors and are independent, we may use small ball probability estimates. We have
By Lemma 4.8, we have for some absolute constant
Let be a random matrix as in Theorem 4.1. Let . Let denote column vectors of matrix . Let with . Then we have
Applying Lemma 4.9 with , 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 into compressible and incompressible vectors, we get
The last two inequalities together imply the result.
To relax the condition of Theorem 4.1 to we should put . Then the value in Lemma 4.8 is at most , and hence we get the bound in (153). This yields the bound in (4). Thus Theorem 4.1 holds with chosen to be of order .
Proof of the main theorem
According to Theorem 4.1 with , we have
According to Lemma .2 with , we have
Let be such that as . A more specific choice will be made later. Consider the potential . We have
where denotes an indicator function of an event and denotes the complement of .
Assuming the conditions of Theorem 4.1, for such that
Applying Cauchy’s inequality, we get, for any ,
Furthermore, since is uniformly distributed in the unit disc and independent of , we may write
Since for any , the function is not decreasing on the interval , we have for ,
Using this inequality, we obtain, for ,
If we choose , then we get
The following bound holds for . Note that for and sufficiently small . Using this inequality, we obtain
Furthermore, inequalities (188), (190), (5) and (196) together imply
We choose and rewrite the last inequality as follows:
If we choose we obtain , then (189) holds and the lemma is proved.
We shall investigate now. We may write
where is the distribution function corresponding to the restriction of the measure to the set . Introduce the notation
Note that, for any , . This implies that
Since the distribution function has a density which is bounded (see Remark 3.1) we obtain
Choose . Inequalities (203) and (53) together imply
From inequalities (204) and (200) it follows that
Furthermore, let and be probability measures supported on the compact set and , respectively, such that
Introduce the logarithmic potential of the measure ,
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 denote the eigenvalues of the matrix ordered via decreasing absolute values, and let denote the singular values of the matrix .
Under condition of Theorem 1.1 for sufficiently large we have
Assume that with , , and . Then there exists some absolute positive constant such that
where .
Let us introduce . Using Chebyshev’s inequality we obtain, for sufficiently large ,
Furthermore, for any value , splitting into the events and , we get
Now choose . Thus, since ,
Taking into account Lemma .1 and inequality (53) we obtain
for some positive constant , thus proving the lemma.
Let . The following inequality holds:
Since the function not decreasing, it follows from inequality (2) that
Let be the empirical spectral measure of the matrix and be the uniform distribution on the disc of radius . Let be the empirical spectral measure of the matrix , where is a random variable which is uniformly distributed on the unit disc. Then the measure is the convolution of the measures and , that is,
Let be a random variable which is uniformly distributed on the set . Let be the eigenvalues of the matrix . Then are eigenvalues of the matrix . Let be denote the Dirac measure. Then
Denote by the distribution of . Then
Denote by the characteristic function of the joint distribution of the real and imaginary parts of ,
If for any there exists , then
The first equality follows immediately from the independence of the random variable and the matrix . Since the first equality implies the second one.
Let and be distribution functions with Stieltjes transforms and , respectively. Assume that . Let have a bounded support and density bounded by some constant . Let and be positive numbers such that
Then there exist some constants depending on and only such that
Let , , be independent complex random variables with and . Assume furthermore that
Then we have, for some positive and ,
First we note, that there exists a positive number such that
Let be a small positive number. For we have
Consider now . 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.