The number of solutions to the alternate matrix equation over a finite field and a q-identity

Journal of Statistical Planning and Inference 94 (2001) 349–358

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The number of solutions to the alternate matrix equation over a *nite *eld and a q-identity
Hongzeng Wei ∗ , Yibin Zhang Department of Mathematics, Hebei Normal College, Shijiazhuang, Hebei 050091, People’s Republic of China

Abstract Let Fq be a *nite *eld with q elements, where q is a power of a prime. In this paper, we *rst correct a counting error for the formula N (K2 ; 0(m) ) occurring in Carlitz (1954. Arch. Math. V, 19 –31). Next, using the geometry of symplectic group over Fq , we have given the numbers of solutions X of rank k and solutions X to equation XAX
= B over Fq , where A and B are alternate matrices of order n, rank 2 and order m, rank 2s, respectively. Finally, an elementary q-identity is obtained from N (K2 ; 0(0) ), and the explicit results for N (Kn; 2 ; Km; 2s ) is represented c 2001 Published by Elsevier Science B.V. by terminating q-hypergeometric series.
MSC: 05E15 Keywords: Alternate matrix equation; Symplectic geometry; q-identity; q-hypergeometric series

1. Introduction Let Fq be a *nite *eld of q = py elements, p a prime. Let Nm×n (A; B; k) and Nm×n (A; B) denote the number of m × n matrices X with rank k and of m × n matrices X such that XAX  = B;

(1)

respectively, where A is an n × n alternate matrix and B an m × m alternate matrix over Fq . Carlitz (1954) and Hodges (1966) have applied the exponential sums to compute the number of Nm×n (A; B) and Nm×n (A; B; k) over a *nite *eld of Ch = 2, respectively. Later, using exponential sums, Buckhiester (1973) studied the above problem over F2y . The paper is organized as follows. In Section 2, the formula Nm×n (A; 0) occurring Carlitz (1954) is corrected by using exponential sums. In Section 3, the above problem


The project supported by NSF of China (19571024). Corresponding author. E-mail address: [email protected] (H. Wei).

∗

c 2001 Published by Elsevier Science B.V. 0378-3758/01/$ - see front matter  PII: S 0 3 7 8 - 3 7 5 8 ( 0 0 ) 0 0 2 6 5 - 2

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will be studied in its full generality, using the geometry of symplectic groups over Fq and some enumerational theorems for subspaces. In Section 4, we obtain an elementary q-identity. Let A and B be two n × n matrices over Fq . If there exists an n × n nonsingular matrix Q over Fq such that QAQ = B, we say that B is cogredient to A. Clearly, without loss of generality, let 26n, then an n × n alternate matrix A of rank 2 is necessarily cogredient to the following normal form 
0 I () K2 = if 2 = n −I () 0 or 
K2 if 2 ¡ n; Kn; 2 = 0(n−2) where =0; 1; 2; : : : ; [n=2], called the index of A. Now, if another m×m alternate matrix B is cogredient to Km; 2s , where 2s6m. Let N (Kn; 2 ; Km; 2s ; k) and N (Kn; 2 ; Km; 2s ) denote the number of m×n matrices X with rank k and of m×n matrices X over Fq such that XKn; 2 X  = Km; 2s ;

(2)

respectively. Then, we have Nm×n (A; B; k) = N (Kn; 2 ; Km; 2s ; k) and

Nm×n (A; B) = N (Kn; 2 ; Km; 2s )

under cogredient transformations. Thus, it is suFcient to determine N (Kn; 2 ; Km; 2s ; k) and N (Kn; 2 ; Km; 2s ). Let P be an m-dimensional subspace in the 2-dimensional row vector space Fq(2) . We use the same letter P to denote a m × 2 matrix representation with rank m of the subspace P. If PK2 P  = 0, then P is called an m-dimensional totally isotropic subspace with respect to K2 . The m-dimensional totally isotropic subspaces exist if and only if m6. 2. A correction for N (K2 ; 0(m) ) Let Fq be a *nite *eld with q = py elements, p an odd prime. For a  ∈ Fq , de*ne e() = e2it()=p ; where t() =  + p + · · · + p

(3) y−1

:

Clearly, for any ;  ∈ Fq , we have t( + ) = t() + t(); e( + ) = e()e();

H. Wei, Y. Zhang / Journal of Statistical Planning and Inference 94 (2001) 349–358

we can deduce that  q  e() = 0 ∈Fq and

 ;∈Fq

if  = 0;

351

(4)

if  = 0

e() = q:

(5)

Furthermore, we have  (m) q 2 if A = 0;  e((AB)) = B 0 if A = 0;

(6)

where ( m2 ) = m(m − 1)=2, A is an m × m alternate matrix, the sum extends over all m × m alternate matrices B, and (AB) denotes the trace of the matrix AB. Let Fq(m) denote an n-dimensional row vector space over Fq . It is known that for all ;  ∈ Fq(m) , fA (; ) = A is a alternate bilinear form on Fq(m) × Fq(m) if and only if A is an m × m alternate matrix. De*ne  T (fA ) = e(fA (; )): (7) ;∈Fq(m)

Then, we have Theorem 1. Let A be an m × m alternate matrix over Fq . (i) If B is cogredient to A; then T (fB ) = T (fA ). (ii) If A is of rank 2r; then T (fA ) = q2(m−r) :

(8)

Proof. If B = PAP  , for a nonsingular matrix P, then  T (fB ) = e(fB (; )) ;∈Fq(m)

= =

 ;∈Fq(m)

 ";#∈Fq(m)

e((P)A(P) ) e("A# )

= T (fA ): Now, suppose that A is of rank 2r, then there exists a nonsingular matrix P such that PAP  = Km; 2r . But r r   fKm; 2r (; ) = Km; 2r  = ai br+i + (−ar+i )bi ; i=1

i=1

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H. Wei, Y. Zhang / Journal of Statistical Planning and Inference 94 (2001) 349–358

where  = (a1 ; a2 ; : : : ; am ),  = (b1 ; b2 ; : : : ; bm ). Hence,   r r    e ai br+i + (−ar+i )bi T (fA ) = i=1

a1 ; :::; am ∈Fq b1 ; :::; bm ∈Fq

=

=

=





r  i=1

a1 ; :::; am ∈Fq b1 ; :::; bm ∈Fq







a2r+1 ; :::; am ∈Fq b2r+1 ; :::; bm ∈Fq

= q2r

 e(ai br+i )

r  i=1

 a2r+1 ; :::; am ∈Fq b2r+1 ; :::; bm ∈Fq

 q

r  i=1

e((−ar+i )bi ) 

r   i=1 ai ;br+i

a2r+1 ; :::; am ∈Fq b2r+1 ; :::; bm ∈Fq



i=1

e(ai br+i ) r  i=1

r   i=1 ar+i; bi

 e((−ar+i )bi )

q

1

= q2(m−r) : Theorem 2. (m)

N (K2 ; 0

2m−( m2 )

)=q

 2r6m

q

−2r(+1)

2r−1 (1 − qm−i ) i=0 : r −2i ) i=1 (1 − q

(9)

 m Proof. By (6), it is clear that e((XK2 X  B)) = q( 2 ) if and only if XK2 X  = 0, B  and B e((XK2 X  B)) = 0, otherwise, where the summation extends over all m × m alternate matrices B. Thus,  m e((XK2 X  B)) = N (K2 ; 0(m) )q( 2 ) : (10) B

X

A simple calculation shows that

   m  2 m  m m      (xi; +k (−bji )xjk ) + (xi; k− bji xjk ) : (XK2 X B) = k=1

j=1 i=1

k=+1

j=1 i=1

(11) Let fB be the alternate bilinear form de*ned by fB (x; y) = xBy for all x; y ∈ Fq(m) . m m Let x = (x1 ; x2 ; : : : ; xm ), y = (y1 ; y2 ; : : : ; ym ). Then fB (x; y) = i=1 j=1 xi bij yj . Thus, (11) becomes (XK2 X  B) = Hence,  B

X

  k=1

f−B (xk ; x+k ) +

e((XK2 X  B))

2  k=+1

fB (xk ; xk− ):

(12)

H. Wei, Y. Zhang / Journal of Statistical Planning and Inference 94 (2001) 349–358

= =

 B

X



  k=1

   B

X k=1

353

 2  e(f−B (xk ; x+k )) e(fB (xk ; xk− )) k=+1

e(f−B (xk ; x+k ))e(fB (x+k ; xk )):

Thus, (10) becomes m

N (K2 ; 0(m) )q( 2 ) =

   B

X k=1

e(f−B+B (xk ; x+k )):

 ), where xk =(x1k ; x2k ; : : : ; xmk ), 16k62. Furthermore, let Let X =(x1 ; x2 ; : : : ; x2

Fq(m) .

indicates a sum extending over all vectors xk in Then (13) becomes      m N (K2 ; 0(m) )q( 2 ) = ··· e(f−B+B (xk ; x+k )) : B

x2

x1

k=1

(13) 

xk

(14)

If B extends over all m × m alternate matrices, then also does −B + B = −2B. Thus, (14) becomes
     m N (K2 ; 0(m) )q( 2 ) = e(f−B+B ("; #)) B k=1

= =






B

 ;

";#

 e(fB (; ))

 (T (fB )) :

(15)

B

Let K(m; 2r) denote the number of all m × m alternate matrices of rank 2r. Then, from Theorem 3:30 in Wan (1993), m (qi − 1) r (16) K(m; 2r) = qr(r−1) i=m−2r+1 2i i=1 (q − 1) and, from (8) and (16), we have  m N (K2 ; 0(m) ) = q−( 2 ) K(m; 2r)(q2(m−r) ) 2r6m

=q

2m−( m2 )

 2r6m

q

r(r−1−2)

m i i=m−2r+1 (q − 1) r : 2i i=1 (q − 1)

Thus, we obtain (9), and correct a counting error occurring in (Carlitz, 1954, Section 6). 3. Determination of N (Kn; 2 ; Km; 2s ; k) and N (Kn; 2 ; Km; 2s ) Clearly, when n = 2, from (2), we have N (K2 ; Km; 2s ) = 0;

if s ¿ 

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H. Wei, Y. Zhang / Journal of Statistical Planning and Inference 94 (2001) 349–358

and N (K2 ; K2s ; 2s) = N (K2 ; K2s ): Theorem 3. Let s6. Then; N (K2 ; K2s ) = qs(2−s)

 

(q2i − 1):

(17)

i=−s+1

Theorem 4. Let ¿0. Then    (m) ( k2 ) m N (K2 ; 0 ; k) = q (q2i − 1); k q i=−k+1 (m)

N (K2 ; 0

)=

min{;  m} k=0

q

( k2 )



m k



 

q i=−k+1

(18)

(q2i − 1);

where the Gaussian coe?cient is de@ned by  (qm − 1)(qm−1 − 1) · · · (qm−k+1 − 1) m = ; k q (qk − 1)(qk−1 − 1) · · · (q − 1)

(19)

0 ¡ k6m;

[ m0 ]q = 1; and [ mk ]q = 0; if k ¿ m. Proof. Let X be an m × 2 matrix of rank k over Fq such that XK2 X  = 0(m) :

(20)

Assume that the rows of X span a subspace of dimension k. Then, k6m and P is a k-dimensional totally isotropic subspace with respect to K2 . Hence, k6. Let v1 ; v2 ; : : : ; vk be a basis of P. Then there is a uniquely determined m × k matrix T of rank k such that   v1    v2    (21) X = T  . :  .   .  vk Conversely, for any m × k matrix T of rank k, de*ne X by (21), then X satis*es (20), and the rows of X span P. By Corollary 3:19 of Wan (1993) the number of k-dimensional totally isotropic subspaces is  2i i=−k+1 (q − 1) : (22) k i i=1 (q − 1) By Lemma 1:5 of Wan (1993), the number of m × k matrices of rank k is equal to m  k q( 2 ) (qi − 1): (23) i=m−k+1

Hence, we have (18) and (19).

H. Wei, Y. Zhang / Journal of Statistical Planning and Inference 94 (2001) 349–358

355

Theorem 5. Let s6. Then N (K2 ; Km; 2s ; k) = N (K2 ; K2s )N (K2(−s) ; 0(m−2s) ; k − 2s);

(24)

N (K2 ; Km; 2s ) = N (K2 ; K2s )N (K2(−s) ; 0(m−2s) ):

(25)

Proof. Let X be an m × 2 matrix of rank k over Fq satisfying XK2 X  = Km; 2s :

(26)

Write X in the block form
 X1 2s X= ; m − 2s X2

(27)

From (26), we deduce X1 K2 X1 = K2s ;

(28)

X1 K2 X2 = 0(2s; m−2s)

(29)

X2 K2 X2 = 0(m−2s) :

(30)

and For any 2s×2 matrix X1 satisfying (28), clearly, we have rank X1 =2s. The 2s rows of X1 span a subspace of dimension 2s, which will also be denoted by X1 if no ambiguity arises. De*ne X1⊥ = {v ∈ Fq(2) | X1 K2 v = 0}:

(31)  X1 K2 X1⊥

X1⊥

Then is a subspace of dimension 2( − s). Clearly, = 0. By (28), X1 ∩ ⊥ X1 = 0. Denote any 2( − s) × 2 matrix over Fq whose rows form a basis of X1⊥ also by X1⊥ , then ( XX⊥1 ) is a 2 × 2 nonsingular matrix and 1


 K2s X1 X1 K2 : (32) =  X1⊥ X1⊥ X1⊥ K2 X1⊥ 

Therefore, X1⊥ K2 X1⊥ is a 2( − s) × 2( − s) nonsingular alternate matrix. We can assume that 

X1⊥ K2 X1⊥ = K2(−s) :

(33)

It is clear that for any (m − 2s) × 2( − s) matrix T satisfying TK2(−s) T  = 0;

(34)

the m × 2 matrix
 X1 TX1⊥ satis*es (26). Conversely, for any m×2 matrix (27) satisfying (26), there is a uniquely determined (m − 2s) × 2( − s) matrix T such that X2 = TX1⊥ and T satis*es (34).

356

H. Wei, Y. Zhang / Journal of Statistical Planning and Inference 94 (2001) 349–358

If rank X = k, then from rank X1 = 2s and X2 = TX1⊥ ⊂ X1⊥ , we deduce rank T = rank X2 = k − 2s. Hence, we have obtained (24) and (25). Theorem 6. Let s6. Then N (Kn; 2 ; Km; 2s ; k) =

=



min{+s;  k} t=max{2s; k−n+2} n−2  i=n−2−k+t+1

q

t(n−2)+( k−t 2 )

m−t k −t

(qi − 1)N (K2 ; km; 2s ; t);

 q

(35)

where 2s6k6min{n −  + s; m};

(36)

N (Kn; 2 ; Km; 2s ) = qm(n−2) N (K2 ; Km; 2s ):

(37)

Proof. Let X be an m × n matrix over Fq satisfying (2). Write X in the block form 2

X = ( X1

n−2 X2 )m ;

(38)

then X1 K2 X1 = Km; 2s :

(39)

which is clearly independent of X2 . Therefore, X2 can be any m × (n − 2) matrix over Fq ; X1 is an m × 2 matrix satisfying (39), and conversely. Thus we obtain (37). Now for any m × n matrix X = (X1 ; X2 ) of rank k over Fq satisfying (2), we assume that rank X1 = t. Therefore, the number of such matrices X1 is equal to N (K2 ; Km; 2s ; t) by (39). Assume that the rows of X1 span a subspace P of dimension t, and X1 has a decomposition
 P X1 = TP = (TS) (40) 0 where t

(T

m−t

S )

is an m × m nonsingular matrix, P is a t × 2 representation matrix with rank t of subspace P. Once X1 has been chosen, to compute formula (35), it is enough to compute the number of matrices X of rank k with the block form (X1 X2 ) = (TS)


2 n−2 P X21 0 X22

t ; m−t

(41)

where P; T , and S are all *xed matrices. Clearly, X21 is any t × (n − 2) matrix, and X22 is a (m − t) × (n − 2) matrix of rank k − t. It is known that the number of m × n

H. Wei, Y. Zhang / Journal of Statistical Planning and Inference 94 (2001) 349–358

matrices over Fq of rank k is

 n m  ( k2 ) q (qi − 1): k q i=n−k+1

357

(42)

Hence, (35) follows. From (24) and (41), we have k − 2s6 − s, i.e., 2s6k6 + s, and k − t6n − 2, respectively. Hence, the conditions max{2s; k − n + 2}6t6min{ + s; k} and (36) are obtained. Remark 1. The proof of Theorem 6 corrects an error about the number of choices of X2 in Hodges (1966). 4. An elementary q-identity First, we have Theorem 7. Let q be any number. Then

 min{;    m} ( k ) m (q2i − 1) q2 k q i=−k+1 k=0 =q

2m−( m2 )

 2r6m

q

−2r(+1)

2r−1 (1 − qm−i ) i=0 : r −2i ) i=1 (1 − q

(43)

Proof. When q is odd, we can obtain equality (43) from formulas (9) and (19) of N (K2 ; 0(m) ). Clearly, both sides of equality (43) are rational fractions in q. Since there are in*nitely many odd prime power q. Therefore, (43) is an identity in q. Next, let us express above formulas in terms of terminating q-hypergeometric series. Theorem 8. Let s6. Then m

N (K2 ; 0(m) ) = q2m−( 2 ) 2 ,0 [q−(1=2)m ; q−(1=2)(m−1) ; q+1 ] m

N (Kn; 2 ; Km; 2s ) = qmn−2s+s(s−1)−( 2 ) 2 ,0 [q

where 2 ,0 [a; b; z]

=

−(1=2)(m−2s)

  i=−s+1

(44)

(q2i − 1)

; q−(1=2)(m−2s−1) ; q−s+1 ]

∞ (a) (b)  r r r z; (q)r r=0

(a)r = (1 − a)(1 − qa) · · · (1 − qr−1 a) if r¿1; (a)0 = 1 and the prime indicates that q is to be replaced by q−2 in the function 2 ,0 .

(45)

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H. Wei, Y. Zhang / Journal of Statistical Planning and Inference 94 (2001) 349–358

Proof. From (9), we deduce (44); from (37), (25), and (44), (45) holds. Remark 2. How to give a direct proof for identity (43) will be an interesting work. 5. Uncited Reference Wan (1995). References Buckhiester, P.G., 1973. Guass sums and the number of solutions to the matrix equation XAX
= 0 over GF(2 ). Acta Arith. 23, 271–278. Carlitz, L., 1954. Representations by skew forms in a *nite *eld. Arch. Math. V, 19–31. Hodges, J.H., 1966. A skew matrix equation over a *nite *eld. Arch. Math. 17, 49–55. Wan, Z., 1993. Geometry of classical groups over *nite *elds. Student literature, Lund. 1993. Wan, Z., 1995. Representations of forms by forms in a *nite *eld. Finite Fields Appl. 2.