On the Smallest Eigenvalue of General correlated Gaussian Matrices
aa r X i v : . [ c s . I T ] D ec On the Smallest Eigenvalue of General correlatedGaussian Matrices
Abla Kammoun,
Member, IEEE and Mohamed-Slim Alouini,
Fellow, IEEE,
Abstract —This paper investigates the behaviour of the spec-trum of generally correlated Gaussian random matrices whosecolumns are zero-mean independent vectors but have differentcorrelations, under the specific regime where the number of theircolumns and that of their rows grow at infinity with the samepace. This work is, in particular, motivated by applications fromstatistical signal processing and wireless communications, wherethis kind of matrices naturally arise. Following the approachproposed in [1], we prove that under some specific conditions, thesmallest singular value of generally correlated Gaussian matricesis almost surely away from zero.
I. I
NTRODUCTION
Let Σ n be a rectangular random matrix of size N × n .The study of the behaviour of the asymptotic spectrum of Σ n when N, n → + ∞ has been investigated in severalworks. As is known, when the elements of Σ n are zero-mean and unit variance independent and identically distributed(i.i.d.) and Nn → c < , the empirical measure of theeigenvalues of n Σ n Σ ∗ n converge weakly to a deterministicprobability distribution which is supported by the interval (cid:2) (1 −√ c ) , (1+ √ c ) (cid:3) [2]. A question which immediatelyarises in connection with this result concerns the asymptoticbehaviour of the extreme singular values. At first sight, onewould expect the smallest and the largest eigenvalues of n Σ n Σ ∗ n to converge to (1 −√ c ) and (1+ √ c ) , respectively.While this statement is correct, it cannot be directly inferredfrom the aforementioned weak convergence result. As a matterof fact, the proof generally requires the use of more advancedtechniques improving the weak convergence result. First find-ings related to these issues can be traced back to the worksof J. Silverstein [3] and S. Geman [4], who provided a rig-orous proof showing that the extreme eigenvalues of n Σ n Σ ∗ n converge in the Gaussian case to the edges of the limitingsupport (1 −√ c ) and (1+ √ c ) . This result was then extendedto the case of non-Gaussian matrices but with independent andidentically distributed entries [5]. The characterization of thelimiting support of Σ n is much more difficult in the case wherethe column entries of Σ n are correlated. Instead of determiningthe exact support, many works focused on establishing thealmost sure absence of eigenvalues of n Σ n Σ ∗ n in any closedinterval outside the support of the limiting distribution. We cancite, for sake of illustration, the work of [6] applying for thesimple-correlated case where the columns of Σ n are correlatedwith the same correlation matrix and that of [1] which dealswith non-centered uncorrelated models.In many applications, this result, though limited, is essential.It can be, for instance, used to efficiently handle randomquantities involving the Gram matrix n Σ n Σ ∗ n or its inverse. In this paper, we consider the generally correlated Gaussianmodel in which the columns of Σ n are zero-mean independentGaussian random vectors but with different correlations. Firstresults related to this model are due to Wagner et al. [7]who characterize the asymptotic behaviour of the limitingdistribution of n Σ n Σ ∗ n . This result was in particular applied tothe analysis of the performance of the regularized-zero forcinglinear precoding technique [7].Since then, this model has known an increasing popularity,mostly spurred by applications in multi-user nultiple-input-single-output (MISO) systems [8], [9] and the very recentrobust signal processing applications [10]. In what follows,we provide two different applications where the general cor-relation Gaussian model arises. a) Multiple Input Single Output Channel: Consider thedownlink of a single-cell system in which a base station (BS)with N antennas serves n users equipped each with a singleantenna each and assume that N < n . The downlink channelvector h k between the BS and the k th user is given by [7]: h k = R k z k . with z k is a standard complex Gaussian vector and matrix R k is essentially function of the richness of the scattering betweenthe BS and the user of interest and as such is specific foreach user. To mitigate inter-user interference, the BS precodesthe transmitted signal by a matrix G which depends on thechannel conditions for all users. Among the used precodingtechniques, we can cite the Zero-forcing (ZF) precoding givenby [11]: G = (cid:18) n HH ∗ (cid:19) − H , where H = [ h , · · · , h n ] . The ZF precoding involves theinversion of the Gram matrix HH ∗ , a step which becomescritical in case the smallest eigenvalue is near zero. In orderto analyze the performance of using the ZF precoding, theregime under which the number of antennas N and the numberof users n increase with the same pace is often assumed. Theperformance of the ZF precoding under this regime has beenstudied in [7], where it has been assumed that the smallesteigenvalue of n HH ∗ is bounded away from zero for all large N and n . Although this assumption holds true for specificcases where all matrices R k are equal, there is no proofsupporting its validity in general. This is the reason why theauthors in [7] opted to add it as an assumption, which is likelyto always hold true and thus is unnecessary. b) Robust Statistics: Consider a temporal series of n vector observations y , · · · , y n of size N × . Assume that the contribution of each y i can be decomposed as the sum ofa useful signal plus an elliptical noise, i.e, y i = s i + x i , (1)where s , · · · , s n are Gaussian independent N × randomGaussian vectors with covariance R and x i is drawn froma Compound Gaussian distribution, i.e, x i = √ τ i z i , (2)where z i are standard complex Gaussian vectors and τ , · · · , τ n are scalar positive-valued random variables. Weconsider the problem of estimating the covarince matrix of x i .In order to mitigate the impact of the heavy-tailed distributednoise, the use of robust covariance estimates known also asrobust scatter estimates has been proven to be a good solution.These are given as the unique solution of the followingequation: ˆ C N = n X i =1 u ( x ∗ i ˆ C − N x i ) x i x ∗ i , (3)where x : u ( x ) is a scalar functional satisfying certainconditions [12]. In a recent submitted work, we prove thatmatrix ˆ C N converges in the operator norm to ˆ S N where ˆ S N is given by: ˆ S N = n X i =1 v ( δ i ) x i x ∗ i , (4)with δ , · · · , δ n are solutions of some fixed point equations[10]. Conditioning on τ i , matrix S N follows the model ofgenerally correlated Gaussian matrices. The proof in [10] relieson the control of the smallest eigenvalue of ˆ S N .Despite its importance, the generally correlated Gaussianmodel has not been extensively explored, most probablybecause of its recent emergence as a major practical model.Several questions related to the behaviour of the eigenvaluesremain unanswered. A major question, illustrated by the twoexamples above, and which triggered our motivation for thiswork, concerns the control of the smallest eigenvalue of theGram matrix n Σ n Σ ∗ n . Knowing that the smallest eigenvaluestay away of zero in the i.i.d case when N < n , one can expectthe same behaviour to hold for the general Gaussian correlatedcase under probably some mild conditions on the correlationmatrices. In this paper, we provide a rigorous proof for thisstatement by essentially building on the techniques developedby [1].II. P
ROBLEM STATEMENT AND REVIEW OF SOME RESULTS
All along the paper, we consider integers n, N, N such that n ≥ N and N ≥ N . We denote by c N the ratio Nn . We makethe following assumptions: Assumption A-1. < lim inf c N ≤ lim sup c N < . (5)The objective of this paper is to provide some interestingproperties of the spectrum of generally correlated Gaussian matrices, i.e matrices whose columns are zero-mean indepen-dent random vectors but have different covariances. Through-out this paper, matrix Σ n represents the complex-valued N × n matrix given by: Σ n = [ ξ , · · · , ξ n ] , (6)where ξ , · · · , ξ n are assumed to satisfy the following as-sumptions: Assumption A-2. ( ξ i ) ni =1 are zero-mean complex Gaussianvectors of size N × with covariance Θ i where ( Θ i ) ni =1 is asequence of N × N matrices verifying: w min , inf N min ≤ i ≤ n λ ( Ω i ) > , (7) w max , sup N max ≤ i ≤ n λ N ( Ω i ) < + ∞ , (8) where Ω i , Θ i Θ ∗ i and λ ( Ω i ) and λ N ( Ω i ) are the smallestand largest eigenvalues of Ω i . We denote in what follows by λ ≤ · · · ≤ λ N theeigenvalues of n Σ n Σ ∗ n . The empirical eigenvalue distributionof n Σ N Σ ∗ N is defined as: ˆ µ N = 1 N N X k =1 δ λ k . (9)In order to characterize the asymptotic behaviour of ˆ µ N , itis in practice quite common to analyze that of its Stieltjestransform (ST). Since the ST of a positive finite measure µ isgiven by: Ψ µ ( z ) = Z R dµ ( λ ) λ − z , the ST of the empirical eigenvalue distribution in (9) can bewritten as: ˆ m N ( z ) = 1 N N X k =1 λ k − z . (10)Denote by Q N ( z ) = (cid:0) n Σ n Σ ∗ n − z I N (cid:1) − . In the parlance ofrandom matrix theory, Q N ( z ) is referred to as the resolventmatrix. From (10), one can easily see that: ˆ m N ( z ) = 1 N tr Q N ( z ) . (11)Relation (11) clearly establishes the link between the resolventmatrix and the ST of the empirical eigenvalue distribution ˆ µ N . It is a fundamental equation that accounts for the keyrole played by the resolvent matrix in the theory of randommatrices. As a matter of fact, the study of the asymptoticbehaviour of the resolvent matrix has provided an importantload of new results concerning different statistical models [13],[14]. The model of generally correlated random matrices hasrecently been studied in [7], where it has been proven that theST of the empirical eigenvalue distribution converges almostsurely to a deterministic function which is the ST of someprobability distribution. More formally, it is well known from[7], that it exists a sequence of deterministic measures µ N such that ˆ µ N − µ N converges weakly to zero almost surely. Measure µ N is characterized through its ST m N ( z ) which isgiven by: m N ( z ) = 1 N tr n n X i =1 Ω i δ i ( z ) − zI N ! − , where δ , · · · , δ n form the unique solutions that are ST of non-negative finite measure of the following system of equations: δ i ( z ) = 1 n tr Ω i n n X j =1 Ω j δ j ( z ) − zI N − for each z ∈ C \ R + .In the following, we denote by T N , the matrix: T N ( z ) = n n X i =1 Ω i δ i ( z ) − zI N ! − , and m N ( z ) = 1 N tr T N ( z ) . As ˆ µ N − µ N converge to zero weakly almost surely, we have: ˆ m N ( z ) − m N ( z ) a.s. → for each z ∈ C \ R + . III. M AIN RESULTS
In this paper, we prove that under Assumptions 1-2, thesmallest eigenvalue of the Gram matrix n Σ n Σ ∗ n stays awayzero almost surely for N large enough. This in particularimplies, that for some ǫ > , ˆ µ N [0 , ǫ ] = 0 for N large enough.Since ˆ µ N − µ N converges weakly to zero, it is not difficult toconvince oneself that one needs to start by showing that thesupport S N of µ N does not contain . In particular, we provethe following result: Theorem 1.
Under Assumption 1 and 2, / ∈ S N . In particular,there exists ǫ > such that: [0 , ǫ ] ∩S N = ∅ . To avoid disrupting the flow of the article, the proof ofTheorem 1 is deferred to Appendix B.Theorem 1 ensures that does not belong to the supportof the deterministic measure µ N . To conclude, it suffices tosupplement this result with a second one, which establishesthat almost surely, there is no eigenvalue of n Σ n Σ ∗ n thatgoes outside the support of S N . This kind of result has alreadybeen shown to hold for other statistical models, by either usingproperties of the ST and bounds on the moments of martingaledifference sequences [15]–[17] or resorting to tools based onGaussian calculus [1]. Since we assume in this paper that Σ n has Gaussian entries, we rather build on the method of[1] which also originates from some of the ideas of [18]. Inparticular, we establish the following result: Theorem 2.
Assume that there exists a positive quantity ǫ > and two real values a, b ∈ R such that for all N large enough: ] a − ǫ, b + ǫ [ ∩S N = ∅ Then, with probability one, no eigenvalue of n Σ n Σ ∗ n appearsin [ a, b ] for all N large enough.Proof: The following proposition will be crucial in orderto prove Theorem 2. It merely quantifies the error that weincur by replacing E N tr Q ( z ) by N tr T N ( z ) . The proofis quite demanding and heavily relies on Gaussian calculustools. It will be detailed in the corpus of the paper, namely insection IV, since we believe that some intermediate results beof independent interest. Proposition 3. ∀ z ∈ C \ R + , we have for N large enough, E (cid:20) N tr Q ( z ) (cid:21) = 1 N tr T N ( z )+ 1 N χ N ( z ) with χ is analytic on C \ R + and satisfies: | χ N ( z ) | ≤ K ( | z | + C ) k P (cid:16) |ℑ z | − (cid:17) (12) for each z ∈ C + where C, K are constants, k is an integerindependent of N and P is a polynomial with positive coeffi-cients independent of N . Proposition 3 will essentially serve to provide asymptoticapproximates of linear statistics of the eigenvalues of the Grammatrix. In fact, with the help of proposition 3, we prove thefollowing result:
Lemma 4.
Let φ be a compactly supported real-valued smoothfunction defined on R , i.e, φ ∈ C ∞ c ( R , R ) . Then , E (cid:20) φ (cid:18) N Σ n Σ ∗ n (cid:19)(cid:21) − Z S N φ ( λ ) dµ N ( λ ) = O (cid:18) N (cid:19) . (13) Proof:
The proof is built around the use of the inversionlemma of ST. Recall that if m is the ST of some finite measure µ , then for any continuous real function φ with compactsupport in R Z R φ ( λ ) µ ( dλ ) = 1 π ℑ (cid:18) lim y ↓ Z R φ ( x ) m ( x + ıy ) dx (cid:19) . We therefore have: E (cid:20) N tr φ ( 1 n Σ n Σ H n ) (cid:21) = 1 π ℑ (cid:18) lim y ↓ Z R φ ( x ) E (cid:20) N tr Q ( x + ıy ) (cid:21) dx (cid:19)Z S N φ ( λ ) dµ N ( λ ) = 1 π ℑ (cid:18) lim y ↓ Z R φ ( x ) E (cid:20) N tr T N ( x + ıy ) (cid:21) dx (cid:19) . By proposition 3, we get: E (cid:20) N tr φ ( 1 n Σ n Σ H n ) (cid:21) − Z S N φ ( λ ) dµ N ( λ )= 1 N π lim y ↓ ℑ "Z R + φ ( x ) χ N ( x + ıy ) dx . Since the function χ N ( z ) satisfies (12), Theorem 6.2 in [19]implies that: lim sup y ↓ (cid:12)(cid:12)(cid:12)(cid:12)Z R φ ( x ) χ N ( x + ıy ) dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ C < + ∞ . If A = P Ni =1 λ i u i u H i is an eigenvalue decomposition of A , then φ ( A ) = P Ni =1 φ ( λ i ) u i u H i . where C is a constant independent of N , thereby establishing(13).We return now to the proof of Theorem 2. With the aboveresults at hand, Theorem 2 can be shown along the same linesas the proof of Theorem 3 in [1]. The details are provided inthe sequel for sake of completeness. Consider ψ ∈ C ∞ c ( R , R ) satisfying ≤ ψ ≤ and: ψ ( λ ) = (cid:26) for λ ∈ [ a, b ]0 for λ ∈ R \ ] a − ǫ, b + ǫ [ . For N large enough, function ψ is zero in the support S N .Therefore, E (cid:20) N ψ (cid:18) n Σ n Σ H n (cid:19)(cid:21) = O (cid:18) N (cid:19) . We need also to prove that the variance of N ψ ( n Σ n Σ H n ) isof order N : var (cid:20) N ψ (cid:18) n Σ n Σ H n (cid:19)(cid:21) = O (cid:18) N (cid:19) . (14)To establish (18), it suffices to resort to the Nash-Poincar´einequality which is stated in Lemma 7 of the next section.Applying Lemma 7, we obtain: var (cid:18) N tr ψ (cid:18) n Σ n Σ H n (cid:19)(cid:19) ≤ n X k =1 N X s =1 N X r =1 ∂ N tr ψ (cid:0) n Σ n Σ H n (cid:1) ∂ξ s,k [ Ω k ] s,r (cid:2) ∂ N tr ψ (cid:0) n Σ n Σ H n (cid:1)(cid:3) ∗ ∂ξ r,k + n X k =1 N X s =1 N X r =1 ∂ N tr ψ (cid:0) n Σ n Σ H n (cid:1) ∂ξ ∗ s,k [ Ω k ] s,r (cid:2) ∂ N tr ψ (cid:0) n Σ n Σ H n (cid:1)(cid:3) ∗ ∂ξ ∗ r,k . (15)By Lemma 4.6 in [19], we have: ∂ (cid:2) N tr ψ (cid:0) n Σ n Σ H n (cid:1)(cid:3) ∂ξ s,k = (cid:20) N n Σ H n ψ ′ (cid:18) n Σ n Σ H n (cid:19)(cid:21) k,s (16) ∂ (cid:2) N tr ψ (cid:0) n Σ n Σ H n (cid:1)(cid:3) ∂ξ ∗ s,k = (cid:20) N n ψ ′ (cid:18) n Σ n Σ H n (cid:19) Σ n (cid:21) s,k . (17)Plugging (16) and (17) into (15), we get: var (cid:20) N ψ (cid:18) n Σ n Σ H n (cid:19)(cid:21) ≤ n X k =1 N n E (cid:20) tr (cid:18) Σ n Σ H n ψ ′ (cid:18) n Σ n Σ H n (cid:19) Ω k ψ ′ (cid:18) n Σ n Σ H n (cid:19)(cid:19)(cid:21) ( a ) ≤ w max n X k =1 N n E (cid:20) tr (cid:18) ψ ′ (cid:18) n Σ n Σ H n (cid:19) Σ n Σ H n ψ ′ (cid:18) n Σ n Σ H n (cid:19)(cid:19)(cid:21) , where ( a ) follows from the fact that tr AB ≤ k A k tr B for A hermitian and B positive definite matrix. Consider h : λ λ (cid:12)(cid:12)(cid:12) ψ ′ ( λ ) (cid:12)(cid:12)(cid:12) . Clearly h belongs to C ∞ c ( R , R ) . We therefore have: E (cid:20) n tr (cid:18) ψ ′ (cid:18) n Σ n Σ H n (cid:19) Σ n Σ H n ψ ′ (cid:18) n Σ n Σ H n (cid:19)(cid:19)(cid:21) = Z S N h ( λ ) dµ N ( λ )+ O (cid:18) N (cid:19) . It is clear that for N large enough, R S N h ( λ ) dµ N ( λ ) = 0 , thusproving: var (cid:18) N ψ (cid:18) n Σ n Σ H n (cid:19)(cid:19) = O (cid:18) N (cid:19) . Applying the classical Markov inequality, we obtain: P (cid:18) N tr ψ (cid:18) n Σ n Σ H n (cid:19)(cid:19) ≤ N / E "(cid:12)(cid:12)(cid:12)(cid:12) N tr ψ (cid:18) n Σ n Σ H n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = N / (cid:12)(cid:12)(cid:12)(cid:12) E (cid:20) N tr ψ (cid:18) n Σ n Σ H n (cid:19)(cid:21)(cid:12)(cid:12)(cid:12)(cid:12) +var (cid:18) N tr ψ (cid:18) n Σ n Σ H n (cid:19)(cid:19)(cid:19) = O (cid:18) N / (cid:19) . Thus, by Borel-Cantelli lemma, for N large enough, N tr ψ (cid:18) n Σ n Σ H n (cid:19) ≤ N / , or equivalently, tr ψ (cid:18) n Σ n Σ H n (cid:19) ≤ N / By definition of function ψ , the number of eigenvalues of theGram matrix n Σ n Σ H n that lies in the in the interval [ a, b ] is upper-bounded by tr ψ (cid:0) n Σ n Σ H n (cid:1) , and is therefore lessthan N / with probability . Since this number has to bean integer, we deduce that it is zero for N large enough. Asa consequence, there is no eigenvalue in [ a, b ] for N largeenough.Gathering the results of Theorem 2 and Theorem 1, we get: Corollary 5.
Assume the setting of Theorem 1. Then, for N large enough, the smallest eigenvalue of Σ n Σ ∗ n is boundedaway from zero. IV. A
PPROXIMATION RULE
This section aims at showing the approximation in propo-sition 3 stating that: E (cid:20) N tr Q ( z ) (cid:21) = 1 N tr T N ( z )+ 1 N χ N ( z ) for N large enough, where χ is analytic on C \ R + and satisfiesinequality (12).As far as generally correlated Gaussian matrices are con-cerned, the convergence of N tr Q N ( z ) to N tr T N ( z ) hasbeen shown to hold in the almost sure sense, [7]. Thisresult directly implies that the empirical eigenvalue distributionconverges weakly to a measure µ N which is characterized byits stieltjes transform m N ( z ) = N tr T N ( z ) . Its importancelies in that it gives us insights on the proportion of eigenvaluesfalling in any interval. But, it does not rule out the possibilityof a o ( n ) proportion of eigenvalues lying outside the limitingsupport of µ N . As it has been shown above, a sufficientcondition that can eliminate this possibility is constituted bythe statement of proposition 3. This statement is already known to hold for other models, mainly the non-centered Gaussianmodel [1]. Its proof for the model of generally correlatedGaussian matrices has not been carried out, to the best ofthe authors’ knowledge.While the proof of proposition 3 relies on the standard use ofGaussian calculus tools, several adaptations to the specificityof the random matrix model are far from being immediate. Tofacilitate the understanding of the highly technical proof, westart by introducing the main key steps. In order to control thedifference N E tr Q N ( z ) − N tr T N ( z ) , we need to introduce,similar to previous works [14], an intermediate deterministicmatrix denoted by R N ( z ) and which writes as: R N ( z ) = n n X k =1 Ω k α k ( z ) − z I N ! − , where α k ( z ) = n tr Ω k E Q ( z ) , k = 1 , · · · , n . With matrix R N ( z ) at hand, we decompose the difference N E tr Q N ( z ) − N tr T N ( z ) as: N E tr Q N ( z ) − N tr T N ( z ) = 1 N E tr Q N ( z ) − N tr R N ( z )+ 1 N tr R N ( z ) − N tr T N ( z ) , N χ ( z )+ 1 N χ ( z ) . This decomposition is quite standard in random matrix theory.While the direct control of the difference N E tr Q N ( z ) − N tr T N ( z ) is complicated, much can be inferred from bothdifferences N E tr Q N ( z ) − N tr R N ( z ) and N tr R N ( z ) − N tr T N ( z ) . In order to prove proposition 3, it suffices toshow that: | χ i ( z ) | ≤ ( | z | + C i ) k i P i (cid:16) |ℑ z | − (cid:17) , i = 1 , , where C i , i = 1 , are positive constants, k i , i = 1 , arepositive integers and P i , i = 1 , are polynomial with positivecoefficients independent of N . In addition to R N ( z ) , we willneed to introduce the following deterministic quantities: ˜ r i = − z (1+ α i ( z )) , i = 1 , · · · , n ˜ R N = diag (˜ r , · · · , ˜ r n ) . It can be easily shown along the same lines of Proposition5.1 of [13] that matrix valued functions R N ( z ) and ˜ R N ( z ) are holomorphic in C \ R + and coincide with the Stieltjestransforms of positive matrix valued probability measurescarried by R + , the mass of which are equal to I . Their spectralnorms are thus bounded by |ℑ z | − . In particular, we have: max (cid:16) k ˜ R N k , k R N k (cid:17) ≤ |ℑ z | − . With these quantities at hand, we are now in position tosequentially control the terms χ ( z ) and χ ( z ) . A. Control of χ ( z ) The control of χ ( z ) will extensively rely on the useof Gaussian calculus tools, namely the Integration by Partformulae and the Nash-Poincar´e inequality. Before delvinginto the core of the proof, we shall recall these tools. Lemma 6 (Integration by Part Lemma) . Let x =[ x , · · · , x N ] T a complex Gaussian vector such that E [ x ] = 0 , E [ xx T ] = 0 and E [ xx ∗ ] = R . If Γ : x Γ( x ) is a C complex function polynomially bounded together with itsderivatives, then: E [ x p Γ( x )] = N X m =1 [ R ] p,m E (cid:20) ∂ Γ( x ) ∂x ∗ m (cid:21) Lemma 7 (Nash-Poincar´e Inequality) . Let x = [ x , · · · , x N ] T a complex Gaussian vector such that E [ x ] = 0 , E [ xx T ] = 0 and E [ xx ∗ ] = R . If Γ : x Γ( x ) is a C complex functionpolynomially bounded together with its derivatives, then, not-ing ∇ x Γ = h ∂ Γ ∂x , · · · , ∂ Γ ∂x M i T and ∇ x ∗ Γ = h ∂ Γ ∂x ∗ , · · · , ∂ Γ ∂x ∗ M i T , var (Γ( x )) ≤ E (cid:2) ∇ x Γ( x ) T R ( ∇ x Γ( x )) ∗ (cid:3) + E (cid:2) ( ∇ x ∗ Γ( x )) ∗ R ∇ x ∗ Γ( x ) (cid:3) . Applying Lemma 7, we will thus get: var (Γ( ξ , · · · , ξ n )) ≤ n X k =1 N X s =1 N X r =1 E (cid:20) ∂ Γ ∂ξ s,k [Ω k ] s,r ∂ Γ ∗ ∂ξ r,k (cid:21) + n X k =1 N X s =1 N X r =1 E " ∂ Γ ∗ ∂ξ ∗ s,k [Ω k ] s,r ∂ Γ ∂ξ ∗ r,k . (18)The application of these tools will require us to computedifferentials of the resolvent matrix with respect to the entriesof Σ n . In particular, we will need in the sequel, the followingdifferentiation formulas: ∂ [ Q ] ℓ,p ∂ξ ∗ m,k = − n [ Q ∂ Σ n Σ ∗ n Q ] ℓ,p ∂ξ ∗ m,k = − n [ Q ξ k e T m Q ] ℓ,p = − n [ Q ξ k ] ℓ [ Q ] m,p . (19)Moreover, we also have: ∂ [ Q ] ℓ,p ∂ξ s,k = − n [ Q ] ℓ,s [ ξ ∗ k Q ] p . (20)The use of the integration by part lemma along with the abovedifferential formulae will allow us to establish the followinglemma: Lemma 8.
Let β i , i = 1 , · · · , n be given by β i = n tr Ω i Q ( z ) . For each z ∈ C + and any deterministic matrix A , it holds that: E tr AQ ( z ) = tr AR ( z ) − z E tr AQΣ n ˜ RBΣ ∗ n R n where B = diag (cid:18) o β , · · · , o β n (cid:19) with o β i = β i − α i . Proof:
From the identity: Q (cid:18) n Σ n Σ ∗ n − z I N (cid:19) = I N we have: z E [ Q ] p,q = E (cid:20) Q Σ n Σ ∗ n n (cid:21) p,q − δ p,q (21) = N X i =1 n X j =1 n E (cid:2) Q p,i ξ i,j ξ ∗ q,j (cid:3) − δ p,q . Using the integration by parts formula in Lemma 6, we have: E (cid:2) Q p,i ξ i,j ξ ∗ q,j (cid:3) = N X m =1 E " [ Ω j ] i,m ∂ξ ∗ q,j [ Q ] p,i ∂ξ ∗ m,j = N X m =1 [ Ω j ] i,m δ m,q E [ Q ] p,i − N X m =1 [ Ω j ] i,m n E h ξ ∗ q,j (cid:2) Q ξ j (cid:3) p [ Q ] m,i i . Summing the above equality over i , we obtain: E h(cid:2) Q ξ j (cid:3) p ξ ∗ q,j i = E [ QΩ j ] p,q − E h β j (cid:2) Q ξ j (cid:3) p ξ ∗ q,j i Plugging o β j = β j − α j into the above equality, we get: E h(cid:2) Q ξ j (cid:3) p ξ ∗ q,j i = E [ QΩ j ] p,q − α j E h ξ ∗ q,j (cid:2) Q ξ j (cid:3) p i − E (cid:20) o β j ξ ∗ q,j (cid:2) Q ξ j (cid:3) p (cid:21) Hence: E h(cid:2) Q ξ j (cid:3) p ξ ∗ q,j i = E " [ QΩ j ] p,q (1+ α j ) − E o β j ξ ∗ q,j (cid:2) Q ξ j (cid:3) p (1+ α j ) Summing over j , we finally get: E (cid:20) QΣ n Σ ∗ n n (cid:21) p,q = E Q n n X j =1 Ω j (1+ α j ) p,q + z E " QΣ n ˜ RBΣ ∗ n n p,q Plugging the above equality into (21), we thus get: E [ z Q ] p,q = E Q n n X j =1 Ω j (1+ α j ) p,q − [ I N ] p,q + z E " QΣ n ˜ RBΣ ∗ n n p,q Therefore, E (cid:2) QR − (cid:3) p,q = [ I N ] p,q − z E " QΣ n ˜ RBΣ ∗ n n p,q thereby proving that: E QR − = I N − z E " QΣ n ˜ RBΣ ∗ n n . As a consequence: E tr AQ = tr AR − zn tr E h AQΣ n ˜ RBΣ ∗ n R i . From Lemma 8, it appears that the control of χ amountsto showing that: z Γ , z E h tr QΣ n ˜ RBΣ ∗ n R i ≤ n ( | z | + C ) k P (cid:16) |ℑ z | − (cid:17) with C , k and P verifying the conditions of proposition 3.The proof relies on the use of the Nash-poincar´e inequality.But before that, we need to further workout quantity Γ bymeans of the Integration by Part formula. We first expand Γ as: Γ = 1 n N X p,q,m =1 n X ℓ =1 E (cid:20) [ Q ] p,q ξ q,ℓ ξ ∗ m,ℓ o β ℓ (cid:21) [ R ] m,p ˜ r ℓ (22)Using the integration by part formula, we have: E (cid:20) [ Q ] p,q ξ q,ℓ ξ ∗ m,ℓ o β ℓ (cid:21) = N X s =1 [ Ω ℓ ] q,s E ∂ [ Q ] p,q ξ ∗ m,ℓ o β ℓ ∂ξ ∗ s,ℓ = N X s =1 [ Ω ℓ ] q,s E (cid:20) [ Q ] p,q o β ℓ (cid:21) δ m,s + N X s =1 [ Ω ℓ ] q,s E [ Q ] p,q ξ ∗ m,ℓ ∂ o β ℓ ∂ξ ∗ s,ℓ − [ Ω ℓ ] q,s n E (cid:20) [ Q ξ ℓ ] p [ Q ] s,q ξ ∗ m,ℓ o β ℓ (cid:21) = − n E (cid:20) [ Ω ℓ Q ] q,q [ Q ξ ℓ ] p ξ ∗ m,ℓ o β ℓ (cid:21) +[ Ω ℓ ] q,m E (cid:20) [ Q ] p,q o β ℓ (cid:21) + N X s =1 [ Ω ℓ ] q,s E [ Q ] p,q ξ ∗ m,ℓ ∂ o β ℓ ∂ξ ∗ s,ℓ Summing the above equation over q , we get: E (cid:20) [ Q ξ ℓ ] p ξ ∗ m,ℓ o β ℓ (cid:21) = − E (cid:20) n tr ( Ω ℓ Q ) [ Q ξ ℓ ] p ξ ∗ m,ℓ o β ℓ (cid:21) + E (cid:20) [ QΩ ℓ ] p,m o β ℓ (cid:21) + N X q =1 N X s =1 [ Ω ℓ ] q,s E [ Q ] p,q ξ ∗ m,ℓ ∂ o β ℓ ∂ξ ∗ s,ℓ Writing n tr Ω ℓ Q as o β ℓ + α ℓ and using the same techniqueas in the proof of Lemma 8, we finally get: E (cid:20) [ Q ξ ℓ ] p ξ ∗ m,ℓ o β ℓ (cid:21) = z ˜ r ℓ E "(cid:18) o β ℓ (cid:19) [ Q ξ ℓ ] p ξ ∗ m,ℓ − z ˜ r ℓ E (cid:20) [ QΩ ℓ ] p,m o β ℓ (cid:21) − N X s,q =1 z ˜ r ℓ [ Ω ℓ ] q,s E [ Q ] p,q ξ ∗ m,ℓ ∂ o β ℓ ∂ξ ∗ s,ℓ (23) Plugging (23) into (22), we finally obtain:
Γ = zn E h tr (cid:16) QΣ n ˜ R B Σ ∗ n R (cid:17)i − zn n X ℓ =1 E (cid:20) o β ℓ tr (cid:16) QΩ ℓ R ˜ R (cid:17)(cid:21) − zn n X ℓ =1 N X s =1 ˜ r ℓ E [ Σ ∗ n RQΩ ℓ ] ℓ,s ∂ o β ℓ ∂ξ ∗ s,ℓ , ∆ − ∆ − ∆ . In the following we will prove that ∆ i satisfies: ∆ i ≤ K i n (cid:16) | z | + ˜ C i (cid:17) ˜ k i ˜ P i ( |ℑ z | − ) for some positive constant ˜ C i , K i , integer k i and polynomial ˜ P i independent of N . This will be sufficient to control χ ( z ) since the underlying polynomials have positive coefficients.Closer scrutiny of the expressions of ∆ i , i = 1 , , , revealsthat they make appear quantities of the form n tr AQ ( z ) with A is a some deterministic matrix. It is thus easy to convinceoneself that controlling the variance of these terms is essential.This will be the goal of the following lemma whose proof isdeferred to Appendix C: Lemma 9.
Let A be a N × N deterministic matrix. Then, wehave for any z ∈ C + , var (cid:18) n tr AQ ( z ) (cid:19) ≤ Cn k A k ( | z | +1) (cid:18) |ℑ z | + 1 |ℑ z | (cid:19) where C , a positive constant and P , a polynomial with positivecoefficients, are independent of N . With Lemma 9 at hand, we are now in position to handlethe terms ∆ i , i = 1 , , . We start by controlling ∆ . For that,consider Σ ( i ) to be the matrix Σ n without its i -th column.Define Q ( i ) the resolvent matrix given by: Q ( i ) = (cid:18) n Σ ( i ) Σ ∗ ( i ) − z I N (cid:19) − and β i, ( i ) = n tr Ω i Q ( i ) . Let o β i, ( i ) = β i, ( i ) − E β i, ( i ) and B ( i ) = diag (cid:18) o β , (1) , · · · , o β n, ( n ) (cid:19) . From the rank-one pertur-bation Lemma [20, Lemma 2.6], we obtain: max ≤ i ≤ n (cid:12)(cid:12)(cid:12)(cid:12) o β i − o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ w max n |ℑ z | Decompose ∆ as: ∆ = zn n X i =1 E " (cid:12)(cid:12)(cid:12)(cid:12) o β i (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) ! h Σ ∗ n R N QΣ n ˜ R N i i,i + zn n X i =1 E "(cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) h Σ ∗ n R N QΣ n ˜ R N i i,i , ∆ , +∆ , . We start by dealing with ∆ , . First, we need to bound the quantity (cid:12)(cid:12)(cid:12)(cid:12) o β i (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) . We have: (cid:12)(cid:12)(cid:12)(cid:12) o β i (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) o β i (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) o β i (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) ≤ N w max n |ℑ z | (cid:12)(cid:12)(cid:12)(cid:12) o β i − o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ N w n |ℑ z | . (24)From (24), ∆ , can be bounded by: ∆ , ≤ | z | n N w |ℑ z | n X i =1 E (cid:12)(cid:12)(cid:12)(cid:12)h Σ ∗ n R N QΣ n ˜ R N i i,i (cid:12)(cid:12)(cid:12)(cid:12) . We need thus to bound E (cid:12)(cid:12)(cid:12)(cid:12)h n Σ ∗ n R N QΣ n ˜ R N i i,i (cid:12)(cid:12)(cid:12)(cid:12) . We have: E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:20) n Σ ∗ n R N QΣ n ˜ R N (cid:21) i,i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = E (cid:20) n ξ ∗ i R N Q ξ i ˜ r i (cid:21) ≤ | ˜ r | i E (cid:20) k R N Q k n ξ ∗ i ξ i (cid:21) ≤ |ℑ z | n tr Ω i ≤ N w max n |ℑ z | and thus: ∆ , ≤ | z | (cid:18) lim sup N Nn (cid:19) w n |ℑ z | . We now move to the control of ∆ , . First, write ∆ , as: ∆ , = zn n X i =1 E "(cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) (cid:2) ξ ∗ i RQ ξ i ˜ r i (cid:3) . Using the relation Q ξ i = Q ( i ) ξ ∗ i n ξ ∗ i Q ξ ∗ i , (25)we obtain: ∆ , ≤ | z | n n X i =1 E "(cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12) ξ ∗ i R N Q ( i ) ξ i ˜ r i (cid:12)(cid:12) n ξ ∗ i Q ξ i ≤ | z | n |ℑ z | n X i =1 E "(cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) ξ ∗ i ξ i . Since β i, ( i ) is independent of ξ i , and thus : ∆ , ≤ | z | n |ℑ z | n X i =1 tr Ω i E | o β i, ( i ) | ≤ N w max | z | n |ℑ z | n X i =1 E | o β i, ( i ) | From Lemma 9, we have: E (cid:12)(cid:12)(cid:12)(cid:12) o β i, ( i ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ w n ( | z | +1) (cid:18) |ℑ z | + 1 |ℑ z | (cid:19) Hence, ∆ , ≤ lim sup Nn w n |ℑ z | ( | z | +1) (cid:18) |ℑ z | + 1 |ℑ z | (cid:19) , Kn ( | z | +1) P ( |ℑ z | − ) , thereby proving the desired result. The control of ∆ relieson the use of the Cauchy-schwartz inequality. We have: ∆ = zn n X ℓ =1 E (cid:20) o β ℓ tr (cid:16) QΩ ℓ R N ˜ R N (cid:17)(cid:21) ≤ | z | n X ℓ =1 q E | o β ℓ | r var 1 n tr QΩ ℓ R N ˜ R N From Lemma 9, we can bound E (cid:12)(cid:12)(cid:12)(cid:12) o β ℓ (cid:12)(cid:12)(cid:12)(cid:12) and var tr QΩ ℓ R N ˜ R N as: E (cid:12)(cid:12)(cid:12)(cid:12) o β ℓ (cid:12)(cid:12)(cid:12)(cid:12) ≤ w n ( | z | +1) (cid:18) |ℑ z | + 1 |ℑ z | (cid:19) var tr 1 n QΩ ℓ R N ˜ R N ≤ w |ℑ z | n ( | z | +1) (cid:18) |ℑ z | + 1 |ℑ z | (cid:19) . Using the fact that √ xy ≤ x + y for positive scalars x, y , wefinally get: | ∆ | ≤ w ( | z | +1) n (cid:18) |ℑ z | + 1 |ℑ z | + 1 |ℑ z | + 1 |ℑ z | (cid:19) , K ( | z | +1) P ( |ℑ z | − ) Finally, we will move to the treatment of ∆ . Recall that ∆ is given by: ∆ = zn n X ℓ =1 N X s =1 ˜ r ℓ E [ Σ ∗ n R N QΩ ℓ ] ℓ,s ∂ o β ℓ ∂ξ ∗ s,ℓ . Using the differentiation formulae in (19), we get: ∂ o β ℓ ∂ξ ∗ s,ℓ = − n [ QΩ ℓ QΣ n ] s,ℓ . Hence, ∆ = − zn n X ℓ =1 N X s =1 ˜ r ℓ E h [ Σ ∗ n R N QΩ ℓ ] ℓ,s [ QΩ ℓ QΣ n ] s,ℓ i = − zn n X ℓ =1 ˜ r ℓ E [ ξ ∗ ℓ R N QΩ ℓ QΩ ℓ Q ξ ℓ ] . The above relation allows us to bound ∆ as: | ∆ | ≤ | z | w n n X ℓ =1 | ˜ r ℓ | k R N k E (cid:2) ξ ∗ ℓ ξ ℓ k Q k (cid:3) ≤ | z | w n |ℑ z | lim sup Nn , K | z | n P ( |ℑ z | − ) . From the obtained bounds for the scalars ∆ i , i = 1 , , , wecan deduce that: | z Γ | ≤ n ( | z | + C ) k P ( |ℑ z | − ) , which is, as mentioned above, the required inequality tocontrol χ . B. Control of χ ( z ) We now move to the control of χ ( z ) given by: χ ( z ) = N tr R N − N tr T N . To this end, we will resort to the resolvent identity : A − − B − = B − ( A − B ) A − for any invertible matrices B and A . We therefore obtain: N tr R N − N tr T N = Nn tr R N n X j =1 Ω j δ j − Ω j α j T = Nn n X j =1 tr( R N Ω j T )( α j − δ j )(1+ α j )(1+ δ j )= Nn n X j =1 z ˜ r j ˜ δ j tr R N Ω j T ( α j − δ j ) , where ˜ δ j = − z (1+ δ j ) . Using property 6 of Lemma 1 in [1],we can easily check that ˜ δ j , j = 1 , · · · , n similar to ˜ r j areStieltjes transforms of probability measures carried by R + .We therefore have: max (cid:16)(cid:12)(cid:12)(cid:12) ˜ δ j (cid:12)(cid:12)(cid:12) , | ˜ r j | (cid:17) ≤ |ℑ z | . Hence, | N tr R N − N tr T N | ≤ | z | N |ℑ z | max ≤ j ≤ n | α j − δ j | . To control χ , it suffices to show that there exists constants C and K , integer k and polynomial P with positive coefficientsand independent of N such that: max ≤ j ≤ n | α j − δ j | ≤ KN ( | z | + C ) k P ( |ℑ z | − ) . This will be the objective of the next derivations in this section.We start by decomposing α j − δ j as: α j − δ j = 1 n tr Ω j E Q − n tr Ω j R + 1 n tr Ω j R − n tr Ω j T = ǫ j ( z )+ 1 n tr Ω j R − n tr Ω j T . The control of ǫ j ( z ) is similar to that of χ ( z ) , the presence ofmatrix Ω j instead of the identity matrix requiring only slightmodifications of the proof. We can thus deduce that: max ≤ j ≤ n | ǫ j | ≤ K ǫ N ( | z | + C ǫ ) k ǫ P ǫ ( |ℑ z | − ) , (26)for some constants K ǫ and C ǫ , integer k ǫ and polynomial P ǫ independent of N . Again, using the resolvent identity as above,we obtain: α j − δ j = ǫ j ( z )+ 1 n n X k =1 ( α k − δ k ) tr Ω j R N Ω k T (1+ α k )(1+ δ k ) . (27)Define α = [ α , · · · , α n ] T , δ = [ δ , · · · , δ n ] T and ǫ =[ ǫ ( z ) , · · · , ǫ n ( z )] . Then (27) writes as: ( I n − A ) ( α − δ ) = ǫ , (28)where A is a n × n matrix with entries: [ A ] j,k = 1 n tr Ω j R N Ω k T (1+ α k )(1+ δ k ) . In order to control the difference vector α − δ , we need first tocheck that I n − A is invertible. For that, notice that by Cauchy-Schwartz inequality: (cid:12)(cid:12)(cid:12) [ A ] j,k (cid:12)(cid:12)(cid:12) ≤ r(cid:12)(cid:12)(cid:12) [ B ] j,k (cid:12)(cid:12)(cid:12)r(cid:12)(cid:12)(cid:12) [ C ] j,k (cid:12)(cid:12)(cid:12) where B and C are n × n matrices with entries: [ B ] j,k = 1 n tr Ω j R N Ω k R N | α k | [ C ] j,k = 1 n tr Ω j TΩ k T | δ k | . It follows from the algebraic lemma proven in Appendix Ethat I n − A is invertible provided that B or C have spectralnorms strictly less than , in which case: (cid:13)(cid:13)(cid:13) ( I n − A ) − (cid:13)(cid:13)(cid:13) ∞ ≤ r(cid:13)(cid:13)(cid:13) ( I n − B ) − (cid:13)(cid:13)(cid:13) ∞ r(cid:13)(cid:13)(cid:13) ( I n − C ) − (cid:13)(cid:13)(cid:13) ∞ . (29)It appears from (29) that one needs to study matrices B and C ,which are at first sight easier to manipulate, mainly becausethey either involve R N or T . This however is not trivial. Westate the result in the following proposition and for sake ofreadability defer the proof to Appendix D. Proposition 10.
Assume that z ∈ C + . Then, Matrix C satisfies ρ ( C ) < . Moreover, (cid:13)(cid:13)(cid:13) ( I n − C ) − (cid:13)(cid:13)(cid:13) ∞ ≤ K ( η + | z | ) |ℑ z | (30) where K and η are some positive constants independentof N . There exists polynomials Q and Q independent of N with positive coefficients such that for N large enoughand z ∈ E N given by E N = (cid:26) z ∈ C + , N Q ( | z | ) Q ( |ℑ z | − ) ≤ (cid:27) we have ρ ( B ) ≤ and: k ( I n − B ) − k ≤ ˜ K (˜ η + | z | ) |ℑ z | . It follows from proposition 10 that the spectral norm of A isstrictly less than . Thus, I n − A is invertible and for z ∈ E N , k ( I n − A ) − k ∞ ≤ (cid:13)(cid:13)(cid:13) ( I n − B ) − (cid:13)(cid:13)(cid:13) ∞ + 12 (cid:13)(cid:13)(cid:13) ( I n − C ) − (cid:13)(cid:13)(cid:13) ∞ ≤ K max ( η + | z | ) |ℑ z | , (31)where K max = max( K, ˜ K ) and η max = max( η, ˜ η ) . Plugging(31) into (28), we obtain: k α − δ k ∞ ≤ K max K ǫ N ( | z | + C ǫ ) k ǫ ( η max + | z | ) P ǫ ( |ℑ z | − ) |ℑ z | , where the right hand side of the above inequality can be putunder the form: K ( C + | z | ) k N P ( |ℑ z | − ) . for K and C positive constants, k integer, and P somepolynomial with positive coefficients. Consider now the casewhere z ∈ C + \E N . We first remark that: | α j − δ j | ≤ | α j | + | δ j | ≤ w max |ℑ z | . Since z / ∈ E N , we therefore have: N Q ( | z | ) Q ( |ℑ z | − ) ≥ . Hence: k α − δ k ∞ ≤ w max |ℑ z | N Q ( | z | ) Q ( |ℑ z | − ) As a consequence, we can find for
C, K constants, k integerand P polynomial with positive coefficients such that: k α − δ k ∞ ≤ KN ( | z | + C ) k P ( |ℑ z | − ) , thereby ending the proof.A PPENDIX AP RELIMINARIES
Many of the results of the appendix part are based on thefollowing key lemmas, which we recall in this section for sakeof clarity.
Lemma 11.
Let A = ( a ℓ,m ) nℓ,m =1 be an n × n real matrix and u and v be two real n × vectors. Assume that the entries of A are positive and that of u and v strictly positive. Assume,furthermore, that the equation: u = Au + v is satisfied. Then, the spectral radius ρ ( A ) of A of A satisfies: ρ ( A ) ≤ − min( v ℓ )max( u ℓ ) < . Lemma 12 (Matrix Inequality) . Let A be a n × n hermitianmatrix. Then, n tr AA ∗ ≥ (cid:12)(cid:12)(cid:12)(cid:12) n tr A (cid:12)(cid:12)(cid:12)(cid:12) with equality only if A is proportional to identity.Proof: Let A = UΛU H be an eiengevalue decompositionof A . Consider λ , · · · , λ n the eigenvalues of A . Then, if thereis i = j such that λ i = λ j , we have due to the strict-convexityof x x : n tr AA ∗ = 1 n n X i =1 λ i > (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n n X i =1 λ i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A PPENDIX BP ROOF OF T HEOREM does not belong to the support S N , we show that it exists ǫ > for which µ N ([0 , x ]) = 0 foreach x ∈ ]0 , ǫ [ . To this end, define function φ : R n + × R + → R n + , with: φ ( x , · · · , x n , z ) = ( φ ( x , · · · , x n , z ) , · · · , φ n ( x , · · · , x n , z )) where φ i : R n + × R + → R + is given by: φ i ( x , · · · , x n , z ) = 1 n tr Ω i n n X k =1 Ω k x k − zI N ! − . We need to show that there exists ℓ , · · · , ℓ n such that: φ i ( ℓ , · · · , ℓ n ,
0) = ℓ i . Let p ∈ N and r p = − p . We will first start by proving thatfor each p , there exists a unique x p , · · · , x pn such that: φ i ( x p , · · · , x pn , r p ) = x pi . For that, it suffices to show that ˜ φ p : R n + → R n + , ( x , · · · , x n ) φ ( x , · · · , x n , r p ) is a standard interfer-ence function. In particular, we need to check that φ satisfythe following properties: • Nonnegativity: For each x , · · · , x n ≥ and each i and p , φ i ( x , · · · , x n , r p ) > . • Monotonicity: For each x ≥ x ′ , · · · , x n ≥ x ′ n , and each i and p , φ i ( x , · · · , x n , r p ) ≥ φ i (cid:16) x ′ , · · · , x ′ n , r p (cid:17) . • Scalability: For each α > , and each i and p , αφ i ( x , · · · , x n , r p ) > φ i ( αx , · · · , αx n , r p ) .The first item is obvious since Ω i are positive definite matri-ces, while the second one follows from the fact that for positivedefinite matrices, A (cid:23) B implies B − (cid:23) A − . Finally, toprove the last item, note that for α > , φ i ( αx , · · · , αx n , r p ) < N tr Ω i n n X k =1 Ω k α (1+ x k ) − r p α I N ! − = αφ i ( x , · · · , x n , r p ) . Therefore, φ i ( αx , · · · , αx n , r p ) > αφ i ( x , · · · , x n , r p ) . According to [21, Theorem 2], ˜ φ p is a standard interferencefunction. To prove that there exists a unique x p , · · · , x pn satisfying: x pi = φ i ( x p , · · · , x pn ) , we need to check that there exits x , · · · , x n such that: x i > φ i ( x , · · · , x n , r p ) . This condition holds true, since φ i ( x , · · · , x n ) ≤ r p , and soincreasing x i to infinity will satisfy the above inequality.Moreover, consider the sequence: x ( t,p ) i = φ i ( x ( t − ,p )1 , · · · , x ( t − ,p ) n ) , i = 1 , · · · , n where x (0 ,p )1 , · · · , x (0 ,p ) n are arbitrary positive reals. Then, x ( t,p ) = (cid:16) x ( t,p )1 , · · · , x ( t,p ) n (cid:17) converge to x p = ( x p , · · · , x pn ) .From this, we can prove that for p ≥ q , we have for each i ∈ { , · · · , n } , x pi ≥ x qi . To this end, we will consider the sequence, x ( t,p ) i = φ i ( x ( t − ,p )1 , · · · , x ( t − ,p ) n ) , i = 1 , · · · , n where x (0 ,p ) i = x qi and will show that for any t , x ( t,p ) i ≥ x qi . We will proceed by induction on t . For t = 0 , the resultobviously holds. Assume that the resuld holds for any k ≤ t ,i.e, x ( k,p ) i ≥ x qi , i = 1 , · · · , n and k ≤ t. And let us prove it for t = k +1 . We have: x ( t +1 ,p ) i = φ i ( x ( t,p )1 , · · · , x ( t,p ) n , r p ) ≥ φ i ( x ( t,p )1 , · · · , x ( t,p ) n , r q ) ( a ) ≥ φ i ( x q , · · · , x qn , r q )= x qi . where ( a ) follows since φ i is increasing in each variable and x ( t,p ) i ≥ x ( q ) i by the induction assumption.We have therefore shown that for p ≥ q , x pi ≥ x qi . As p tends to infinity, x pi will converge to a limit ℓ i ∈ R + ∪{ + ∞} . Assume that for i ∈ { , · · · , n } , ℓ i = + ∞ . Then,one can easily see, that necessarily, ℓ i = + ∞ for any i ∈{ , · · · , n } . We will prove now, that the case of ℓ i = + ∞ forall i = 1 , · · · , n cannot hold. For this observe that: n X i =1 x pi x pi = n X i =1 n tr Ω i x pi n n X k =1 Ω k x pk + r p I N ! − ≤ N. Let x p min = min ≤ i ≤ n x pi . We have thus: x min x min ≤ Nn or equivalently: x min ≤ Nn − Nn . which is contradiction with the fact that ℓ i = + ∞ for all i .Recall now that: φ i ( x p , · · · , x pn , r p ) = x pi . Taking the limit in p , we thus get that: φ i ( ℓ , · · · , ℓ n ,
0) = ℓ i , or equivalently: n tr Ω i n n X k =1 Ω k ℓ k ! − = ℓ i . The Jakobian matrix corresponding to ˜ φ ∞ : R n + → R n + :( x , · · · , x n ) φ ( x , · · · , x n , at x i = ℓ i , i = 1 , · · · , n , isgiven by: [ J ] i,m = 1 n tr Ω i n n X k =1 Ω k ℓ k ! Ω m (1+ ℓ m ) n n X r =1 Ω r ℓ r ! − Let u = [1+ ℓ , · · · , ℓ n ] T and v = [ ℓ , · · · , ℓ n ] T . Then,after simple calculations, one can show that: Ju = v . The entries of J , u and v are strictly positive. A directapplication of Lemma 11 in section A implies that: ρ ( J ) ≤ − min ≤ i ≤ n ℓ i ≤ i ≤ n ℓ i < . thereby showing that I n − J is invertible. Hence, the implicitfunction theorem implies that there exists an open disk at zerowith radius η > , i.e D (0 , η ) and unique analytic functions ϕ , · · · , ϕ n defined in D (0 , η ) such that: φ i ( ϕ ( z ) , · · · , ϕ n ( z ) , z ) = ϕ i ( z ) and ϕ i (0) = ℓ i , i = 1 , · · · , n. On the other hand, one can show that there exists ǫ > suchthat ϕ i ( t ) is real valued and strictly positive for any t ∈ [ − ǫ, ǫ ] .Indeed, writing ℑ ϕ i ( t ) as: ℑ ϕ i ( t ) = 12 ı n tr Ω i n n X k =1 Ω k ϕ k ( t ) − t I N ! − − n tr Ω i n n X k =1 Ω k ϕ ∗ k ( t ) − t I N ! − = 1 n tr Ω i n n X k =1 Ω k ϕ k ( t ) − t I N ! − × n n X k =1 Ω k ℑ ( ϕ k ( t )) | ϕ k ( t ) | ! n n X k =1 Ω k ϕ k ( t ) − t I N ! − . Therefore, the vector g t = [ ℑ ( ϕ ( t )) , · · · , ℑ ( ϕ n ( t ))] T issolution of the following system of equations: g t = J t g t . As t ρ ( J t ) is continuous, and since for t = 0 , ρ ( J t ) = ρ ( J ) < , there exists ǫ > such that: ρ ( J t ) < for every t ∈ [ − ǫ, ǫ ] . Therefore, g t = 0 . Furthermore, sinceat t = 0 , ϕ i (0) = ℓ i > , we can futher assume that ǫ is chosen such that ϕ i ( t ) is real-valued and strictly positivefor any t ∈ [ − ǫ, ǫ ] . From [7, Theorem 1], we know that for t < , δ ( t ) , · · · , δ n ( t ) are the unique non-negative pointwisesolutions of the following system of equations δ i ( t ) = φ i ( δ ( t ) , · · · , δ n ( t ) , t ) , thereby implying that: δ i ( t ) = ϕ i ( t ) for any t ∈ [ − ǫ, . Since, the set of functionals δ ( t ) , · · · , δ n ( t ) and ϕ ( t ) , · · · , ϕ n ( t ) are holomorphic on D (0 , ǫ ) \ { [0 , ǫ [ } and coincide on a set of values with anaccumulation point, they must coincide on the whole domaineof analicity, namely D (0 , ǫ ) \ { [0 , ǫ [ } .Let m be given by: m = 1 N tr n n X k =1 Ω k ϕ k ( z ) − z I N ! − . Obviously m is analytic on D (0 , ǫ ) and satisfies: m ( z ) = m N ( z ) for all z ∈ D (0 , ǫ ) \ { [0 , ǫ [ } . We recall that for ≤ x < ǫ , µ N ([0 , x ]) can be expressed as: µ N ([0 , x ]) = 1 π lim y → ,y> Z x ℑ ( m N ( s + ıy )) ds. Therefore, µ N ([0 , x ]) = 1 π lim y → ,y> Z x ℑ ( m ( s + ıy )) ds As m is holomorphic on D (0 , ǫ ) , the dominated convergencetheorem implies that: π lim y → ,y> Z x ℑ ( m ( s + ıy )) ds = 1 π Z x ℑ ( m ( s )) ds = 0 since m ( s ) ∈ R for s ∈ [0 , x ] . Thus, we establish that µ N ([0 , x ]) = 0 . A PPENDIX CP ROOF OF LEMMA β A = n tr AQ ( z ) .We then have: var( β A ( z )) ≤ n X k =1 N X s =1 N X r =1 n E h [ Σ ∗ n QAQ ] k,s [ Ω k ] s,r [ Q ∗ A ∗ Q ∗ Σ n ] r,k i + n X k =1 N X s =1 N X r =1 n E h [ Q ∗ A ∗ Q ∗ Σ n ] k,s [ Ω k ] s,r [ Σ ∗ n QAQ ] r,k i = n X k =1 n E h [ Σ ∗ n QAQΩ k Q ∗ A ∗ Q ∗ Σ n ] k,k i + n X k =1 n E h [ Q ∗ A ∗ Q ∗ Σ n Ω k Σ ∗ n QAQ ] k,k i . Since Ω k (cid:22) w max I N with w max = sup N max ≤ k ≤ n k Ω k k ,we have: var( β A )( z ) ≤ w max n tr (cid:18) QAQQ ∗ A ∗ Q ∗ Σ n Σ ∗ n n (cid:19) + w max n tr (cid:18) Q ∗ A ∗ Q ∗ Σ n Σ ∗ n n QAQ (cid:19) . Using the resolvent identity: Q ( z ) Σ n Σ ∗ n n = Σ n Σ ∗ n n Q ( z ) = I N + z Q ( z ) , and the inequality k Q ( z ) k ≤ |ℑ ( z ) | , we obtain: var( β A ( z )) ≤ w max k A k n |ℑ ( z ) | + | z ||ℑ ( z ) | ! ≤ w max k A k n ( | z | +1) |ℑ ( z ) | + 1 |ℑ ( z ) | ! . A PPENDIX DP ROOF OF PROPOSITION µ N is tight. To this end, we willfollow the same steps as in [13, Lemma C1]. Observe that: Z + ∞ λµ N ( dλ ) = lim y → + ∞ ℜ [ − ıy ( ıym N ( ıy )+1)]= lim y → + ∞ ℜ (cid:20) − ıy (cid:18) ıy N tr T N ( ıy )+1 (cid:19)(cid:21) . (32)On the other hand: T N ( ıy ) n n X k =1 Ω k δ k ( ıy ) − ıy I N ! = I N . Therefore,
1+ 1 N tr ıy T N ( ıy ) = 1 n n X k =1 N tr Ω k T N ( ıy )1+ δ k ( ıy )= 1 n n X k =1 c N δ k ( ıy )1+ δ k ( ıy ) . (33)Plugging (33) into (32), we finally get: Z + ∞ λµ N ( dλ ) = lim y → + ∞ n c N n X k =1 ℜ [ − ıyδ k ( ıy )] | δ k ( iy ) | . Since δ k are Stieltjes transforms of finite positive measures,we have: lim y → + ∞ | δ k ( ıy ) | = 0 Moreover, we have lim y → + ∞ − ıyδ k ( ıy ) = n tr Ω k , therebyestablishing that: sup N Z + ∞ λµ N ( dλ ) < + ∞ . The tightness of the sequence µ N follows directly from theabove inequality. In the same way, we can also show that thesequence of measures corresponding to the Stieltjes transforms N tr R is also tight. These two results will be of fundamentalimportance in the proof of proposition 10.We now return to the proof of proposition10: a) Proof of proposition 10-1): The proof is based on theuse of Lemma 11 in section A. For that, we need to find alinear system involving matrix C . For z ∈ C + , we have: ℑ ( δ j ) = 12 ın (tr Ω j T − tr Ω j T H )= 1 n tr Ω j T n n X k =1 Ω k ℑ ( δ k ) | δ k | + ℑ ( z ) I N ! T H = 1 n n X k =1 tr Ω j TΩ k T H | δ k | ℑ ( δ k )+ ℑ ( z ) 1 n tr Ω j TT H . Let I δ and c be the n × vectors given by: I δ = [ ℑ ( δ ) , · · · , ℑ ( δ n )] T c = (cid:20) n tr Ω TT H , · · · , n tr Ω n TT H (cid:21) T , Then: I δ = C I δ + ℑ ( z ) c . Since ℑ ( δ j ) > for all j and ℑ z > and C , c have positiveentries, we get from Lemma 11, (cid:13)(cid:13)(cid:13) ( I n − C ) − (cid:13)(cid:13)(cid:13) ∞ ≤ max ≤ j ≤ n ℑ δ j ℑ z min ≤ j ≤ n n tr Ω j TT H ≤ w max |ℑ z | min ≤ j ≤ n n tr Ω j TT H , where the second inequality follows from the fact that max ≤ j ≤ n ℑ δ j ≤ max ≤ j ≤ n | δ j | ≤ w max ℑ z . Using the inequal-ity n tr AB ≥ λ ( A ) n tr B for A and B hermitian positivedefinite matrices with λ ( A ) the smallest eigenvalue of A , weget: (cid:13)(cid:13)(cid:13) ( I n − C ) − (cid:13)(cid:13)(cid:13) ∞ ≤ w max |ℑ z | w min 1 n tr TT H . (34)In order to obtain a lower bound on N tr TT H , we first remarkthat by the Jensen inequality in Lemma 12: N tr TT H ≥ (cid:12)(cid:12) N tr T (cid:12)(cid:12) = | m N ( z ) | ≥ ℑ ( m N ( z )) . As ( µ N ) N ≥ is tight,it exists η > for which µ N ([ η, + ∞ )) ≤ for all N and assuch: µ N ([0 , η ]) ≥ . As a consequence, ℑ ( m N ( z )) = ℑ ( z ) Z + ∞ dµ N ( λ ) | λ − z | > Z η ℑ ( z ) dµ N ( λ )2( η + | z | ) µ N ([0 , η ]) ≥ ℑ ( z )4( η + | z | ) . (35)Plugging (35) into (34), we finally get (30). b) Proof of proposition 10-2): The proof is similar tothat of the first statement. We first decompose α j as: α j = α j − n tr Ω j R + 1 n tr Ω j R = ǫ j + 1 n tr Ω j R . Hence, ℑ ( α j ) = ℑ ( ǫ j ( z ))+ ℑ (cid:18) n tr Ω j R (cid:19) . Using the same kind of calculations as above, we thus get: ℑ ( α j ) = ℑ ( ǫ j )+ 1 n n X k =1 tr Ω j RΩ k R H ℑ α k | α k ( z ) | + ℑ ( z ) 1 n tr Ω j RR H . (36)In order to determine a subset of C + on which ℑ ( z ) n tr Ω j RR H + ℑ ( ǫ j ( z )) > , we evaluate a lower boundof n tr Ω j RR H . We have by the Jensen inequality in Lemma12: n tr Ω j RR ∗ ≥ w min (cid:12)(cid:12)(cid:12)(cid:12) n tr R (cid:12)(cid:12)(cid:12)(cid:12) = w min (cid:18) Nn (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) N tr R (cid:12)(cid:12)(cid:12)(cid:12) . From the discussion in the beginning of this section, we knowthat the sequence of measures corresponding to the Stieltjestransforms N tr R is tight. Hence, there exists ˜ η such that: ℑ (cid:18) N tr R (cid:19) ≥ ℑ z (cid:16) ˜ η + | z | (cid:17) . Hence, n tr Ω j RR ∗ ≥ w min (cid:18) Nn (cid:19) |ℑ z | (cid:16) ˜ η + | z | (cid:17) . On the other hand, from (26), we recall that: | ǫ j ( z ) | ≤ K ǫ N ( | z | + C ǫ ) k ǫ P ǫ ( |ℑ z | − ) . Consider E N, the set given by: E N, = ( z ∈ C + , w min (cid:0) Nn (cid:1) |ℑ z | η + | z | ) − K ǫ N ( | z | + C ǫ ) k ǫ P ǫ ( |ℑ z | − ) > (cid:27) Then, as before, using the fact that for z ∈ E N, (36) can becast into a linear system of equations involving positive-entriesmatrix and vectors, we deduce that ρ ( B ) < and: (cid:13)(cid:13)(cid:13) ( I n − B ) − (cid:13)(cid:13)(cid:13) ∞ ≤ max ≤ j ≤ n α jw min N n |ℑ z | η + | z | ) − K ǫ N ( | z | + C ǫ ) k ǫ P ǫ ( |ℑ z | − ) ≤ w min N n |ℑ z | η + | z | ) (cid:16) − N Q ( | z | ) Q ( |ℑ z | − ) (cid:17) , where Q and Q are polynomials with positive coefficients.Take E N as the set defined by: E N = (cid:26) z ∈ C + , N Q ( | z | ) Q (cid:0) ℑ z − (cid:1) ≤ (cid:27) . Obviously E N ⊆ E N, , and for all z ∈ E N , we get: (cid:13)(cid:13)(cid:13) ( I n − B ) − (cid:13)(cid:13)(cid:13) ∞ ≤ n (cid:0) ˜ η + | z | (cid:1) w min N |ℑ z | . A PPENDIX EA LINEAR ALGEBRAIC RESULT
Finally, we finish the Appendix part with a linear algebraiclemma which we need in our derivation and can be ofindependent interest.
Lemma 13.
Let B and C be n × n matrices with non-negativeentries. Let A be a n × n matrix satisfying: (cid:12)(cid:12)(cid:12) [ A ] i,j (cid:12)(cid:12)(cid:12) ≤ q [ B ] i,j q [ C ] i,j . (37) Then, ρ ( A ) ≤ p ρ ( B ) p ρ ( C ) . If furthermore max( ρ ( A ) , ρ ( B )) < , then ρ ( A ) < and: (cid:13)(cid:13)(cid:13) ( I n − A ) − (cid:13)(cid:13)(cid:13) ∞ ≤ r(cid:13)(cid:13)(cid:13) ( I n − B ) − (cid:13)(cid:13)(cid:13) ∞ r(cid:13)(cid:13)(cid:13) ( I n − C ) − (cid:13)(cid:13)(cid:13) ∞ Proof:
We start by proving that ρ ( A ) ≤ p ρ ( B ) p ρ ( C ) .For that, consider ˜ A , the matrix given by: h ˜ A i i,j = q [ B ] i,j q [ C ] i,j Consider | A | the matrix such that [ | A | ] i,j = (cid:12)(cid:12)(cid:12) [ A ] i,j (cid:12)(cid:12)(cid:12) . Then, ρ ( | A | ) ≤ ρ ( ˜ A ) . Recall, that for any matrix D , ρ ( D ) = lim k → + ∞ k D k k k ∞ . From the above convergence, we have: h ˜ A k i i,j = n X i , ··· ,i k − h ˜ A i i,i h ˜ A i i ,i · · · h ˜ A i i k − ,j = X ≤ i , ··· ,i k − ≤ n q [ B ] i,i [ B ] i ,i · · · [ B ] i k − ,j × q [ C ] i,i [ C ] i ,i · · · [ C ] i k − ,j ≤ s X ≤ i , ··· ,i k − ≤ n [ B ] i,i [ B ] i ,i · · · [ B ] i k − ,j s X ≤ i , ··· ,i k − ≤ n [ C ] i,i [ C ] i ,i · · · [ C ] i k − ,j = q [ B k ] i,j q [ C k ] i,j . With this inequality at hand, we are now in position to bound k ˜ A k k ∞ . We have: (cid:13)(cid:13)(cid:13) ˜ A k (cid:13)(cid:13)(cid:13) ∞ = max ≤ i ≤ n n X j =1 h ˜ A k i i,j ≤ max ≤ i ≤ n n X j =1 (cid:2) B k (cid:3) i,j (cid:2) C k (cid:3) i,j ≤ max ≤ i ≤ n vuut n X j =1 [ B k ] i,j vuut n X j =1 [ C k ] i,j ≤ q k B k k ∞ q k C k k ∞ . We therefore have: ρ ( ˜ A ) = lim k → + ∞ (cid:13)(cid:13)(cid:13) ˜ A k (cid:13)(cid:13)(cid:13) k ∞ ≤ lim k → + ∞ (cid:13)(cid:13) B k (cid:13)(cid:13) k ∞ (cid:13)(cid:13) C k (cid:13)(cid:13) k ∞ = p ρ ( B ) p ρ ( C ) . Therefore, ρ ( ˜ A ) < and thus, ρ ( A ) < if max( ρ ( C ) , ρ ( B )) < . In this case, I n − A is invertible andalso are I n − B and I n − C . Since ( I n − A ) − = P + ∞ k =0 A k .for any ≤ i ≤ n , we have: n X j =1 (cid:12)(cid:12)(cid:12)(cid:12)h ( I n − A ) − i i,j (cid:12)(cid:12)(cid:12)(cid:12) ≤ + ∞ X k =0 n X j =1 (cid:12)(cid:12)(cid:12)(cid:2) A k (cid:3) i,j (cid:12)(cid:12)(cid:12) ≤ ∞ X k =0 n X j =1 h | A | k i i,j ≤ + ∞ X k =0 n X j =1 q [ B k ] i,j rh | C | k i i,j ≤ + ∞ X k =0 vuut n X j =1 [ B k ] i,j vuut n X j =1 [ C k ] i,j ≤ vuut + ∞ X k =0 n X j =1 [ B k ] i,j vuut + ∞ X k =0 n X j =1 [ C k ] i,j ≤ r(cid:13)(cid:13)(cid:13) ( I n − B ) − (cid:13)(cid:13)(cid:13) ∞ r(cid:13)(cid:13)(cid:13) ( I n − C ) − (cid:13)(cid:13)(cid:13) ∞ . As a consequence, we have: (cid:13)(cid:13)(cid:13) ( I n − A ) − (cid:13)(cid:13)(cid:13) ∞ ≤ r(cid:13)(cid:13)(cid:13) ( I n − B ) − (cid:13)(cid:13)(cid:13) ∞ r(cid:13)(cid:13)(cid:13) ( I n − C ) − (cid:13)(cid:13)(cid:13) ∞ . R EFERENCES[1] P. Vallet, P. Loubaton, and X. Mestre, “Improved subspace estimationfor multivariate observations of high dimension: the deterministic signalscase,”
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