The Logarithmic Law of Random Determinant 3

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an bn 1 an 1 Dn =  − −  . ······  b1 a1      Here an,...,a1,bn 1,...,b1 is a collection of independent variables such that ai { − } ∼ χi,bj χj for i =1,...,n,j =1,...,n 1. Such a tridiagonal form is well known for Gaussian∼ matrix. One can refer to Dumitriu− and Edelman [4], for instance. Then appar- ently one has

n 2 T d 2 det An = det AnAn = ai , i=1 Y which implies that log det An can be represented by a sum of n independent random variables. Then by the| elementary| properties of χ2 distributions, the following CLT can be obtained

log det A (1/2) log(n 1)! d | n|− − N(0, 1). (1.1) (1/2) logn −→

For details, one can see Rouaultp [18] or the Appendix of Costello and Vu [2] for instance. Like most of topics in the Random Matrix Theory, one may ask whether there is a “universal” phenomenon for the CLT of log det An under general distribution assumption. The best result on this problem was given| by Girko| in [9] (one can also refer to Girko’s books [7] and [8] or his paper [6] for his former results on this topic), where the author only required the existence of the (4+ δ)th moment of the entries for some positive δ. Girko named (1.1) as “the logarithmic law” for random determinant. In Girko [9], using an elegant “method of perpendiculars” and combining the classical CLT for martingale, Girko claimed (1.1) is universal under the moment assumption mentioned above. How- ever, though the proof route of Girko [9] is clear and quite original, it seems the proof is not complete and several parts are lack of mathematical rigour. Recently, Nguyen and Vu [14] provided a transparent proof of (1.1) for the general distribution case, under much stronger moment assumption in the sense that for all t> 0,

P( a t) C exp( tC2 ) (1.2) | ij |≥ ≤ 1 − with some positive constants C1, C2 independent of i,j,n. Obviously, (1.2) implies the existence of the moment of any order. The basic framework of the proof in Nguyen and The logarithmic law of random determinant 3

Vu [14] is similar to Girko’s method of perpendiculars, which will be introduced in the next section. However, in order to provide a transparent proof, the authors of Nguyen and Vu [14] inserted a lot of new ingredients. Moreover, some unrigorous steps in Girko [9] can be ﬁxed by the methods provided in Nguyen and Vu [14]. One may ﬁnd that Nguyen 1/3+o(1) and Vu [14] also provided a convergence rate of the logarithmic law as log− n, which is nearly optimal. In this paper, also relying on the basic strategy of Girko’s method of perpendiculars, we will provide a complete and rigorous proof under a weaker moment condition. More precisely, we only require the existence of 4th moment of the matrix entries. Our main result is

Theorem 1.1. Let An = (aij )n,n be a square random matrices, where aij , 1 i, j n is a collection of independent real random variables with common mean{ 0 and≤ variance≤ } 1. Moreover, we assume E 4 sup max aij < . 1 i,j n ∞ n ≤ ≤ Then we have the logarithmic law for det A : as n tends to inﬁnity, | n| log det A (1/2) log(n 1)! d | n|− − N(0, 1). (1/2) logn −→

Our paper is organized as follows.p In Section 2, we will sketch the main idea of Girko’s method of perpendiculars and the proof route of Theorem 1.1. In Sections 3, 4 and 5, we will present the details of the proof with the aid of some additional lemmas, whose proofs will be given in the Appendix. Throughout the paper, the notation such as C,K will be used to denote some positive constants independent of n, whose values may diﬀer from line to line. We use and k·k2 op to represent the Euclidean norm of a vector and the operator norm of a matrix k·krespectively as usual.

2. Girko’s method of perpendiculars

In this section, we will sketch the main framework of Girko’s method of perpendiculars, which was also pursued by Nguyen and Vu in the recent work [14]. To state this method rigorously, we need the following proposition whose proof will be given in Appendix B.

Proposition 2.1. For the matrix An defined in Theorem 1.1, we can find a modified matrix An′ = (aij′ )n,n satisfying the assumptions in Theorem 1.1 such that

P all square submatrices of A′ are invertible = 1 (2.1) { n } and

P log det A log det A′ = o( log n) =1 o(1). (2.2) { | n|− | n| } − p 4 Z. Bao, G. Pan and W. Zhou

Remark 2.2. The construction of the modified matrix An′ can be found in Nguyen and 2 1/2 Vu [14]. The strategy is to set a′ := (1 ε ) a + εθ , where θ , 1 i, j n is a ij − ij ij { ij ≤ ≤ } collection of independent bounded continuous random variables with common mean zero and variance one and is independent of An. By choosing ε to be extremely small, say Kn n− for large enough constant K> 0, it was shown in Nguyen and Vu [14] that (2.2) holds. The proof in Nguyen and Vu [14] relies on a lower bound estimate on the smallest singular value of square random matrices provided in Theorem 2.1 of Tao and Vu [20]. To adapt to our condition, we will use Theorem 4.1 of Götze and Tikhomirov [11] instead (by choosing pn = 1 in Götze and Tikhomirov [11]). For convenience of the reader, we sketch the proof of Proposition 2.1 in Appendix B. The proof is just a slight modification of that in Nguyen and Vu [14] under our setting and assumptions.

Therefore, with the aid of Proposition 2.1, we can always work under the following assumption.

Assumption C0. We assume that An = (aij )n,n is a square random matrix, where aij , 1 i, j n is a collection of independent real random variables with common mean { ≤ ≤ } E 4 zero and variance one. Besides, supn max1 i,j n aij < . Moreover, all square subma- ≤ ≤ ∞ trices of An are invertible with probability one.

The starting point of the method of perpendiculars is the elementary fact that the magnitude of the determinant of n real vectors in n dimensions is equal to the volume of the parallelepiped spanned by those vectors. Therefore, by the basic “base times height” formula, one can represent det An by the products of n perpendiculars. To make it more precise, we introduce some| notations| at first. T In the sequel, we will use ak to denote the kth row of An. And let A(k) be the k n T × rectangular matrix formed by the first k rows of An. Particularly, one has A(1) = a1 and A(n) = An. Moreover, we use the notation Vi to denote the subspace generated by the first i rows of An and Pi = (pjk(i))n,n to denote the projection matrix onto the space T T V ⊥. Let γ be the distance from a to V for 1 i n 1. And we set γ = a . i i+1 i+1 i ≤ ≤ − 1 k 1 k2 Then by the “base times height” formula, we can write

n 1 2 − 2 det An = γi+1. (2.3) i=0 Y T Observe that γi+1 is the norm of the projection of ai+1 onto Vi⊥. Thus we also have

γ2 = aT P a , 1 i n 1. (2.4) i+1 i+1 i i+1 ≤ ≤ −

Moreover, by the deﬁnition of Vi and Assumption C0 one has that with probability one T A(i)A(i) is invertible and

T T 1 P = I A (A A )− A . (2.5) i n − (i) (i) (i) (i) The logarithmic law of random determinant 5

Then a direct consequence of the deﬁnition of γi+1 is

E γ2 P = tr P = n i, 0 i n 1. { i+1| i} i − ≤ ≤ − It follows from (2.3) that

n 1 − 2 2 log det An = log γi+1, i=0 X which yields

n 1 − γ2 log det A2 log(n 1)! = log i+1 + log n. (2.6) n − − n i i=0 X − Now we set γ2 (n i) X := X = i+1 − − . i+1 n,i+1 n i − And we write γ2 X2 log i+1 = X i+1 + R , (2.7) n i i+1 − 2 i+1 − where X2 R := log(1 + X ) X i+1 . i+1 i+1 − i+1 − 2 Then by (2.6), one can write

log det A2 log(n 1)! n − − √2 log n (2.8) n 1 n 1 2 n 1 1 − 1 − Xi+1 1 − = Xi+1 log n + Ri+1. √2 logn − √2 log n 2 − √2 log n i=0 i=0 ! i=0 X X X Crudely speaking, the main route is to prove that the first term of (2.8) weakly con- verges to the standard Gaussian distribution and the remaining two terms tend to zero in probability. Let i be the σ-algebra generated by the first i rows of An, by definition we have E

E X =0. { i+1|Ei} Thus X ,...,X is a martingale difference sequence with respect to the filtration ∅ 1 n ⊂ 1 n 1. In Girko [9], under the assumption of the existence of (4 + δ)th moments E ⊂···⊂E − 6 Z. Bao, G. Pan and W. Zhou of the matrix entries for some δ> 0, Girko used the CLT for martingales to show the first term of (2.8) is asymptotically Gaussian. He also showed that the last term of (2.8) is asymptotically negligible. However, some steps in the proofs of these two parts are lack of mathematical rigour. Moreover, we do not find the discussion of the second term of (2.8) in Girko’s original proof in [9]. Recently, Nguyen and Vu provided a complete proof under the assumption that the distributions of the matrix entries satisfy (1.2), thus with finite moments of all orders. In the following sections, we will also adopt the representation (2.7) and the theory on the weak convergence of martingales. It will be clear that the proof will rely on some approximations of Xi+1 and Ri+1. However, these approximations are i-dependent and the large i case turns out to be badly approximated. To see this, we can take the Gaussian case for example. Note that 2 2 when the entries are standard Gaussian, γi+1 χn i, thus ∼ − 1/2 3/2 X = ((n i)− ), R = ((n i)− ) i+1 O − i+1 O − with high probability. Especially, when n i is (1), the main term X and the neg- − O i+1 ligible term R are comparable to be (1). Such a fact will be an obstacle if we use i+1 O crude estimations for Xi+1 and Ri+1 for general distribution case. This is explained such as follows. When we estimate the last term of (2.8), a basic strategy is to use the Taylor expansion of log(1 + Xi+1) to gain a relatively small remainder Ri+1, which requires X 1 c for some small positive constant c. However, as we mentioned above, when | i+1|≤ − n i is too small, such a bound is hard to be guaranteed since Xi+1 = (1) with high probability,− especially under the assumption of the 4th moment. Fortunately,O if all the T 2 ai+1’s are Gaussian for large i n s1 for some positive number s1, γi+1 are independent 2 ≥ − χ variables for all i n s1 even if A(n s1) is generally distributed. Such an explicit distribution information≥ can− be used to deal− with the large i part. Therefore, in [9] Girko proposed to replace some rows by Gaussian ones and prove the logarithmic law for the matrix after replacement, and then recover the result to the original one by a comparison procedure. Such a strategy was also used in Nguyen and Vu [14]. To pursue this idea, we set s = log3a n 1 ⌊ ⌋ for some sufficiently large positive constant a. Our proof route can be split into the following five steps. (i)

n s − 1 X i=0 i+1 d N(0, 1). √2 log n −→ P (ii)

n s1 2 − X /2 log n P i=0 i+1 − 0. √2 log n −→ P The logarithmic law of random determinant 7

(iii)

n s − 1 R P i=0 i+1 0. √2 log n −→ P

(iv) If the last s1 rows of An are Gaussian, then

n 1 2 i=−n s log(γi+1/(n i)) P − 1 − 0. P √2 log n −→

(v) Let Bn be a random matrix satisfying the basic Assumption C0 and diﬀering from An only in the last s1 rows. Then one has

sup P det An x P det Bn x 0. x | {| |≤ }− {| |≤ }| −→

We will prove (i) and (ii) together in Section 3, and prove (iii) and (iv) in Section 4. Section 5 is devoted to the proof of (v).

3. Convergence issue on the martingale diﬀerence sequence

In this section, we will prove the statements (i) and (ii). The arguments for both two parts heavily rely on the fact that Xi, 1 i n is a martingale diﬀerence sequence. In order to prove (i), we will use{ the following≤ ≤ } classical CLT for martingales, which can be found in the book of Hall and Heyde [12], for instance.

Proposition 3.1. Let S , , 1 i k ,n 1 be a zero-mean, square-integrable { ni Fni ≤ ≤ n ≥ } martingale array with diﬀerences Zni. Suppose that

P max Zni 0, i | | −→ P Z2 1. ni −→ i X E 2 Moreover, (maxi Zni) is bounded in n. Then we have

S d N(0, 1). nkn −→

Now we use the above proposition to prove (i). Let kn = n s1, Zni = Xi/√2 log n. Thus, it suﬃces to show that −

1 P max Xi+1 0, (3.1) √ 0 i n s1 log n ≤ ≤ − | | −→ 8 Z. Bao, G. Pan and W. Zhou

n s1 1 − P X2 1, (3.2) 2 log n i+1 −→ i=0 X and

n s − 1 1 E 2 1 E 2 max Xi+1 Xi+1 C (3.3) log n i ≤ log n ≤ i=0 X for some positive constant C independent of n. To verify (3.1) it suﬃces to show the following lemma.

Lemma 3.2. Under the Assumption C0, we have for any constant ε> 0

n s − 1 1 P Xi+1 ε 0 √log n| |≥ −→ i=0 X as n tends to inﬁnity.

Proof. Below we use the notation 1 Q = [q (i)] =: P . i jk n,n n i i − Thus by deﬁnition and the fact that tr Qi = 1 we have

X = aT Q a 1= q (i)(a2 1)+ q (i)a a . i+1 i+1 i i+1 − kk i+1,k − uv i+1,u i+1,v k u=v X X6 Now we introduce the quantities

U = q (i)(a2 1), V = q (i)a a . i+1 kk i+1,k − i+1 uv i+1,u i+1,v k u=v X X6 Obviously

X U + V . | i+1| ≤ | i+1| | i+1| Then it is elementary to see

n s − 1 1 ε P Ui+1 0 √log n| |≥ 2 −→ i=0 X The logarithmic law of random determinant 9 and

n s − 1 1 ε P Vi+1 0, √log n| |≥ 2 −→ i=0 X which can be implied by

n s 1 − 1 EU 2 0, (3.4) log n i+1 −→ i=0 X and

n s 1 − 1 EV 4 0. (3.5) 2 i+1 −→ log n i=0 X First, we verify (3.4). In the sequel, we set

20a s = n log− n . 2 ⌊ ⌋ By deﬁnition, we see

2 EU 2 = E q (i)(a2 1) CE q2 (i). i+1 kk i+1,k − ≤ kk Xk Xk Using the basic fact 0 p (i) 1 one has 0 q (i) 1/(n i). Taking this fact and ≤ kk ≤ ≤ kk ≤ − tr Qi = 1 into account we have

n s n s 1 − 1 1 − 1 EU 2 C E q2 (i) log n i+1 ≤ log n kk i=0 i=0 X X Xk n s n s 1 − 2 1 1 − 1 C + E max qkk(i) ≤ log n n i log n k i=0 i=n s ! X − X− 2 n s 1 − 1 log log n C E max qkk(i)+ (3.6) ≤ log n k O log n i=n s2 X− n s 1 − 1 1 log log n C E max pkk(i)+ ≤ log n n i k O log n i=n s2 X− − log log n C max E max pkk(i)+ . ≤ n s2 i n s1 k O log n − ≤ ≤ − Now we need the following crucial technical lemma which will be used repeatedly in the sequel. 10 Z. Bao, G. Pan and W. Zhou

Lemma 3.3. Under the above notation, we have

8a max E max pkk(i) C log− n (3.7) n s i n 1 k − 2≤ ≤ − ≤ for some positive constant C.

The proof of Lemma 3.3 will be given later. Now we proceed to the proof of (3.4) by assuming Lemma 3.3. Note that (3.4) follows from (3.6) and (3.7) immediately. Hence, it remains to show (3.5). We need the following simple deviation lemma for the quadratic form, whose proof is quite elementary and will be given in Appendix A.

Lemma 3.4. Suppose xi,i =1,...,n are independent real random variables with com- l mon mean zero and variance 1. Moreover, we assume maxi E xi νl. Let Mn = (mij )n,n be a nonnegative deﬁnite matrix which is deterministic. The|n we| ≤ have

n 4 E 2 4 2 2 miixi tr Mn C(ν8 tr Mn + (ν4 tr Mn) ) (3.8) − ≤ i=1 X and

4 E m x x Cν2(tr M 2)2 (3.9) uv u v ≤ 4 n u=v X6 for some positive constant C.

Note that by (3.9) one has

1 EV 4 CE(tr Q2)2 = C , i+1 ≤ i (n i)2 − which implies that

n s − 1 1 4 2 3a EV = (log− − n). (3.10) 2 i+1 O 4 log n i=0 X Thus (3.5) holds. Then Lemma 3.2 follows from (3.4) and (3.5) immediately. Thus (3.1) is veriﬁed.

Now we prove Lemma 3.3.

Proof of Lemma 3.3. We denote the jth column of A(i) by bj (i) and use the notation A(i,j) to denote the matrix induced from A(i) by deleting the jth column bj (i). Moreover, The logarithmic law of random determinant 11

1/6 we set the positive parameter α = αn := n− . By (2.5), we have

T T 1 p (i)=1 b (i) (A A )− b (i) kk − k (i) (i) k T T T 1 =1 b (i) (A A + b (i)b (i) )− b (i) − k (i,k) (i,k) k k k T T T 1 1 b (i) (A A + nαI + b (i)b (i) )− b (i) ≤ − k (i,k) (i,k) i k k k T T 1 1 =(1+ bk(i) (A(i,k)A(i,k) + nαIi)− bk(i))− , where in the last step we used the Sherman–Morrison formula

1 T 1 T 1 1 M − bk(i)bk(i) M − b b − (M + k(i) k(i) ) = M − T 1 (3.11) − 1+ bk(i) M − bk(i) for k k invertible matrix M. Let × 1 1 1 − 1 − G (α)= A AT + αI , G (α)= A AT + αI . (i,k) n (i,k) (i,k) i (i) n (i) (i) i Then one has

1 1 − p (i) 1+ b (i)T G (α)b (i) . kk ≤ n k (i,k) k Hence, to verify (3.7) we only need to show

1 E 1 T − 8a max 1+ bk(i) G(i,k)(α)bk(i) C log− n, n s2 i n 1. (3.12) k n ≤ − ≤ ≤ −

It is apparent that G(i)(α) and G(i,k)(α) are positive-deﬁnite and

1 G (α) , G (α) α− . k (i) kop k (i,k) kop ≤ Moreover, we have

tr G (α) tr G (α) | (i) − (i,k) | 1 1 1 1 − 1 − = tr A AT + b (i)b (i)T + αI tr A AT + αI (3.13) n (i,k) (i,k) n k k i − n (i,k) (i,k) i

(1/n)b (i)T G (α)2b (i) k (i,k) k 1 = T α− , 1+(1/n)bk(i) G(i,k)(α)bk(i) ≤ where in the second step above we used the Sherman–Morrison formula (3.11) again. Now we set

χ(i)= 1 (1/n)tr G (α) log10a n , { (i) ≥ } 12 Z. Bao, G. Pan and W. Zhou and we denote the (u, v)th entry of G(i,k)(α) by G(i,k)(u, v) below. Moreover, for ease of presentation, when there is no confusion, we will omit the parameter α from the notation G(i,k)(α) and G(i)(α). Then we have for some small constant 0 <ε< 1/2,

1 E 1 T − max 1+ bk(i) G(i,k)bk(i) k n 1 P 1 10a E 1 T − tr G(i) log n + χ(i)max 1+ bk(i) G(i,k)bk(i) ≤ n ≤ k n 1 i − P 1 10a E 1 2 tr G(i) log n + χ(i)max 1+ G(i,k)(j, j)ajk ε ≤ n ≤ k n − j=1 ! X P 1 + max G(i,k)(u, v)aukavk ε n 1 k n ≥ ≤ ≤ 1 u=v i ≤X6 ≤ 1 − P 1 10a E a 1 tr G(i) log n + χ(i)max 1 + log− n tr G(i,k) ε ≤ n ≤ k · n − (3.14) n i 2 a 1 10a + C P G (j, j)a < log− n tr G , tr G log n (i,k) jk · (i,k) n (i) ≥ (j=1 ) Xk=1 X P 1 + max G(i,k)(u, v)aukavk ε n 1 k n ≥ ≤ ≤ 1 u=v i ≤X6 ≤ 1 1 10a a 1 − P tr G log n + Eχ(i) 1 + log− n tr G 2ε ≤ n (i) ≤ · n (i) − n i 2 a 1 10a + C P G (j, j)a < log− n tr G , tr G log n (i,k) jk (i,k) n (i) ≥ (j=1 ) Xk=1 X P 1 + max G(i,k)(u, v)aukavk ε , n 1 k n ≥ ≤ ≤ 1 u=v i ≤X6 ≤

where in the above last inequality, we used (3.13). Below we will estimate (3.14) term by term. To this end, we need the following two lemmas, whose proofs will be given in Appendix A.

Lemma 3.5. Under the assumption of Theorem 1.1, for n s2 i n 1, we have for 1/6 − ≤ ≤ − α = n−

1 1/6 E tr G (α) = s (α)+ (n− ) n (i) i O The logarithmic law of random determinant 13 and

1 1/3 Var tr G (α) = (n− ), n (i) O where

2 1 i i i − s (α)=2 α +1 + α + 1 +4α . i − n − n n s With the aid of Lemma 3.5, we can estimate the ﬁrst term of (3.14) as follows. Note that by deﬁnition s (α) 1 log20a n for n s i n 1, we have i ≥ 10 − 2 ≤ ≤ − 1 1 1 1 P tr G log10a n P tr G E tr G log20a n n (i) ≤ ≤ n (i) − n (i) ≥ 20

40a 1 1/3 C log − n Var tr G = o(n− ). ≤ n (i) For the second term of (3.14), with the deﬁnition of χ(i), obviously one has

1 a 1 − 8a Eχ(i) 1 + log− n tr G 2ε C log− n. · n (i) − ≤ Now we deal with the third term of (3.14). We set

aˆjk Eaˆjk a aˆjk = ajk1 ajk log n , a˜jk = − . {| |≤ } Var aˆ { jk} Since G (j, j)’s are positive anda ˆ2 a2 one has p (i,k) jk ≤ jk

n i 2 a 1 10a P G (j, j)a < log− n tr G , tr G log n (i,k) jk · (i,k) n (i) ≥ (j=1 ) Xk=1 X n 2 a 1 10a P G (j, j)ˆa < log− n tr G , tr G log n . ≤ (i,k) jk · (i,k) n (i) ≥ j Xk=1 X Moreover, by the assumption sup max Ea4 < it is easy to derive that n ij ij ∞ 3a 2a Eaˆ = (log− n), Var aˆ =1+ (log− n). jk O { jk} O Consequently,

a a˜ =ˆa + (log− n), jk jk O 14 Z. Bao, G. Pan and W. Zhou which implies

2 2 2a 2 a a˜ 2â + (log− n) 2â + log− n jk ≤ jk O ≤ jk for sufficiently large n. Therefore, we have

n 2 a 1 10a P G (j, j)a < log− n tr G , tr G log n (i,k) jk · (i,k) n (i) ≥ j Xk=1 X n 2 a 1 10a P G (j, j)â < log− n tr G , tr G log n ≤ (i,k) jk · (i,k) n (i) ≥ j Xk=1 X n 2 a 1 10a P G (j, j)ã < 3 log− n tr G , tr G log n ≤ (i,k) jk · (i,k) n (i) ≥ j Xk=1 X n 1 1 P G (j, j)ã2 tr G tr G , tr G log10a n ≤ (i,k) jk − (i,k) ≥ 2 (i,k) n (i) ≥ k=1 j X X n 1 1 P G (j, j)ã2 tr G log10a n ≤ n (i,k) jk − (i,k) ≥ 4 k=1 j X X n 40a 4 4a 4 2 2 C log− n n− E[log n tr G +(tr G ) ] ≤ · · (i,k) i,k kX=1 1 o . ≤ nα4 In the fourth inequality we used the fact (3.13) and in the fifth inequality we used (3.8) and the fact E a˜ 4+t = (logta n) for any t 0, which is easy to see from the definition | ij | O ≥ ofa ˜ij . Now we begin to deal with the last term of (3.14). Note that by (3.9)

P 1 max G(i,k)(u, v)aukavk ε n 1 k n ≥ ≤ ≤ u=v X6 n 1 P G (u, v)a a ε ≤ n (i,k) uk vk ≥ k=1 u=v X X6 n 4 4 4 ε− n− E G (u, v)a a ≤ (i,k) uk vk k=1 u=v X X6 n 4 4 2 2 Cε− n− E(tr G ) ≤ (i,k) Xk=1 The logarithmic law of random determinant 15

1 . ≤ O nα4 Therefore, (3.12) follows from the above estimates, so does (3.7). Hence, we complete the proof.

Now we come to deal with (3.2) and (3.3). Note that (3.2) can be implied by (ii) directly. Thus we will prove the statement (ii) and (3.3) below. We reformulate them as the following lemma and then prove it.

Lemma 3.6. Under the Assumption C0, one has

n s1 2 − X 2 log n P i=0 i+1 − 0, (3.15) √log n −→ P and

n s 1 − 1 EX2 =1+o(1). (3.16) 2 log n i+1 i=0 X Proof. We begin with (3.15). We split the proof of (3.15) into two steps.

n s1 2 2 − (X E X i ) P i=0 i+1 − { i+1|E } 0, (3.17) √log n −→ P and

n s1 2 − E X i 2 logn P i=0 { i+1|E }− 0. (3.18) √log n −→ P Observe that

n 2 E 2 E Xi+1 i = qjk(i)ai+1,j ai+1,k 1 i { |E } ( − ! E ) j,k=1 X (3.19) 2 = + q (i)2(E a 4 3). n i kk | i+1,k| − − Xk Therefore, to verify (3.18), we only need

n s − 1 1 2 E qkk(i) 0. (3.20) √log n −→ i=0 X Xk 16 Z. Bao, G. Pan and W. Zhou

Note that from the proof of (3.4) we can get the following estimate directly.

n s − 1 1 2 log log n E qkk(i) = . (3.21) √log n O √log n i=0 X Xk Thus it suﬃces to show (3.17). By elementary calculations, we have X2 E X2 i+1 − { i+1|Ei} = 2 q (i)(a2 1)+2 q (i)q (i)(a2 a2 1) − uu i+1,u − uu vv i+1,u i+1,v − u u=v X X6 +2 q (i)2(a2 a2 1) uv i+1,u i+1,v − u=v X6

+2 qu1v1 (i)qu2v2 (i)ai+1,u1 ai+1,v1 ai+1,u2 ai+1,v2 u =v ,u =v 16 1 26 2 u ,v X= u =v { 1 1}6 { 26 2}

+ q (i)2(a4 Ea4 )+2 q (i)(a2 1) q (i)a a uu i+1,u − i+1,u uu i+1,u − uv i+1,u i+1,v u u u=v X X X6 =:2W1(i)+2W2(i), where

W (i)= q (i)(a2 1)+ q (i)q (i)(a2 a2 1) 1 − uu i+1,u − uu vv i+1,u i+1,v − i u=v X X6 + q (i)2(a2 a2 1) uv i+1,u i+1,v − u=v X6

+ qu1v1 (i)qu2v2 (i)ai+1,u1 ai+1,v1 ai+1,u2 ai+1,v2 , u =v ,u =v 16 1 26 2 u ,v X= u =v { 1 1}6 { 26 2} and 1 W (i)= q (i)2(a4 Ea4 ) 2 2 uu i+1,u − i+1,u u X + q (i)(a2 1) q (i)a a . uu i+1,u − uv i+1,u i+1,v u u=v X X6 We split the issue to show

s1 s1 1 P 1 P W1(i) 0, W2(i) 0. (3.22) √log n −→ √log n −→ i=0 i=0 X X The logarithmic law of random determinant 17

First, we deal with the second statement of (3.22). Note that

n s 1 − 1 E W2(i) √log n | | i=0 X n s − 1 1 2 C E quu(i) ≤ √log n i=0 u X X (3.23) n s1 2 1/2 2 1/2 − 1 2 + C E quu(i)(a 1) E quv(i)ai+1,uai+1,v √log n i+1,u − i=0 u u=v X X X6

n s1 n s1 1/2 1/2 − − 1 2 1 1 2 C E quu(i) + C E quu(i) . ≤ √log n √log n n i i=0 u i=0 u X X X − X By (3.21), we have that the ﬁrst term of (3.23) is of the order of (log log n/√log n). For O the second term, by using (3.7) we have

n s1 1/2 1/2 − 1 1 2 E quu(i) √log n n i i=0 u X − X n s2 n s1 1/2 − − 1 1 1 1 2 + E quu(i) ≤ √log n n i √log n n i i=0 i=n s2 u X − X− − X n s 1 − 1 1 1/2 log log n E max puu(i) + ≤ √log n n i u O √log n i=n s2 X− − log log n = . O √log n Now we consider the ﬁrst term of (3.22). It is easy to see

E W (i) =0, E W (i)W (j) =0, i = j, { 1 } { 1 1 } 6 which yields

2 s1 s1 1 1 2 E W1(i) = EW1(i) √log n log n i=0 ! i=0 X X n s n s C − 1 C − 1 E q (i)2 + E q (i)2q (i)q (i) ≤ log n uu log n uu vv ww i=0 u i=0 u=v,u=w X X X 6 X6 (3.24) 18 Z. Bao, G. Pan and W. Zhou

n s C − 1 + E q (i)2q (i)2 log n uv uw i=0 u=v,u=w X 6 X6 n s C − 1 + E q (i)q (i)q (i)q (i) . log n | u1v1 u1v2 u2v1 u2v2 | i=0 u =v ,u =v X 16 X1 26 2 The estimation of (3.24) is elementary but somewhat tedious. In fact, one can ﬁnd the estimate towards every term of (3.24) in Nguyen and Vu [14] (see the estimation of n s1 Var( i=0− Yi+1) in Section 6 of Nguyen and Vu [14]). Here we omit the details and claim the following estimation P s 2 1 1 log log n E W1(i) = . √log n O log n i=0 ! X Then (3.17) follows, so does (3.15). Moreover, it is easy to see that (3.16) holds by combining (3.19) and (3.20). Therefore, Lemma 3.6 is proved.

Thus we complete the proof of (i) and (ii).

4. Negligible parts (iii) and (iv)

In this section, we will prove the statements (iii) and (iv). We start with (iii). The following elementary but crucial lemma will be needed.

a/2 Lemma 4.1. By the deﬁnitions above, if X 1 + log− n, one has i+1 ≥− R C(U 2 + V 2+δ) log log n | i+1|≤ i+1 | i+1| for any 0 δ 1. Here C := C(a,δ) is a positive constant only depends on a and δ. ≤ ≤ 1 Proof. We split the discussion into three cases. Choose some small constant 0 <ε< 10 a/2 (say) and consider the three cases X 1 ε, X > 1 ε and 1 + log− n | i+1|≤ − i+1 − − ≤ Xi+1 < 1+ ε separately. For the ﬁrst case, we can use the elementary Taylor expansion to see that−

R C X 3 C U + V 2+δ | i+1|≤ | i+1| ≤ | i+1 i+1| for any 0 δ 1. If U 1 or V 1, we will immediately get ≤ ≤ | i+1|≥ | i+1|≥ R C(U 2 + V 2+δ) | i+1|≤ i+1 | i+1| since U + V < 1. If both U and V are less than 1, we have | i+1 i+1| | i+1| | i+1| R C( U 2+δ + V 2+δ) C(U 2 + V 2+δ). | i+1|≤ | i+1| | i+1| ≤ i+1 | i+1| The logarithmic law of random determinant 19

Now we come to deal with the second case. When X > 1 ε, obviously one has i+1 − R C(U + V )2. | i+1|≤ i+1 i+1 Then it is elementary to see that we always can ﬁnd some positive constant C such that

R C(U 2 + V 2+δ) | i+1|≤ i+1 | i+1| 1 ε since max Ui+1, Vi+1 > 2 2 . Finally, we deal with the last case. Note that when c{ } − 1 + log− n X < 1+ ε, we have − ≤ i+1 − R C log log n. | i+1|≤ Moreover, it is obvious that we have max U , V > 1 ε . Consequently, one has {| i+1| | i+1|} 2 − 2 R C(U 2 + V 2+δ) log log n. | i+1|≤ i+1 | i+1| In conclusion, we completed the proof.

n s1 The next lemma is devoted to bounding the probability of the event i=0− Xi+1 < a { 1 + log− n . − } S

Lemma 4.2. Under the Assumption C0, we have

n s1 − a/2 P X < 1 + log− n 0 { i+1 − } −→ i=0 X as n tends to inﬁnity.

Proof. Note that

a/2 P X < 1 + log− n { i+1 − } T a/2 = P a Q a < log− n { i+1 i i+1 } 2 a/2 1 a/2 P q (i)a < 2 log− n + P q (i)a a log− n . ≤ kk i+1,k uv i+1,u i+1,v ≥ 2 k u=v X X6

Now we recall the deﬁnition

aî+1,k Eaî+1,k a aî+1,k = ai+1,k1 ai ,k log n , a˜i+1,k = − . {| +1 |≤ } Var aˆ { i+1,k} A similar discussion as that in the last section yields p

2 a/2 2 a/2 P q (i)a < 2 log− n P q (i)˜a < C log− n kk i+1,k ≤ kk i+1,k Xk Xk 20 Z. Bao, G. Pan and W. Zhou

1 P q (i)˜a2 1 ≤ kk i+1,k − ≥ 2 k X 4a 2 C(log n tr Q4 + (tr Q 2) ) ≤ · i i 4a 3 2 C(log n (n i)− + (n i)− ). ≤ · − − Here in the third inequality, we used (3.8) again. Moreover, by (3.9), we obtain

1 a/2 2 2a P q (i)a a log− n C(n i)− log n. uv i+1,u i+1,v ≥ 2 ≤ − u=v X6

Thus ﬁnally, we have

n s1 − a/2 P X < 1 + log− n { i+1 − } i=0 X n s n s − 1 − 1 4a 3 2a 2 C log n (n i)− + log n (n i)− ≤ − − i=0 i=0 ! X X a C log− n. ≤ Therefore, we complete the proof.

Combining Lemmas 4.1 with 4.2, one has with probability 1 o(1), − R C(U 2 + V 2+δ) log log n, 0 i n s . | i+1|≤ i+1 i+1 ≤ ≤ − 1 Thus to show (iii), it suﬃces to verify that

n s − 1 log log n 2 2+δ P (U + Vi+1 ) 0, √log n i+1 | | −→ i=0 X which can be implied by

n s − 1 log log n 2 2+δ E(U + Vi+1 ) 0. (4.1) √log n i+1 | | −→ i=0 X Note that n s n s − 1 − 1 EU 2 C E q (i)2 = (log log n). (4.2) i+1 ≤ jj O i=0 i=0 j X X X Moreover, we have

n s n s n s − 1 − 1 − 1 2+δ 4 (2+δ)/4 1 δ/2 E V (EV ) C (n i)− − = o(1). (4.3) | i+1| ≤ i+1 ≤ − i=0 i=0 i=0 X X X The logarithmic law of random determinant 21

Then (4.1) follows from (4.2) and (4.3) immediately. Thus, we completed the proof of (iii). It remains to show (iv) in this section. The proof is quite elementary owing to the fact 2 2 2 that γi+1,i = n s1,...,n 1 is an independent sequence and γi+1 χn i. One may refer{ to Section 7− of Nguyen− and} Vu [14] for instance. By using the Laplace∼ − transform trick, Nguyen and Vu [14] showed that for 0

log(γ2 /(n i)) P i+1 − < log1/2+c n = o(exp( logc/2 n)), √2 logn − − n s1 i n 1 − X≤ ≤ − which implies (v) immediately.

5. A replacement issue: Proof of (v)

In this section, we present the proof for (v). In other words, we shall replace the last s1 rows of Gaussian entries by generally distributed entries. We will need the following classical Berry–Esseen bound for sum of independent random variables. For instance, one can refer to Theorem 5.4 of Petrov [16].

Lemma 5.1. Let Z1,...,Zm be independent real random variables such that EZj =0 and E Z 3 < , j =1,...,n. Assume that | j | ∞ m m 2 2 2 3/2 3 σ = EZ ,D = σ ,L = D− E Z . j j m j m m | j | i=1 j=1 X X Then there exists a constant C> 0 such that

m P 1/2 sup Dm− Zj x Φ(x) CLm. x ≤ ! − ≤ j=1 X

Our strategy is to replace one row at each step, and derive the difference between the distributions of the logarithms of the magnitudes of two adjacent determinants. Hence, it suffices to compare two matrices with only one different row. Noting that since the magnitude of a determinant is invariant under swapping of two rows, thus without loss of generality, we only need to compare two random matrices An = (aij )n,n and A¯n =(¯aij )n,n satisfying Assumption C0 such that they only differ in the last row. More precisely, we T T assume that aij =¯aij , 1 i n 1, 1 j n and an and ¯an are independent. Here we T T ≤ ≤ − ≤ ≤ ¯ use an and ¯an to denote the nth row of An and An, respectively as above. Below we use the notation αni to denote the cofactor of ani. It is elementary that

n det An = ankαnk, kX=1 22 Z. Bao, G. Pan and W. Zhou and n det A¯n = a¯nkαnk. kX=1 Now we set

∆= α2 + + α2 . n1 ··· nn q Consider the quantities

det A n α det A¯ n α n = a nk , n = a¯ nk . ∆ nk ∆ ∆ nk ∆ Xk=1 Xk=1 By using Lemma 5.1, we obtain

n 3 P det An αni sup x n 1 Φ(x) C | 3| . x ∆ ≤ E − − ≤ ∆ i=1 X

Therefore, we have det A sup P n x Φ(x) ∆ ≤ − x

det An = sup E P x n 1 Φ(x) x ∆ ≤ E − − n 3 αnk CE | | . ≤ ∆3 Xk=1 For simplicity, we will brieﬂy denote bk(n 1) and A(n 1,k) by bk and Ank respectively in the sequel. Then by the deﬁnitions of− cofactors and− the Cauchy–Binet formula, we have

2 2 αnk det Ank T T T 1 b b − = T = det Ank(AnkAnk + k k ) Ank ∆ det A(n 1)A(n 1) − − 1 T T 1 1 T T 1 1 = det(In 1 + A− bkbk (Ank)− )− =(1+ bk (AnkAnk)− bk)− . − nk Moreover, one always has

n 3 E αnk E αnk E T T 1 1/2 | | max | | = max (1+ bk (AnkAnk)− bk)− . ∆3 ≤ k=1,...,n ∆ k=1,...,n Xk=1 Recall the deﬁnition 1 1 1 T − 1 T − G(n 1,k)(α)= AnkAnk + αIn 1 , G(n 1)(α)= A(n 1)A(n 1) + αIn 1 . − n − − n − − − The logarithmic law of random determinant 23

By using (3.12), we obtain

E T T 1 1/2 max (1+ bk (AnkAnk)− bk)− k=1,...,n

1/2 E 1 T − max 1+ bk G(n 1,k)(α)bk ≤ k=1,...,n n − 1 1/2 E 1 T − max 1+ bk G(n 1,k)(α)bk ≤ k=1,...,n n − 4a (log− n). ≤ O Thus we have

det An 4a sup P x Φ(x) C log− n, ∆ ≤ − ≤ x

and ¯ det An 4a sup P x Φ(x) C log− n. ∆ ≤ − ≤ x

Consequently, we have ¯ det An det An 4a sup P x P x C log− n, ∆ ≤ − ∆ ≤ ≤ x

which implies

2 ¯2 log det An log(n 1)! log det An log(n 1)! 4a sup P − − x P − − x C log− n. x √2 log n ≤ − √2 logn ≤ ≤

Then after s = log3a n steps of replacing, we can ﬁnally recover the logarithmic law to 1 ⌊ ⌋ general distribution case. Thus we completed the proof.

Appendix A

In this appendix, we provide the proof of Lemma 3.5 and Lemma 3.4. The proof is intrinsically the same as the counterpart in Bai [1]. For convenience of the reader, we sketch it below. For ease of notation, we represent Lemma 3.5 as follows.

Lemma A.1. Let X = (xij )p n be a random matrix, where n s1 p n and xij , 1 × − ≤ ≤ { ≤ i p, 1 j n is a collection of real independent random variables with means zero and ≤ ≤ ≤ } 4 1 T 1 variances 1. Moreover, we assume sup max Ex < . Let G(α) = ( XX + α)− , n i,j ij ∞ n 24 Z. Bao, G. Pan and W. Zhou

1/6 we have for α = n−

1 1/6 E tr G(α) = s (α)+ (n− ) n p O and

1 1/3 Var tr G(α) = (n− ), n O where

2 1 i i i − s (α)=2 α +1 + α + 1 +4α . p − n − n n s Proof. For convenience, we set 1 r (α)= E tr G(α) p p p 1E 1 = T 2 T T 1 T . p (1/n)xkxk + α (1/n )xkX(k) ((1/n)X(k)X (k)+ αIp 1)− X(k)xk Xk=1 − − Here xk is the kth row of X and X(k) is the submatrix of X by deleting its kth row. Set yp = p/n. Then we write

p 1 1 rp(α)= E p εk +1+ α yp + ypαrp(α) k=1 − X (A.1) 1 = + δ, 1+ α y + y αr (α) − p p p where n 1 1 2 1 T 1 T − T εk = (xkj 1)+ yp ypαrp(α) xkX (k) X(k)X (k)+ αIp 1 X(k)xk , n − − − n2 n − j=1 X and p 1 εk δ = δp = E −p (1 + α yp + ypαrp(α))(1 + α yp + ypαrp(α)+ εk) Xk=1 − − p 1 E εk = 2 (A.2) −p (1 + α yp + ypαrp(α)) kX=1 − p 2 1 E εk 2 . − p (1 + α yp + ypαrp(α)) (1 + α yp + ypαrp(α)+ εk) kX=1 − − The logarithmic law of random determinant 25

From (A.1), we can get

1 2 rp(α)= ( (1 + α yp ypαδ) +4ypα (1 + α yp ypαδ)). (A.3) 2ypα − − − − − q It is not diﬃcult to see that

1 1 α− . (A.4) 1+ α y + y αr (α)+ ε ≤ | − p p p k| By (A.2) and (A.4), we can get

p 1 1 2 2 δ ( Eε + α− Eε )(1+ α y + y αr (α)) . (A.5) | |≤ p | k| k − p p p Xk=1 First, note that

1 1 1 1 T − 1 T Eεk = E yp ypα tr G(α) tr X(k)X (k)+ αIp 1 X(k)X (k) | | − n − n2 n − n

1 T 1 T 1 = α E(tr(XX + αIp)− tr(X(k)X (k)+ αIp 1)) + (A.6) n| − − | n 1 = . O n E 2 Next, we come to estimate εk. Note that C Eε2 = Var ε + (Eε )2 + T + T , (A.7) k { k} k ≤ n 1 2 where

1 E 1 T 1 T − T T1 = xkX (k) X(k)X (k)+ αIp 1 X(k)xk n2 n −

1 2 E(k) 1 T 1 T − T xkX (k) X(k)X (k)+ αIp 1 X(k)xk , − n2 n −

and 2 1 1 2 α 1 T − 1 T − T2 = E tr X(k)X (k)+ αIp 1 E tr X(k)X (k)+ αIp 1 . (A.8) n2 n − − n −

(k) Here E represents the conditional expectation given xij ,i = k . Let { 6 } 1 1 T 1 T − Γk = (γij (k)) = X (k) X(k)X (k)+ αIp 1 X(k). n n − 26 Z. Bao, G. Pan and W. Zhou

Then one has C C T E tr Γ2 . (A.9) 1 ≤ n2 k ≤ nα2 Let E be the conditional expectation given x , d +1 i p, 1 j n . Deﬁne d { ij ≤ ≤ ≤ ≤ } 1 1 1 T − 1 T − γd(k)= Ed 1 tr X(k)X (k)+ αIp 1 Ed tr X(k)X (k)+ αIp 1 − n − − n − = Ed 1σd(k) Edσd(k), d =1, 2, . . ., p, − − where

1 1 1 T − 1 T − σd(k)=tr X(k)X (k)+ αIp 1 tr X(k, d)X (k, d)+ αIp 2 . n − − n − Noting that

1 σ (k) α− , | d |≤ one has

p C C T E γ2(k) . (A.10) 2 ≤ n2 | d |≤ nα2 Xd=1 Combining (A.7)–(A.10) we can get

C Eε2 . (A.11) k ≤ nα2 Substituting (A.6), (A.11) and the basic fact

2 2 (1+ α y + yαr (α)) α− − p p ≤ into (A.5) one has

C δ . | |≤ nα5 Then by (A.3) one has

1 2 1 5 rp(α)= ( (1 + α yp) +4ypα (1 + α yp)) + (n− α− ) 2ypα − − − O q 2 1 1 5 = 2( (1 + α y ) +4y α +(1+ α y ))− + (n− α− ). − p p − p O q The logarithmic law of random determinant 27

Moreover, similar to the estimate towards (A.8), we can get that 1 C Var tr G(α) . n ≤ nα4 Therefore, we can complete the proof.

Now let us prove Lemma 3.4.

Proof of Lemma 3.4. For the diagonal part, we have

n 4 E 2 E 2 2 2 2 mii(xi 1) = mjj mkkmuumvv (xjj 1)(xkk 1)(xuu 1)(xvv 1) − − − − − i=1 j,k,u,v X X

C ν m4 + ν2 m2 m2 ≤ 8 jj 4 jj kk j j=k X X6 C(ν tr M 4 + (ν tr M 2)2). ≤ 8 n 4 n For the oﬀ-diagonal part, we have

4 4 4 E E muvxuxv = muivi xui xvi u=v ui=vi,i=1,...,4 i=1 i=1 X6 6 X Y Y

C ν2 m4 + ν2ν m2 m m + ν4 m m m m ≤ 4 uv 3 2 uv ur rv 2 uv vr rw wu u=v u,v,r u,v,r,w X6 X X Cν2((tr M 2)2 + tr M 2 (tr M 4)1/2 + tr M 4) ≤ 4 n n · n n Cν2(tr M 2)2. ≤ 4 n Thus, we may complete the proof of Lemma 3.4.

Appendix B

In this appendix, we state the proof of Proposition 2.1. We use the idea in Nguyen and Vu [14]. First, we need the following lemma derived from Theorem 4.1 in G¨otze and Tikhomirov [11]. Let sn(Wn) sn 1(Wn) s1(Wn) be the ordered singular values of an n n matrix W . ≤ − ≤···≤ × n Lemma B.1. Under the assumptions of Theorem 1.1, there exist some positive constants c,C,L such that

L cn C√log n P s (A ) n− ; s (A ) n e− + . { n n ≤ 1 n ≤ }≤ √n 28 Z. Bao, G. Pan and W. Zhou

Remark B.2. It is easy to get Lemma B.1 from Theorem 4.1 of G¨otze and Tikhomirov [11] by choosing pn =1. We also remark here Theorem 4.1 of G¨otze and Tikhomirov [11] are stated for more general case under weaker moment assumption. For convenience, we just restate it under our setting.

Now by the result of Latala[ 13], it is easy to see under the assumption of Theorem 1.1,

E E 2 E 2 E 4 s1(An) C max a + max a + 4 a C√n. ≤ i ij j ij ij ≤ j i ij sX sX sX Therefore, one has E s1(An) 1/2 P s (A ) n = Cn− . { 1 n ≥ }≤ n Together with Lemma B.1 we obtain

L √log n P s (A ) n− =1 . (B.1) { n n ≥ } − O √n Now let θ follow the uniform distribution on the interval [ √3, √3] independent of 0 − An. Let θij , 1 i, j n be independent copies of θ0. And we set An′ = (aij′ ), where 2 1/≤2 ≤ (100+2L)n a′ = (1 ε ) a +ε θ . Here we choose ε = n− (say). Writing Θ = (θ ) , ij − n ij n ij n n ij n,n then by Weyl’s inequality, one has

2 1/2 (99+2L)n s (A′ ) (1 ε ) s (A ) ε Θ Cn− . | i n − − n i n |≤ nk nkop ≤ Therefore, by (B.1) we have with probability 1 √log n − √n n n 2 n/2 (99+L)n n det A′ = s (A′ )=(1 ε ) (1+ (n− )) s (A )=(1+o(1)) det A , | n| i n − n O i n | n| i=1 i=1 Y Y which implies (2.2). Moreover, by the construction, since θ0 is a continuous variable, it is obvious that (2.1) holds. Thus, we may complete the proof.

Acknowledgements

Z.G. Bao was supported in part by NSFC Grant 11371317,NSFC Grant 11101362, ZJNSF Grant R6090034 and SRFDP Grant 20100101110001; G.M. Pan was supported in part by the Ministry of Education, Singapore, under Grant # ARC 14/11; W. Zhou was supported in part by the Ministry of Education, Singapore, under Grant # ARC 14/11, and by a Grant R-155-000-131-112 at the National University of Singapore. The logarithmic law of random determinant 29 References

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Received May 2013 and revised January 2014