HADAMARD MATRICES AND THE SPECTRUM OF QUADRATIC SYMMETRIC POLYNOMIALS OVER FINITE FIELDS

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1 HADAMARD MATRICES AND THE SPECTRUM OF QUADRATIC SYMMETRIC POLYNOMIALS OVER FINITE FIELDS FRANCIS N CASTRO AND LUIS A MEDINA Abstract In this article, we resent a beautiful connection between Hadamard matrices and exonential sums of quadratic symmetric olynomials over Galois fields This connection aears when the recursive nature of these sequences is analyzed We calculate the sectrum for the Hadamard matrices that dominate these recurrences The eigenvalues deend on the Legendre symbol and the quadratic Gauss sum over finite field extensions In articular, these formulas allow us to calculate closed formulas for the exonential sums over Galois field of quadratic symmetric olynomials Finally, in the articular case of finite extensions of the binary field, we show that the corresonding Hadamard matrix is a ermutation away from a classical construction of these matrices 1 Introduction Boolean functions are functions from the vector sace F n 2 to F 2 where F 2 reresents the binary field Alications of these beautiful combinatorial objects to comuter science fields such as coding theory, crytograhy and information theory have made them a source of active research Moreover, due to memory restrictions of current technology efficient imlementations of these functions is an area of secial interest Efficient imlementations, in the most general sense, is a very hard roblem However, some classes like the class of symmetric Boolean functions and the class of rotation symmetric Boolean functions are excellent candidates for efficient imlementations These functions are art of ongoing research In some alications related to crytograhy it is imortant for Boolean functions to be balanced A balanced Boolean function is one for which the number of zeros and the number of ones are equal in its truth table Let F (X be a Boolean function List the elements of F n 2 in lexicograhic order and label them as x 0 (0, 0,, 0, x 1 (0, 0,, 1 and so on The vector (F (x 0, F (x 1,, F (x 2n 1 is called the truth table of F Balancedness of Boolean functions are usually studied from the oint of view of Hamming weights or from the oint of view of exonential sums The Hamming weight of a Boolean function F, usually denoted by wt(f, is the number of 1 s in the truth table of F Thus, a Boolean function F is balanced if and only if wt(f 2 n 1 Weights of symmetric Boolean functions are somewhat understood For instance, it is known since the 90 s that weights of symmetric Boolean functions are linear recursive with integer coefficients [4, 5] On the other hand, the study of weights of rotations symmetric Boolean functions is becoming an active area of research [3, 13, 15, 27, 28] Similar to the symmetric case, it had been observed that weights of cubic rotation symmetric Boolean functions are linear recursive [3, 13] This romted the question if the same holds for any degree An answer was given by Cusick [12] when he showed that weights of any rotation symmetric Boolean function satisfy linear recurrences with integer coefficients The exonential sum of an n-variable Boolean function F (X is defined as (11 S(F ( 1 F (x x F n 2 Observe that a Boolean function F (X is balanced if and only if S(F 0 This oint of view is also a very active area of research For some examles, lease refer to [5 7, 9, 11, 19, 21, 24] Moreover, both oint of views are equivalent and are linked by the equation (12 wt(f 2n S(F 2 Date: December 14, Mathematics Subject Classification 05B20, 05E05, 11T23 Key words and hrases Hadamard matrices, Walsh transform, recurrences, exonential sums 1

2 2 FRANCIS N CASTRO AND LUIS A MEDINA Cusick s result about the linear recursive behavior of exonential sums of rotation symmetric Boolean functions was generalized in [8] to the general setting of exonential sums over finite fields Consider the Galois field F q where q r with rime and r 1 The exonential sum over F q of a function F : F q F q is given by (13 S Fq (F x F n q e 2πi Tr Fq /F (F (x, where i 1 and Tr Fq/F reresents the field trace from F q to F In [8] it was roved that exonential sums over Galois fields of rotation symmetric olynomials are also linear recurrent with integer coefficients Therefore, the recursive nature of these sequences is not secial to the binary field The authors of [8] also roved that exonential sums over Galois fields of symmetric olynomials are linear recurrent with integer coefficients Moreover, a beautiful relation between Hadamard matrices and exonential sums of quadratic symmetric olynomials over the rime fields F ( rime was found This link is the main focus of this article where we show that the same holds true when quadratic symmetric olynomials are considered over any Galois fields F q The connection to Hadamard matrices emerges when the roblem of finding linear recurrences for these exonential sums is considered This sounds technical, but in reality, once the roer framework is established, it is a very straight forward roblem In fact, the main idea to find such recurrences is to transform the roblem into a linear system and to comute an annihilator for the matrix associated to it An annihilator for a matrix A is simly a olynomial (X satisfied by A, ie a olynomial (X for which (A 0 The characteristic olynomial of A is such an examle We show that the matrices that dominates the recurrences associated to quadratic symmetric olynomials over any Galois field are all Hadamard matrices Moreover, revious constructions of Hadamard matrices re-emerge when the equivalent class of the matrix associated to quadratic symmetric olynomials is considered As just mentioned, in this article we show that the matrices that dominate the linear recursive nature of exonential sums of quadratic symmetric olynomials over F q are all Hadamard (regardless of q In the articular case when q is odd we comute the eigenvalues and eigenvectors for the corresonding Hadamard matrices These formulas deend on the Legendre symbol and quadratic Gauss sum over F q Moreover, these formulas for the eigenvalues allow us to comute closed formulas for the corresonding exonential sums This aer is divided as follows Next section is a short survey of Hadamard matrices This survey is included for comleteness uroses The exert reader may ski it Some reliminaries and examles that exose the relation between Hadamard matrices and exonential sums of quadratic symmetric olynomials are the subject of Section 3 In the fourth and final section we showed that the matrices related to the recursions considered in this article are indeed Hadamard We also comute the sectrum for these matrices when q is odd and rovide closed formulas for their corresonding exonential sums These closed formulas deend on the Legendre symbol and the quadratic Gauss sum These behavior is in line with the classical results in number theory where Gauss sums aear (naturally in the general theory of exonential sums However, as far we know, this is a new result Finally, in the articular case of finite extensions of the binary field, we show that the corresonding Hadamard matrix is a ermutation away from a classical construction of these matrices 2 Hadamard matrices: a short survey Hadamard matrices, named after the great mathematician Jacques Hadamard, are fascinating combinatorial objects with alications to many scientific areas Some examles include statistics, modern communications systems, error-correcting codes and crytograhy In the articular case of error-correcting codes, they are an imortant ingredient in the famous Reed-Muller code and Walsh-Hadamard code One of the most aealing asects of Hadamard matrices is the simlicity of its formal definition: a square matrix H of order n is a Hadamard matrix if all its entries are either 1 or 1 and it satisfies (21 HH T ni n,

3 QUADRATIC SYMMETRIC POLYNOMIALS AND HADAMARD MATRICES 3 where I n is the n n identity matrix Some simle examles of Hadamard matrices are (22 ( 1, ( , U to equivalence, these are the three smallest examles Two Hadamard matrices H and H are said to be equivalent if H can be obtained from H by interchanging rows or columns, or by multilying rows or columns by 1 U to equivalence, there is a unique Hadamard matrix of orders 1, 2, 4, 8, and 12 There are 5 inequivalent matrices of order 16 In contrast, there are millions (13,710,027 to be exact of inequivalent matrices of order 32 [18] See sequence A in [25] There exist no n n Hadamard matrices for n / {1, 2, 4k k N} In fact, it has been conjectured that a Hadamard matrix of order n exists if and only if n 1, 2, 4k for k a natural number This is known as Hadamard s Conjecture It remains an oen question, but it is widely believed to be true One of the earliest chamions of Hadamard matrices (and erhas, the first one was James Joseh Sylvester In [29] (1867, Sylvester rovided a construction of a Hadamard matrix of order 2 r for every non-negative integer r Let H be a Hadamard matrix of order n He noticed that the artition matrix (23 ( H H H H is also Hadamard Therefore, defining H 1 (1 and alying (23 reeatedly lead to the construction of the Hadamard matrix H2 r 1 H (24 H 2 r 2 r 1 H 2 r 1 H 2 r 1 In fact, the three examles in (22 are just H 1, H 2 and H 4 Sylvester s construction can be written in terms of Kronecker roducts If A and B are matrices, their Kronecker roduct A B is the matrix M constructed by relacing each entry A i,j in A by A i,j B Therefore, in terms of Kronecker roducts, Sylvester s construction is (25 H 2 r H 2 H 2 r 1 Sylvester s construction has been generalized Secifically, if H n, H m are Hadamard matrices of orders n and m (res, then their Kronecker roduct H n H m is an Hadamard matrix of order nm See [26] for more information The existence of Hadamard matrices for orders different than a ower of two was established by Hadamard [16] when he constructed Hadamard matrices of order 12 and 20 Other constructions of Hadamard matrices include the famous Paley Construction [23] and Williamson Construction [31] In fact, the literature of Hadamard matrices and their constructions is quite extensive Some examles are [2, 10, 17, 20, 30] The smallest order of the form 4k for which a Hadamard matrix cannot be constructed by a combination of known methods and for which no Hadamard matrix is known is 668 There are various ways to generalize the concet of Hadamard matrices The one that we use in this article is the concet of comlex Hadamard Matrices An n n matrix H is called a comlex Hadamard matrix if all its entries H ij are comlex numbers with H ij 1 (unimodularity and (26 HH T ni n, where H reresents the matrix obtained by comlex conjugation of all entries of H One of the most notable differences between comlex Hadamard matrices and real Hadamard matrices is their existence While the order n of a real Hadamard matrix is necessarily 1, 2 or 4m, with m a natural number, comlex Hadamard matrices exist for any natural n The classical examle is the n n Discrete

4 4 FRANCIS N CASTRO AND LUIS A MEDINA Fourier Transform matrix W n (DFT, which is defined by (27 W n 1 n ω ω 2 ω 3 ω n 1 1 ω 2 ω 4 ω 6 ω 2(n 1 1 ω 3 ω 6 ω 9 ω 3(n 1 1 ω n 1 ω 2(n 1 ω 3(n 1 ω (n 1(n 1, where ω e 2πi/n is a rimitive nth root of unity Observe that if F n nw n, then (28 F n F n T nin Thus, F n is a comlex Hadamard matrix of order n Two comlex Hadamard matrices H and H are called equivalent if there exist diagonal unitary matrices D 1 and D 2 and ermutation matrices P 1 and P 2 such that (29 H D 1 P 1 HP 2 D 2 An n n comlex Hadamard matrix is called dehased when its first column and first row are 1 and 1 T (res where 1 is the column vector of dimension n whose entries are all 1 s For examle, the rescaled Discrete Fourier Transform matrix F n is dehased Every comlex Hadamard matrix is equivalent to a dehased one This is actually very simle to see if the Hadamard matrix H has first row equal to 1 T Given an n n comlex Hadamard matrix H (H jk whose first row is equal to 1 T, define Deh(H as (210 Deh(H D H H, where H2, (211 D H 0 0 H3, Hn,1 1 Observe that the result of this matrix multilication is equal to the matrix obtained from H by multilying each row of H by the inverse of its first entry Clearly, Deh(H is dehased and (212 Deh(H Deh(H T D H H H T D H T D H ni n D H T nin Thus, Deh(H is comlex Hadamard and, according to definition (29, equivalent to H Remark 21 Following what is now a not so good habit in the culture of mathematicians, we abuse notation Deh(H For now on we use Deh(H to reresent a dehased Hadamard matrix for which H is equivalent to, even though we defined Deh(H for Hadamard matrices with first row equal to 1 T In the next section we exose a connection between Hadamard matrices and exonential sums of quadratic symmetric olynomials over Galois fields This relationshi flourishes when the recursive nature of this sequences is exlored 3 A connection to Hadamard matrices As mentioned in the introduction, we resent a beautiful relation between Hadamard matrices and exonential sums of quadratic symmetric olynomials over Galois fields Let e n,k denotes the elementary symmetric olynomial in n variables of degree k This olynomial is formed by adding together all distinct roducts of k distinct variables For examle, (31 e 4,3 X 1 X 2 X 3 + X 1 X 4 X 3 + X 2 X 4 X 3 + X 1 X 2 X 4

5 QUADRATIC SYMMETRIC POLYNOMIALS AND HADAMARD MATRICES 5 Consider the Galois field F q where q r with rime and r 1 Recall that the exonential sum of a function F : F n q F q is defined by (32 S Fq (F x F n q e 2πi Tr Fq /F (F (x, where Tr Fq/F reresents the field trace function from F q to F The field trace function can be exlicitly defined as r 1 (33 Tr Fq/F (α α i, with arithmetic done in F q Similar to the Boolean case, a function F : F n q F q is balanced if its values are uniformly distributed That is, if F takes each value of F q exactly q n 1 times As in the case of Boolean functions, if a function F : F n q F q is balanced, then S Fq (F is 0 The converse is not necessarily true, however it holds if q In [8], Castro, Chaman, Medina and Seúlveda showed that exonential sums of elementary symmetric olynomials and linear combinations of them are linear recurrent with integer coefficients In rincile, this imlies that exonential sums of this tye of functions can be comuted raidly The argument resented in [8] is actually quite simle: find a recursive generating set for S Fq (e n,k That is, find a set of sequences such that for some integer l, i0 {{a 1 (n}, {a 2 (n},, {a s (n}}, S Fq (e n,k s c j a j (n l where c j s are constants, and, for each 1 j 0 s and every n, one has a j0 (n j1 s d j0,j a j (n 1, j1 where d j0,j s are constant Once this is done, then to find a linear recurrence satisfied by {S Fq (e n,k }, it is enough to find an annihilating olynomial Q(X for the matrix A (d ij A olynomial Q(X is said to be an annihilating olynomial for a n n matrix A if Q(A O, where O reresents the n n matrix whose entries are all 0 Note that the sequence {S Fq (e n,k } will satisfy the linear recurrence with characteristic olynomial Q(X because it is a linear combination of the {a j (n} Let us illustrate the roblem with a detailed examle Examle 31 Consider the irreducible olynomial f(x X 3 + X + 1 in F 2 [X] Make the identification (34 F 8 F 2 [X]/(f(X, and let α [X], ie α is the equivalence class of X This imlies that α 3 α + 1 Now order F 8 using the lexicograhic order, that is, (35 F 8 { 0, 1, α, α + 1, α 2, α 2 + 1, α 2 + α, α 2 + α + 1 } {β 0, β 1, β 2, β 3, β 4, β 5, β 6, β 7 } Consider the sequence {S F8 (e n,2 } Observe that if we let X n assume the value β j, then e n,2 gets transformed to (36 e n 1,2 + β j e n 1,1 Thus, letting X n assume all values in the field leads to

6 6 FRANCIS N CASTRO AND LUIS A MEDINA (37 S F8 (e n,2 ( 1 Tr F 8 /F 2 (e n,2(x x F n 8 7 j0 7 j0 (x 1,,x n 1 F n 1 8 x F n 1 8 ( 1 Tr F 8 /F 2 (e n,2(x 1,,x n 1,β j ( 1 Tr F 8 /F 2 (e n 1,2(x+β je n 1,1(x 7 S F8 (e n 1,2 + β j e n 1,1 j0 Define a βj (n S F8 (e n,2 +β j e n,1 Observe that if we show that every {a βj (n} satisfies a common linear recurrence, then {S F8 (e n,2 } will satisfy such recurrence Consider the olynomial e n,2 + β l e n,1 As before, letting X n assume the value β j transforms e n,2 + β l e n,1 into (38 e n 1,2 + (β l + β j e n 1,1 + β l β j Therefore, letting X n assume all values in the field leads to 7 (39 a βl (n ( 1 Tr F 8 /F 2 (β l β j a βl +β j (n 1 Using the values (310 j0 Tr F8/F 2 (β j 0, for j 0, 2, 4, 6 Tr F8/F 2 (β j 1, for j 1, 3, 5, 7, equation (39 can be written in matrix form as a β0 (n a β1 (n a β2 (n (311 a β3 (n a β4 (n a β5 (n a β6 (n a β7 (n a β0 (n 1 a β1 (n 1 a β2 (n 1 a β3 (n 1 a β4 (n 1 a β5 (n 1 a β6 (n 1 a β7 (n 1 Denote by M 8 the matrix in (311 If we wish to find a linear recurrence with constant coefficients satisfied by {S F8 (e n,2 }, then it is enough to find an annihilating olynomial for M 8 That is because every {a βj (n} will satisfy the linear recurrence whose characteristic olynomial is given by this annihilating olynomial A simle, but erhas tedious, calculation shows that the minimal olynomial of M 8 is q 8 (X X Thus, {S F8 (e n,2 } satisfies the linear recurrence whose characteristic olynomial is q 8 (X Exlicitly, if we define (312 x 2 S F8 (e 2,2 8 x 3 S F8 (e 3,2 0 x 4 S F8 (e 4,2 64 x 5 S F8 (e 5,2 512 x n 64x n 4, for n 6, then S F8 (e n,2 x n In fact, using this recurrence we know that the first few values of {S F8 (e n,2 } n 2 are 8, 0, 64, 512, 512, 0, 4096, 32768, 32768, 0, , This calculation is done almost instantly For examle, using an old comuter (whose features are not to of the art from one of the authors, it took 0116 seconds to calculate S F8 (e 50000,2 which is a number with

7 QUADRATIC SYMMETRIC POLYNOMIALS AND HADAMARD MATRICES 7 22, 578 digits Also, this recurrence imlies that e n,2 is balanced over F 8 if and only if n 4m 1 for m a ositive integer This coincides with a conjecture given in [1] This concludes the examle In [8], the sequence {S F (e n,2 }, for rime, was studied in some details In articular, it was observed that this sequence is linear recurrent and that its recursive nature is dominated by the matrix e 2( 1πi 1 e 2πi e 4πi e 2( 2πi e 2 2( 2πi e 2 2( 1πi 1 e 4πi e 2 2( 3πi (313 M e 2 3( 3πi e 2 3( 2πi e 2 3( 1πi 1 e 2 2( 3πi e 2( 1πi e 2 2( 1πi e 2 3( 1πi e 2 4( 1πi 1 The matrix M turns out to be a very interesting mathematical object First, it can be obtained from the Fourier Discrete Transform Matrix by relacing its j-row r j by RT C j 1 (r j, where RT C is the rotate through carry function (314 RT C(a 1, a 2,, a n (a n, a 1, a 2,, a n 1 and RT C m reresents m iterations of RT C Second, its eigenvalues are given by 2 λ g(1; ζ sa2, where is the Legendre symbol, g(a; is the quadratic Gauss sum mod, s ( 1/2, and ζ e 2πi/ a And third, the matrix M is comlex Hadamard The fact that all the matrices associated to {S F (e n,2 } for rime are Hadamard leads to question if this holds for matrices associated to {S Fq (e n,2 } when q is a ower of a rime In order to try to answer this, let us revisit M 8, the matrix in Examle 31 The reader can check using her favorite comuter algebra system that this matrix is indeed real Hadamard Of course, there is, u to equivalence, only one Hadamard matrix of order 8 This means that the Sylvester matrix H 8 and M 8 are equivalent Indeed, Deh(M 8 can be constructed from H 8 by alying the ermutation σ (1 8(2 7(3 4(5 6 to the columns of H 8 Notation 32 Let A be an n n matrix and let σ S n, where S n reresents the symmetric grou of n symbols The exression σ(a reresents the matrix that can be constructed from A by alying the ermutation σ to its columns Let us do another examle to see if the attern continues Examle 33 Construct F 16 by identifying (315 F 16 F 2 [X]/(f(X, where f(x X 4 + X + 1 in F 2 [X] is irreducible Let α [X], then α 4 α + 1 Order F 16 using the lexicograhic order, that is, (316 F 16 { 0, 1, α, α + 1, α 2,, α 3 + α 2 + α + 1 } {β 0, β 1, β 2, β 3, β 4,, β 15 } Consider the sequence {S F16 (e n,2 } and roceed as in Examle 31 This rocess roduces a matrix M 16 of order 16 with entries ±1 As in the revious examle, the matrix M 16 turns out to be (real Hadamard It was mentioned in the revious section that there are 5 inequivalent Hadamard matrices of order 16, one of them being the Sylvester matrix H 16 It is natural to ask to which of these 5 matrices our Hadamard matrix is equivalent to The answer is to the Sylvester matrix H 16! Indeed, Deh(M 16 can be constructed from H 16 by alying the ermutation σ ( (2 8( ( to the columns of H 16 In other words, Deh(M 16 σ(h 16 Another examle, but with q odd

8 8 FRANCIS N CASTRO AND LUIS A MEDINA Examle 34 Similar to Examle 31, the identification F 9 F 3 [X]/(f(X, where f(x X is irreducible in F 3 [X] roduces the matrix e 2iπ 3 1 e 2iπ 3 1 e 2iπ 3 1 e 2iπ 3 e 2iπ 3 1 e 2iπ 3 1 e 2iπ 3 1 e 2iπ e 2iπ 3 M 9 1 e 2iπ 3 1 e 2iπ e 2iπ 3 1 e 2iπ 3 1 e 2iπ 3 e 2iπ e 2iπ 3 1 e 2iπ 3 1 e 2iπ 3 1 e 2iπ 3 1 e 2iπ 3 1 It can be verified that this matrix is indeed comlex Hadamard Moreover, its eigenvalues are all of the form g(1; 3 2 ζ b 3 where b F 3, g(a; is the quadratic Gauss sum mod and ζ 3 e 2πi/3 Thus, a behavior similar to the one for S F (e n,2 seems to hold for q r This is exlored in the next section where a formula for the eigenvalues of M q when q is odd is rovided The reader is encouraged to exeriment and convince herself that all matrices M q s associated to {S Fq (e n,2 }, for q r, aear to be comlex Hadamard Moreover, in the case when q 2 r is a ower of 2, the matrix seems to be real Hadamard and its dehased form is a ermutation of the columns of the Sylvester matrix H 2 r Think of this as if the dehased form of our matrix is a ermutation away from the Sylvester matrix In the next section we rove these facts 4 The recursive nature of quadratic symmetric olynomials and Hadamard matrices In this section we rove the assertions resented in the revious section That is, all matrices M q s associated to {S Fq (e n,2 } are comlex Hadamard and in the case when q 2 r, the roduced matrix is equivalent to H 2 r because its Deh(M 2 r is a ermutation away from it We also find exlicit formulas for the eigenvalues of our matrices, which in turns allows us to rovide closed formulas for S Fq (e n,2 Let α F q be a rimitive element, that is, F q {0, 1, α, α 2,, α q 2 } Similar to Examle 31, one has q 2 (41 S Fq (e n,2 S Fq (e n 1,2 + S Fq (e n 1,2 + α j e n 1,1 Therefore, if we define (42 a β (n S Fq (e n,2 + βe n,1, for β F q, then (43 S Fq (e n,2 β F q a β (n 1 j0 Thus, the recursive nature of {S Fq (e n,2 } is dominated by the sequences {a β (n} Again, similar to Examle 31, one has (44 a γ (n β F q e 2πi Tr(γβ a γ+β (n 1, where Tr Tr Fq/F is the field trace Therefore, if we let a 0 (n a 1 (n a α (n (45 a(n a α 2(n, a α q 2(n then (44 can be written as a(n M q a(n 1 for a suitable matrix M q

9 QUADRATIC SYMMETRIC POLYNOMIALS AND HADAMARD MATRICES 9 Theorem 41 Let q r with rime The matrix M q is a q q comlex Hadamard matrix Proof Define λ β (x x + β Then the matrix M q is given by R 0 (q R 1 (q (46 M q R α (q R α q 2(q where R β (q, for β F q, is the row vector ( (47 R β (r e 2πi Tr(βλ β(0 e 2πi Tr(βλ β(1 e 2πi Tr(βλ β(α e 2πi Tr(βλ β(α q 2 The row R 0 (q is simly the row vector whose entries are all 1 because Tr(0 0 Now, for any a F, we have (48 {α F q : Tr(α a} q r 1 This, together with the fact that the function α j λ α j (X is a ermutation of F q, imlies that in row R α j (q with j 0, each number in the set } {1, e 2πi 4πi 2( 1πi, e,, e aears the same amount of times It is clear that (49 Also, if j 0, then R 0 (q R 0 (q q R α j (q R α j (q q, for j 0 (410 R 0 (q R α j (q 0, because, as mentioned before, all entries of R 0 (q are 1 s and each number in the set } {1, e 2πi 4πi 2( 1πi, e,, e aears the same amount of times in R α j (q Suose now that 0 l, j q 2 and l j The linearity of the field trace function imlies (411 R α l(q R α j (q e 2πi (Tr(αl λ α l (β Tr(α j λ α j (β β F q e 2πi Tr(αl λ α l (β α j λ α j (β β F q However, note that if f : F q F q is defined as (412 f(x α l λ α l(x α j λ α j (X, then (413 f(x α l λ α l(x α j λ α j (X α l (X α l α j (X α j (α l α j X + α 2l α 2j But α is a rimitive element of F q and l j, therefore f(x is injective This imlies that (414 R α l(q R α j (q e 2πi Tr(αl λ α l (β α j λ α j (β β F q γ F q e 2πi Tr(γ However, as mentioned before, for any a F one has (415 {α F q : Tr(α a} q r 1

10 10 FRANCIS N CASTRO AND LUIS A MEDINA Thus, (416 R α l(q R α j (q e 2πi Tr(γ γ F q 1 q r 1 e 2πij 0, because e 2πi/ is a root of 1 + X + X X 1 Equations (49, (410 and (416 imly j0 (417 M q M q T qiq where I q is the q q identity matrix Thus, M q is Hadamard This concludes the roof In [8], a beautiful formula for the eigenvalues of M was established Their formula included the numbertheoretical quadratic Gauss sum mod and the Legendre symbol In articular, they show the following theorem Theorem 42 [8, Th 54] Let C( be set of eigenvalues of M were is an odd rime Let ζ e 2πi/ Then λ C( if and only if 2 (418 λ g(1; ζ sa2, where g(a; is the quadratic Gauss sum mod and s ( 1/2 In articular, C( ( + 1/2 Moreover, in the roof of Theorem 42, the authors of [8] showed that M is diagonalizable A consequence of this, and which was not stated in [8], is the following result Theorem 43 Let be an odd rime Let (419 λ j ( 2 g(1; ζ sa2 j, for 1 j ( + 1/2 be all the different eigenvalues in C( Then, (420 for some suitable constants c j ( S F (e n,2 (+1/2 j1 (+1/2 j1 c j (λ j ( n c j ( Proof Since the matrix M is diagonalizable, then the olynomial µ (X (+1/2 j1 n 2 g(1; n ζ sna2 j (X λ j ( is the minimal olynomial for M But then the sequence {S F (e n,2 } satisfies the linear recurrence whose characteristic olynomial is µ (X The result now follows from the theory of linear recurrences It is natural to ask if something similar haens for the matrices M q As mentioned in the revious section (see Examle 34, the eigenvalues of M 9 are all of the form g(1; 3 2 ζ3 b where b F 3, g(a; is the quadratic Gauss sum mod and ζ 3 e 2πi/3 A similar calculation imlies that the eigenvalues of M 25 are all of the form g(1; 5 2 ζ5 b and the ones for M 27 and M 125 are of the form 2 2 g(1; 3 3 ζ3 b and g(1; 5 3 ζ b 3 5 5, resectively In fact, we have the following result

11 QUADRATIC SYMMETRIC POLYNOMIALS AND HADAMARD MATRICES 11 Theorem 44 Suose that q r with odd rime and r > 1 Let C(q be the eigenvalues of M q Let ζ e 2πi/ Then λ C(q if and only if 2 (421 λ j (q g(1; F q ζ j F q r 2 ( 1 r 1 g(1; r ζ j, where j F, g(a; F q and g(a; are the quadratic Gauss sum in F q and F (res, and ( 2/F q and ( 2/ are the Legendre s symbol in F q and F (res In articular, C(q Proof Let F q {α 0, α 1,, α q 1 } with α 0 0 Let Tr Tr Fq/F We will rove that the λ j (q s are all the eigenvalues of M q by finding their corresonding eigenvectors Define v αa (q as the column vector whose k entry, for 0 k q 1, is The j entry of M q v αa (q is However, observe that (422 k0 ζ str((α k α a 2 q 1 ζ Tr(αj(α k α j+str((α k α a 2 Tr (α j (α k α j + str ( (α k α a 2 Tr ( α j (α k α j + s(α k α a 2 Tr ( α j α k αj 2 + sαk 2 2sα a α k + sαa 2 Tr ( 2sα j α k + 2sαj 2 + sαk 2 2sα a α k + sαa 2 This imlies that the j entry of M q v αa (q is k0 Tr ( s(α k α a α j 2 + sα 2 j 2sα a α j q 1 ζ Tr(s(α k α a α j 2 +sα 2 2sαaαj j g(s; F q ζ Tr(s(αj αa2 sα 2 a Observe that this is g(s; F q ζ Tr( sα2 a times the j entry in v αa (q Therefore, v αa (q is an eigenvector with eigenvalue g(s; F q ζ Tr( sα2 a s g(1; F q ζ Tr( sα2 a 2 g(1; F q ζ str(α2 a F q F q The claim that the eigenvalues of M q are all of the form r 2 2 g(1; F q ζ b ( 1 r 1 g(1; r ζ b, for b F F q follows from the fact that the function f : F q F given by f(α a str(α a is surjective and the classical formulas r a a (423 and g(1; F q ( 1 r 1 g(1; r F q The linear indeendence of the vectors v αa (q, for 0 a q 1 (see next roosition comletes the roof In order for the roof of Theorem 44 to be comlete, we need to show that the vectors v αa (q, for 0 a q 1, are linearly indeendent However, this is a consequence of the following (beautiful result Proosition 45 The matrix A q, whose a-th row is given by v αa (q T, is comlex Hadamard, ie A q A q T qi q where I q is the q q identity matrix In articular, the vectors v αa (q, for 0 a q 1, are linearly indeendent

12 12 FRANCIS N CASTRO AND LUIS A MEDINA Proof Let Tr Tr Fq/F By definition, ( v αj (q T ζ str((α0 αj2, ζ str((α1 αj2,, ζ str((αq 1 αj2 It is clear that v αj (q T v αj (q T q On the other hand, suose that 0 k, j q 1 and k j Then, v αk (q T v αj (q T q 1 m0 q 1 m0 ζ str((αm α k 2 str((α m α j 2 ζ str(2(αj α kα m+α 2 k α2 j However, the function f(x 2s(α j α k X + α 2 k α2 j is a bijection of F q Thus, v αk (q T v αj (q T q 1 m0 ζ str(2(αj α kα m+α 2 k α2 j β F q ζ Tr(β 0 We conclude that A q is comlex Hadamard Observe that Theorem 44 and Proosition 45 imly that, for q r with odd rime and r > 1, the minimal olynomial for the sequence {S Fq (e n,2 } is In articular, we have the following corollary 1 µ q (X (X λ j (q j0 Corollary 46 Let q r with odd rime and r > 1 Then, (424 for some suitable constants c j (q S Fq (σ m,2 for m 2, 3,, + 1 S Fq (e n,2 1 c j (qλ j (q n j0 1 n 2 c j (q g(1; F q n ζ jn j0 F q F q n 2 1 g(1; F q n c j (qζ jn j0 Moreover, the constants c j (q s deend on the λ k (q s and the values Proof The first claim is consequence of Theorem 44, Proosition 45 and the theory of linear recurrences The second claim follows by solving the linear system λ 0 (q 2 λ 1 (q 2 λ 1 (q 2 c 0 (q S Fq (e 2,2 λ 0 (q 3 λ 1 (q 3 λ 1 (q 3 c 1 (q S Fq (e 3,2 λ 0 (q +1 λ 1 (q +1 λ 1 (q +1 c 1 (q S Fq (e +1,2 In ractice, the calculation of the values S Fq (e m,2 for m 2, 3,, + 1 might not be an easy task

13 QUADRATIC SYMMETRIC POLYNOMIALS AND HADAMARD MATRICES 13 Examle 47 Consider F 9 F 3 [X]/(X Let α [X] Observe that 2 1 and g(1; F q 3 Therefore, Using the values and solving the corresonding linear system we find F 3 S F9 (e n,2 3 n c 0 (9 + c 1 (9e 2πin 3 + c 2 (9e 4πin 3 S F9 (e 2,2 9 S F9 (e 3,2 27 S F9 (e 4,2 243, c 0 (9 5 3, c 1(9 2 3 e 4πi 3, c2 (9 2 3 e 2πi 3 We conclude that (425 (426 S F9 (e n,2 3 n e 2iπ(n e 2iπ(2n+1 3 { 3 n n 0, 2 mod 3 3 n+1 n 1 mod 3 A similar argument roduces the identities (427 S F25 (e n,2 ( 1 n 5 n 1 6e (2n+3iπ 5 + 6e 3(2n+3iπ 5 + 6e (4n+1iπ 5 + 6e (8n+7iπ 5 1 { ( 5 n n 1 mod 5 ( 5 n+1 n ( 1 mod 5 S F27 (e n,2 ( i n (3n 1 4e 1 6 iπ(4n 1 + 2e 1 6 iπ(8n 5 + 3i ( n 3i 3 n 0 mod 3 ( 3i 3 n+1 n 1 mod 3 ( 3i 3 n ( n 2 mod 3 S F81 (e n,2 ( 1 n 3 9 n 1 10e π 3 i(2n e π 3 i(4n 1 7 { ( 9 n n 1 mod 3 ( 9 n+1 n 1 mod 3 Note that in all these examles S Fq (e n,2 0 for every n This is evidence of a conjecture resented in [1] Examle 48 The value of S F3 r (e n,2 is given by (428 S F3 r (e n,2 c 0 λ n 0 + c 1 λ n 1 + c 2 λ n 2, where λ j λ j (3 r ( 1 r 1 (i 3 r e 2πij 3, c j c j (3 r S F 3 r (e 2,2 λ [1+j] λ [2+j] S F3 r (e 3,2 ( λ [1+j] + λ [2+j] + SF3 r (e 4,2 λ [j] 2 ( λ [j] λ [1+j] ( λ[j] λ [2+j], and [a] reresents the unique integer l {0, 1, 2} such that a l mod 3 Theorem 44 and Corollary 46 do not aly when q even However, not everything is lost for that case In the revious section we observed that if q is even, then the matrix Deh(M q seems to be a ermutation away from the Sylvester matrix Theorem 49 Deh(M 2 r is a ermutation away from the Sylvester matrix H 2 r

14 14 FRANCIS N CASTRO AND LUIS A MEDINA Proof Let q r with a rime and r a ositive integer Let f : F q C be a function and set ζ e 2πi/ The Walsh transform of f, denoted by ˆf, is defined by (429 ˆf(α f(βζ Tr(αβ, β F q where Tr Tr Fq/F (see [22] for more details Let α F q be a rimitive element, that is, F q {0, 1, α, α 2,, α q 2 } The Walsh transform can be written in matrix form as (430 ˆf(0 ˆf(1 ˆf(α ˆf(α q 2 ζ Tr(0 0 ζ Tr(1 0 ζ Tr(α 0 ζ Tr(αq 2 0 ζ Tr(0 1 ζ Tr(1 1 ζ Tr(α 1 ζ Tr(αq 2 1 ζ Tr(0 α ζ Tr(0 αq 2 ζ Tr(1 α ζ Tr(1 αq 2 ζ Tr(α α ζ Tr(α αq 2 ζ Tr(αq 2 α ζ Tr(αq 2 α q 2 f(0 f(1 f(α f(α q 2 The matrix in equation (430 is known as the Walsh Transform Matrix and it is usually denoted by W q The careful reader robably noticed that this is the same notation used for the Discrete Fourier Matrix That is because the Walsh Transform matrix of order q for q rime coincides (uon rescaling with the Fourier Transform Matrix of the same order It is known that when q 2 r, W q is equal, u to a ermutation of columns, to the matrix obtained by letting W 1 (1, 1 1 (431 W W 2 r W 2 W 2 r 1 Observe that if σ (1 2, then W 2 σ(h 2 Moreover, (432 H and W Thus, if σ (1 4(2 3, then W 4 σ(h 4 In general, if σ is the ermutation σ (1 2 r (2 2 r 1 (2 r 1 2 r 1 + 1, then W 2 r σ(h 2 r Therefore, the Walsh transform matrix W 2 r is a ermutation away from the Sylvester matrix H 2 r (in fact, in other contexts, the Walsh transform matrix W 2 r is the Sylvester matrix Let us revisit our matrix M q Observe that if we index our matrix by the elements of F q, then the (β, γ entry of M q is (433 e 2πi Tr(βλ β(γ ζ Tr(βλ β(γ ζ Tr(β(γ β This imlies that the (β, γ entry of Deh(M q is (434 ζ Tr(βλ β(γ ζ Tr(βλ β(0 ζ Tr(βλ β(γ Tr(βλ β (0 ζ Tr(βλ β(γ βλ β (0 ζ Tr(β(γ β β(0 β ζ Tr(βγ In other words, Deh(M q W q This imlies that Deh(M 2 r is a ermutation away from H 2 r This concludes the roof

15 QUADRATIC SYMMETRIC POLYNOMIALS AND HADAMARD MATRICES 15 5 Concluding remarks We show that the recursive behavior of exonential sums of quadratic symmetric olynomials over Galois fields are dominated by Hadamard matrices We also comuted the sectrum of such matrices when working over finite fields extensions of F for odd rime This result allowed us to rovide closed formulas for the corresonding exonential sums It would be nice if similar results can be found for the associated matrices of symmetric olynomials of higher degree References [1] R A Arce-Nazario, F N Castro, O E González, L A Medina and I M Rubio New families of balanced symmetric functions and a generalization of Cuscik, Li and P Stǎnicǎ Designs, Codes and Crytograhy, 2017, DOI: /s [2] LD Baumert, M Hall Jr Hadamard matrices of the Williamson tye Math Com, 19, , 1965 [3] M L Bileschi, TW Cusick and D Padgett Weights of Boolean cubic monomial rotation symmetric functions Crytogr Commun, 4, , 2012 [4] J Cai, F Green and T Thierauf On the correlation of symmetric functions Math Systems Theory, 29, , 1996 [5] F N Castro and L A Medina Linear Recurrences and Asymtotic Behavior of Exonential Sums of Symmetric Boolean Functions Elec J Combinatorics, 18:#P8, 2011 [6] F N Castro and L A Medina Asymtotic Behavior of Perturbations of Symmetric Functions Annals of Combinatorics, 18: , 2014 [7] F N Castro and L A Medina Modular eriodicity of exonential sums of symmetric Boolean functions Discrete Al Math 217, , 2017 [8] F N Castro, R Chaman, L A Medina, and L B Seúlveda Recursions associated to traezoid, symmetric and rotation symmetric functions over Galois fields arxiv: , 2017 [9] F N Castro, O E González and L A Medina Diohantine equations with binomial coefficients and erturbations of symmetric Boolean functions IEEE Trans Inf Theory, 2017, DOI /TIT [10] R Craigen and H Kharaghani A combined aroach to the construction of Hadamard matrices Australas J Combin, 13, , 1996 [11] T W Cusick Hamming weights of symmetric Boolean functions Discrete Al Math 215, 14 19, 2016 [12] T W Cusick Weight recursions for any rotation symmetric Boolean functions arxiv: [mathco] [13] T W Cusick and B Johns Recursion orders for weights of Boolean cubic rotation symmetric functions Discr Al Math, 186, 1 6, 2015 [14] T W Cusick, Y Li, and P Stǎnicǎ Balanced Symmetric Functions over GF ( IEEE Trans on Information Theory 5, , 2008 [15] TW Cusick and P Stǎnicǎ Fast evaluation, weights and nonlinearity of rotation symmetric functions Discr Math, 258, , 2002 [16] J Hadamard Résolution d une question relative aux déterminants Bulletin des Sciences Mathématiques, 17, , 1893 [17] H Kharaghani A construction for Hadamard matrices Discrete Math, 120, , 1993 [18] H Kharaghani and B Tayfeh-Rezaie Hadamard matrices of order 32 J Combin Des, 21(5, , 2013 [19] M Kolountzakis, R J Liton, E Markakis, A Metha and N K Vishnoi On the Fourier Sectrum of Symmetric Boolean Functions Combinatorica, 29, , 2009 [20] M Miyamoto A Construction of Hadamard Matrices J Combinatorial Theory, Ser A, 57, , 1991 [21] O Moreno and C J Moreno Imrovement of the Chevalley-Warning and the Ax-Katz theorems Amer J Math, 117, , 1995 [22] G L Mullen and D Panario Handbook of Finite Fields Taylor & Francis, 2013 [23] Raymond EAC Paley On Orthogonal Matrices Journal of Mathematics and Physics, 12, , 1933

16 16 FRANCIS N CASTRO AND LUIS A MEDINA [24] A Shilka and A Tal On the Minimal Fourier Degree of Symmetric Boolean Functions Combinatorica, 88, , 2014 [25] N J Sloane The On-Line Encycloedia of Integer Sequences htt://oeisorg/ [26] Douglas R Stinson Combinatorial Designs: Constructions and Analysis Sringer-Verlag, New York, 2004 [27] P Stǎnicǎ and S Maitra Rotation Symmetric Boolean Functions Count and Crytograhic Proerties Discr Al Math, 156, , 2008 [28] P Stǎnicǎ, S Maitra and J Clark Results on Rotation Symmetric Bent and Correlation Immune Boolean Functions Fast Software Encrytion, FSE 2004, Lecture Notes in Comuter Science, 3017, SringerVerlag, 2004 [29] J J Sylvester Thoughts on Orthogonal Matrices, Simultaneous Sign Successions, and Tessellated Pavements in Two or More Colours, with Alications to Newton s Rule, Ornamental Tile-Work, and the Theory of Numbers Phil Mag, 34, , 1867 [30] JS Wallis On Hadamard Matrices J Combinatorial Theory, Ser A, 18, , 1975 [31] J Williamson Hadamard s Determinant Theorem and the Sum of Four Squares Duke Math J, 11, 65 81, 1944 Deartment of Mathematics, University of Puerto Rico, San Juan, PR address: franciscastr@gmailcom Deartment of Mathematics, University of Puerto Rico, San Juan, PR address: luismedina17@uredu