The Birkhoff theorem for unitary matrices of prime-power dimension
aa r X i v : . [ m a t h - ph ] D ec The Birkhoff theoremfor unitary matrices of prime-power dimension
Alexis De Vos and Stijn De BaerdemackerDecember 24, 2018
Abstract
The unitary Birkhoff theorem states that any unitary matrix withall row sums and all column sums equal unity can be decomposed as aweighted sum of permutation matrices, such that both the sum of theweights and the sum of the squared moduli of the weights are equalto unity. If the dimension n of the unitary matrix equals a power of aprime p , i.e. if n = p w , then the Birkhoff decomposition does not needall n ! possible permutation matrices, as the epicirculant permutationmatrices suffice. This group of permutation matrices is isomorphicto the general affine group GA( w, p ) of order only p w ( p w − p w − p ) ... ( p w − p w − ) ≪ ( p w )!. Let D( n ) be the semigroup of n × n doubly stochastic matrices; let P( n ) bethe group of n × n permutation matrices. Birkhoff [1] has demonstrated Theorem 1
Every D( n ) matrix D can be written D = X σ c σ P σ with all P σ ∈ P( n ) and the weights c σ real, satisfying both ≤ c σ ≤ and P σ c σ = 1 . The question arises whether a similar theorem holds for matrices from theunitary group U( n ). This question is discussed by De Baerdemacker et al.[2] [3]. For this purpose, the subgroup XU( n ) of U( n ) is introduced [4][5]. It consists of all U( n ) matrices with all line sums (i.e. all row sumsand all column sums) equal to 1. Whereas U( n ) is an n -dimensional Lie1roup, the group XU( n ) is only ( n − -dimensional. A unitary Birkhofftheorem has been proved for XU( n ) matrices [2] [3]. Remarkable is the factthat the case n = p with p an arbitrary prime [3] has been treated in avery different way from the case where n is an arbitrary integer [2]. As aresult, the decomposition, tailored to prime numbers [3], can be restrictedto n terms, whereas the general case [2] leads to a summation over all n !(or at least over n ! /
2) permutation matrices, albeit with a large number ofdegrees of freedom. In the present paper, we will treat the two cases ina unified way. Moreover, the unified approach will be applied to the case n = p w , i.e. n equal to an arbitrary power w of an arbitrary prime p .In general, the Birkhoff theorem for unitary matrices is easily proved asfollows. Let G( n ) be a finite subgroup of XU( n ). Lemma 1
If an XU( n ) matrix X can be written X = X σ c σ G σ with all G σ ∈ G( n ), then the weights c σ satisfy P σ c σ = 1 . The proof is trivial: all line sums of G σ equal unity; therefore, all line sumsof the matrix c σ G σ equal c σ and thus all line sums of the matrix P σ c σ G σ are equal to P σ c σ . As all line sums of X are equal to 1, we thus need P σ c σ = 1. Lemma 2
If every XU( n ) matrix X can be written X = X σ a σ G σ with all G σ ∈ G( n ), then there exists a decomposition X = X σ b σ G σ , such that not only P σ b σ = 1 , but also P σ | b σ | = 1 . This fact follows from the Klappenecker–R¨otteler theorem [6]. n ) Remark 1
For sake of convenience, in the present paper, the rows andcolums of a matrix are not numbered starting from 1, but instead startingfrom 0. Thus the upper-left entry of any m × m square matrix A is A , andits lower-right entry is A m − ,m − .
2e recall that the group XU( n ) is an ( n − -dimensional subgroup ofthe n -dimensional unitary group U( n ). Any member X of XU( n ) can bewritten X = T (cid:18) U (cid:19) T − , (1)where U is a member of U( n −
1) and where the constant unitary matrix T is1 / √ n times a dephased complex Hadamard matrix [7]. Thus (1) constitutesa 1-to-1 mapping between X and U . Because of T j, = T ,k = 1 / √ n , (2)eqn (1) leads to X k,l = 1 n + n − X r =1 n − X s =1 T k,r U r − ,s − ( T − ) s,l . With T being unitary, i.e. with T − = T † , this becomes X k,l = 1 n + n − X r =1 n − X s =1 U r − ,s − T k,r T l,s . We thus can write the matrix X as a sum of 1 + ( n − matrices: X = W + 1 n n − X r =1 n − X s =1 U r − ,s − M r,s , (3)where W is the n × n van der Waerden matrix, i.e. the doubly stochasticmatrix with all entries equal to n , and where M r,s is an n × n matrix definedby ( M r,s ) k,l = n T k,r T l,s . (4)The labels r and s of the matrix M r,s run from 1 to n −
1, in contrast tothe indices k and l of its entries, which run from 0 to n −
1. We thus have( n − such matrices, each having n entries. Each entry of the matrix M r,s equals the leftmost entry of its row times the uppermost entry of itscolumn. Taking into account (2), one indeed easily checks( M r,s ) ,l ( M r,s ) k, = ( M r,s ) k,l . (5)Both the first row and the first column of M r,s equal a line of the Hadamardmatrix T (up to complex conjugation and up to the factor √ n ):( M r,s ) ,l = √ n T l,s ( M r,s ) k, = √ n T k,r . (6)3ecause T is 1 / √ n times a Hadamard matrix, we have | T l,s | = 1 / √ n and | T k,r | = 1 / √ n , such that | ( M r,s ) ,l | = 1 and | ( M r,s ) k, | = 1, and thus,because of (5), we conclude that all entries ( M r,s ) k,l have unit modulus. In the present section, we consider an arbitrary doubly transitive group G( n )of n × n permutation matrices. We denote by N the order of the group.We generalize the ideas and computations in Reference [2], where G( n ) isequal to the group P( n ) of all n × n permutation matrices, thus G( n ) beingisomorphic to the symmetric group S n and N being equal to n !.In the next three sections, we will apply the Lemmas 1 and 2 to threedifferent choices of G( n ): • In case of arbitrary n , we choose the group of all n × n permutationmatrices (i.e. a group isomorphic to the symmetric group S n ). SeeSection 4. • In case of n equal to some prime p , we choose the group of all n × n supercirculant permutation matrices (i.e. a group isomorphic to asemidirect-product group C n : C n − ). See Section 5. • In case of n equal to some power w of some prime p (i.e. equal to p w ),we choose the group of all n × n epicirculant permutation matrices(i.e. a group isomorphic to the general affine group GA( w, p )). SeeSection 6.The meaning of the words ‘supercirculant’ and ‘epicirculant’ will be madeclear below. The mentioned groups are doubly transitive, as it is knownthat the symmetric group S n is n -transitive, the alternating group A n is( n − C n , which is only 1-transitive.In each of the three cases, we will prove below that every XU( n ) matrix X can be written as X = X σ c σ G σ (7)with all G σ member of the appropriate group G( n ). Because of Lemmas 1and 2, we are then allowed to put the case that both P σ c σ = 1 and P σ | c σ | = 1. For the explicit computation of the weights c σ , we notethat the G( n ) matrices form an n -dimensional reducible representation of4ome abstract group G . We assume that G has µ different irreducible repre-sentations. According to Lemma (29.1) of [9], because G is 2-transitive, the n -dimensional natural representation is the sum of the 1-dimensional trivialrepresentation and an ( n − µ irreducible repre-sentations of G : U ( ν ) = X σ c σ D ( ν ) σ , (8)where ν is the label of the irrep (0 ≤ ν ≤ µ − D ( ν ) σ is the ν thirreducible representation of G σ , and U ( ν ) is an appropriate n ν × n ν unitarymatrix, with a special mentioning for ν = 0 anf ν = 1 (see further). Here, n ν is the dimension of the ν th representation. We have µ such matrixequations (8). Each matrix eqn constitutes n ν scalar equations. We thushave a total of P µ − ν =0 n ν = N scalar equations with N unknowns c σ : X σ c σ ( D ( ν ) ( σ )) k,l = ( U ( ν ) ) k,l . Solution of this set of equations is: c σ = 1 N X ν n ν n ν − X i =0 n ν − X j =0 ( D ( ν ) ( σ )) i,j ( U ( ν ) ( σ )) i,j = 1 N X ν n ν Tr (cid:16) D ( ν ) ( σ ) † U ( ν ) ( σ ) (cid:17) . (9)We choose for ν = 0 the trivial representation, i.e. the 1-dimensionalirreducible representation with all characters equal to 1. We choose for ν = 1 the standard representation, i.e. the ( n − P σ : P σ = T (cid:18) D (1) ( σ ) (cid:19) T − and thus (cid:18) D (1) ( σ ) (cid:19) = T − P σ T .
In (9), the matrix U (0) ( σ ) equals the 1 × U (1) ( σ ) equals the ( n − × ( n −
1) lower-right block of (cid:18) U (cid:19) = T − XT . U ( ν ) ( σ ) with 2 ≤ ν ≤ µ −
1, we are allowed tochoose any unitary matrix of the right dimension n ν . This usually allows alarge number of degrees of freedom. Here, we propose two different strategiesto take advantage of this freedom. For each matrix U ( ν ) ( σ ) with 2 ≤ ν ≤ µ −
1, we choose the n ν × n ν unitmatrix. Then (9) becomes c σ = 1 N [ n Tr (cid:16) D (0) † ( σ ) (cid:17) + n Tr (cid:16) D (1) † ( σ ) U (cid:17) + µ − X ν =2 n ν Tr (cid:16) D ( ν ) † ( σ ) (cid:17) ] . (10)We take advantage of Shur’s orthogonality relation: X ν n ν Tr (cid:16) D ( ν ) † ( σ ) (cid:17) = X ν n ν Tr (cid:16) D ( ν ) † ( σ ) D ( ν ) ( ǫ ) (cid:17) = δ σ N , where ǫ is the trivial identity permutation and where δ ǫ = 1 while δ σ = 0 if σ = ǫ . Because moreover D (1) † ( σ ) = D (1) ( σ − ) and n = n −
1, we obtainthe explicit expression for the weight: c σ = δ σ + n − N Tr (cid:16) D (1) ( σ − ) U (cid:17) − n − N χ (1) ( σ − ) . (11)The number χ ( ν ) ( G ) denotes the character of the element G of the group G according to the ν th representation. It is equal to Tr( D ( ν ) ( G )). In partic-ular, we have Tr( D (1) ( G )) = Tr( G ) − The second strategy is only applicable if the group G has an anti-standardirreducible representation, non-equivalent to the standard representation.The anti-standard representation, which we will assign the label ν = 2 (ifit exists), has the same characters as the standard representation (withlabel ν = 1), except for a factor − N ≥ n − . (12)As in the first strategy, we again choose the 1 × U (0) ( σ )and the ( n − × ( n −
1) matrix U for U (1) ( σ ). However, in this second6trategy, we also choose the matrix U for each matrix U (2) ( σ ). For eachmatrix U ( ν ) ( σ ) with 3 ≤ ν ≤ µ −
1, we choose the n ν × n ν unit matrix.Then (9) becomes c σ = 1 N [ n Tr (cid:16) D (0) † ( σ ) (cid:17) + n Tr (cid:16) D (1) † ( σ ) (cid:17) + n Tr (cid:16) D (2) † ( σ ) (cid:17) + µ − X ν =3 n ν Tr (cid:16) D ( ν ) † ( σ ) (cid:17) ] . (13)Again taking advantage of Shur’s orthogonality relation and n = n = n − c σ = δ σ + n − N Tr (cid:0) D (1) ( σ − ) U (cid:1) − n − N χ (1) ( σ − ) if σ even= 0 if σ odd . (14)In the second strategy, the group G ∩ A n thus takes over the role of G and N/ N . n Lemma 3
Every XU( n ) matrix X can be written X = X σ c σ P σ with all P σ ∈ P( n ). The proof is provided by [3], by means of induction on n . Combining Lem-mas 1, 2, and 3 leads to the unitary Birkhoff theorem: Theorem 2
Every XU( n ) matrix X can be written X = X σ c σ P σ with all P σ ∈ P( n ), such that both P σ c σ = 1 and P σ | c σ | = 1 . We can apply result (11) with N = n !. The only possible values of χ (1) areTr(P σ ) − − , , , , ..., n −
1, with exception of n − .2 Second strategy The character tables of the groups S and S show no anti-standard repre-sentation. For n >
3, the group S n has an anti-standard representation. Inthis case, we can apply result (14) with N = n !. The restriction n > N = n ! is fulfilled neither if n = 2 nor if n = 3. n = p We call an n × n matrix A supercirculant iff each row k equals row k − x positions to the right. Thus A k,l = A k − ,l − x , where addition andsubtraction are modulo n . We equivalently may write A ,a = A k,a + kx . We call x the pitch of the matrix. If x = 1, then the supercirculant matrixis called circulant; if x = n −
1, then the supercirculant matrix is calledanticirculant.If p denotes a prime, then the p × p supercirculant permutation matricesare denoted S a,x , where x is the pitch and a (called the shift) is the columnwith the unit entry in the upper row (i.e. row 0). The unit entries of such p × p permutation matrix thus are located at the p positions (0 , a ), (1 , a + x ),(2 , a + 2 x ), ..., and ( p − , a + ( p − x ), where sums are to be taken modulo p .Because x and p are co-prime, the consecutive columns with a 1, i.e. thecolumns a , a + x , a + 2 x , ..., and a + ( p − x , are all different.If n equals some prime p , then we choose for the p × p Hadamard matrix T of Section 2 the p × p discrete Fourier transform F , with entries F k,l = 1 √ p ω kl , where ω is equal to the p th root of unity. Thus (4) becomes( M r,s ) k,l = ω kr − ls . From [3], we know that M can be written as a weighted sum of p supercir-culant permutation matrices: M r,s = p − X a =0 ( M r,s ) ,a S a,x ( r,s ) , (15)8here the pitch x of the matrix S a,x is a function of r and s . Indeed, thecondition ( M r,s ) k,a + kx = ( M r,s ) ,a yields kr − ( a + kx ) s = − as and thus r − xs = 0. Thus x has to satisfy the eqn sx = r mod p . This eqn has one solution: x = rs − mod p , where s − is the inverse of s modulo p . As p is prime, each non-zero integerhas exactly one inverse. With ( M r,s ) ,a = ω − as , we finally obtain M r,s = p − X a =0 ω − as S a,rs − . The supercirculant p × p permutation matrices form a group S( p ), sub-group of P( p ) (proof in Appendix A), isomorphic to the semidirect productof the cyclic group of order p and the multiplicative group of integers mod-ulo p . The group thus is isomorphic to the semidirect product of two cyclicgroups: C p : C p − , a non-Abelian group of order p ( p − Lemma 4 If n is prime, then every XU( n ) matrix X can be written X = X σ c σ S σ with all S σ ∈ S( n ). The proof is as follows. If n is a prime p , then all matrices M r,s are super-circulant with a pitch x = rs − modulo p . Also the van der Waerden matrix W is supercirculant, as it is circulant: W = n − X a =0 n S a, . Hence, according to (3), X is a weighted sum of supercirculant permutationmatrices.Combining Lemmas 1, 2, and 4 leads to9 heorem 3 If n is prime, then every XU( n ) matrix X can be written X = X σ c σ S σ with all S σ ∈ S( n ), such that both P σ c σ = 1 and P σ | c σ | = 1 . We can apply result (11) with N = p ( p − χ (1) are −
1, 0, and p −
1, as demonstrated in Appendix B. Thus we find a unitaryBirkhoff decomposition with only p ( p −
1) terms. For a prime exceeding 3,this number is substantially smaller than the number p ! / p terms. The group S(2), isomorphic to the cyclic group C , has only two irreduciblerepresentations: the trivial one and the standard one. Also the group S( n )with n equal to an odd prime p , has no inequivalent anti-standard represen-tation. Indeed, because all odd supercirculant permutations have non-unitpitch (see Appendix C) and thus have unit trace (see Appendix B) andhence have zero character χ (1) , all characters of the anti-standard represen-tation equal the corresponding characters of the standard representation.Therefore, the standard and anti-standard representations are equivalent.We conclude that we cannot apply the second strategy of Subsection 3.2.The absence of any inequivalent anti-standard representation is no surprise,as N = n ( n −
1) does not satisfy (12). n = p w For n = p w with arbitrary positive w , we can choose for T of Section 2 theKronecker product of w small (i.e. p × p ) Fourier matrices F : T = F ⊗ F ⊗ ... ⊗ F = F ⊗ w . The n × n matrix T has following entries: T a,b = 1 √ n ω f ( a,b ) , f ( x, y ) is the sum of the ditwise product of the p -ary numbers x and y : f ( x, y ) = X j x j y j mod p . As a consequence, we have( M r,s ) k,l = ω f ( k,r ) − f ( s,l ) . (16)Among the n entries of this matrix, n /p are equal to 1, n /p are equalto ω , ..., and n /p are equal to ω p − . Remark 2
For sake of convenience, below, the rows and the colums of amatrix will sometimes be pointed at, not by a number, but instead by a vector.This will allow matrix computations for the row and column numbers. Forthis purpose, any number z = z + z p + z p ... + z w − p w − has an associatedboldfaced w × vector z = ( z , z , z , ..., z w − ) T , consisting of the w dits ofthe number z . We call a matrix A epicirculant if row k equals row 0, ‘shifted to theright’ according to A , a = A k , a + xk , where a is the w × a and where x is a w × w matrix called the pitch matrix, consisting of w entries, all ∈ { , , ..., p − } . A matrix of the form (16) is automatically epicirculant.It is a weighted sum of epicirculant permutation matrices E : we have M r,s = p − X a =0 ( M r,s ) ,a E a , x ( r,s ) . (17)Here, x is an appropriate w × w pitch matrix, depending on r and s . Proofis in Appendix D. We note that vector a and matrix x constitute a pair,fully specifying an affine transformation [10].If n is a prime power, say n = p w , then the epicirculant p w × p w permuta-tion matrices form a group E( n ), subgroup of P( n ) (proof in Appendix E),isomorphic to the general affine group GA( w, p ), a semidirect product ofthe direct product of cyclic groups of order p and the general linear groupGL( w, p ): GA( w, p ) = C wp : GL( w, p )of order p w ( p w − p w − p )( p w − p ) ... ( p w − p w − ) . (18)11e note that GA( w, p ) is a maximal subgroup of the symmetric group S p w (O’Nan–Scott theorem) [11].Each of the w subgroups C p consists of p matrices, each a Kroneckerproduct with a total of w factors: I ⊗ I ⊗ ... ⊗ I ⊗ M ⊗ I ⊗ ... ⊗ I , where I denotes the p × p unit matrix and M a p × p circulant permutationmatrix S a, . Lemma 5 If n is a prime power, then every XU( n ) matrix X can be written X = X σ c σ E σ with all E σ ∈ E( n ). The proof is as follows. If n is a prime power p w , then all matrices M r,s areepicirculant with an invertible pitch matrix x . Also the van der Waerdenmatrix W is epicirculant, as it is circulant: W = n − X a =0 n E a , , where the pitch matrix denotes the w × w unit matrix. Hence, accordingto (3), X is a weighted sum of epicirculant permutation matrices.Combining Lemmas 1, 2, and 5 leads to Theorem 4 If n is a prime power, then every XU( n ) matrix X can bewritten X = X σ c σ E σ with all E σ ∈ E( n ), such that both P σ c σ = 1 and P σ | c σ | = 1 . We can apply result (11) with N given by (18). The only possible valuesof χ (1) are −
1, 0, p − p − p −
1, ..., and p w −
1, as demonstrated inAppendix F. 12 .2 Second strategy
For w > p >
2, the general affine groups have, besides the stan-dard representation, also an inequivalent anti-standard representation. Fora proof, it suffices to point to a single example of an odd epicirculant per-mutation matrix with trace different from unity. We choose the p w × p w matrix E = I ⊗ I ⊗ ... ⊗ I ⊗ M , i.e. the Kronecker product of w − I (i.e. the p × p unit matrix)and the p × p supercirculant matrix M = S ,q . The w × w pitch matrixassociated with E is the diagonal matrix diag( q, , , ..., A ⊗ B ) = [ Det( A ) ]dim ( B ) [ Det( B ) ]dim ( A ) . (19)Therefore, we have Det( E ) = Det( M ) ( p w − ) . We choose the number q suchthat Det( M ) = − E ) = −
1. This is always possible. Suf-fice it to choose q equal to g ( p ), where g is a generator of the modulo p multiplication group [12]. Unfortunately, there is no algorithm known forfinding such generator except brute force [13]. Nevertheless, we can provethat Det( S ,g ( p ) ) = −
1, without a priori knowing the value of g ( p ): seeAppendix C.On the other hand, we have Tr( E ) = p w − Tr( M ) = p w − p w − .Because w >
1, we have Tr( E ) > χ (1) >
0. We thus concludethat we can apply result (14) with N according to (18).The above reasoning is not valid for p = 2, because, in that case,Det( M ) = − E ) = −
1. For the case p = 2, wewill prove that all 2 w × w epicirculant matrices are even permutations. Forthis purpose, it is sufficient to demonstrate that all group generators areeven. From reversible computing [14] [15] [16], it is known that the groupGA( w,
2) is generated by following matrices: A = I ⊗ I ⊗ ... ⊗ I ⊗ (cid:18) (cid:19) ⊗ I ⊗ I ⊗ ... ⊗ IB = I ⊗ I ⊗ ... ⊗ I ⊗ ⊗ I ⊗ I ⊗ ... ⊗ I = I ⊗ I ⊗ ... ⊗ I ⊗ ⊗ I ⊗ I ⊗ ... ⊗ I , with a total of w − A ) or w − B and C ) factors I . In the contextof computing, these matrices represent NOT gates, respectively controlled
NOT gates. Applying (19), we have:Det( A ) = [ Det (cid:18) (cid:19) ] ( p w − ) = ( − w − = 1Det( B ) = [ Det ] ( p w − ) = ( − w − = 1Det( C ) = [ Det ] ( p w − ) = ( − w − = 1 , except if w = 2. Thus, for w >
2, all members of GA( w,
2) represent evenpermutations and the second strategy (Subsection 3.2) is not applicable.This leaves us with the case p = 2 and w = 2. The epicirculant matricesform a group E(4) isomorphic to the symmetric group S . As stated in Sec-tion 4.2, the second strategy is applicable. The results on the applicabilityof the second strategy are summarized in Table 1.Table 1: Applicability of the second strategy for the Birkhoff decompositionof an XU( n ) matrix with n = p w . p = 2 p ≥ w = 1 no no w = 2 yes yes w ≥ According to [2], every unit-linesum n × n unitary matrix can be decomposedas a weighted sum of the n × n permutation matrices, such that both the sum14f the weights and the sum of the squared moduli of the weights equal unity.Such Birkhoff sum contains n ! terms. In the present paper, we demonstratethe following: • If n ≥
4, then n ! / • If n = p w with p an arbitrary prime and w an arbitrary integer, then p w ( p w − p w − )( p w − p w − ) ... ( p w − p )( p w −
1) suffice. • If n = p w with p an arbitrary odd prime and w an integer ≥
2, then p w ( p w − p w − )( p w − p w − ) ... ( p w − p )( p w − / n × n unit-linesum unitary matrix. n n
12 13 14 15 16 17239,500,800 156 43,589,145,600 653,837,184,000 322,560 272The case of n equal to the product of two different primes is left forfurther investigation. References [1] G. Birkhoff, “Tres observaciones sobre el algebra lineal”,
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Journal ofAlgebra , vol. 111 (1987), pp. 365-383.[12] mathworld.wolfram.com/ModuloMultiplicationGroup.html (2018).[13] K. Conrad, “Cyclicity of (Z/( p )) × ”, (2018).[14] T. Beth and M. R¨otteler, “Quantum algorithms: applicable algebra andquantum physics”, In: G. Alber, T. Beth, M. Horodecki, P. Horodecki,R. Horodecki, M. R¨otteler, H. Weinfurter, R. Werner, and A. Zeilinger,“Quantum information”, Springer Verlag, Berlin (2001), pp. 96-150.[15] K. Patel, I. Markov, and J. Hayes, “Optimal synthesis of linearreversible circuits”, Quantum Information and Computation , vol. 8(2008), pp. 282-294.[16] A. De Vos, “Reversible computing”, Wiley–VCH, Weinheim (2010).16
The group of supercirculant permutation ma-trices
The supercirculant n × n permutation matrices form a group. Indeed, theproduct of two such matrices (say S a,x and S b,y ) yields a third such matrix.In order to prove this fact, we compute the matrix entry at position ( u, v ):( S a,x S b,y ) u,v = X f ( S a,x ) u,f ( S b,y ) f,v = X f δ f, a + ux δ v, b + fy = δ v, b +( a + ux ) y = δ v, b + ay + uxy = ( S b + ay, xy ) u,v and hence S a,x S b,y = S b + ay, xy . (20)If n is a prime p , each non-zero number x has an inverse number x − .Applying (20), we find S a,x S − ax − , x − = S , . The right-hand side being the p × p unit matrix, the result proves that eachsupercirculant matrix has an inverse matrix that also is supercirculant:( S a,x ) − = S − ax − , x − . We conclude by considering two applications of eqn(20): • choosing x = y = 1 leads to S a, S b, = S a + b, illustrating that the p matrices S a, are isomorphic to the additionmodulo p ; • choosing a = b = 0 leads to S ,x S ,y = S , xy illustrating that the p − S ,x are isomorphic to the multi-plication modulo p .Each supercirculant matrix can be decomposed as the product of a zero-shiftmatrix and a unit-pitch matrix: S a,x = S ,x S a, = S ax − , S ,x . The trace of a supercirculant permutation ma-trix
We compute the trace of the supercirculant permutation matrix S a,x :Tr( S a,x ) = X u ( S a,x ) u,u = X u δ u, a + ux . If the eqn u (1 − x ) = a is fulfilled, then the corresponding number u points to a unit entry in position( u, u ) of the matrix S a,x . We notice: • If x = 1, then u = a (1 − x ) − is the one and only solution; • if x = 1 and a = 0, then the eqn has no solution u ; • if x = 1 and a = 0, then u may have any value from { , , , ...,p − } .Thus we conclude: • Tr( S a,x ) = 1, if x = 1, • Tr( S a, ) = 0, if a = 0, and • Tr( S , ) = p . C The determinant of a supercirculant permuta-tion matrix
As mentioned in Appendix A, each supercirculant matrix can be decomposedas follows: S a,x = S ,x S a, . Hence: Det( S a,x ) = Det( S ,x ) Det( S a, ) . We have S a, = ( S , ) a and therefore Det( S a, ) = (Det( S , )) a . If p is odd,then Det( S , ) = 1, such that Det( S a, ) = 1. In other words: for oddprimes, all of the p circulant permutation matrices have unit determinant.The situation is different for the p − S ,x . Half of them haveunit determinant and half of them have determinant equal to −
1. In order18o prove this fact, the key observation is the fact that the cyclic groupis Abelian; so there exists a similarity transformation that diagonalizes all matrices S ,x . We now prove that the following matrix F serves our purpose: F u,v = u = v = 00 if u = 0 and v = 00 if u = 0 and v = 0 ω vϕ ( u ) √ p − if u = 0 and v = 0 , where ω = exp( πip − ) is the ( p −
1) th root of unity, and the function ϕ ( a )gives the ‘position’ of the number a in the cyclic group C p − (multiplicativegroup modulo p ), as a power of the (a priori unknown) generator g , i.e. a = g ϕ ( a ) . From this definition, the following interesting properties of ϕ can be deduced: ϕ (1) = 0 ϕ ( g ) = 1 ϕ ( ab ) = ϕ ( a ) + ϕ ( b ) . These properties are key in the following derivation. We compute the simi-larity transformation given by F † S ,x F . Because both F and S ,x are blockdiagonal with a single 1 in the upper-left corner, we only need to computethe lower-right part:( F † S ,x F ) u,v = p − X k =1 p − X l =1 F k,u ( S ,x ) k,l F l,v = 1 p − p − X k =1 p − X l =1 ω − uϕ ( k ) δ l,xk ω vϕ ( l ) = 1 p − p − X k =1 ω − uϕ ( k )+ vϕ ( xk ) = 1 p − p − X k =1 ω − uϕ ( k )+ vϕ ( x )+ vϕ ( k ) = ω vϕ ( x ) δ u,v . This result leads to two conclusions:19
By choosing x = 1, we find that ( F † F ) u,v = δ u,v and thus that F isunitary. • By choosing x arbitrary, we find that the matrix S ,x has the eigen-values ω vϕ ( x ) plus an additional 1 from the upper-left matrix block.The determinant is just the product of all eigenvalues:Det( S ,x ) = p − Y v =1 ω vϕ ( x ) = ω ϕ ( x ) P p − v =1 v = ω ϕ ( x ) p ( p − = e πip − ϕ ( x ) p ( p − = e πiϕ ( x ) p . Now, if p is an odd prime, then e πip = −
1, such that Det( S ,x ) = ( − ϕ ( x ) ,which proves that the sign of the determinant of S ,x alternates in the chainof successive elements of C p − . More in particular, the position of x = g always is ϕ ( g ) = 1, so we have Det( S ,g ) = − S a, and S ,x are only validfor odd primes p . If p is even, i.e. if p = 2, then there exist only twosupercirculant matrices S , = (cid:0) (cid:1) , with determinant equal to 1, and S , = (cid:0) (cid:1) , with determinant equal to − D The pitch matrix
In (17), the epicirculant matrix E a , x needs a unit entry in position ( k , a + xk )if ( M r,s ) k , a + xk = ( M r,s ) , a implying f ( k, r ) − f ( s, a + X u X v x u,v k v p u ) = f (0 , r ) − f ( s, a )or X j k j r j − X j s j a j + ( X u X v x u,v k v p u ) j ! = − X s j a j and thus X j s j X v x j,v k v = X j k j r j or X v k v X j x j,v s j = X v k v r v X v k v ( X j x j,v s j − r v ) = 0 . We fulfil this condition by the set of w non-coupled eqns X j s j x j,v = r v . (21)For each eqn, we expect p w − solutions (as we can choose w − w dits x j,v arbitrarily from { , , ..., p − } ). However, many solutions have tobe rejected. Indeed, each column of the matrix E a , x in (17) should containone and only one unit entry. For this purpose, it is necessary and sufficientthat the matrix x is invertible. Proof is as follows. We require that for anytwo different row numbers ( k ′ = k ) the unit entry of the permutation matrixis in another column: a + xk ′ = a + xk and thus x ( k ′ − k ) = . This requires that for any non-zero number K wehave xK = . This, in turn, requires that the rows of x are linearly independent and thusthat the matrix x is invertible.We now prove that, for any pair ( r, s ), the set (21) has at least one ac-ceptable solution, i.e. a solution such that the matrix x is invertible. Indeed: • Because both r and s are non-zero, at least one dit r u is non-zero andat least one dit s j is non-zero. Let r α be the least-significant non-zerodit of r ; let s β be the least-significant non-zero dit of s . • We choose all dits x j,v = 0, except the dits x v,v , x β,v , and x α,β . Thuseqns (21) become s v x v,v + s β x β,v = r v mod p if v = βs α x α,β + s β x β,β = r β mod p . (22) • For v = α and v = β , we choose x v,v = 1. Further we choose x α,α = 0and x α,β = 1. Thus eqns (22) become s β x β,v = r v − s v mod p if v = α and v = βs β x β,α = r α mod p (23) s β x β,β = r β − s α mod p which lead to a single solution set x β,v .21he resulting pitch matrix x consists of a non-zero diagonal, one non-zero row, and one extra unit entry. E.g. for w = 7, α = 2, and β = 4, wehave: x , x , x , x , x , x , x , . We note that here Det( x ) equals x , . In general, we haveDet( x ) = ± x β,α = ± r α s − β . Because Det( x ) = 0, we have that x is invertible. E The group of epicirculant permutation matrices
The epicirculant permutation matrices form a group. An arbitrary entry(at location ( k , l )) of such matrix E a , x is δ l , a + xk . The product of two suchmatrices yields a third such matrix. Indeed:( E a , x E b , y ) u,v = X f ( E a , x ) u,f ( E b , y ) f,v = X f δ f , a + xu δ v , b + yf = δ v , b + ya + yxu = ( E b + ya , yx ) u,v and hence E a , x E b , y = E b + ya , yx . Straightforward application of this result leads to E a , x E − x − a , x − = E , . The right-hand side being the p w × p w unit matrix, the result proves thateach epicirculant matrix has an inverse matrix that also is epicirculant:( E a , x ) − = E − x − a , x − . a and a matrix with unit pitch matrix x : E a , x = E , x E a , = E x − a , E , x . F The trace of an epicirculant permutation matrix
We compute the trace of the epicirculant permutation matrix E a , x :Tr( E a , x ) = X u ( E a , x ) u,u = X u δ u , a + xu . If the eqn ( − x ) u = a is fulfilled, then the corresponding number u points to a unit entry in position( u, u ) of the matrix E a , x . Here, denotes the w × w unit matrix. We notice: • If ( − x ) is invertible, then u = ( − x ) − a is the one and only solu-tion; • if ( − x ) = and a = , then the eqn has no solutions u ; • if ( − x ) = and a = , then u may have any value from { , , , ...,p w − } ; • if ( − x ) is neither invertible nor zero, then ( − x ) has rank λ with1 ≤ λ ≤ w − u can have as many values as there are solutionsof the eqn ( − x ) u = , i.e. as the size of the kernel of ( − x ), i.e. p w − λ .Thus we conclude: • Tr( E a , ) = 0, if a = , • Tr( E , ) = p w , and • Tr( E a , x ) = p w − λ , if ( − x ) has rank λ6