DSpace at VNU: On the permanents of matrices with restricted entries over finite fields tài liệu, giáo án, bài giảng , l...
Trang 1ON THE PERMANENTS OF MATRICES WITH RESTRICTED
LE ANH VINH†
Abstract For a prime power q, we study the distribution of permanents of n × n matrices over
a finite field Fq of q elements We show that if A is a sufficiently large subset of F q, then the set
of permanents ofn × n matrices with entries in A covers all (or almost all) of F ∗
q Whenq = p is a
prime, andA is a subinterval of [0, p − 1] of cardinality |A| p 1/2logp, we show that the number
of matrices with entries inA having permanent t is asymptotically close to the expected value.
Key words permanents, matrices with restricted entries, finite fields AMS subject classifications 15A15, 15B33, 11L07
DOI 10.1137/110835050
1 Introduction Throughout this paper, let q = p r , where p is an odd prime and r is a positive integer Let F q be a finite field of q elements The prime base
field Fp ofFq may then be naturally identified withZp =Z/pZ Let M be an n × n matrix Two basic parameters of M are its determinant
Det(M ) :=
sgn(σ)
n
i=1
and its permanent
Per(M ) :=
n
i=1
The distribution of the determinant of matrices with entries in a finite field Fq
has been studied by various researchers Suppose that the ground fieldFq is fixed and
M = M n is a random n × n matrix with entries chosen independently from F q If the entries are chosen uniformly fromFq, then it is well known that
i1
(1− q −i ) as n → ∞.
It is interesting that (1.1) is quite robust Specifically, Kahn and Koml´os [9] proved
a strong necessary and sufficient condition for (1.1)
Theorem 1.1 (see [9]) Let M n be a random n × n matrix with entries chosen according to some fixed nondegenerate probability distribution μ on F q Then (1.1) holds if and only if the support of μ is not contained in any proper affine subfield
of Fq
∗Received by the editors May 23, 2011; accepted for publication (in revised form) April 17, 2012;
published electronically July 19, 2012 This research was supported by Vietnam National University,
Hanoi under the project Some problems on matrices over finite fields An extended abstract of this work appeared in Proceedings of the European Conference on Combinatorics, Graph Theory and
Applications (EuroComb 2009), Electron Notes Discrete Math 34, Elsevier Sci B.V., Amsterdam,
2009, pp 519–523 [13].
http://www.siam.org/journals/sidma/26-3/83505.html
†University of Education, Vietnam National University, Hanoi, Vietnam (vinhla@vnu.edu.vn).
997
Trang 2An extension of the uniform limit to random matrices with μ depending on n was
considered by Kovalenko, Leviskaya, and Savchuk [10] They proved that the standard
limit (1.1) holds under the condition that the entries m ij of M are independent and Pr(m ij = α) > (log n + ω(1))/n for all α ∈ F q The behavior of the nullity of M n for
1− μ(0) close to log n/n and μ(α) = (1 − μ(0))/(q − 1) for α = 0 was also studied by
Bl¨omer, Karp, and Welzl [3]
Another direction is to fix the dimension of matrices For an integer number n
and a subset E ⊆ F n
q , let M n(E) denote the set of n × n matrices with rows in E.
For any t ∈ F q , let D n(E; t) be the number of n × n matrices in M n(E) having
determinant t Ahmadi and Shparlinski [1] studied some natural classes of matrices
over finite fieldsFp of p elements with components in a given subinterval [−H, H] ⊆
[−(p − 1)/2, (p − 1)/2] They showed that
; t) = (1 + o(1)) (2H + 1)
p
if t ∈ F ∗ q and H q 3/4 In the case n = 2, the lower bound can be improved to
H q 1/2+ε for any constant ε > 0.
Covert et al [4] studied this problem in a more general setting A subsetE ⊆ F n
called a product-like set if|H d ∩ E| |E| d/n for any d-dimensional subspace H d ⊂ F n
Covert et al [4] showed that
D3(E; t) = (1 + o(1)) |E|3
q
if t ∈ F ∗ q and E ⊂ F3
q is a product-like set of cardinality |E| q 15/8 Using the
geometry incidence machinery developed in [4], and some properties of nonsingular matrices, the author [12] obtained the following result for higher dimensional cases
(n ≥ 4):
D n(A n ; t) = (1 + o(1)) |A| n2
q
if t ∈ F ∗ q andA ⊆ F q of cardinality|A| q d
2d−1
On the other hand, little has been known about the permanent The only known uniform limit similar to (1.1) for the permanent is due to Lyapkov and Sevast’yanov
[11] They proved that the permanent of a random n × m matrix M nmwith elements fromFp and independent rows has the limit distribution of the form
lim
n→∞ Pr(Per(M nm ) = k) = ρ m δ k0+ (1− ρ m )/p, k ∈ F p ,
where δ k0 is the Kronecker delta Here, as usual, the permanent Per(A) of an n × m matrix A is defined as the sum of all possible products of n coefficients of A chosen
such that no two of the coefficients are taken from the same row nor from the same column The purpose of this paper is to study the distribution of the permanent when
the dimension of matrices is fixed For any t ∈ F q and E ⊂ F d
q , let P n(E; t) be the
number of n × n matrices with rows in E having determinant t We are also interested
in the set of all permanents, P n(E) = {Per(M) : M ∈ M n(E)} The first result of this
paper is the following
Trang 3Theorem 1.2 Suppose that q is an odd prime power and gcd(q, n) = 1.
(a) If E ∩ (F ∗
q)n = ∅, and |E| > cq n+1
2 , thenF∗
(b) If A ⊆ F q of cardinality |A| q 2/3 , then for each t ∈ F ∗ q ,
P3(A3; t) = (1 + o(1)) |A|9
q .
We have an immediate corollary of Theorem 1.2
Corollary 1.3 Suppose that q is an odd prime power and gcd(q, n) = 1.
(a) If E ⊂ F n
q of cardinality |E| > nq n−1 , thenF∗
(b) If A ⊂ F q of cardinality |A| q1 + 1
2n , then F∗
Note that the bound in the first part of Corollary 1.3 is tight up to a factor of n.
For example,|{x ∈ F n
q : x1= 0}| = q n−1 and P n({x ∈ F n
q : x1 = 0}) = 0 When E is
a product-like set, we can get a positive proportion of the permanents under a weaker assumption
Theorem 1.4 Suppose that q is an odd prime power and gcd(q, n) = 1 If
E ⊂ F n
q is a product-like set of cardinality |E| q n2
2n−1 , then
|P n(E)| (1 + o(1))q.
In the special caseE = A × · · · × A, we have the following corollary.
Corollary 1.5 Suppose that q is an odd prime power and gcd(q, n) = 1 If
A ⊆ F q of cardinality |A| q1+2n−11 , then |P n(A n)| (1 + o(1))q.
Furthermore, if we restrict our study to matrices over a finite fieldFp of p elements (p is a prime) with components in a given interval, we obtain a stronger result.
Theorem 1.6 Suppose that q = p is a prime, and entries of M are chosen from
a given interval I := [a + 1, a + b] ⊆ F p , where
b
p 1/2 log p → ∞, p → ∞;
then
P n(I n ; t) = (1 + o(1)) b
p for any t ∈ F p
As we will see in the proof of Theorem 1.6, the method can be applied to a large variety of multihomogeneous forms In particular, we have a similar result for the determinant
Theorem 1.7 Suppose that q = p is a prime and entries m ij of M are chosen from a given interval I := [a + 1, a + b] ⊆ F p with
b
p 1/2 log p → ∞, p → ∞;
then
D n(I n ; t) = (1 + o(1)) b
p for any t ∈ F p
Trang 4This result improves [1, Theorem 11] and [12, Theorem 1.5] The proof of Theo-rem 1.7 (and the statement for multihomogeneous forms) is left for interested readers
Throughout the paper, the implied constants in the symbols o and may depend
on integer parameter n We recall that the notation U V is equivalent to the assertion that the inequality U c|V | holds for some constant c > 0.
2 Some estimates.
2.1 Incidence estimates Let f be a complex-valued function on F n q The
Fourier transform of f on F n q with respect to a nontrivial principal additive character
ψ on F q is given by
ˆ
x∈F n q
f (x)ψ(−x · m).
One of our main tools is a two-set version of the geometric incidence theorem developed
by Hart et al in [8] (see also [7] for an earlier version and [4] for a functional version
of this theorem) Note that going from a one-set formulation in the proof of Theorem 2.1 in [8] to a two-set formulation is just a matter of inserting a different letter in a couple of places
Theorem 2.1 (see [8, Theorem 2.1]) For any E, F ⊂ F n
q , let
ν t(E, F) =
x·y=t E(x)F(y),
where here and throughout the paper, E(x) denotes the characteristic function of E.
(a) Then
(2.1) ν t(E, F) = |E||F|q −1 + R t(E, F),
where
|R t(E, F)|2 |E||F|q n−1 ,
if t = 0, and
|R0(E, F)|2 |E||F|q n
(b) Moreover, if (0, , 0) ∈ E, then
ν t2(E, F) |E|2|F|2q −1+|E|q 2n−1
k=(0, ,0)
| ˆ F(k)|2|E ∩ l k |, where lk={tk : t ∈ F ∗
2.2 Arithmetic estimates The congruence
where p is a large prime, arises in many problems of number theory (see, for example, [2, 5] and the references therein) Let L i , N i, 1 i 4, be integers with 0 L i <
L i + N i < p Denote by J(λ) the number of solutions of congruence (2.3) in the box
(2.4) L i+ 1 x i L i + N i (1 i 4).
Ayyad, Cochrane, and Zheng [2] proved that
(2.5) J(0) = N1N2N3N4
N1N2N3N4log2p).
Trang 5It has been shown by Garcia [6], in a personal communication, that a similar estimate
holds for J(λ) when λ = 0.
Theorem 2.2 (see [6]) The following asymptotic formula holds for any λ ∈ F q :
(2.6) J(λ) = N1N2N3N4
N1N2N3N4log2
p).
Proof Writing J(λ) in terms of character sums, we have
p − 1
χ
χ(x ∗1x ∗2(λ − x3x4)),
where χ runs over the set of multiplicative characters modulo p, and the range for the variables in summations over x1, x2, x3, and x4 is defined by (2.4) In addition,
x ∗ denotes the inverse multiplicative of x (mod p) Separating the principal character
χ = χ0, we have
J(λ) = N1N2N3N4
By the Cauchy–Schwarz inequality, we have
p − 1
χ(x ∗1x ∗2)
χ(λ − x3x4)
⎛
⎝ 1
p − 1
χ(x ∗1x ∗2)
2⎞
⎠
1/2⎛
⎝ 1
p − 1
χ(λ − x3x4)
2⎞
⎠
1/2
= J1− N2N2
p − 1
1/2
J2− N2N2
p − 1
1/2
,
(2.7)
where, J1and J2denote, respectively, the number of solutions of the congruence
x1x2≡ x
1x 2(mod p), L1+ 1 x1, x 1 L1+ N1, L2+ 1 x2, x 2 L2+ N2;
λ − x3x4≡ λ − x
3x 4(mod p), L3+ 1 x3, x 3 L3+ N3, L4+ 1 x4, x 4 L4+ N4.
It follows from (2.5) that
J1− N2N2
p − 1 1N2log
2p,
(2.8)
J2− N2N2
p − 1 3N4log
2
p.
(2.9)
The theorem follows immediately from (2.7), (2.8), and (2.9)
Corollary 2.3 For any L1, L2, L3, L4, N1, N2, N3, N4 with
N i
p 1/2 log p → ∞, p → ∞, let J(λ) be the number of solutions of
x1x4+ x2x3= λ,
where x i ∈ [L i +1, L i +N i ] (mod p) Then J(λ) = (1+o(1)) N1N2N3N4
q for any λ ∈ F p Proof If [L i + 1, L i + N i ] (mod p) is not a subinterval of [1, p − 1], we break
it into two disjoint subintervals and an element 0 Applying Theorem 2.2 for these subintervals, and adding up the results, we obtain the corollary
Trang 62.3 Estimation of matrices with small per-rank Let P n(2)(A n) be the
number of n × n matrices M with entries in A such that all (n − 1) × (n − 1) minors
of M have permanent 0.
Lemma 2.4 For any fixed n 2 and A ⊂ F q , we have
P n(2)(A n ) = O(|A| n2−4 ).
Proof The proof proceeds by induction The base case n = 2 is trivial Suppose
that the statement holds for n − 1 We show that it also holds for n For any n × n matrix M = (x ij)1i,jn with entries in A, let M st (2 s, t n) be the submatrix obtained from M by deleting the first row, the sth row, the first column, and the tth column Suppose that all (n − 1) × (n − 1) minors of M have permanent 0 We have
two cases
(1) Suppose that Per(M st)= 0 for some 2 s, t n Without loss of generality,
we assume that Per(M22)= 0 For 1 i, j 2, let M ij be the (n − 1) × (n − 1) minor of M obtained by deleting the ith column and the jth row Let i1= 3− i and
j1= 3− j; then
Per(M ij ) = x i1j1Per(M22) + α for some α does not depend on x i1j1 Therefore, if we fix M 22with Per(M22)= 0, there
is at most one choice of x i1j1 such that Per(M ij) = 0 Hence, this case contributes at mostn−1
2 |A| n2−4 = O(|A| n2−4 ) matrices whose (n − 1) × (n − 1) minors have zero
permanent
(2) Suppose that Per(M st) = 0 for all 2 s, t n From the induction
hypoth-esis, this case contributes at most|A| 2(n−1)+1 P n−1(2) (A n−1 ) = O(|A| 2n−1+(n−1)2−4) =
O(|A| n2−4 ) matrices whose (n − 1) × (n − 1) minors have zero permanent.
Putting these two cases together, we complete the proof of the lemma
3 Proof of Theorem 1.2 (a) Fix an a = (a1, , a n)∈ E ∩ (F ∗
q)n For any
x = (x1, , x n), andy = (y1, , y n)∈ E, let M(a; x, y) denote the matrix whose
rows arex, y and (n − 2) a’s Letting 1 := (1, , 1), x/a := (x1/a1, , x n /a n), and
y/a := (y1/a1, , y n /a n), we have
Per(M (a; x, y)) =
n
i=1
a i Per(M (1; x/a, y/a))
=
n
i=1
a i
n
i=1
x i
a i
j=i
y j
a j
Set
E1:={(x i /a i)n i=1 : (x1, , x n)∈ E} ,
(3.1)
E2:=
⎧
⎨
⎩
⎛
j=i
y i /a i
⎞
⎠
n
i=1
: (y1, , y n)∈ E
⎫
⎬
⎭. (3.2)
It is clear that |E1| = |E2| = |E| (as gcd(n, q) = 1) From (2.1), for any t ∈ F ∗
have
(3.3) ν t(E1, E2) = q −1 |E1||E2| + O q d−1 |E1||E2|
= q −1 |E|2+ O(q (d−1)/2 |E|).
Trang 7This follows that F∗
q ⊂ {Per(M(a; x, y)) : x, y ∈ E} if |E| > cq (d+1)/2 for some large
constant c > 0.
(b) For anya, x, y ∈ A3, let M (a, x, y) denote the 3 × 3 matrix whose rows are
a, x, and y We have three cases.
(1) Suppose thata ∈ (A ∩ F ∗
q)3 Similarly as above, for a fixeda, we have
|{Per(a, x, y) = t : x, y ∈ A3}| = |A|6
q + O(q|A|
3) = (1 + o(1)) |A|6
q .
Since|(A ∩ F ∗
q)3| = (1 + o(1))|A|3,
(3.4) |{Per(a, x, y) = t : a ∈ (A ∩ F ∗
q)3, x, y ∈ A3}| = (1 + o(1)) |A|9
q .
(2) Suppose thata = (0, a2, a3) for some a2, a3∈ A We have
Per(a, x, y) = x1(a3y2+ a2y3) + y1(a3x2+ a2x3).
Therefore, for any choice of a2, a3, x2, x3, y1, y2, y3, there is at most one possibility of
x1 such that Per(a, x, y) = t (t ∈ F ∗
q) This implies that (3.5) |{Per(a, x, y) = t : a = (0, a2, a3), x, y ∈ A3}| = O(|A|7).
(3) Suppose thata = (0, 0, a3) for some a3∈ A We have
Per(a, x, y) = a3(x1y3+ x2y1).
Therefore, for any choice of x1, y3, x2, y1, there is at most one possibility of a3 such that Per(a, x, y) = t (t ∈ F ∗
q) This implies that (3.6) |{Per(a, x, y) = t : a = (0, 0, a3), x, y ∈ A3}| = O(|A|6).
Putting (3.4), (3.5), and (3.6) together, we have
P3(A3; t) = (1 + o(1)) |A|9
q + O(|A|
7) = (1 + o(1)) |A|9
q
if|A| q 2/3 This completes the proof of Theorem 1.2.
4 Proof of Theorem 1.4 Since E is a product-like set, E ∩ (F ∗
q)n = ∅ Fix
ana ∈ E ∩ (F ∗
q)n Define E1 and E2 as in (3.1) and (3.2), respectively Similarly to section 3, we have
(4.1) P n(E; t) |{x · y : x ∈ E1, y ∈ E2}|.
Removing (0, , 0) from E does not increase the lower bound of P n(E; t), so we may
assume that (0, , 0) ∈ E From (2.2), we have
ν t2(E1, E2) |E1|2|E2|2q −1+|E1|q 2n−1
k=(0, ,0)
| E2(k)|2|E1∩ l k |.
Trang 8SinceE is product-like, so is E1 (note thatE1 is obtained fromE by scaling) Hence,
|E1∩ l k | |E1| 1/n It follows that
ν t2(E1, E2) |E1|2|E2|2q −1 + O
|E1|1+ 1
n q 2n−1
k
| E2(k)|2
=|E1|2|E2|2q −1 + O
|E1|1+ 1
n q n−1
y
|E2(y)|2
=|E|4q −1 + O(|E|2+1n q n−1 ),
(4.2)
where the second line follows from Plancherel’s theorem, and the last line follows from
|E1| = |E2| = |E| Applying the Cauchy–Schwarz inequality, we have
|E|4=|E1|2|E2|2=
⎛
v t(E1, E2)
⎞
⎠
2
|{x · y : x ∈ E1, y ∈ E2}|
ν t2(E1, E2).
(4.3)
It follows from (4.2) and (4.3) that
(4.4) |{x · y : x ∈ E1, y ∈ E2}| |E|4
|E|4q −1 + O(|E|2+n1q n−1) =
q
1 + O q n
Theorem 1.4 now follows immediately from (4.1) and (4.4)
5 Proof of Theorem 1.6 The proof proceeds by induction The case n = 2
follows from Theorem 2.2 Let M = (x ij)1i,j3 be a 3× 3 matrix with entries in
[a + 1, a + b] We have two cases.
(1) If x33= 0, then we can write
Per(M ) = x33 x11+x13x31
x33 x22+x23x32
x33
+ x12+x13x32
x33 x21+x23x31
x33
−2 x13x31x23x32
Hence, Per(M ) = t if and only if
x11+x13x31
x33 x22+x23x32
x33
+ x12+x13x32
x33 x21+x23x31
x33
= tx33− 2x13x31x23x32
Since
x ij+x i3 x 3j
x33 ∈
a + x i3 x 3j
x33 , a + x i3 x 3j
x33 + b
(mod p), 1 i, j 2,
it follows from Corollary 2.3 that, for any fixed a i3 , a 3j ∈ A, 1 ≤ i, j ≤ 3, there are
(1 + o(1))b4/p choices of x11, x12, x21, and x22such that Per(M ) = t Therefore, this
case contributes
(1 + o(1))(b − I(0))b4b4/p = (1 + o(1))b9/p
matrices having permanent t, where I(0) = 1 if 0 ∈ I and I(0) = 0 otherwise.
Trang 9(2) If x33= 0, then
Per(M ) = x32(x11x23+ x21x13) + x31(x22x23+ x13x22).
If x11= 0, there are at most b7choices of the remaining entries such that Per(M ) = t.
If x11 = 0, there is at most one choice of x23 such that x11x23+ x21x13 = 0 If
x11x23+ x21x13= 0, there is at most one choice of x32 such that Per(M ) = t Thus, this case contributes at most 3b7 matrices having permanent t.
Since b2 p log2p, putting these two cases together, we have
P3(I3; t) = (1 + o(1))b9/p.
Suppose that the statement holds for n − 2 2 We show that it also holds for n For any n × n matrix M = (x ij)1i,jn with entries in [a, a + b], let M be the
submatrix obtained from M by deleting the first two rows and two columns, and let
M st (3 s, t n) be the submatrix obtained from M by deleting the first row, the second row, the sth row, the first column, the second column, and the tth column Choose entries of M randomly from the interval [a + 1, a + b] We have three cases (A) Suppose that per(M ) = 0 For any choice of x 1i , x 2i , x j1 , x j2 , x ij (3 i,
j n), we can write
Per(M ) = Per(M )(x11x22+ x12x21) + v11x11+ v12x12+ v21x21+ v22x22+ v for some fixed v11, v12, v21, v22, and v (depending on chosen entries of M ) Hence, Per(M ) = t if and only if
x11+ v22
Per(M ) x22+ v11
Per(M )
+ x12+ v21
Per(M ) x21+ v21
Per(M )
Per(M )− v11v22+ v21v12
(Per(M ))2 .
Since
x ij+ v ij
Per(M ) ∈
a + v ij
Per(M ), a + v ij
Per(M )+ b
(mod p),
it follows from Corollary 2.3 that there are (1 + o(1))b4/p choices for x11, x12, x21, x22.
Therefore, this case contributes
(b (n−2)2− P n−2(I n−2 ; 0))b 4(n−2) (1 + o(1)) b
4
p = (1 + o(1))
b n2 p
−(1 + o(1))P n−2(I n−2; 0)b 4n−4
p
matrices having permanent t.
(B) Suppose that Per(M ) = 0 and Per(M st)= 0 for some 3 s, t n Without
loss of generality, we may assume that Per(M33)= 0 For any choice of x 1i , x 2i , x i1 , x i1
(3 i n), we can write
Per(M ) = v11x11+ v12x12+ v21x21+ v22x22+ v, where v11, v12, v21, v22, and v depend on x ij , i, j ≥ 3 By factoring Per(M ) by rows,
we have
(5.1) v11= x23(x32Per(M33) + α) + β,
Trang 10where α does not depend on x23, x32 and β does not depend on x23 Now, suppose
that Per(M ) = t For any fix x 1i , x i1 , x 2j , x j2 (i 3, j 4), there is at most one possibility of x32 such that x32Per(M33) + α = 0 Besides, if x32Per(M33) + α = 0, there is at most one possibility of x23 such that v11 = 0 Hence, there are at most
2b possibilities of (x23, x32) such that v11 = 0 From (5.1), if v11 = 0, there are at
most b3 possibilities of (x11, x12, x21, x22) such that Per(M ) = t Therefore, this case
contributes at most
(P n−2(I n−2; 0)− P n−2(2) (I n−2 ))(2b 4(n−2)−1 b4+ b 4(n−2) b3)
= 3b 4n−5 (P n−2(I n−2; 0)− P n−2(2) (I n−2))
matrices having permanent t.
(C) Suppose that Per(M st) = 0 for all 3 s, t n It is trivial that there are at most b 4(n−2)+4 choices of a 1i , a 2i , a i1 , a i2, 1≤ i ≤ n, such that Per(M) = t Therefore,
this cases contributes at most
b 4n−4 P n−2(2) (I n−2)
matrices having permanent t.
Putting all three cases together, we have
P n(I n ; t) = (1 + o(1)) b
p + O P n−2(I n−2; 0) b 4n−4
4n−5
+ O(b 4n−4 P n−2(2) (I n−2 )).
(5.2)
From the induction hypothesis,
(5.3) P n−2(I n−2 ; 0) = O(b (n−2)2/p),
and from Lemma 2.3,
(5.4) P n−2(2) (I n−2 ) = O(b (n−2)2−4 ).
It follows immediately from (5.2), (5.3), and (5.4) that
P n(I n ; t) = (1 + o(1)) b
p
whenever b/p 1/2 log p → ∞, completing the proof of the theorem.
Acknowledgment The author would like to thank Victor Garcia for kindly
granting his permission to present Theorem 2.2 here
REFERENCES
[1] O Ahmadi and I E Shparlinski, Distribution of matrices with restricted entries over finite
fields, Indag Math (N.S.), 18 (2007), pp 327–337.
[2] A Ayyad, T Cochrane, and Zh Zheng, The congruence x1x2 ≡ x3x4 (mod p), the equation x1x2=x3 x4, and mean values of character sums, J Number Theory, 59 (1996), pp 398–
413.
[3] J Bl¨ omer, R Karp, and E Welzl, The rank of sparse random matrices over finite fields,
Random Structures Algorithms, 10 (1997), pp 407–419.
... n2−4 ) matrices whose (n − 1) × (n − 1) minors have zero permanent.Putting these two cases together, we complete the proof of the lemma
3 Proof of Theorem 1.2 (a)...
[1] O Ahmadi and I E Shparlinski, Distribution of matrices with restricted entries over finite< /small>
fields, Indag Math (N.S.), 18 (2007), pp 327–337.
[2]... x11= 0, there are at most b7choices of the remaining entries such that Per(M ) = t.
If x11 = 0, there is at most one choice of x23