Quadratic Gauss sums on matrices

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1 Linear Algebra and its Applications 384 (2004) Quadratic Gauss sums on matrices Mitsuru Kuroda 410, Haitsu-hosha, 61, Nojiri-cho, Sakai, Osaka , Japan Received 13 March 2003; accepted 27 January 2004 Submitted by R.A. Brualdi Dedicated to Professor Yoshio Mimura on his 60th birthday Abstract Gauss sums over a finite field are generalized to ones over a matrix ring, related to numbers of solutions of diagonal matrix equations. A similar relation of quadratic Gauss sums of two types is obtained. We also consider quadratic diagonal matrix equations with scalar coefficients Elsevier Inc. All rights reserved. AMS classification: 11L03; 11T24 Keywords: Character sum; Gauss sum; Matrix equation; Quadratic form 1. Introduction Let F = F q be a finite field of order q = p f (p: a prime). Our interest is a Gauss sum: g s (a) = ( ) 2πi exp p tr F (ax s ), x F where tr F denotes the trace map of field extension F/F p. Let e a denote an additive character of F : e a (x) = exp(2πi tr F (ax)/p), thatis g s (a) = e a (x s ) = e 1 (ax s ). x F x F address: mkuroda@ab.mbn.or.jp (M. Kuroda) /$ - see front matter 2004 Elsevier Inc. All rights reserved. doi: /j.laa

2 188 M. Kuroda / Linear Algebra and its Applications 384 (2004) When a = 1weomitsubscript1andsoe(x) = e 1 (x) andalsoputg s = g s (1). Let R=R(n) = M n (F ) for n 1. We define a Gauss sum on matrices as follows: G s (A) = ( ) 2πi exp p tr F (tr(ax s )), X R where the second tr in the right hand side is the trace map on matrices. Porter evaluated quadratic Gauss sum G 2 (A) on matrices for A M 2 (F ) (but the result should be corrected) in [7], with the intention of applying them to the problem of finding numbers of solutions of matrix equations [7, Introduction]. In case p/= 2, a relation between quadratic Gauss sums on matrices and quadratic forms Q A = tr(ax 2 ) was shown [4,5] and we can easily evaluate G 2 (A) for A M 3 (F ) by the last proposition in [4]. In this paper, we give an explicit formula G 2 (A) for A M n (F ) and an analogous relation of quadratic Gauss sums of two types by direct computation in Section 3. Before that we briefly discuss a connection between Gauss sums on matrices and the number of solutions of matrix equations, especially diagonal matrix equations, over a finite field in Section 2, which was essentially done in [4, Introduction]. In Section 4, we consider quadratic diagonal matrix equations with scalar coefficients. 2. Gauss sums and number of solutions of matrix equations In this section we give a connection between Gauss sums on matrices and the number of solutions of matrix equation: f(x 1,X 2,...,X m ) = B. (1) Fix A Rand define a map e A :R C by ( ) 2πi e A (X) = exp p tr F (tr(ax)). We easily see that e A is an additive character of a finite abelian group R with respect to addition. Moreover, we can show that ˆR ={e A A R},where ˆR is a character group of R. This equality can be proved in the same way as in case n = 1andwe only need an additional fact that a nonzero linear form represents every element in F. And hence we have the following proposition by the orthogonality relation for characters. Proposition 1. Let N(f = B),whichwefollowin [1,2], denotethenumberofsolutions of the matrix equation (1) and define G(f = B) = e A (f (X 1,...,X m ))e A ( B). X i,a R Then we have R N(f = B) = G(f = B).

3 M. Kuroda / Linear Algebra and its Applications 384 (2004) From now on, we sometimes omit subscript I = I n, the identity matrix of size n, in the notation of additive characters and Gauss sums on matrices as described in case n = 1 in Section 1. For example, G s means G s (I) = X R e I (X s ) = X R e(xs ). Now, we consider diagonal matrix equations: f(x 1,X 2,...,X m ) = C 1 X s C 2X s C mx s m m = B, B R. (2) In this situation, by an easy computation, we can express G(f = B) as a character sum including Gauss sums on matrices, or G(f = B) = m G si (AC i )e A ( B). (3) A R i=1 Now, we introduce an equivalence relation on R by similarity (over F ). That is, A and A are equivalent if there exists a regular matrix P in R such that A = P 1 AP. Lemma 1. Let A, A R. If A is equivalent to A, then we have G s (A) = G s (A ) for all s. Proof. We have X R e I (A X s ) = X R e(a(p XP 1 ) s ) = Y R e(ay s ). Let R 1, R 2,...,R h be equivalence classes with respect to similarity: R=R 1 R h, R i R j = if i/= j,anda i R i their representatives for i = 1, 2,...,h. Now, we further assume that coefficient matrices of Eq. (2) are scalar. Note that only scalar matrices commute with all regular matrices. By (3) and Lemma 1, we have the following theorem. Note that if B is similar to B we have N(f = B) = N(f = B ), since we are assuming that C i s are scalar and so there is a one-to-one correspondence between the solutions. Theorem 1. Let C i s be scalar matrices and consider the matrix equation (2). Then we have h m G(f (X 1,...,X m ) = B) = G si (A j C i ) e A ( B). A R j j=1 i=1 In case B = O n = O, the zero matrix (of size n), we have G(f = O) = h j=1 i=1 m G si (A j C i ) R j. Moreover, if B is similar to B we have G(f = B) = G(f = B ).

4 190 M. Kuroda / Linear Algebra and its Applications 384 (2004) Quadratic Gauss sums on matrices In this section we consider quadratic Gauss sums on matrices, which have a similar relation with quadratic Gauss sums over a finite field F. Throughout this section we assume that the characteristic of the basic field F is not equal to 2. Fix A R(n) and consider the quadratic form Q A = tr(ax 2 ) with n 2 variables. Let V A denote the quadratic space corresponding to Q A. In consideration of quadratic Gauss sums on matrices, we have the following fundamental relation [5, Appendix A]. Proposition 2. G 2 (A) = G 2 (A ) if and only if V A =VA. We use an exponential notation for repeated quadratic spaces, that is, if U is a quadratic space, U n means an orthogonal sum of n copies of U : U n =U U U. If V A could be determined for given A R, then G 2 (A) can be easily computed by the following lemma [5, Lemma A1]. Lemma 2. Let g 2 = g 2 (1) be a quadratic Gauss sum (over F), H a hyperbolic plane and ι A = 0 or 1. Then, for nonzero A R(n) we can express V A =H r A 1 ι A d A 0 s A, n 2 = 2r A + ι A s A (4) for some d A F q /(F q )2 and hence we have G 2 (A) = q r A+s A ϱ(d A )g ι A+1 2, where ϱ is the quadratic character on F q. Further s A = dim rad V A in (4) can be determined by the following lemma [6, Corollary to Proposition 3]. Lemma 3. Let A [A 1,A 2,...,A t ] (notation in [4]) be a Jordan splitting of A R over the algebraic closure of F, α i be the eigenvalue of A i for i = 1,...,t.Then s A = dim rad V A = min {size(a i ), size(a j )}, where the sum is taken over 1 i t and 1 j t such that α i + α j = 0. Note that if A is given, then n, s A, r A, ι A in (4) can be determined successively. Therefore, we have only to determine d A in (4), which differs from the discriminant of a maximal regular subspace of V A by (a multiple) ( 1) r A, to determine the structure of V A, that is, to determine quadratic Gauss sum G 2 (A). And we have the following structure theorem of quadratic space V A which was shown in [6, (only one) Theorem]. Note that we can consider the quadratic space V A for the basic field being arbitrary.

5 M. Kuroda / Linear Algebra and its Applications 384 (2004) Lemma 4. Let F be a field of characteristic different from 2. Write the characteristic polynomial f A (x) of A Ras a product of powers of irreducible polynomials which are relatively prime: f A (x) = x d0 s f i (x) di i=1 t g j (x) e j, (5) j=1 where f i (x) is a polynomial of x 2 and g j (x) is not so. Let B j be the companion matrix associated with g j (x). Then V A e =W 1 B 1 W e t B t H O, (6) where H is a hyperbolic space, O is the radical of V A and dim O is given by Lemma 3, and W Bj is a regular subspace of V A depending on B j with dim W Bj = deg g j (x) and disc W Bj = det B j for each j. Now, we determine quadratic Gauss sums on matrices explicitly. Theorem 2. Let A be in R(n), and write the characteristic polynomial f A of A as in (5), and put D A (x) = g j (x) e j and α A = deg(d A (x)) in (5). Then quadratic Gauss sums on matrices are given as follows. (i) G 2 (O) = q n2. (ii) If A is nonzero, then we have G 2 (A) = q (n2 +s A ι A 1)/2 ϱ(d A )g ι A+1 2, where s A is given by Lemma 3 and d A = ( 1) n2 s A (1+ι A +α A )/2 β. D A (β)=0 Here, the product is taken over all roots β of D A (x) (with multiplicity). Proof. (i) We trivially have G s (O n ) = q n2 for all s. (ii) By (6), we can express V A =WA H rad V A, (7) where W A is regular, dim W A = α A, and disc W A = D A (β)=0 β where the product is over all roots β of D A (x) (with multiplicity). Let dim H = 2h A in (7). By (4) and (7), we have n 2 = 2r A + ι A s A = α A + 2h A + s A. (8) By comparison of both discriminants of regular parts in (4) and (7), we have ( 1) r A d A = ( 1) h A β. D A (β)=0

6 192 M. Kuroda / Linear Algebra and its Applications 384 (2004) Hence d A = ( 1) r A+h A β. By (8) and then applying Lemma 2, the proof is completed. Example 1. Let a,b F with ( a + b/= ) 0andγ a nonsquare element in( F. Suppose A M 4 (F ) is similar to,whereb M ) B O2 0 1 O 2 C 2 (F ) is similar to γ 0 ( ) a 0 and C M 2 (F ) is similar to. Then the characteristic polynomial of A is 0 b given by f A (x) = (x 2 γ )(x a)(x b), where the right hand side is a product of irreducible polynomials in ( F [x]. Let ) F denote an algebraic closure of F. Then, we β 0 have that B is similar to over F with β 0 β 2 = γ and β F. By Lemma 3, we have s A = s B and s A = = 2. Since 4 2 s A 0 (mod 2), wehaveι A = 1. And we have α A = 2andd A = ab, sinced A (x) = (x a)(x b). Therefore, we have G 2 (A) = q 8 ϱ(ab)g 2 2 = ϱ( ab)q9. In a finite field F we have a relation between Gauss sums of two types. Let R = R (n) = GL n (F ) and G ϱ (A) = ϱ(det X)e(AX) = ϱ(det X)e(AX). X R X R When n = 1 we use small letter g instead of G as in Section 1 with Gauss sums on matrices and also put G ϱ = G ϱ (I). This is just a convention in notation. Now, in a finite field case, we have a relation g 2 (a) = g ϱ (a) if a/= 0. We will show G ϱ (A) = G 2 (A) if V A is regular, or equivalently if s A = 0. Note that for nonzero a F the corresponding uniary quadratic space V a is regular. We will show the equality by direct computations of G 2 and G ϱ. Lemma 5. G ϱ (I n ) = G 2 (I n ) = q n(n 1)/2 g n 2. Proof. Since V I = 1 n H n(n 1)/2, G 2 can be easily seen. G ϱ is known due to Eichler and others. Apply Theorem 4.1 in [3]. Lemma 6. If a quadratic space V A for A R(n) is regular, then its discriminant is given by disc V A = ( 1) n(n 1)/2 det A. Proof. This is easily seen from Theorem 5 in [4].

7 Proposition 3. We have M. Kuroda / Linear Algebra and its Applications 384 (2004) (i) G ϱ (A) = ϱ(det A)G ϱ if A R. (ii) G 2 (A) = ϱ(det A)G 2 if and only if s A = 0. Proof. (i) For A with det A/= 0wehave ϱ(det X)e(AX)= ϱ((det A) 2 det X)e(AX) X R X R =ϱ(det A) ϱ(det Y)e(Y). Y R (ii) Suppose s A = 0, or equivalently V A is regular. By Lemma 6 we have disc V A = ( 1) n(n 1)/2 det A = det A disc V I. By Theorem 2, we have d A =( 1) (n2 ι A 1)/2 disc V A =( 1) (n2 ι I 1)/2 det A disc V I = det A d I, since s A = s I = 0andι A = ι I. Therefore, we have G 2 (A) = q (n2 1 s A )/2 ϱ(d A )g ι A+1 2 = ϱ(det A)G 2. The converse is clear by Lemma 2 and by consideration of the absolute values of both sides. By the proposition and Lemmas 5, 6, we have the following. Theorem 3. G 2 (A) = G ϱ (A) if s A = 0. Remark. ( ) If A R and s A /= 0, then G 2 (A) /= G ϱ (A). For example, if A = 1 0,wehaveG (A) = q 3 and G ϱ (A) = q 2. This can be easily seen by consideration of the absolute values. Note that G 2 (A) is minimal if and only if s A = Quadratic diagonal matrix equations with scalar coefficients In this section we consider the quadratic diagonal 2 2 matrix equation with scalar coefficients: f(x 1,...,X m ) = a 1 X a mxm 2 = B, B R(2), a i F, (9)

8 194 M. Kuroda / Linear Algebra and its Applications 384 (2004) over a finite field of characteristic different from 2. In [5], the problem of finding the number N(f = B) of solutions of Eq. (9) was reduced to ones of finding N(g = B) and N(h = B),whereg = X X2 m (for any m), and h = X Xm δx2 m for even m and a nonsquare element δ F,andN(h = O) was evaluated [5, Proposition 1]. N(g = B) is known [4, Proposition 7] (Proposition 7, (3) in [4] is incorrect. See [5, Correction]). Here we consider this problem. But first, we consider the quadratic diagonal matrix equation with scalar coefficients: f(x 1,...,X m ) = a 1 X a mxm 2 = B, B R(n), a i F, (10) over a finite field of characteristic different from 2 and establish an analogous theorem to Theorem 1, which can be considered as a special case of Theorem 1. Throughout this section we assume that F is a finite field of order q with 2 /= 0. In Section 2, we introduced an equivalence relation on R(n) by similarity. In thissection, we introduce another equivalence relation on R: A A if the quadratic Gauss sums are equal, that is, if G 2 (A) = G 2 (A ). Let M 1,...,M h be the equivalence classes with respect to this equivalence relation and A i s representatives of M i s. Note that each M i is a disjoint union of some equivalence classes R j s with respect to similarity by Lemma 1. Lemma 7. Let t A denote the number of 0 s as eigenvalues of A R, that is,x t A fully divides the characteristic polynomial f A (x) of A. Then, we have t A s A (mod 2). ( ) B O Proof. Let be a Jordan canonical form of A over F, where the characteristic polynomial of B is f B = x t A O C and det C/= 0. By Theorem 8 in [4], we have s A s B (mod 2) since s C is even by Lemma 3. By considering Jordan canonical form of B and then applying Lemma 3, we have s B t A (mod 2). Lemma 8. For A, B R(n), the following assertions are equivalent. (1) n 2 s A n 2 s B (mod 2). (2) ι A = ι B. (3) s A s B (mod 2). (4) t A t B (mod 2). (5) α A α B (mod 2). By Lemma 8 and the formula for d A in Theorem 2, we have the following lemma. Lemma 9. Let A, B in R. If A B, that is, if G 2 (A) = G 2 (B), then we have G 2 (aa) = G 2 (ab) for a F. By using Lemma 9, we have the following theorem in the same way as Theorem 1 is obtained.

9 M. Kuroda / Linear Algebra and its Applications 384 (2004) Theorem 4. Consider Eq. (10). Then, we have G(f (X 1,...,X m ) = B) = In case B = O, we have G(f = O) = h j=1 i=1 h j=1 i=1 m G 2 (a i A j ) M j. Moreover, if B is similar to B we have G(f = B) = G(f = B ). m G 2 (a i A j ) A M j e A ( B). Now, we return to the problem of finding the number N(f = B) of solutions of Eq. (9). From now on, we fix R=R(2). There are six equivalence classes and they are given as follows: M 1 ={O 2 }, M 2 ={A R A/= O, tr(a) = 0}, M 3 ={A R det(a) = 0, tr(a) (F ) 2 }, M 4 ={A R det(a) = 0, tr(a) δ(f ) 2 }, M 5 ={A R tr(a) /= 0, det(a) (F ) 2 }, M 6 ={A R tr(a) /= 0, det(a) δ(f ) 2 }, where δ denotes a nonsquare element in F [4]. The following lemma is easily seen from Theorem 6 in [4]. Lemma 10. For A R, put T = tr(a) and D = det(a). Then we have q 4 if A = O, q G 2 (A) = 3 if A/= O and T = 0, q 2 ϱ(t )g 2 if T /= 0 and D = 0, q 2 ϱ( D) if TD /= 0. Hence we have, for a F, { G2 (A) if a δ(f G 2 (aa) = ) 2,T /= 0 and D = 0, G 2 (A) otherwise. Character sums A M j e A ( B) in Theorem 4 (h = 6) are given as follows, which were already used to obtain Proposition 7 in [4].

10 196 M. Kuroda / Linear Algebra and its Applications 384 (2004) Lemma 11. The character sums A M j e A ( B) are given by the following. (I) When B = O, we have (1) A M 1 e A ( B) = 1, (2) A M 2 e A ( B) = q 3 1, (3) A M 3 e A ( B) = 1 2 q(q2 1), (4) A M 4 e A ( B) = 1 2 q(q2 1), (5) A M 5 e A ( B) = 1 2 q(q 1)(q2 q 1 ϱ( 1)), (6) A M 6 e A ( B) = 1 2 q(q 1)(q2 q 1 + ϱ( 1)). (II) When B = λi, λ F, we have (1) A M 1 e A ( B) = 1, (2) A M 2 e A ( B) = q 3 1, (3) A M 3 e A ( B) = 1 2 q(q + 1)(ϱ( λ)g 2 1), (4) A M 4 e A ( B) = 1 2 q(q + 1)(ϱ( λ)g 2 + 1), (5) A M 5 e A ( B) = 1 2 q(q2 q 1 ϱ( 1)), (6) A M 6 e A ( B) = 1 2 q(q2 q 1 + ϱ( 1)). ( ) 0 1 (III) When B is similar to, put r(b) = k D T 2 +Tk+D=0 ϱ(k), then we have k F (1) A M 1 e A ( B) = 1, (2) A M 2 e A ( B) = 1, (3) A M 3 e A ( B) = q 2 {(1 ϱ(d)2 )q + g 2 r(b) ϱ(t 2 4D) 1}, (4) A M 4 e A ( B) = q 2 {(1 ϱ(d)2 )q g 2 r(b) ϱ(t 2 4D) 1}, (5) A M 5 e A ( B) = 1 2 q{(ϱ( D) + ϱ(d)2 ϱ( 1) + ϱ( 1)ϱ(T 2 4D) 2 1)q + ϱ(t 2 4D) + ϱ( 1) + 1}, (6) A M 6 e A ( B) = 1 2 q{(ϱ( D) ϱ(d)2 ϱ( 1) + ϱ( 1)ϱ(T 2 4D) 2 + 1)q ϱ(t 2 4D) + ϱ( 1) 1}. ( ) 0 1 Remark. To obtain Proposition 7 in [4], we assumed that B = in case D T (III) of the lemma. But the character sums A M j e A ( B)(j = 1,...,6) are invariant under similarity and we have A M j e A ( B) = A M j e A ( P 1 BP) for a regular matrix P R, since each M j is a disjoint union of some R i s and e A ( P 1 BP) = e( AP 1 BP) = e( PAP 1 B) A M j A M j A M j by definition of additive characters e A.Wealsohaver(B) = r(p 1 BP) since k in ϱ(k) is an eigenvalue of B which belongs to F. By using Theorem 4, we can evaluate N(f = B) in the same way as N(g = B) is done, where g = X X2 m. But an easier way is to compute

11 M. Kuroda / Linear Algebra and its Applications 384 (2004) N(f = B) N(g = B). Now, we can find the number N(f = B) of solutions of Eq. (9). Note that G(f = B) = q 4 N(f = B). We compute G(f = B) G(g = B). Suppose that a 1,...,a l (F ) 2 and a l+1,...,a m δ(f ) 2 (0 l m) in Eq. (9). Then we have G(f = B) G(g = B) = (q 2 g 2 ) l ( q 2 g 2 ) m l e A ( B) + ( q 2 g 2 ) l (q 2 g 2 ) m l e A ( B) A M 3 A M 4 (q 2 g 2 ) m e A ( B) + ( q 2 g 2 ) m e A ( B) A M 3 A M 4 = (( 1) m l 1)q 2m g2 m e A ( B) A M 3 +(( 1) l ( 1) m )q 2m g2 m A M 4 e A ( B). Therefore, we have the following proposition. Proposition 4. Notations as above. For m 1 and any B R(2), we have a relation q 4 (N(f = B) N(g = B)) = (( 1) m l 1)q 2m g2 m e A ( B) A M 3 + (( 1) l ( 1) m )q 2m g m 2 A M 4 e A ( B), where A M 3 e A ( B) and A M 4 e A ( B) are the character sums in Lemma 11 and l is the number of square elements of coefficients a i s in Eq. (9). Correction. Eq. 7 in [5, Lemma 3] is incorrect (Table 1 is correct and it is the author s careless mistake). The correct formula should be ( 1,1) 1 = 1 4 q(q 1)3. (Condition ( 1, 1) below the summation symbol is incorrect and Eq. 6 in [5, Lemma 3] is correct.) References [1] B.C. Berndt, R.J. Evans, K.S. Williams, Gauss and Jacobi Sums, A Wiley-Interscience Publication, New York, [2] K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, second ed., Springerverlag, New York, [3] D.S. Kim, Gauss sums for general and special linear groups over a finite field, Arch. Math. 69 (1997) [4] M. Kuroda, A quadratic form tr (AX 2 ) and its application, Linear Algebra Appl. 266 (1997)

12 198 M. Kuroda / Linear Algebra and its Applications 384 (2004) [5] M. Kuroda, On Rosenfeld s problem, Linear Algebra Appl. 295 (1999) [6] M. Kuroda, Y. Mimura, Gauss sums and quadratic forms on matrices, Far East J. Math. Sci. 2 (2000) [7] A.D. Porter, An exponential sum in a finite field, Publ. Math. (Debrecen) 20 (1973)

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