Integer Sequences and Matrices Over Finite Fields

Transcription

1 Integer Sequences and Matrices Over Finite Fields arxiv:math/ v [mathco] 2 Jun 2006 Kent E Morrison Department of Mathematics California Polytechnic State University San Luis Obispo, CA kmorriso@calpolyedu February, 2006 Abstract In this expository article we collect the integer sequences that count several different types of matrices over finite fields and provide references to the Online Encyclopedia of Integer Sequences (OEIS) Section contains the sequences, their generating functions, and examples Section 2 contains the proofs of the formulas for the coefficients and the generating functions of those sequences if the proofs are not easily available in the literature The cycle index for matrices is an essential ingredient in most of the derivations 2000 Mathematics Subject Classification: Primary 05A5, 5A33; Secondary 05A0, 05A30, B65, B73 Keywords: Integer sequences, Matrices over finite fields, Cycle index for matrices Notation q a prime power F q field with q elements GL n (q) group of n n invertible matrices over F q M n (q) algebra of n n matrices over F q (q) order of the group GL n (q) order of GL n (q) with q understood ν d number of irreducible monic polynomials of degree d over F q

2 Sequences Section OEIS All matrices A00246 Invertible matrices 2 A Subspaces, q-binomial coefficients 3 A A A Splittings (direct sum decompositions) 4 q-stirling numbers, q-bell numbers 4 Flags of subspaces 5 A005329, A Linear binary codes 6 A02266, A07683 Matrices by rank 7 Linear derangements 8, 22 A Projective derangements 8, 22 Diagonalizable matrices 9, 23 Projections 0, 24 A Solutions of A k = I, 25 A05378, 722, 725 A A A A Nilpotent matrices 2 A Cyclic(regular) matrices 3, 26 Semi-simple matrices 4, 27 Separable matrices 5, 28 Conjugacy classes 6, 29 A070933, A The Sequences References to the OEIS are accurate as of February, 2006 Sequences mentioned in this article may have been added since then and entries in the OEIS may have been modified n n matrices over F q Over the field F q the number of n n matrices is q n2 For q = 2 this sequence is A00246 of the OEIS, indexed from n = 0 The terms for 0 n 4 are 2 Invertible matrices,2,6,52,65536 The number of invertible n n matrices is given by (q) := GL n (q) = (q n )(q n q)(q n q 2 ) (q n q n ) 2

3 We will use in place of (q) unless there is eed to be explicit about the base field It is convenient to define 0 to be For q = 2 the sequence is A002884, and from 0 to 5 the terms are,,6,68,2060, A second formula for is = (q ) n q (n 2) [n]q!, where [n] q! := [n] q [n ] q [2] q [] q and [i] q := +q+q 2 + +q i are the q-analogs of n! and i Also, [i] q is the number of points in the projective space of dimension i over F q A third formula is = ( ) n q (n 2) (q;q)n, where (a;q) n is defined for n > 0 by n (a;q) n := ( aq j ) j=0 Construct a random n n matrix over F q by choosing the entries independently and uniformly from F q Then /q n2 is the probability that the matrix is invertible, and this probability has a limit as n lim n = ( q ) q n2 r r Forq = 2thelimit is Forq = 3thelimitis05602 Asq theprobability of being invertible goes to 3 Subspaces The number of k-dimensional subspaces of a vector space of dimension n over F q is given by the Gaussian binomial coefficient (also called the q-binomial coefficient) ( ) n = (qn )(q n q)(q n q 2 ) (q n q k ) k (q k )(q k q)(q k q 2 ) (q k q k ) q Also, we define ( ) 0 = Other useful expressions for the Gaussian binomial coefficients are 0 q ( ) n = k q [n] q! [k] q![n k] q!, which shows the q-analog nature of the Gaussian coefficients, and ( ) n = k k k q k(n k), q which comes from the transitive action of GL n (q) on the set of subspaces of dimension k, the denominator being the order of the subgroup stabilizing one of the subspaces 3

4 Entries A02266 A02288 are the triangles of Gaussian binomial coefficients for q = 2 to q = 24 Note that when q is not a prime power the formula does not count subspaces For q = 2 the rows from n = 0 to 6 are For the Gaussian binomial coefficients we have the q-binomial theorem [2] (+q i t) = ( ) q 2) n (k t k k q i n 0 k n Summing over k from 0 to n for a fixed n gives the total number of subspaces Sequences A0066 A00622 and A0595 A0527 correspond to the values q = 2,,8 and q = 9,,24 (Same warning applies to those q that are not prime powers) For q = 2 and n = 0,,8 the sequence begins 4 Splittings,2,5,6,67,374,2825,2922,4799 Let { } n be the number of direct sum splittings of an n-dimensional vector space into k k q non-trivial subspaces without regard to the order among the subspaces These numbers are q-analogs of the Stirling numbers of the second kind, which count the number of partitions of an n-set into k non-empty subsets, and the notation follows that of Knuth for the Stirling numbers Then it is easy to see that { } n k q = k! n + +n k =n n i k It is not difficult to verify that the triangle { } n satisfies the two variable generating function k q identity + n { } ( n u n t k = exp t ) u r k n k= q r r For q = 2 the entries for n 6 and k n are

5 Let b n be the total number of non-trivial splittings of an n-dimensional vector space: n { } n b n = k k= The b n are analogs of the Bell numbers counting the number of partitions of finite sets Then we have the formula for the generating function for the b n, + ( ) u n u r b n = exp n n r r For q = 2 the values of b n for n =,,6 are q, 4, 57, 292, 54045, Bender and Goldman [] first wrote down the generating function for the q-bell numbers, which, along with the q-stirling numbers, can be put into the context of q-exponential families [0] 5 Flags A flag of length k in a vector space V is an increasing sequence of subspaces {0} = V 0 V V 2 V k = V If dimv = n, then a complete flag is a flag of length n Necessarily, dimv i = i The number of complete flags of an n-dimensional vector space over F q is ( ) ( ) ( ) n n = [n] q [n ] q [] q = [n] q! q q q The reason is that having selected V,,V i, the number of choices for V i+ is ( ) n i Alternatively, the general linear group acts transitively on the set of complete flags The stabilizer q subgroup of the standard flag is the subgroup of upper triangular matrices, whose order is (q ) n q 2) (n Hence, the number of complete flags is (q ) n q k) = [n] q! (n Define [0] q! = The sequence [n] q! for q = 2 is A The terms for n = 0,,2,,8 are,, 3, 2, 35, 9765, 6595, , By looking up the phrase q-factorial numbers in the OEIS one can find the sequences [n] q! for q 3 The triangle of q-factorial numbers is A The q-multinomial coefficient ( ) n := n n 2 n k q [n] q! [n ] q![n 2 ] q! [n k ] q!, where n = n + +n k, is the number of flags of length k in an n-dimensional space such that dimv i = n + +n i 5

6 6 Linear binary codes An[n,k] linear binarycodeis ak-dimensional subspace ofthespace off n 2 Thus, the number of [n,k] linear binary codes is the Gaussian binomial coefficient ( ) n from section 3 and is k 2 given by the triangle A02266 Two linear binary codes are equivalent (or isometric) if there is a permutation matrix P mapping one subspace to the other The number of equivalence classes of [n,k] codes is given by the triangle A07683 The early entries are identical with the corresponding binomial coefficients, but that does not hold in general The rows for n = 0 to 6 are For linear codes over other finite fields, the notion of equivalence uses matrices P having exactly one non-zero entry in each row and column These matrices, as linear maps, preserve the Hamming distance between vectors They form a group isomorphic to the wreath product of the multiplicative group of F q with S n There are more than a dozen sequences and triangles associated to linear binary codes in the OEIS They are listed in the short index of the OEIS under Codes 7 Matrices by rank The number of m n matrices of rank k over F q is ( ) m (q n )(q n q) (q n q k ) k q ) = ( n k (q m )(q m q) (q m q k ) q = (qm )(q m q) (q m q k ) (q n )(q n q) (q n q k ) (q k )(q k q) (q k q k ) (Tojustifythefirstlineoftheformulanotethatthenumberofk-dimensionalsubspacesofF m q to serve as the column space of a rank k matrix is ( ) m Identify the column space with the k q image of the associated linear map from F n q to Fm q There are (qn )(q n q) (q n q k ) surjective linear maps from F n q to that k-dimensional image The second line follows by transposing) Define the triangle r(n,k) to be the number of n n matrices of rank k over F q and r(0,0) := Thus, ( (q n )(q n q) (q n q k ) ) 2 r(n,k) = (q k )(q k q) (q k q k ) 6

7 For q = 2 the entries from r(0,0) to r(5,5) are Linear and projective derangements A matrix is a linear derangement if it is invertible and does not fix any non-zero vector Such a matrix is characterized as not having 0 or as an eigenvalue Let e n be the number of linear derangements and define e 0 = Then e n satisfies the recursion e n = e n (q n )q n +( ) n q n(n )/2 For q = 2, the sequence is A (with offset 2) and the first few terms beginning with e 0 are,0,2,48,5824, , The sequence can be obtained from the generating function + e n u n = ( uq ) n n u r r The proof of this is in section 2 The asymptotic probability that an invertible matrix is a linear derangement is e n lim = ( q ) n r n r The asymptotic probability that a square matrix is a linear derangement is lim n e n = ( q ) 2 q n2 r r A matrix over F q is a projective derangement if the induced map on projective space has no fixed points Equivalently, this means that the matrix has no eigenvalues in F q If d n is the number of such n n matrices, then d n u n = u r ( u q r ) q For q = 2 the notions of projective and linear derangement are the same, and the corresponding sequence is given above For q = 3 the sequence has initial terms from n = to 5 0, 8, 3456, ,

8 Since two matrices that differ by a scalar multiple induce the same map on projective space, the number of maps that are projective derangements is d n /(q ) The asymptotic probability that a random invertible matrix is a projective derangement is the limit d n lim = ( ) q n q r r The asymptotic probability that a random n n matrix is a projective derangement is ( ) q q r lim n d n q n2 = r As q, this probability goes to /e, the same value as the asymptotic probability that a random permutation is a derangement 9 Diagonalizable matrices In this section let d n be the number of diagonalizable n n matrices over F q Then + ( ) q d n u n u m = n n m 0 m It follows that For q = 2 this simplifies to The sequence for d to d 8 is d n = n + +n q=n d n = n i=0 q i i For q = 3 the five initial terms are 2, 8, 58, 802, 20834, 05586, , For arbitrary q, we can easily find d 2, d 2 = 3, 39, 209, 4753, n + +n q=2 = q ( q 2 = q4 q 2 +2q 2 8 q ) 2

9 0 Projections A projection is a matrix P such that P 2 = P Let p n be the number of n n projections Then + ( ) 2 p n u n u m = n n m 0 m It follows that p n = 0 i n i i In the sum the term / i i is the number of projections of rank i Projections are also characterized as diagonalizable matrices having eigenvalues 0 or Thus, for q = 2 the diagonalizable matrices are precisely the projections, and so we get the same sequence as in section 9 From p to p 8 the sequence is 2, 8, 58, 802, 20834, 05586, , For q = 3 the map P P +I gives a bijection from the set of projections to the set of diagonalizable matrices with eigenvalues or 2 Such matrices are precisely the solutions of X 2 = I The sequence is given in the OEIS by A From p to p 7 the sequence is, 2, 4, 236, 2692, , , There is a bijection between the set of n n projections and the direct sum splittings F n q = V W The projection P corresponds to the splitting ImP KerP Note that V W is regarded as different from W V Hence, } because { } n 2 q as the same as W V From 3 we have { n p n = 2+2 2, q is the number of splittings into two proper subspaces, with V W regarded ( ) n = k q k k q k(n k), showing the relationship between the number of subspaces of dimension k and the number of projections of rank k From it we see that t there are q k(n k) complementary subspaces for a fixed subspace of dimension k Solutions of A k = I Let be the number of n n matrices A satisfying A 2 = I Such matrices correspond to group homomorphisms from the cyclic group of order 2 to GL n (q) In characteristic other than two, the generating function for the is u n = 9 ( m 0 ) 2 u m, m

10 and so = n i=0 i i These matrices are those which are diagonalizable and have only and for eigenvalues If we fix two distinct elements λ and λ 2 in the base field F q, then also gives the number of diagonalizable matrices having eigenvalues λ and λ 2 Taking the eigenvalues to be 0 and gives the set of projections, and so we see that the number of projections is the same as the number of solutions to A 2 = I, when q is not a power of 2 The proof in 24 for the number of projections is essentially the proof for the more general case The sequence A gives the number of solutions over F 3 In characteristic two, A 2 = I does not imply that A is diagonalizable and the previous formula does not hold We have the formula = q i(2n 3i) i 2i 0 i n/2 to be proved in 25 Because A 2 = I is equivalent to (A+I) 2 = 0, the formula for also counts the number of n n nilpotent matrices N such that N 2 = 0 For q = 2 the sequence is A Although there appears to be an error in the formula given in the OEIS entry, the initial terms of the sequence are correct For n =,,8 they are, 4, 22, 36, 6976, , , For q = 4 the sequence is A and the initial terms for n =,,7 are, 6, 36, 6966, , , More generally, let k be a positive integer not divisible by p, where q is a power of p We consider the solutions of A k = I and let be the number of n n solutions with coefficients in F q Now z k factors into a product of distinct irreducible polynomials z k = φ (z)φ 2 (z) φ r (z) Let d i = degφ i Then the generating function for the is u n = r i= m 0 q md i m (q d i ) Note that counts the homormorphisms from the cyclic group of order k to GL n (q) The generating function given here is a special case of the generating function given by Chigira, Takegahara, and Yoshida [3] for the sequence whose nth term is the number of homomorphisms from a finite group G to GL n (q) under the assumption that the characteristic of F q does not divide the order of G Now we consider some specific examples Let k = 3 and let q be a power of 2 Then z 3 = z 3 + = (z +)(z 2 +z +) 0

11 is the irreducible factorization Thus, d = and d 2 = 2 The generating function is + ( )( ) u n u m u 2m = n n m 0 m m 0 m (q 2 ) With q = 2 we get the sequence A with initial terms for n =,,8, 3, 57, 233, 75393, , , With q = 4 we get the sequence A with initial terms for n =,,5 3, 63, 8739, , For the next example, let k = 8 and let q be a power of 3 Then z 8 = (z )(z +)φ 3 (z)φ 4 (z)φ 5 (z), wherethelastthreefactorsallhavedegree2 Thus, r = 5,d = d 2 =,andd 3 = d 4 = d 5 = 2 The generating function is given by ( ) 2 ( ) 3 u n u m u 2m = n n m 0 m m 0 m (q 2 ) For q = 3 this gives the sequence A beginning for n =,,5 2, 32, 4448, 38628, These are the sequences in the OEIS for small values of k and q 2 Nilpotent matrices k q = 2 q = 3 q = 4 2 A A A A A A A05378 A A A A A A05377 A05385 A A A A A A A A A A A A A Fine and Herstein [5] proved that the number of nilpotent n n matrices is q n(n ), and Gerstenhaber [7] simplified the proof Recently Crabb has given an even more accessible proof [4] For q = 2 this is sequence A with initial terms from n = to n = 6,, 4, 64, 4096, ,

12 3 Cyclic matrices A matrix A is cyclic if there exists a vector v such that {A i v i = 0,,2,} spans the underlying vector space (The term regular is also used ) An equivalent description is that the minimal and characteristic polynomials of A are the same Let be the number of cyclic matrices over F q The generating function factors as and can be put into the form u n = d u n = u ( + d u d q d (u/q) d ) νd, ( ) u d νd +, q d (q d ) where ν d is thenumber of irreducible, monic polynomials of degree d over F q The generating function can be extracted from the proof of Theorem in [6], which we present in section 2 For q = 2 the sequence from a to a 7 is 2, 4, 42, 50832, , , Wall [3] proved that the probability that an n n matrix is cyclic has a limit ( lim = ) ( q ) n q n2 q 5 r r 3 Fulman [6] also has a proof For q = 2 this limit is Semi-simple matrices A matrix A is semi-simple if it diagonalizes over the algebraic closure of the base field Let be the number of semi-simple n n matrices over F q Then the generating function has a factorization + u n = ( + ) u jd νd, n n d j j (q d ) where j (q d ) = GL j (q d ) For q = 2 the sequence from a to a 7 is 2, 0, 28, 25426, , , Separable matrices A matrix is separable if it is both cyclic and semi-simple, which is equivalent to having a characteristic polynomial that is square-free Let be the number of separable n n matrices over F q Then the generating function factors u n = d 2 (+ ud q d ) νd

13 This can also be factored as u n = u For q = 2 the sequence from a to a 7 is d ( ) + ud (u d νd ) q d (q d ) 2, 8, 60, 22272, , , The number of conjugacy classes of separable n n matrices is q n q n for n 2 and q for n = This is proved by Neumann and Praeger [, Lemma 32] by showing that the number of square-free monic polynomials ofdegree n isq n q n There is anatural bijection between the set of conjugacy classes of separable matrices and the set of square-free monic polynomials obtained by associating to each conjugacy class the characteristic polynomial of the matrices in that class 6 Conjugacy classes Let be the number of conjugacy classes of n n matrices over F q The generating function for this sequence is + u n = qu r n r The number of conjugacy classes grows like q n In fact, lim n q = ( ), n q r r which is the reciprocal of the limiting probability that a matrix is invertible See section 2 For q = 2 the sequence is A The initial terms for n =,,0 are 2,6,4,34,74,66,350,746,546,3206 For q = 3 the initial terms for n =,,0 are 3,2,39,29,399,245,3783,54,34734,04754 Let b n be the number of conjugacy classes in the general linear group GL n (q), the group of n n invertible matrices over F q Then + n 0 b n u n = r For q = 2 the sequence is A00695, which starts u r qu r,3,6,4,27,60,7,246,490,002 3

14 For q = 3 the sequence is A006952, which starts 2,8,24,78,232,720,252,6528,9578,58944 For q = 4,5,7 the sequences are A04934, A04935, and A04936 The number of conjugacy classes is asymptotic to q n : b n lim n q = n Hence, in the limit the ratio of the number of conjugacy classes of invertible matrices to the number of conjugacy classes of all matrices is the same as the limiting probability that a matrix is invertible That is, b n lim = ( q ) n a r n r Sequence A07073 gives the size of the largest conjugacy class in GL n (2) Starting with n = the initial terms are, 3, 56, 3360, , The minimal order of the centralizers of elements in GL n (2) is given by the quotient of by the nth term in this sequence The resulting sequence for n =,0 is This sequence is A in the OEIS 2 Selected Proofs,2,3,6,2,2,42,84,47,294 2 The cycle index and generating functions In sections 6-3 we make heavy use of generating functions of the form u n, where the sequence counts some class of n n matrices These generating functions come from the cycle index for matrices that was first defined by Kung [9] and later extended by Stong [2] We follow Fulman s notation in [6] The cycle index for conjugation action of GL n (q) on M n (q) is a polynomial in the indeterminates x φ,λ, where φ ranges over the set Φ of monic irreducible polynomials over F q and λ ranges over the partitions of the positive integers First we recall that the conjugacy class of a matrix A is determined by the isomorphism type of the associated F q [z]-module on the vector space F n q in which the action of z is that of A This module is isomorphic to a direct sum k i= l i j= F q [z]/(φ λ i,j i ) 4

15 whereφ,,φ k aredistinctmonicirreduciblepolynomials; foreachi, λ i, λ i,2 λ i,li is a partition of n i = j λ i,j Since A is n n, then n = i n idegφ i = i,j λ i,jdegφ i Let λ i denote the partition of n i given by the λ i,j and define λ i = n i The conjugacy class of A in M n (q) is determined by the data consisting of the finite list of distinct monic irreducible polynomials φ,,φ k and the partitions λ,,λ k The cycle index is defined to be x φ,λφ (A), A M n(q) φ Φ where λ φ (A) is the partition associated to φ in the conjugacy class data for A If φ does not occur in the polynomials associated to A, then λ φ (A) is the empty partition, and we define x φ,λφ (A) = We construct the generating function for the cycle index u n A M n(q) φ Φ x φ,λφ (A) This generating function has a formal factorization over the monic irreducible polynomials For the proof we refer to the paper of Stong [2, p 70] Lemma + n= u n A M n(q) x φ,λφ (A) = φ φ x φ,λ u λ degφ λ c φ (λ) wherec φ (λ) is the orderof the group ofmoduleautomorphismsof the F q [z]-module j F q[z]/(φ λ j ) Lemma 2 Fix a monic irreducible φ Then λ u λ degφ c φ (λ) = r ( udegφ q rdegφ ) Proof We use the formula for c φ (λ) proved by Kung [9, Lemma 2, p 46] Let λ be a partition of n and let b i be the number of parts of size i Let d i = b + 2b ib i + i(b i+ + +b n ) Then c φ (λ) = i b i k= (q d idegφ q (d i k)degφ ) We see from this formula that c φ (λ) is a function of λ and q degφ Therefore it is sufficient to prove the lemma for a single polynomial of degree, say φ(z) = z, and then to replace u by u degφ and q by q degφ So, now we will prove that for φ(z) = z, λ u λ c φ (λ) = r 5 ( u q r )

16 We split the left side into an outer sum over n and an inner sum over the partitions of n λ u λ c φ (λ) = + n λ =n c φ (λ) u n c φ (λ) To evaluate the inner sum we note that λ =n matrices, which is q n(n ) from the theorem of Fine and Herstein [5] Therefore, the u n coefficient of the left side is λ =n c φ (λ) = q n ( )( ) q q n is the number of n n nilpotent For the right side we use a formula of Euler, which is a special case of Cauchy s identity and a limiting case of the q-binomial theorem A convenient reference is the book of Hardy and Wright [8, Theorem 349, p 280] For a <, y <, the coefficient of in the infinite product ( ay r ) is r y n ( y)( y 2 )( y n ) Let a = u and y = q to see that the un coefficient of ( uq ) r is r q n ( q )( q n ) Therefore, the coefficients of u n of the left and right sides are equal and so we have proved that for φ(z) = z, u λ c φ (λ) = ( u ) q r λ r Lemma 3 Let Φ be a subset of the irreducible monic polynomials Let be the number of n n matrices whose conjugacy class data involves only polynomials φ Φ Then u n = φ Φ λ u n = φ Φ r u λ degφ c φ (λ) ( udegφ q rdegφ ) 6

17 Proof Using Lemma we set x φ,λ = or 0 according to whether φ Φ or not Then on the left side the inner sum is over the matrices A that do not have factors of φ Φ in their characteristic polynomials, and so the coefficient of u n is simply the number of such n n matrices This gives the first equality in the statement of the lemma u n = φ Φ λ u λ degφ c φ (λ) Then from Lemma 2 we get the second equality Lemma 4 u = φ z λ u = φ z r u λ degφ c φ (λ) ( udegφ q rdegφ ) Proof Using Lemma 3 we let Φ be the complement of the polynomial φ(z) = z The matrices not having factors of z in their characteristic polynomials are the invertible matraices Thus, =, and so the left side is un We can generalize the first part of Lemma 3 to allow for conjugacy class data inwhich the allowed partitions vary with the polynomials The proof follows immediately from Lemma Lemma 5 For each monic, irreducible polynomial φ let L φ be a subset of all partitions of the positive integers Let be the number of n n matrices A such that λ φ (A) L φ for all φ Then u n = φ u λ degφ c φ (λ) λ L φ Lemma 6 u = φ ( udegφ ) q degφ = d ( ud ) νd q d Proof Unique factorization in the ring of polynomials F q [z] means that each of the q n monic polynomials of degree n has a unique factorization as a product of monic irreducible polynomials This implies the factorization of the generating function q n u n = u kdegφ n 0 φ k 0 The left side and the inner sum on the right are geometric series, and so qu = φ u degφ 7

18 Grouping the factors on the right according to degree we get qu = ( ) νd u d d Substituing u/q for u gives u = ( ) νd (u/q) d d The lemma follows by taking reciprocals The final lemma below is used to evaluate limiting probabilities such as lim n / The proof is straightforward and omitted Lemma 7 If n 0 α n u n = u F(u) where F(u) is analytic and the series for F() is convergent, then lim α n = F() n Now with these lemmas we are ready to prove several results stated in 22 Linear and Projective Derangements In this section we prove the results given in 8 Linear derangements are matrices with no eigenvalues of 0 or, which means that their characteristic polynomials do not have factors of z or z Theorem 8 Let e n be the number of n n linear derangements Then + e n u n = ( uq ) n n u r r Proof In Lemma 3 let Φ = Φ\{z,z } to see that e n u n = φ Φ r ( udegφ q rdegφ ) On the right side multiply and divide by the product corresponding to φ(z) = z, which is ( u ), q r to see that e n u n = φ z r r ( udegφ 8 q rdegφ ) r ( uq r )

19 Then use Lemma 4 to give e n u n = u r ( uq r ) From the generating function we can derive a recursive formula for e n Corollary 9 e n = e n (q n )q n +( ) n q n(n )/2, e 0 = Proof From Theorem 8 it follows that e n / is the sum of the u i coefficients of ( u/q r ) r for i = 0,,,n Now the u i coefficient is ( ) i q r +r 2 + +r i r <r 2 < <r i By induction one can easily show that this coefficient is Therefore Next, e n = + ( ) i (q i )(q i ) (q ) i n ( ) i (q i )(q i ) (q ) e n = e n ( ) n + (q n ) (q ) Making use of the formula for and and canceling where possible we see that e n = e n (q n )q n +( ) n q n(n )/2 We present the next result because the sequence e n for q = 2 is given in the OEIS as 2 n(n )/2 where satisfies a recursive formula Corollary 0 Let Then satisfies the recursion = e n q n(n )/2 = (q n )+( ) n, a 0 = Proof The proof follows immediately from the recursive formula for e n 9

20 Corollary The asymptotic probability that an invertible matrix is a linear derangement is e n lim = ( q ) n r n r The asymptotic that any matrix is a linear derangement is ( ) 2 q r lim n e n q n2 = r Proof We use Lemma 7 for the first statement Then lim n e n e n = lim q n2 n q ( n2 )( e n = lim n lim n ) q n2 We have just computed the first limit on the right and the second limit (from 2) is the same The proof of Theorem 8 easily adapts to give the generating function for the number of projective derangements Theorem 2 Let d n be the number of n n projective derangements Then + d n u n = ( u ) q n n u q r Proof In this case Φ = Φ \ {z a a F q } As in the proof of Theorem 8 this time we multiply and divide by the products corresponding to all the linear polynomials except φ(z) = z There are q of these and so we get e n u n = φ z r r ( udegφ q rdegφ ) r ( u q r ) q Use Lemma 4 to finish the proof It would be interesting to find a recursive formula for the d n analagous to that given in Corollary 9 for the e n 23 Diagonalizable matrices Theorem 3 Let d n be the number of diagonalizable n n matrices Then + ( ) q d n u n u m = n 0 n m 0 m and d n = n + +n q=n q 20

21 Proof We use Lemma 5 Diagonalizable matrices have conjugacy class data that only involves the linear polynomials φ(z) = z a, a F q and partitions λ φ (A) that are either empty or have the form These partitions are indexed by non-negative integer For φ(z) = z a and for λ the partition consisting of m s, the corresponding m m matrix in the canonical form is the diagonal matrix with a on the main diagonal The centralizer subgroup of this matrix is thefull general linear group GL m (q) andso c φ (λ) = m Then from Lemma 5 we see that + n 0 d n u n = = u m a F q m 0 m ( From this the formula for d n follows immediately It should be noted that the formula for d n follows directly from the knowledge of the centralizer subgroup of a diagonalizable matrix and does not require the full machinery of the cycle index generating function Also, there is an alternative approach to this problem given in [0] 24 Projections m 0 u m m ) q Theorem 4 Let p n be the number of n n projections Then + n 0 p n u n = ( m 0 ) 2 u m m and p n = 0 i n i i Proof Since projections are diagonalizable matrices with eigenvalues restricted to be in the set {0,}, it follows that the product in the generating function for the cycle index is over the two polynomials z and z and the partitions are restricted as described in the previous theorem Thus, we see that + n 0 p n u n = = u m z,z m 0 m ( m 0 u m m ) 2 The formula for p n follows immediately 2

22 25 Solutions of A k = I Theorem 5 Let be the number of n n matrices A satisfying A 2 = I, and assume that the base field F q has characteristic two Then = 0 i n/2 q i(2n 3i) i 2i Proof The rational canonical form for A is a direct sum of companion matrices for z and (z ) 2 No other polynomials occur Thus, the partition λ φ (A) for φ(z) = z consists of b repetitions of and b 2 repetitions of 2, where b +2b 2 = n From Lemma 5 we see that u n = b,b 2 u b+2b2 c z (b), where b = (b,b 2 ) denotes the partition To compute c z (b) we refer to the formula stated in the proof of Lemma 2 In that formula we have d = b +b 2 and d 2 = b +2b 2, and then c z (b) = 2 b i (q d i q di k ) i= k= This becomes c z (b) = b k= (q d q d k ) b 2 k= (q d 2 q d 2 k ) Now let i = d k in the first product and let i = d 2 k in the second product, so that c z (b) = d i=b 2 (q d q i ) d 2 i=d 2 b 2 (q d2 q i ) From each factor of the first product we pull out a factor of q b 2 and from each factor of the second product we remove a factor of q d 2 b 2 This gives Then c z (b) = q b b 2 b q (d 2 b 2 )b 2 b2 = b +2b 2 =n c z (b), where we note that d 2 = n and b = n 2b 2 Therefore, = b +2b 2 =n q b 2(3n 2b 2) b b2, which is equivalent to the formula to be proved 22

23 Theorem 6 Assume that k is a positive integer relatively prime to q Let z k = φ (z)φ 2 (z) φ r (z) be the factorization of z k over F q into distinct irreducible polynomials with d i = degφ i, and let be the number of n n matrices over F q that are solutions of A k = I Then u n = r i= m 0 q md i m (q d i ) Proof The rational canonical form of a matrix A satisfying A k = I is a direct sum of any number of copies of the companion matrices of the φ i Thus, with Lemma 5 the product is taken over the φ i for i =,,r and the subset L φi is the same for all i and consists of the partitionsinwhich allpartsare Hence, L φi canbeidentified withthenon-negativeintegers m = 0,,2, For the partition λ given by m s, the value of c φi (λ) is the order of the groupof automorphisms of the F q [z]-module F q [z]/(φ m i ), but this module is thedirect sum of m copies of the extension field of degree d i over F q Therefore, the group of automorphisms is the general linear group GL m (q d i ), and so c φi (λ) = m (q d i ) 26 Cyclic matrices Theorem 7 Let be the number of cyclic n n matrices Then + u n = ( + u d n n q d (u/q) d d ) νd and u n = u d ( ) u d νd + q d (q d ) Proof In terms of the conjugacy class data a matrix is cyclic if λ φ has at most one part for each φ Thus, L φ = {,,2,}, where m means the partition of m having just one part For these partitions we have c φ ( ) = and c φ (m) = q mdegφ q (m )degφ Let be the number of cyclic matrices of size n n Beginning with Lemma 5 we find the generating 23

24 function u n = φ u λ degφ c L φ (λ) φ ( u mdegφ = + q mdegφ q (m )degφ φ m = ( + ) u md νd q md q (m )d d m = ( ) + u md νd q d q (m )d d m = ( + ) u d νd q d (u/q) d d ) For the second part, we use Lemma 6 to get + u n = ) νd ( ud n n u q d d d Combine the products and simplify to finish the proof ( + q d ) u d νd (u/q) d 27 Semi-simple matrices Theorem 8 Let be the number of n n semi-simple matrices over F q Then + u n = ( + ) u jd νd n n d j j (q d ) Proof In order for a matrix to be semi-simple the partition associated to a polynomial φ has no parts greater than Thus, for all φ, L φ = {,, 2,, j,}, where j means the partition of j consisting of j copies of Then c φ ( j ) is the order of the automorphism group of the F q [z]-module which is the direct sum of j copies of F q [z]/(φ), which can be identified with (F q d) j, where d = degφ Furthermore, the automorphisms of the module (F q [z]/(φ)) j can be identified with the F q d-linear automorphisms of (F q d) j This group of automorphisms is GL j (q d ), and so c φ ( j ) = j (q d ) From Lemma 5 we find + u n = u λ degφ n n c φ L φ (λ) φ ( ) u jdegφ = + φ j j (q d ) = ( + ) u jd νd d j j (q d ) 24

25 28 Separable matrices Theorem 9 Let be the number of separable n n matrices over F q Then + u n = ) νd (+ ud n n q d d and u n = u d ( ) + ud (u d νd ) q d (q d ) Proof For a matrix to be separable the allowed partitions are either empty or the unique partition of So the sum over L φ in Lemma 5 is a sum of two terms For λ =, and for λ =, Therefore, u λ degφ c φ (λ) u λ degφ c φ (λ) u n = φ =, = udegφ q degφ u λ degφ c φ (λ) L φ (+ udegφ ) = φ q degφ = d ) νd (+ ud q d For the second statement of the theorem multiply the right side in the line above by ) νd ( ud, u q d which is equal to by Lemma 6 Then combine the products d 29 Conjugacy classes Theorem 20 Let be the number of conjugacy classes of n n matrices Then the ordinary generating function for the sequence { } is given by u n = r qu r 25

26 Proof Recall the ordinary power series generating function for partitions factors as the infinite product n u n 0p n = u r, r where p n is the number of partitions of the integer n A conjugacy class of n n matrices is uniquely specified by the choice of a partition λ φ for each monic irreducible polynomial φ such that φ λ φ degφ = n Consider the infinite product over φ p n u ndegφ φ n 0 The coefficient of u n is a sum of terms of the form p n u n degφ p nk u n kdegφ k where n = nk degφ k Therefore, the u n -coefficient is the number of conjugacy classes of n n matrices Then u n = φ = φ p n u ndegφ n 0 r = r φ = r d u rdegφ u rdegφ ) νd ( u rd Then by substituting qu r for u in Lemma 6 and inverting we see that ( ) νd = u rd qu r d Therefore u n = r qu r References [] Edward A Bender and Jay R Goldman, Enumerative uses of generating functions, Indiana Univ Math J 20 (970/97), [2] Peter J Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge University Press, Cambridge, 994 [3] Naoki Chigira, Yugen Takegahara, and Tomoyuki Yoshida, On the number of homomorphisms from a finite group to a general linear group, J Algebra 232 (2000),

27 [4] M C Crabb, Counting nilpotent endomorphisms, Finite Fields Appl 2 (2006), 5 54 [5] N J Fine and I N Herstein, The probability that a matrix be nilpotent, Illinois J Math 2 (958), [6] Jason Fulman, Random matrix theory over finite fields, Bull Amer Math Soc (NS) 39 (2002), 5 85 (electronic) [7] Murray Gerstenhaber, On the number of nilpotent matrices with coefficients in a finite field, Illinois J Math 5 (96), [8] G H Hardy and E M Wright, An Introduction to the Theory of Numbers, fifth ed, The Clarendon Press Oxford University Press, New York, 979 [9] Joseph P S Kung, The cycle structure of a linear transformation over a finite field, Linear Algebra Appl 36 (98), 4 55 [0] Kent E Morrison, q-exponential families, Electron J Combin (2004), Research Paper 36, pp (electronic) [] Peter M Neumann and Cheryl E Praeger, Cyclic matrices over finite fields, J London Math Soc 52 (995), [2] Richard Stong, Some asymptotic results on finite vector spaces, Adv in Appl Math 9 (988), [3] G E Wall, Counting cyclic and separable matrices over a finite field, Bull Austral Math Soc 60 (999), Sequences considered: A00246, A002820, A002884, A005329, A0066 A00622, A0595 A0527, A02266 A02288, A05378, A053722, A053725, A A053777, A053763, A A053849, A05385 A053857, A A053863, A069777, A070933, A07683, A