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Electron. J. Probab. 18 (2013), no. 29, 1–13.
ISSN: 1083-6489 DOI: 10.1214/EJP.v18-2103
On the expectation of the norm of random matrices
with non-identically distributed entries
Stiene Riemer∗
Carsten Schütt†
Abstract
Let Xi,j , i, j = 1, ..., n, be independent, not necessarily identically distributed random
variables with finite first moments. We show that the norm of the random matrix
(Xi,j )n
i,j=1 is up to a logarithmic factor of the order of
(Xi,j )n
E max (Xi,j )n
j=1 2 .
j=1 2 + E max
i=1,...,n
i=1,...,n
This extends (and improves in most cases) the previous results of Seginer and Latała.
Keywords: Random Matrix; Largest Singular Value; Orlicz Norm.
AMS MSC 2010: 60B20.
Submitted to EJP on June 26, 2012, final version accepted on February 8, 2013.
1
Introduction and Notation
We study the order of magnitude of the expectation of the largest singular value, i.e.
the norm of random matrices with independent entries
E (ai,j gi,j )ni,j=1 2→2 ,
where ai,j ∈ R, i, j = 1, . . . , n, gi,j , i, j = 1, . . . , n, are standard Gaussian random variables and k k2→2 the operator norm on `n
2 . There are two cases with a complete answer.
Chevet [2] showed for matrices satisfying ai,j = ai bj that the expectation is proportional
to
kak2 kbk∞ + kak∞ kbk2 ,
where kak2 denotes the Euclidean norm of a = (a1 , . . . , an ) and kak∞ = max1≤i≤n |ai |.
For diagonal matrices with diagonal elements d1 , . . . , dn the expectation of the norm
is of theq
order of the Orlicz norm k(d1 , . . . , dn )kM where the Orlicz function is given by
Rs
1
M (s) = π2 0 e− 2t2 dt [4]. This Orlicz norm is up to a logarithm of n equal to the norm
max1≤i≤n |di |.
∗ Christian-Albrechts Universität, Kiel, Germany. E-mail: riemer@math.uni-kiel.de
† Christian-Albrechts Universität, Kiel, Germany. E-mail: schuett@math.uni-kiel.de
Norm of random matrices
These two cases are of very different structure and seem to present essentially what
might occur concerning the structure of matrices. This leads us to conjecture that the
expectation for arbitrary matrices is up to a logarithmic factor equal to
max (ai,j )nj=1 2 + max k(ai,j )ni=1 k2 .
(1.1)
j=1,...,n
i=1,...,n
Latała [5] showed for arbitrary matrices
E (ai,j gi,j )ni,j=1 2→2 ≤ C max (ai,j )nj=1 2 + max k(ai,j )ni=1 k2 + (ai,j )ni,j=1 4 .
j=1,...,n
i=1,...,n
n,m
Seginer [12] showed for any n × m random matrix (Xi,j )i,j=1 of independent identically
distributed random variables
n,m n
m E (Xi,j )i,j=1 2→2 ≤ c E max (Xi,j )j=1 2 + E max k(Xi,j )i=1 k2 .
1≤j≤n
1≤i≤n
The behavior of the smallest singular value has been determined in [3, 7, 9, 10, 14].
Theorem 1.1. There is a constant c > 0 such that for all ai,j ∈ R, i, j = 1, ..., n and all
independent standard Gaussian random variables gi,j , i, j = 1, ..., n,
E (ai,j gi,j )ni,j=1 2→2
!! (ai,j )n i,j=1 1
n
(ai,j gi,j )nj=1 + E
≤ c ln e E
max
max
k(a
g
)
k
.
i,j
i,j
i=1 2
2
(ai,j )n i=1,...,n
j=1,...,n
i,j=1 ∞
The norm ratio appearing in the logarithm is at most n2 . In the same way we prove
Theorem 1.1 we can show the similar formula
E (ai,j gi,j )ni,j=1 2→2
!! (ai,j )n i,j=1 1
(ai,j )nj=1 + max k(ai,j )ni=1 k .
max
≤ c ln e 2
2
(ai,j )n i=1,...,n
j=1,...,n
i,j=1 ∞
The proof of this formula is essentially the same as the proof of Theorem 1.1. Through
out the proof we use the expression max (ai,j )n
j=1 2 instead of the expression E
i=1,...,n
max (ai,j gi,j )nj=1 2 .
i=1,...,n
The inequality of Theorem 1.1 is generalized to arbitrary random variables as in [5].
Theorem 1.2. Let Xi,j , i, j = 1, ..., n, be independent, mean zero random variables.
Then
2
E (Xi,j )ni,j=1 2→2 ≤ c (ln(en)) E max (Xi,j )nj=1 2 + E max k(Xi,j )ni=1 k2 .
i=1,...,n
j=1,...,n
On the other hand,
E
max (ai,j gi,j )nj=1 2
i=1,...,n
+E
max k(ai,j gi,j )ni=1 k2
j=1,...,n
(1.2)
is obviously smaller than E (ai,j gi,j )n
i,j=1 2→2 . We show that the expression (1.2) is
equivalent to the Musielak-Orlicz norm of the vector (1, . . . , 1), where the Orlicz functions are given through the coefficients ai,j , i, j = 1, . . . , n. Our formula (Theorem 3.1)
enables us to estimate from below the expectation of the operator norm in many cases
efficiently.
EJP 18 (2013), paper 29.
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Norm of random matrices
Moreover, we do not know of any matrix where the expectation of the norm is not of
the same order as (1.2).
A convex function M : [0, ∞) → [0, ∞) with M (0) = 0 that is not identically 0 is called
an Orlicz function. Let M be an Orlicz function and x ∈ Rn then the Orlicz norm of x,
kxkM , is defined by
(
)
n
X
|xi |
t > 0
M
≤1
t
kxkM = inf
.
i=1
We say that two Orlicz functions M and N are equivalent (M ∼ N ) if there are strictly
positive constants c1 and c2 such that for all s ≥ 0
M (c1 s) ≤ N (s) ≤ M (c2 s).
If two Orlicz functions are equivalent, so are their norms: For all x ∈ Rn
c1 kxkM ≤ kxkN ≤ c2 kxkM .
In addition, let Mi , i = 1, ..., n, be Orlicz functions and let x ∈ Rn then the MusielakOrlicz norm of x, kxk(Mi )i , is defined by
)
n
X
|xi |
t > 0
Mi
≤1 .
t
(
kxk(Mi )i = inf
i=1
2
The upper estimate
In this section we are going to prove the upper estimate. We require the following
known lemma. In a more general form see e.g. ([13], Lemma 10).
l
Lemma 2.1. Let x
(l)
(l)
ε1 xπ(1) , ..., εn xπ(n)
(l)
=
1
√
n−l
z }| { z }| {
(1, ..., 1, 0, ..., 0), l = 1, ..., n, and let BT be the convex hull of
l
, where εi = ±1, i = 1, ..., n, and π denote permutations of {1, ..., n}.
Let k kT be the norm on Rn whose unit ball is BT . Then, for all x ∈ Rn
kxk2 ≤ kxkT ≤
p
ln(en)kxk2 .
∗
Proof. The left hand inequality is obvious. Let x ∈ Rn . Then x∗1√
, . . . , x√
n denotes the
decreasing rearrangement of the numbers |x1 |, . . . , |xn |. Let ak = k − k − 1 for k =
1, . . . , n. Then, for all x ∈ Rn
n
X
kxkT =
√
√
x∗k ( k − k − 1).
k=1
Since
√
k−
√
k−1≤
kxkT ≤
√1
k
n √
X
√
| k − k − 1|2
! 12
k=1
kxk2 ≤
n
X
1
k
! 21
kxk2 ≤
p
ln(en)kxk2 .
k=1
We denote
STn−1
=
x = (x1 , ..., xn ) ∈ S
n−1
∃i = 1, ..., n j = 1, ..., n|xj = ± √1 = i ,
i EJP 18 (2013), paper 29.
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Norm of random matrices
the set of extremal points of BT . Then by our previous lemma we have
kAk2→2 = sup kAxk2 ≤
p
x∈S n−1
ln(en) sup kAxk2 .
(2.1)
n−1
x∈ST
We use now the concentration of sums of independent gaussian random variables
Pn
i=1 gi zi in a Banach space ([6], formula (2.35)): For all t > 0
X=
t2
P{kXk ≥ EkXk + t} ≤ exp −
2σ(X)2
,
(2.2)
where
σ(X) = sup
kξk∗ =1
n
X
! 21
|ξ(zi )|2
.
(2.3)
i=1
Applying (2.2) to

EkGxk2 ≤ 
n
X
 21
a2ij x2j 
and

 21
n
X
σ(Gx) = max 
a2ij x2j  ,
i=1...,n
i,j=1
j=1
we immediately get the following lemma.
Lemma 2.2. For all i, j = 1, ..., n let ai,j ∈ R, let gi,j be independent standard gaussians,
!
G = (ai,j gi,j )ni,j=1 and let x ∈ B2n . For all β ≥ 1 and all x with max
i=1,...,n
n
P
j=1
a2i,j x2j
> 0 we
have
P kGxk2 > β E max (ai,j gi,j )nj=1 2 + E max k(aij gij )ni=1 k2
j=1,...,n
i=1,...,n



! 12 2 



n
P 2 2



β E max (ai,j gi,j )nj=1 + E max k(ai,j gi,j )ni=1 k
 


−
a
x


i,j
j
2
2


i=1,...,n
j=1,...,n

 1
i,j=1
!
.
≤ exp −


2
n
P






max
a2i,j x2j




i=1,...,n j=1




Proposition 2.1. For all i, j = 1, ..., n let ai,j ∈ R, let gi,j be independent standard
pπ
gaussian random variables and let G = (ai,j gi,j )n
i,j=1 . For all β with β ≥
2
p
P kGk2→2 > β ln(en) E max (ai,j gi,j )nj=1 2 + E max k(ai,j gi,j )ni=1 k2
i=1,...,n
≤
n
X
2
exp l ln(2n) − l
l=1
β
π
.
Furthermore, we get for β =
P kGk2→2 >
≤
√
j=1,...,n
p
3π ln(2n)
n n
3π ln(en) E max (ai,j gi,j )j=1 2 + E max k(ai,j gi,j )i=1 k2
i=1,...,n
j=1,...,n
1
.
n2
EJP 18 (2013), paper 29.
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Norm of random matrices
n
P
Proof. We shall apply Lemma 2.2. We may assume that max
i=1,...,n
j=1
!
a2i,j x2j
> 0 for all
x ∈ B2n \ {0}. By (2.1)
kGk2→2 ≤
p
ln(en) sup kGxk2 .
n−1
x∈ST
Therefore, for β ∈ R>0 , we have
n
n max k(ai,j gi,j )i=1 k2
P kGk2→2 > β ln(en) E max (ai,j gi,j )j=1 2 + E
j=1,...,n
i=1,...,n
!
n
n
max k(ai,j gi,j )i=1 k
≤P
sup kGxk > β E max (aij gij )j=1 + E
.
p
2
n−1
x∈ST
2
i=1,...,n
2
j=1,...,n
(l)
For all l = 1, ..., n let Ml be the set of x(l) ∈ STn−1 , such that xj
j = 1, ..., n. Now we apply Lemma 2.2 and get
∈ {0, ± √1l } for all
p
n n
P kGk2→2 > β ln(en) E max (ai,j gi,j )j=1 2 + E max k(ai,j gi,j )i=1 k2
j=1,...,n
i=1,...,n
(
)
n
n
≤P
sup kGxk > β E max (aij gij )j=1 + E max k(ai,j gi,j )i=1 k
2
n−1
x∈ST
≤
n
X
X
2
i=1,...,n
2
j=1,...,n
(l) n n
P Gx > β E max (aij gij )j=1 2 + E max k(ai,j gi,j )i=1 k2
j=1,...,n
i=1,...,n
2
l=1 x(l) ∈Ml






 


n n

E
max
(a
g
)
+
E
max
k(a
g
)
k
n
ij ij j=1 2
i,j i,j i=1 2
 1 
X
X
i=1,...,n
j=1,...,n

≤
exp − · l β
1

2

2 

l=1 x(l) ∈Ml


n
P




max 
a2ij 


i=1,...,n
(l)
j∈{k|xk 6=0}
 12 2 



P



1
2


√

a


i,j
l

 
i=1 j∈{k|x(l) 6=0}

k
.
−
 12 

 

 
n
P


max 
a2i,j   


i=1,...,n

(l)
j∈{k|x 6=0}

n
P
k
We have

n
1 X
√ 
l i=1
 12
X

a2ij  ≤
(l)
j∈{k|xk 6=0}
max
(l)
j∈{k|xk 6=0}
k(aij )ni=1 k2 ≤ max k(aij )ni=1 k2
j=1,...,n
and by triangle inequality
E max k(ai,j gi,j )ni=1 k2
j=1,...,n
≥
≥
max E
j=1,...,n
max
j=1,...,n
n
X
! 12
|ai,j gi,j |2
(2.4)
i=1
n
X
! 21
2
2
|ai,j | (E|gi,j |)
i=1
EJP 18 (2013), paper 29.
r
=
2
max k(ai,j )ni=1 k2 .
π j=1,...,n
ejp.ejpecp.org
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Norm of random matrices
Therefore, we have for all β with β ≥
pπ
2

n
1 X
βE max k(ai,j gi,j )ni=1 k2 − √ 
j=1,...,n
l i=1
 12
X

a2i,j  ≥ 0.
(l)
j∈{k|xk 6=0}
Thus
p
P kGk2→2 > β ln(en) E max (ai,j gi,j )nj=1 2 + E max k(ai,j gi,j )ni=1 k2
j=1,...,n
i=1,...,n

2 







 








n


(a
g
)
E
max
n
i,j
i,j




j=1 2
X X
i=1,...,n
1 

≤
exp − · l β

 12   .


2 

l=1 x(l) ∈Ml
 



n
P



 max 
2   


a


i,j


i=1,...,n
(l)
j∈{k|xk 6=0}
Again, by (2.4) we have for all β with β ≥
P kGk2→2
≤
i=1,...,n
exp −l
P kGk2→2
≤
p
n
X
j=1,...,n
2
2l nl exp −l
l=1
2
exp l ln(2n) − l
We choose β =
≤
β
π
l=1
n
X
2
l=1 x(l) ∈Ml
=
2
p
> β ln(en) E max (ai,j gi,j )nj=1 2 + E max k(ai,j gi,j )ni=1 k2
n
X
X
n
X
pπ
β
π
β
π
.
β2
π .
3π ln(2n), thus such that 3 ln(2n) =
Then
p
n n
> 3π ln(2n) ln(en) E max (aij gij )j=1 2 + E max k(aij gij )i=1 k2
i=1,...,n
exp (−2l ln(2n)) =
n X
l=1
l=1
1
4n2
l
≤
1
n2
j=1,...,n
∞ X
l=1
1
4
l
≤
1
.
n2
Therefore
P kGk2→2 >
≤
√
n n
3π ln(en) E max (aij gij )j=1 2 + E max k(aij gij )i=1 k2
i=1,...,n
j=1,...,n
1
.
n2
Proposition 2.2. Let ai,j ∈ R, i, j = 1, ..., n and gi,j , i, j = 1, ..., n, be independent
standard Gaussian random variables, then
E (ai,j gi,j )ni,j=1 2→2
r
π √
≤
+ 3π ln(en)
E max (ai,j gi,j )nj=1 2 + E max k(ai,j gi,j )ni=1 k2 .
i=1,...,n
j=1,...,n
2
EJP 18 (2013), paper 29.
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Norm of random matrices
Proof. We divide the estimate of E (ai,j gi,j )n
i,j=1 2→2
into two parts. Let M be set of
all points with
(ai,j gi,j )ni,j=1 2→2
√
≤ 3π ln(en) E max (ai,j gi,j )nj=1 2 + E max k(ai,j gi,j )ni=1 k2 .
j=1,...,n
i=1,...,n
Clearly,
E (ai,j gi,j )ni,j=1 2→2 χM
√
n n
≤ 3π ln(en) E max (ai,j gi,j )j=1 2 + E max k(ai,j gi,j )i=1 k2 .
j=1,...,n
i=1,...,n
Furthermore, by Cauchy-Schwarz inequality and Proposition 2.1 we get
p
2 21
E (ai,j gi,j )ni,j=1 2→2 χM c ≤ P (M c ) E (ai,j gi,j )ni,j=1 2→2


 12
 12
n
n
1
X
X
1
1
2
2
E (ai,j gi,j )ni,j=1 HS
≤
= 
|ai,j |2  ≤ max 
|ai,j |2  .
1≤i≤n
n
n i,j=1
j=1
As in (2.4)
max (aij )nj=1 2 + max k(aij )ni=1 k2
i=1,...,n
j=1,...,n
r
π
E max (aij gij )nj=1 2 + max k(aij gij )ni=1 k2 .
≤
j=1,...,n
i=1,...,n
2
Altogether, this yields
rπ n
n n
E (ai,j gi,j )i,j=1 2→2 χM c ≤
E max (ai,j gi,j )j=1 2 + max k(ai,j gi,j )i=1 k2 .
j=1,...,n
i=1,...,n
2
Summing up, we get
E (ai,j gi,j )ni,j=1 2→2
r
π √
n n
≤
+ 3π ln(en)
E max (ai,j gi,j )j=1 2 + E max k(ai,j gi,j )i=1 k2 .
i=1,...,n
j=1,...,n
2
Proof. (Theorem 1.1) W.l.o.g. we assume 0 ≤ ai,j ≤ 1, i, j = 1, ..., n and that there is a
coordinate that equals 1. For all i, j = 1, ..., n and k ∈ N we define
aki,j
=
1
2k
0
, if 21k < ai,j ≤
, else.
1
2k−1
k
k
n
Let G = (ai,j gi,j )n
of
i,j=1 and G = (ai,j gi,j )i,j=1 . Wedenote byφ(k) the
number
nonzero
k n
n
γ
n
entries in (ai,j )i,j=1 and we choose γ such that (ai,j )i,j=1 1 = 2 (ai,j )i,j=1 ∞ . Thus,
we get φ(k) 21k =
n
P
i,j=1
aki,j ≤ 2γ and therefore φ(k) ≤ 2k+γ . Therefore, the non-zero
entries of Gk are contained in a submatrix of size 2k+γ × 2k+γ . Taking this into account
and applying Proposition 2.2 to Gk
E Gk 2→2
r
π √
≤
+ 3π ln(e2k+γ ) E max (aki,j gi,j )nj=1 2 + max (aki,j gi,j )ni=1 2 .
i=1,...,n
j=1,...,n
2
EJP 18 (2013), paper 29.
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Norm of random matrices
Since
r
π √
13 31
+ 3π ln(e2k+γ ) ≤
+ (1 + ln(2k+γ )) ≤ 8(k + γ)
2
10 10
we get

 12
n
X
k
γ
k
|aki,j gi,j |2  ≤ 8(k + γ)2 2 − 2 .
E G 2→2 ≤ 8(k + γ) E
i,j=1
Therefore,
∞
X k+γ
X
k
E Gk 2→2 ≤ 8
≤ 16
k
k .
4
4
k≥2γ 2
k≥2γ
k=1 2
X
Since one of the coordinates of the matrix is 1
E G1 2→2 ≥
∞
Z
r
|g|dt =
−∞
2
.
π
Therefore, there is a constant c such that
E kGk2→2
X
k
G ≤ 2E k≤2γ +2
2→2
The matrix
X
EkGk k2→2
k>2γ
X
k
G ≤ cE k≤2γ .
2→2
k
P
k≤2γ
G has at most
X
φ(k) ≤
k≤2γ
X
!3
(ai,j )n i,j=1 1
(ai,j )n 2γ+k ≤ 23γ+1 ≤ 2
(2.5)
i,j=1 ∞
k≤2γ
k
entries that are different from 0. Therefore, all nonzero entries of
k≤2γ G are contained in a square submatrix having less than (2.5) rows and columns. We may apply
Proposition 2.2 and get with a proper constant c
P


!3 
r
(ai,j )n √
π
i,j=1 1
 ×
E (kGk2→2 ) ≤c 
+ 3π ln e · 2 (ai,j )n 2
i,j=1 ∞



n n
X
X



aki,j gi,j  aki,j gi,j 
E  max max + j=1,...,n
i=1,...,n k≤2γ
k≤2γ
j=1 2
3


.
i=1 2
The lower estimate
Theorem 3.1. For all i, j = 1, ..., n let ai,j ∈ R≥0 and gi,j be independent standard
−1
gaussians. For all i = 1, ..., n and all s with 0 ≤ s < (ai,j )n
j=1 2

Ni (s) = s max ai,j exp −
j=1,...,n
s2

1

max a2i,j
j=1,...,n
−1
and for all s ≥ (ai,j )n
j=1 2
Ni (s) =
max ai,j
j=1,...,n
(ai,j )n j=1 2
 
(ai,j )n 2
j=1 2
 + 3 (ai,j )nj=1 exp −
2
2
max ai,j
e
j=1,...,n
EJP 18 (2013), paper 29.
1
s− (ai,j )n j=1 2
!
.
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Norm of random matrices
ej , j = 1, ..., n, are defined in the same way for the transposed matrix
Moreover, N
ej , j = 1, . . . , n, are Orlicz functions and
(ai,j )nj,i=1 . Then Ni , i = 1, . . . , n and N
c1 k(1)ni=1 k(Ni )i + (1)nj=1 (N
fj )j
max k(ai,j gi,j )ni=1 k2
≤ E max (ai,j gi,j )nj=1 2 + E
j=1,...,n
i=1,...,n
n n
≤ c2 k(1)i=1 k(Ni )i + (1)j=1 (N
,
f)
j j
where c1 and c2 are absolute constants.
The following example is an immediate consequence of Theorem 3.1.
Toeplitz matrices.
It covers
Example 3.2. Let A be a n × n-matrix such that for all i, = 1 . . . , n and k = 1, . . . , n

n
X

1

2
|ai,j |
2
=
j=1
1
2
n
X
|aj,k |
2
j=1
and
max |ai,j | = max |aj,k |.
1≤j≤n
1≤j≤n
Then
E
max
i=1,.....,n
(ai,j gi,j )n j=1 2
+E
max
j=1,.....,n
k(ai,j gi,j )n
i=1 k2


1


2
n
 X

√
2
|a1,j |  , ln n max |a1,j | .
∼ max 


1≤j≤n
 j=1

We associate to a random variable X a Orlicz function M by
Z
s
Z
|X|dPdt.
M (s) =
0
(3.1)
1
t ≤|X|
We have
Zs Z
|X|dPdt
M (s) =
0
Zs
=
1
t ≤|X|


Z∞
1
1

+ P (|X| ≥ u) du dt.
 P |X| ≥
t
t
0
(3.2)
1
t
Lemma 3.3. There are strictly positive constants c1 and c2 such that for all n ∈ N, all
independent random variables X1 , ..., Xn with finite first moments and for all x ∈ Rn
c1 kxk(Mi )i ≤ E max |xi Xi | ≤ c2 kxk(Mi )i ,
1≤i≤n
where M1 , ..., Mn are Orlicz functions that are associated to the random variables (3.1).
Lemma 3.3 is a generalization of the same result for identically distributed random
variables [4]. It can be generalized from the `∞ -norm to Orlicz norms.
We use the fact [11] that for all x > 0
√
1 2
2π
√
e− 2 x ≤
2
(π − 1)x + x + 2π
q Z
2
π
∞
1
2
e− 2 s ds ≤
x
EJP 18 (2013), paper 29.
q
2 1 − 21 x2
.
π xe
(3.3)
ejp.ejpecp.org
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Norm of random matrices
Proof. (Theorem 3.1) We apply Lemma 3.3 to the random variables

Xi = 
n
X
 12
|ai,j gi,j |2 
i = 1, . . . , n.
j=1
Now, it is enough to show that Mi ∼ Ni for all i = 1, . . . , n. We have two cases.
12 −1
Pn
2
2
E
a
g
. There are constants c1 , c2 > 0 such
j=1 i,j i,j
! 12
n
P
2
2
that for all u with u > 2E
ai,j gi,j
1
2
We consider first s <
j=1

u
max a2i,j
exp −c21



 12
n
X


2 
 ≤ P 
a2i,j gi,j
≥ u ≤ exp −c22

2
j=1
j=1,...,n

u2
.
max a2i,j
(3.4)
j=1,...,n
The right-hand side inequality follows from (2.2). The left-hand side inequality follows
from
n
X
2
2
a2i,j gi,j
≥ a2i,1 gi,1
.
j=1
Since
1
t
! 21
n
P
> 2E
j=1
Mi (s) =
for 0 < t < s, we can apply (3.4). Therefore,

 


 21
 12

Z∞
n
n

X
X
1 
1


2
2 
2
2 
ai,j gi,j
P
≥  + P
ai,j gi,j
≥ u du dt


t

t
j=1
j=1

Zs 

0
≤
2
a2i,j gi,j
1
t

Zs 
1

t
0


exp −
t2
c22
+
max a2i,j
j=1,...,n
Z∞
1
t



u2
 du dt.
exp −c22

max a2i,j

j=1,...,n


By (3.3)
Mi (s) ≤

Zs 





2
2
1
t
c
c
2
2
2
+
 dt.
−
exp − 2
max
a
exp
i,j
t
t max a2i,j
2c22 j=1,...,n
t2 max a2i,j 
j=1,...,n
0
Since
1
t
> 2E
! 21
n
P
j=1
2
a2i,j gi,j
≥
j=1,...,n
q 2 n π (ai,j )j=1 2 , we get
1
1
t
1
+ 2 max a2i,j ≤ + 2
t
2c2 j=1,...,n
t
2c2
r
π
1
c
(ai,j )nj=1 2 ≤ ,
n
2
2 (ai,j )j=1
t
2
where c is an absolute constant. Altogether, we get
Zs
Mi (s) ≤
0




Z∞
2
2 2
c
c
c
c
u
2
2
 dt =
 du.
exp − 2
exp −
t
t max a2i,j
u
max a2i,j
j=1,...,n
1
s
j=1,...,n
Passing to a new constant c2 and using (3.3) we get for all s with 0 ≤ s <
Z∞
Mi (s) ≤ c
1
s
1
2
12 −1
Pn
2
2
E
a
g
j=1 i,j i,j




2 2
2
1
c
u
s
c
2
2
 du ≤ ( max ai,j ) exp −
 . (3.5)
exp −
u
max a2i,j
c2 j=1,...,n
s2 max a2i,j
j=1,...,n
j=1,...,n
EJP 18 (2013), paper 29.
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Norm of random matrices
From this and the definition of Ni we get that there is a constant c such that for all s
with 0 ≤ s <
1
2
12 −1
Pn
2
2
E
j=1 ai,j gi,j
Mi (s) ≤ Ni (cs).
Indeed, the inequality follows immediately from (3.5) provided that
c2
2
− 12
P
n
j=1
|ai,j |2
12
P
n
2
2
,
E
j=1 ai,j gi,j
If
Moreover,
1
2
P
n
j=1
|ai,j |2
− 12
P
n
j=1
12 
P
n
2
2
2
g
c
(E
a
)2  √2π
j=1 i,j i,j
 2
.
≤
12 exp −
4 max a2i,j
c2

j=1,...,n
c2 E
≤
2
a2i,j gi,j
√
j=1,...,n
√
!
!
2π
n
−1
+ 1 k(ai,j )j=1 k
.
c2
2π
≤ Ni
c2
Therefore, with a universal constant c the inequality Mi (s) ≤ Ni (cs) also holds for those
values of s. The inverse inequality is treated in the same way.

Now we consider s with s ≥
1
2
E
n
P
j=1
! 12 −1
2
a2i,j gij

and denote α = E
n
P
j=1
! 21
2
a2ij gij
.
The following holds
Mi (s) =
=
 



 12
 12

Z∞
n
n

X
X
1
1 


2
2 
2
2 
ai,j gi,j
≥  + P
≥ u du dt
P
ai,j gi,j


t

t
j=1
j=1

Zs 

0
1
t


 

 12
 12

Z∞
n
n

X
X
1



2
2 
2 2
ai,j gi,j
P
≥  + P
≥ u du dt
aij gij


t
t

j=1
j=1

1
Z2α 
1
0
1
t
+



 
 12
 12

Z∞
n
n

X
X
1



2 
2 
a2i,j gi,j
a2i,j gi,j
P 
≥  + P 
≥ u du dt


t
t

j=1
j=1

Zs 
1
1
2α
1
t
1
The first summand equals Mi ( 2α
). Therefore, by (3.5) the first summand is of the order
! 12 2 

2
E
 
a2i,j gi,j


max ai,j


j=1
j=1,...,n
 2

−c2
.
! 12 exp 
2


max
a
i,j
n
P


j=1,...,n
2
2


E
ai,j gi,j


n
P
j=1
We estimate the second summand. The second summand is less than or equal to

 12 

 21

 12

Zs
n
n
n

X
X
X
1
2
2 
2 
+ E
a2ij gij
dt ≤ 3E 
a2i,j gi,j
dt ≤ 3E 
a2i,j gi,j
s.


t

j=1
j=1
j=1
1

Zs 

1
2α
.
q
12 −1
Pn
2
2
2
≤ s ≤ 12 E
a
g
then,
by
(3.5)
and
j=1 i,j i,j
π max1≤j≤n ai,j ≤
2 max ai,j
Mi (s) ≤
s
c2
2α
EJP 18 (2013), paper 29.
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Norm of random matrices

Therefore, with a universal constant c we have for all s with s ≥

Mi (s) ≤ (c − 1)s 
n
X
1
2
n
P
E
j=1

Mi (s) ≥
n
X

2
α
1
t≤
n
P
j=1
!1


 21
n
X
|ai,j |2  − 1 ≤ Ni (cs).
|ai,j |2  ≤ cs 
j=1
Now, we sketch a proof of a lower estimate. By (3.1), for all s with s ≥
Z
2
a2i,j gij
 12
j=1
Zs
! 12 −1
 21
2 
dPdt ≥
a2i,j gi,j
j=1
Zs
2
α
2
2
a2i,j gi,j

 12
n
X
2

dPdt.
a2i,j gij
Z
α
2≤
n
P
j=1
2
α
!1
j=1
2
2
a2i,j gi,j
By the definition of α
Mi (s) ≥
1
2
Zs

 12
n
X
2 
dt
E
a2i,j gi,j
j=1
2
α

 
 12 −1
n
n
X

1
 X 2 2   
2  
a2i,j gi,j
ai,j gi,j
E
s
−
2
= E

 


2
j=1
j=1

 21


 21
n
1 X 2 2 
ai,j gi,j
= E
s − 1.
2
j=1
The rest is done as in the case of the upper estimate.
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