Three Enumeration Problems for a Congruence

Three Enumeration Problems for a Congruence Equation (Author’s July 2014 expanded version)

Gregory L. Wilson BerrieHill Research Corporation Dayton, Ohio 45459 [email protected]

MSC 2010 Classifications: 11P83, 15A15, 05A17, 05A05

Acknowledgement In appreciation of guidance provided by my Ph.D. advisor Morris Newman.

Abstract The number of ways of selecting k parts from a given set of integers so that the sum of the parts is congruent to r modulo n is sought. In the three problems considered, the first places no restriction on the parts, the second requires that the parts be non-decreasing, and the third requires that they be strictly increasing. The solutions are obtained by substituting the nth roots of unity into an appropriate generating function and applying a well known transformation. Applications of each enumeration problem are given that address the cycle structure of permutation matrices that arise from generalized matrix functions. Keywords: Ramanujan sum, generalized matrix function, signed permutation matrix

1

1 1.1

Enumerations Unrestricted parts

The argument for the first enumeration problem is quite simple yet it illustrates the basic technique used in all three problems. This technique can be found in reference [1]. Theorem 1. The number of ways of selecting k parts x1 , x2 , . . . , xk from the integers 1, 2, . . . , n so that k  xi ≡ r mod n i=1 k−1

is given by n

.

Proof. Denote the number of solutions (xl , x2 , . . . , xk ) by N and consider the generating polynomial  n  n   n     f (z) = z x1 z x2 · · · z xk x1 =1

 n−1 k  = zk zi  = z

k

i=0

1 − zn 1−z

x2 =1

xk =1

k (for z = 1).

It is apparent that N is the sum of the coefficients of z tn+r , t = 0, 1, 2, . . . in f . Since   n if n | m, m ζ = 0 if n  m, n ζ =1

the sum being over all nth roots of unity, it follows that N =

1  −r ζ f (ζ). n ζ n =1



k

If ζ is an nth root of unity, f (ζ) = lim z z→ζ

k

1 − zn 1−z

 =

1 nk = nk−1 and the proof is complete. Thus, N = n

2

nk 0

if if

ζ = 1, ζ = 1

(1)

1.2

Non-decreasing parts

This section and the next involve the Ramanujan sum  μ(d/Δ) φ(d) s(d, m) = ζm = φ(d/Δ) where ζ runs over all primitive dth roots of 1, μ is the M¨obius function, φ is Euler’s function, and Δ is the greatest common divisor of m and d. The latter expression for the Ramanujan sum is due to O. Holder ([5]). Theorem 2. The number of ways of selecting k parts x1 , x2 , . . . , xk from 0, 1, 2, ..., n such that k  xi ≡ r mod n, 0 ≤ x1 ≤ x2 ≤ · · · ≤ xk ≤ n i=1

is given by

  1  n/d + k/d s(d, r), n/d n

(2)

d|n

where   denotes the greatest integer function and the summation extends over all positive divisors of n. Proof. The number N of solutions described above is also the number of k + 1-tuples (x1 , x2 , . . . , xk+1 ) satisfying k+1 

xi ≡ r mod n,

0 ≤ x1 ≤ x2 · · · ≤ xk+1 = n

i=1

Making the transformation x1 = yk+1 x2 = yk+1 + yk .. . xk+1 = yk+1 + yk + · · · + y1 , we see that N is the number of (k + 1)-tuples (y1 , y2 , . . . , yk+1) with non-negative coordinates satisfying k+1 

i yi ≡ r mod n,

i=1

k+1 

yi = n.

i=1

Consider the generating function ⎞ ⎛  ∞   ∞ ∞    w y1 z y1 w 2y2 z y2 · · · ⎝ w (k+1)yk+1 z yk+1 ⎠ . f (w, z) = y1 =0

y2 =0

yk+1 =0

3

We see that N is the sum of the coefficients of w tn+r z n , t = 0, 1, 2, . . . in f . Since f (w, z) = (1 − wz)−1 (1 − w 2 z)−1 · · · (1 − w k+1z)−1 it follows that f (w, wz) = (1 − w 2 z)−1 · · · (1 − w k+1z)−1 (1 − w k+2z)−1 (1 − wz) f (w, z) = (1 − w k+2z) and we obtain (1 − w k+2 z)f (w, wz) = (1 − wz)f (w, z). Writing f (w, z) =

∞ 

gi z i

i=0

where gi is a polynomial in w, we have (1 − w

k+2

z)

∞ 

gi w z = (1 − wz) i i

i=0

Thus,

∞ 

gi w z − i i

i=0

∞ 

gi w

∞ 

gi z i .

i=0

i+k+2 i+1

z

=

i=0

∞ 

gi z −

i=0

i

∞ 

gi wz i+1 .

i=0

Equating coefficients we have (for i ≥ 1) gi w i − gi−1 w i+k+1 = gi − w gi−1 . which gives

w(1 − w i+k ) gi−1 , gi = 1 − wi Since g0 = 1, this becomes gi

i = 1, 2, . . . .

i 1 − w k+m = w 1 − wm m=1 i

(3)

an empty product being 1. Therefore, N is the sum of the coefficients of w tn+r , t = 0, 1, . . . in the polynomial gn

n k 1 − w k+m 1 − w n+m n = w = w . 1 − wm 1 − wm m=1 m=1 n

The last equality can be obtained by inserting or cancelling the appropriate factors according to the cases k ≥ n and k < n respectively. 4

Using Eq. (1) we obtain N =

1  −r ζ gn (ζ). n ζ n =1

or equivalently 1   −r ζ gn (ζ). n

N =

d|n ◦(ζ)=d

The inner summation is taken over all nth roots of unity of order d, that is, all primitive dth roots of unity. Let d be a positive divisor of n, t an nth root of unity of order d, and m an integer such that 1 ≤ m ≤ k. If d does not divide m then lim

w→ζ

1 − w n+m = 1, 1 − wm

(d  m).

Otherwise, an application of ’Hˆopital’s rule yields lim

w→ζ

1 − w n+m n+m , m = 1−w m

(d | m).

Therefore, ζ

Thus,

−r

k/d m+n sd + n −r = ζ gn (ζ) = ζ m sd s=1 d|m 1≤m≤k   k/d s + n/d n/d + k/d −r −r = ζ = ζ . n/d s s=1 n−r

    1  n/d + k/d 1  n/d + k/d  −r ζ = N = s(d, r). n/d n/d n n ◦(ζ)=d

d|n

d|n

Corollary The number of solutions of k 

xi ≡ r mod n,

1 ≤ x1 ≤ x2 ≤ · · · ≤ xk ≤ n

i=1

is given by

  1  n/d + k/d − 1 s(d, r), k/d n d|g

5

g = gcd(n, k).

(4)

Proof. Let M denote the number of such solutions. It follows from Theorem 2 that     1  n/d + k/d 1  n/d + (k − 1)/d M = s(d, r) − s(d, r) n/d n/d n n d|n d|n      1 n/d + k/d n/d + (k − 1)/d − s(d, r). = n/d n/d n d|n

Since

 k/d =

we have

(k − 1)/d + 1 (k − 1)/d

if if

d | k, d  k,

    1  n/d + k/d n/d + k/d − 1 M = − s(d, r), n/d n/d n d|n, d|k   1  n/d + k/d − 1 s(d, r), = n/d − 1 n d|g    1 n/d + k/d − 1 = s(d, r), k/d n d|g

which completes the proof.

1.3

Strictly increasing parts

In what follows we assume k ≤ n and denote the GCD of n and k by g. Also, for integers t and  with t ≥ 0 and  > 0, we denote the number of k-tuples (x1 , x2 , . . . , xk ) satisfying k 

xi ≡ t mod ,

1 ≤ x1 < x2 · · · < xk ≤ n

i=1

by N(t, ). Lemma 1. If 0 ≤ t1 , t2 < n and t1 ≡ t2 mod g, then N(t1 , n) = N(t2 , n). Proof. Choose integers α and β such that t2 − t1 = αk + βn. Let S be the set of k-tuples (y1 , y2, . . . , yk ) with integer coordinates satisfying k 

yi ≡ t2 mod n,

α + 1 ≤ y1 < y2 < · · · < yk ≤ α + n.

i=1

6

Let S  be the set of k-tuples enumerated by N(t1 , n). If (xl , x2 , . . . , xk ) ∈ S  , then for some integer q k  xi = qn + t1 i=1

and it follows that k 

(xi + α) = qn + t2 − βn ≡ t2 mod n.

i=1

Therefore, a map from S  into S is given by (x1 , x2 , . . . , xk ) −→ (xl + α, x2 + α, . . . , xk + α). Since this map is bijective, S  and S have the same number of elements. It remains to show that S  and S have the same number of elements where S  is the set of k-tuples enumerated by N(t2 , n). Let (y1 , y2 , . . . , yk ) ∈ S. For m = 1, 2, . . . , k, let rm be the remainder obtained upon division of ym by n, unless the remainder is 0 in which case rm = n. Since each pair of integers between α + 1 and α + n (inclusive) are incongruent modulo n, the rm are distinct and can therefore be arranged in increasing order yielding an element in S  . The result now follows from the fact that the mapping from S into S  specified above is a bijection. The enumeration provided in the next theorem is derived in [6] using a different approach. It was discovered independently by the author. Theorem 3. The number of k-tuples (x1 , x2 , . . . , xk ) satisfying k 

xi ≡ r mod n,

1 ≤ x1 < x2 < · · · < xk ≤ n

i=1

is given by     1 1  n/d s d, k(k + 1) − r . N(r, n) = k/d n 2 d|g

Proof. Since 

n/g−1

N(r, g) =

N(jg + r, n),

j=0

It follows from the Lemma 1 that N(r, g) = n g N(r, n), or equivalently N(r, n) = 7

g N(r, g). n

(5)

N(r, g) is also the number of (k + 1)-tuples (x1 , x2 , . . . , xk+1 ) satisfying k+1 

xi ≡ (r + 1) mod g,

1 ≤ x1 < x2 < · · · < xk < xk+1 = n + 1.

i=1

This in turn is the number of (k + 1)-tuples (y1 , y2 , . . . , yk+1) with positive integer coordinates such that k+1 k+1   i yi ≡ (r + 1) mod g, yi = n + 1. i=1

i=1

Define  f (w, z) =

∞ 

 w y1 z y1

y1 =1



∞ 

w 2y2 z y2

⎛ ···⎝

y2 =1

= w (k+1)(k+2)/2 z k+1

k+1

∞ 

⎞ w (k+1)yk+1 z yk+1 ⎠

yk+1 =1

(1 − w m z)−1 .

m=1

It follows that N(r, g) is the sum of the coefficients of w tg+r+1 z n+1 , t = 0, 1, 2, . . . in f . Taking i = n − k in (3), we see that the coefficient of z n+1 in f is given by the polynomial w (k+1)(k+2)/2 w n−k

n−k

k 1 − w k+m 1 − w m+n−k k(k+1)/2+n+1 = w 1 − wm 1 − wm m=1 m=1

Using (1) (where n is replaced by g) and arguing as in Theorem 2, it follows that N(r, g) =

1  k(k+1)/2−r ζ h(ζ) g ζ g =1

where h is the polynomial given by k 1 − w m+n−k h(w) = . m 1 − w m=1

Let ζ be a gth root of unity of order d and m be an integer such that 1 ≤ m ≤ k. Since ζ is an nth and a kth root of unity we have ⎧ ⎨ m+n−k 1 if d  m 1−w = lim w→ζ ⎩ m + n − k if d | m. 1 − wm m

8

It follows that k/d m+n−k sd + n − k = h(ζ) = m sd s=1 d|m 1≤m≤k     k/d s + (n − k)/d (n − k)/d + k/d n/d = = k/d . = k/d s s=1

Collecting the above results we have g N(r, n) = N(r, g) n 1  k(k+1)/2−r ζ h(ζ) = n ζ g =1   1  n/d  k(k+1)/2−r ζ = k/d n d|g ◦(ζ)=d    1 1 n/d = s(d, k(k + 1) − r) k/d n 2 d|g

and the proof is complete. It is interesting to note that (4) and (5) become very simple when it is assumed that n and k are relatively prime. The enumeration formulas in this case are given respectively by     1 n+k−1 1 n and . k n n k

2 2.1

Applications Generalized matrix functions

Three main generalized matrix functions known as the tensor power ([3]), induced ([4]), and compound ([4]) functions are now considered. When these functions are applied to an n × n cyclic permutation matrix P , the result is again a permutation matrix, or in the case of the compound, a signed permutation matrix. A characteristic root of such a matrix is an nth root of unity whose multiplicity reveals information about its cycle structure. For convenience to the reader, we define each of the three generalized matrices below. For a positive integer k and an n × n matrix A with complex entries, the following notation is used: Tk,n (A) = kth tensor power of A Lk,n (A) = kth induced matrix of A Ck,n (A) = kth compound of A (k ≤ n). 9

Defining the tensor product of A with itself by ⎛ a1,1 A a1,2 A · · · a1,n A ⎜ a2,1 A a2,2 A · · · a2,n A ⎜ A ⊗ A = ⎜ .. .. .. ⎝ . . . an,1 A an,2 A · · · an,n A

⎞ ⎟ ⎟ ⎟, ⎠

the kth tensor power of A is the nk × nk matrix given by Tk,n (A) =

k 

A.

i=1

Let x = (x1 , x2 , . . . , xn )t where the xi are indeterminates and let y = Ax. Let k be an arbitrary positive integer and define a set of integer sequences of length n by Γ = {(γ1, γ2 , . . . , γn ) | γi ≥ 0 and

n 

γi = k}.

i=1

The sequences in Γ are listed lexicographically so that γ precedes δ if the first nonzero entry of γ − δ is positive. Let X denote the products xγ11 xγ22 · · · xγnn listed as a column vector as γ runs through the elements of Γ in lexicographical order and similarly for Y . This makes X and Y vectors of length N where   n+k−1 N = . k Since y = Ax, each entry of Y is a linear combination of the entries of X. The kth induced matrix Lk,n (A) is the N × N matrix determined by Y = Lk,n (A)X. The kth compound matrix Ck,n (A) consists of determinants of k × k sub-matrices of A. Its size is M × M where   n M = . k Let α and β denote subsequences of length k taken from (1, 2, . . . , n) and let A[α, β] denotes the k × k sub-matrix of A whose rows are identified by α and whose columns are identified by β. Denoting all such sequences by α1 , α2 , . . . , αM and β1 , β2 , . . . , βM , listed in lexicographical order, Ck,n (A) is the matrix whose (i, j) entry is det(A[αi , βj ]). Some pertinent properties are Tk,n (AB) Lk,n (AB) Ck,n (AB) Ck,n (At )

= = = =

Tk,n (A) Tk,n(B) Lk,n (A) Lk,n (B) Ck,n (A) Ck,n (B) (Ck,n (A))t

(6) (7) (8) (9)

where A and B are any n × n matrices with entries in the complex field and t denotes transpose. 10

2.2

Cycle structure of the tensor power matrix

If α1 , α2 , . . . , αn are the characteristic roots of an n × n matrix A then the characteristic roots of Tk,n (A) are the nk products α i1 α i2 · · · α ik ,

1 ≤ i1 , i2 , . . . , ik ≤ n [3]

Since the characteristic roots of P are the nth roots of unity, it follows that the characteristic roots of the permutation matrix Tk,n (P ) are also nth roots of unity. Let ζ be a primitive nth root of unity and let r be an integer such that 0 ≤ r < n. Then, the multiplicity of ζ r is the number of k-tuples (i1 , i2 , . . . , ik ) such that k 

im ≡ r mod n,

1 ≤ i1 , i2 , . . . ik ≤ n

i=1

which by Theorem 1 is nk−1 . Since Tk,n (P ) is a permutation matrix, it is similar to a direct sum of cyclic permutation matrices ˙ P2 + ˙ ··· + ˙ Pt Tk,n (P ) ∼ P1 + and this similarity can be effected by a permutation matrix. For i = 1, 2, . . . , t, the characteristic roots of Pi are the mi th roots of unity where mi is the order of Pi . Therefore, 1 appears once and only once as a characteristic root of each Pi . By the previous remarks it follows that t = nk−1 . Equivalently stated, the number of cycles that appear in the disjoint cycle decomposition of Tk,n (P ) is nk−1 . Using (6) we have (Tk,n (P ))n = Tk,n (P n ) = Tk,n (In ) = IN ,

N = nk

where Iq denotes the q ×q identity matrix. It follows that mi | n for each i = 1, 2, . . . , nk−1 . Since k−1 n  mi = nk , i=1 k−1

mi = n for i = 1, 2, . . . , n

2.3

and we have shown that Tk,n (P ) consists of nk−1 n-cycles.

Cycle structure of the induced matrix

We now examine the cycle structure of the kth induced matrix of P . For an n × n matrix A with roots α1 , α2 , . . . , αn the characteristic roots of Lk,n (A) are given by  characteristic  n+k−1 the products k α i1 α i2 · · · α ik ,

1 ≤ i1 ≤ i2 ≤ · · · ≤ ik ≤ n [2].

11

Therefore, the characteristic roots of Lk,n (P ) are nth roots of unity. Let ζ be a primitive nth root of unity and r an integer such that 0 ≤ r < n. The corollary of Theorem 2 implies that the multiplicity of ζ r is given by   1  n/d + k/d − 1 Nr = s(d, r), g = gcd(n, k). k/d n d|g

Arguing as before, the number of cycles that appear in the disjoint cycle decomposition of Lk,n (P ) is   1  n/d + k/d − 1 φ(d) N0 = k/d n d|g

The matrix Lk,n (P ) is similar to a direct sum of N0 cyclic permutation matrices ˙ P2 + ˙ ··· + ˙ P N0 Lk,n (P ) ∼ P1 + and the similarity can be effected by a permutation matrix. Let mi be the order of Pi , i = 1, 2, . . . , N0 . Using property (7) we have   n+k−1 n n . (Lk,n (P )) = Lk,n (P ) = Lk,n (In ) = IN , N = k It follows that mi | n for i = 1, 2, . . . , N0 . Let ζ be a primitive nth root of unity. Since the characteristic roots of each Pi are precisely the mi th roots of unity, ζ is a characteristic root of Pi if and only if mi = n. Furthermore, ζ can occur at most once as a characteristic root of Pi . Therefore, the number of Pi of order n is the same as the number of Pi having ζ as a characteristic root which is in turn   1  n/d + k/d − 1 N1 = μ(d). k/d n d|g

To summarize, there are N0 cycles that appear in the disjoint cycle decomposition of Lk,n (P ). The length of each cycle divides n and the number of cycles of length n is N1 . Now let d∗ be the largest divisor of n such that d∗ < n. Let ξ be a primitive d∗ th root of unity. The number of Pi with order d∗ is the same as the number of Pi that have ξ as a characteristic root minus the number of Pi of order n. Let ζ be a primitive nth root of ∗ unity and let r ∗ be an integer such that 0 ≤ r∗ < n and ξ = ζ r . Then, the number of d∗ -cycles appearing in the disjoint cycle decomposition of Lk,n (P ) is Nr∗ −N1 . Proceeding in this fashion, it is clear that the cycle structure of Lk,n (P ) can be described in terms of the quantities N0 , N1 , . . . , Nn−1 . As a simple example, let n and k be relatively prime so that g = 1. Then, N0

1 = n



 n+k−1 , k

N0 

 mi =

i=1

12

 n+k−1 , k

and mi | n.

It follows that mi = N0 for 1 ≤ i ≤ N0 so Lk,n (P ) consists of N0 cycles, each of length n. For a non-trivial example, let n = pq where p and q are distinct primes and k is arbitrary. Therefore, mi ∈ {1, p, q, n}. Let ζ be a primitive nth root of unity so that ζ p is a primitive qth root of unity. Therefore ζ p occurs as a characteristic root of Pi if and only if mi ∈ {q, n}. Since N1 is the number of n-cycles in Lk,n (P ), it follows that Np − N1 is the number of q-cycles in Lk,n (P ). Similarly, ζ q is a primitive pth root of unity which occurs as a characteristic root of Pi if and only if mi ∈ {p, n}. Therefore, the number of p-cycles is Nq − N1 . The remaining cycles are fixed points. In summary, if n is a product of distinct primes p and q, the cycle structure of Lk,n (P ) is given by Number Number Number Number

of of of of

n cycles q cycles p cycles fixed pts

= = = =

N1 , Np − N1 , Nq − N1 , N0 + N1 − Np − Nq .

Taking g = p, 1 [N s(1, r) + M s(p, r)] n 1 [N + M(−1 + p δp,r )] , = n

Nr =

where



N =

 n+k−1 , k

 M =

r = 0, 1, . . . , n − 1,

q + k/p − 1 k/p



 ,

δp,r =

1 if p | r . 0 if p  r

Therefore, N0 = Np =

1 (N − M + p M), n

N1 = Nq =

1 (N − M), n

so that Lk,n (P ) consists of (N − M)/n cycles of length n and M/q cycles of length q.

2.4

Signed permutation matrices

In this section, we establish some relevant facts concerning signed permutation matrices. By definition, a signed permutation matrix is a permutation matrix but with some of the 1’s changed to -1’s. Formally, an n × n matrix Q is a signed permutation matrix if its (i, j) entry is given by i δi,σ(j) for some σ ∈ Sn where i = ±1 and δi,j is Kronecker’s function. Lemma 2. Let Q be an n × n signed permutation matrix with (i, j) entry i δi,σ(j) . Then.  n 

k In = ±In Qn = k==1

13

. Proof. The (i, j) entry of Q2 is n 

i δi,σ(k) k δk,σ(j) = i σ−1 (i) δσ−1 (i),σ(j) .

k=1

Consequently, the (i, j) entry of Q3 is n 

i δi,σ(k) k σ−1 (k) δσ−1 (k),σ(j) = i σ−1 (i) σ−2 (i) δσ−2 (i),σ(j) .

k=1

By induction, the (i, j) entry of Qn is 

i σ−1 (i) σ−2 (i) · · · σ−(n−1) (i) δσ−(n−1) (i),σ(j) =

n



k

δi,j

k==1

Therefore,

 Qn =



n

k

In = ±In .

k==1

Theorem 4. Let Q be an n×n signed cyclic permutation matrix with Qn = In . Then, the characteristic roots of Q are precisely the nth roots of unity. Proof. Its enough to show that every nth root of unity is a characteristic root of Q. Therefore, let ζ be an nth root of unity and let the (i, j) entry of Q be i δi,σ(j) where σ ∈ Sn is an n cycle and i = ±1. We will prove that ζ is a characteristic root of Q by constructing a nonzero complex vector x = (x1 , x2 , . . . , xn )t such that Qx = ζx. Let x1 be any nonzero complex number and for k = 1, 2, . . . , n − 1 define   k

γ m−1 (1) ζ k x1 , γ = σ −1 . (10) xγ k (1) = m=1 n Since σ and hence γ are n cycles, nthis defines x2 , x3 , . . . , xn in some order. Since Q = In , it follows from Lemma 2 that i=1 i = 1, implying that (10) holds when k = n as well. Since the ith entry of Qx is n 

i δi,σ(j) xj = i xγ(i) ,

γ = σ −1 ,

j=1

it remains to show that i xγ(i) = ζxi for i = 1, 2, . . . , n.

14

For each i, 1 ≤ i ≤ n, there exists a unique k with 1 ≤ k ≤ n such that i = γ k−1 (1). Thus,

i xγ(i) = i xγ k (1)  k  = i

γ m−1 (1) ζ k x1 m=1

= i γ k−1 (1) ζ

 k−1



γ m−1 (1)

ζ k−1 x1

(an empty product being 1)

m=1

= 2i ζ xγ k−1 (1) = ζ xi . This concludes the proof. Theorem 5. Let Q be an n × n signed cyclic permutation matrix with Qn = −In . Then, the characteristic roots of Q are given by ζ, ζ 3, . . . , ζ 2n−1 where ζ is a primitive 2nth root of unity. Proof. Given j with 1 ≤ j ≤ n, we wish to show that ζ 2j−1 is a characteristic root of Q by constructing an eigenvector x belonging to ζ 2j−1. Let x1 be any nonzero complex number. For k = 1, 2, . . . , n − 1 define  k  xγ k (1) =

γ m−1 (1) ζ (2j−1)k x1 , γ = σ −1 . (11) m=1

order. It also consistently reproduces This defines complex numbers x2 , x3 , . . . , xn in some n x1 when k = n because ζ = −1 and by Lemma 2, nm=1 m = −1. As in Theorem 4, given an integer i with 1 ≤ i ≤ n, let k be the unique integer with 1 ≤ k ≤ n such that i = γ k−1 (1). Then,

i xγ(i) = i xγ k (1)  k  = i

γ m−1 (1) ζ (2j−1)k x1 m=1

= i γ k−1 (1) ζ

2j−1

 k−1



γ m−1 (1)

ζ (2j−1)(k−1) x1

m=1

=

2i

ζ

2j−1

xγ k−1 (1) = ζ 2j−1 xi .

Thus, as in Theorem 4, this proves Qx = ζ 2j−1x and concludes the proof.

2.5

Cycle structure of the compound matrix

For an application of Theorem 3, consider the kth compound matrix function where 1 ≤ k ≤ n. Applying (8) and (9) to P and P t and using the fact that   n Ck,n (In ) = IN , N = k 15

it follows immediately that Ck,n (P ) (Ck,n(P ))t = (Ck,n (P ))t Ck,n (P ) = IN . Denoting the i, j entry of Ck,n(P ) by ci,j we have N 

c2i,m

=

m=1

N 

c2m,j = 1,

1 ≤ i, j ≤ N.

m=1

Since every entry of Ck,n (P ) is an element of {−1, 0, 1}, it follows that Ck,n (P ) is a signed permutation matrix. That is, Ck,n (P ) becomes a permutation matrix when every -1 is replaced with a 1. When k = 1, Ck,n (P ) = P so C1 (P ) is absent of -1 entries. For k = n, Ck,n (P ) = (−1)n+1 for any n × n cyclic permutation matrix P . When 1 < k < n, we can expect negative entries appearing in Ck,n (P ) for most cyclic permutations P regardless of the evenness or oddness of n and k.1 If the characteristic roots of an n × n complex matrix A are α1 , α2 , . . . , αn then the characteristic roots of Ck,n(A) are given by the N products α i1 α i2 · · · α ik ,

1 ≤ i1 < i2 < · · · < ik ≤ n [4]

Therefore, a characteristic root of Ck,n (P ) is an nth root of unity. From Theorem 3, it follows that if ζ is a primitive nth root of unity and r is an integer with 0 ≤ r < n then the multiplicity of ζ r as a characteristic root of Ck,n (P ) is     1  n/d 1 N(r, n) = s d, k(k + 1) − r . k/d n 2 d|g

where g is the greatest common divisor of n and k. Even though Ck,n(P ) is a signed permutation matrix, information about its cycle structure can be obtained from knowledge of the multiplicities of its characteristic roots. Write ˙ Q2 + ˙ ··· + ˙ Qt Ck,n (P ) ∼ Q1 + where the Qi are signed cyclic permutation matrices and the similarity is effected by a permutation matrix. Let Qi = diag( 1 , 2 , . . . , mi )Pi where Pi is the cyclic permutation matrix corresponding to Qi , mi is the order of Pi , and

j = ±1 for j = 1, 2, . . . , mi . By Lemma 2, m  mi mi i mi = diag

j ,

j , . . . ,

j = ±Imi Qi j=1

j=1

j=1

i If Qm = −Imi , the order of the subgroup of mi × mi complex matrices generated by i i Qi is 2mi . From (8) it follows that Qni = Imi which implies 2mi | n. Therefore Qm i = −Imi mi necessitates n being even. It follows that if n is odd, Qi = Imi for i = 1, 2, . . . , t.

1

The original version of this article published in 1983 has this wrong.

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2.5.1

The case of n odd

If n is odd, the order of Qi is mi and by Theorem 4, the characteristic roots of Qi are precisely the mi th roots of unity. This is the key condition for connecting multiplicities of characteristic roots with cycle structure. Because of this condition, all of the previous analysis for determining the cycle structure of the induced matrix applies to the compound matrix, the only difference being the particular values of the multiplicities involved. For example, the number of cycles of Ck,n (P ) is N(0, n), the length of each cycle divides n, and the number of n cycles is N(1, n). If n is the product of two distinct odd primes p and q, k = g = p, and ζ is a primitive nth root of unity, multiplicities of ζ r as a characteristic root of Ck,n (P ) are given by      1 1 1 n q N(r, n) = s(1, p(p + 1) − r) + s(p, p(p + 1) − r) p 1 n 2 2     1 1 if p | r n . + q(−1 + p δp,r ) , r = 0, 1, . . . , n − 1, δp,r = = 0 if p  r p n Using the analysis given for the induced matrix, the cycle structure of Ck,n (P ) is    n 1 Number of n cycles = N(1, n) = n −q , p Number of q cycles = N(p, n) − N(1, n) = 1, Number of p cycles = N(q, n) − N(1, n) = 0, Number of fixed pts = N(0, n) + N(1, n) − N(p, n) − N(q, n) = 0. 2.5.2

The case of n even

For the case of n even, it is generally not true that N(0, n) enumerates the cycles appearing in the disjoint cycle decomposition of Ck,n (P ). For example, taking n = 4 and k = 2 produces N(0, 4) = 1 while each of the 6 cyclic permutations of S4 yield compound matrices made up of a transposition and a 4-cycle (see Table 1). It is however still true i that the length of each cycle divides n. Since Qm i = ±Imi , the order of Qi is either mi or 2mi . Either way, mi | n. Going back to the example n = 4 and k = 2, the enumeration formula correctly identifies the characteristic roots of C2,4 (P ) as ±1 (each with multiplicity 1) and ±i (each with multiplicity 2). However, the cycle structure for C2,4 (P ) cannot be determined from this information alone. This is evident from example matrices that have these same characteristic roots with the same multiplicities but yield a cycle structure not represented by C2,4 (P ) for any 4 × 4 cyclic permutation matrix P . One such example is       0 1 0 −1 0 1 Q1 = , Q2 = , Q3 = . 1 0 1 0 −1 0

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To the author’s knowledge, determining the cycle structure of Ck,n(P ) for an n × n cyclic permutation matrix P when n is even remains an open problem. However, knowing the characteristic roots with their multiplicities does significantly limit the possibilities. Consequently, the author envisions them playing a significant role in determining cycle structure for the case of n even. In particular, it is envisioned that Theorem 5 will play a role. Table 1: Cycle structure of Ck,n (P ) n k=1 k=2 k=3 k=4 k=5 k=6 k=7 1 1 2 2 1 3 3 3 1 4 4 2,4 4 1 5 5 5(2) 5(2) 5 1 6 6 3,6(2) 2,6(3) 3,6(2) 6 1 7 7 7(3) 7(5) 7(5) 7(3) 7 1 8 8 4,8(3) 8(7) 2,4,8(8) 8(7) 4,8(3) 8 9 9 9(4) 3,9(9) 9(14) 9(14) 3,9(9) 9(4) 10 10 5,10(4) 10(12) 5(2),10(20) 2,10(25) 5(2),10(20) 10(12) 11 11 11(5) 11(15) 11(30) 11(42) 11(42) 11(30) 12 12 6,12(5) 4,12(18) 3,6(2),12(40) 12(66) 2,4,6(3),12(75) 12(66) 13 13 13(6) 13(22) 13(55) 13(99) 13(132) 13(132) 14 14 7,14(6) 14(26) 7(3),14(70) 14(143) 7(5),14(212) 2,14(245) 15 15 15(7) 5,15(30) 15(91) 3,15(200) 5(2),15(333) 15(429)

In Table 1, the entry corresponding to k = 6 and n = 12 means that C6,12 (P ) consists of a transposition, a cycle of length 4, three cycles of length 6, and 75 cycles of length 12. Numerical evidence suggests that the cycle structure of Ck,n (P ) is independent of the particular n × n cyclic permutation matrix P . It also suggests that the cycle structures of Ck,n (P ) and Cn−k,n (P ) are identical. Table 1 was created by a computer program using a brute force algorithm.

References [1] R. A. Brualdi and M. Newman, An Enumeration Problem for a Congruence Equation, J. Res. Nat. Bur. Stand. (U.S.), Math Sci. Vol. 74B, No. 1, Jan.-Mar. 1970. [2] M. Marcus and M. Newman, Inequalities for the Permanent Function, Annals of Math., Vol. 75, No. 1, Jan. 1962. [3] C. MacDuffee, The Theory of Matrices, Chelsea Pub. Co., New York, 1946, pp. 81-88. 18

[4] H. Ryser, Compound and Induced Matrices in Combinatorial Analysis, Proc. of 10th Sym. in Appl. Math. of A.M.S., Vol. 10, 1958, pp. 149-167. [5] 0. H¨older, Prace Mat.- Fiz, 43, 1936, pp. 13-23. [6] K. G. Ramanuthan, Some applications of Ramanujan’s Trigonometrical Sum Cm (n), Proc. Indian Acad. Sci., Sect. A 20, 1944, pp.62-69.

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