The number of terms in the permanent and the determinant of a generic circulant matrix

6 Pages·0.11 MB·English

by  Hugh Thomas

THE NUMBER OF TERMS IN THE PERMANENT AND THE
DETERMINANT OF A GENERIC CIRCULANT MATRIX
3
0 Hugh Thomas
0
2
September 2002

n
a
J Abstract. LetA=(aij)bethegenericn×ncirculantmatrixgivenbyaij =xi+j,

7 withsubscripts on xinterpreted mod n. Define d(n) (resp. p(n))tobe thenumber
oftermsinthedeterminant(resp. permanent)ofA. Thefunctionp(n)iswell-known

] andhasseveralcombinatorialinterpretations. Thefunctiond(n),ontheotherhand,
O hasnotbeenstudiedpreviously. Weshowthatwhennisaprimepower,d(n)=p(n).
C
.
h
t
a Introduction
m
A square matrix is said to be a circulant matrix if its rows are successive cyclic
[
permutations of the first row. Thus, the matrix A = (a ) with a = x , sub-
ij ij i+j
1 scripts on x being interpreted mod n, is a generic circulant matrix.
v
Ifweexpanddet(A), weobtainapolynomialinthex . We defined(n)to bethe
8 i
numberoftermsinthis polynomialafterliketermshavebeencombined. Similarly,
4
0 we define p(n) to be the number of terms in per(A), the permanent of A.
1 The functionp(n)wasstudiedinBrualdiandNewman[1],where itwaspointed
0 out that the main result of Hall [3] shows that p(n) coincides with the number of
3
solutions to
0
/
h
y +2y +···+ny ≡0 (mod n)
t 1 2 n
a
(1) y +···+y =n
m 1 n
:
v in non-negative integers. Using this formulation, they showed by a generating
i function argument that
X
r
a 1 n 2d−1
(2) p(n)= φ( ) .
n d d
Xd|n (cid:18) (cid:19)
Setting w = n y , and rewriting (1) in terms of the w , we see that p(n)
i j=i+1 j i
is the number of non-increasing(n−1)-tuples (w ,...,w ) with n≥w ≥···≥
1 n−1 1
P
w ≥ 0, such that w ≡ 0 (mod n). If we let L be the (n−1)-dimensional
n−1 i i
lattice consistingof(n−1)-tuples ofintegerswhosesumis divisible by n,then this
P
expression for p(n) amounts to the number of points from L lying in the simplex
with vertices (0,0,...,0), (n,0,0,...,0), (n,n,0,...,0), ..., (n,n,...,n).
TypesetbyAMS-TEX
1
2 HUGH THOMAS
There is another combinatorial interpretation for p(n), as follows. Consider all
possible necklaces consisting of n white beads and n black beads, where two neck-
laces are considered equivalent if they differ by a cyclic permutation. A straight-
forwardPo´lyacountingargumentshowsthatthe numberofsuchnecklacesis given
by the right-hand side of (2), and thus that the number of necklaces equals the
number of terms in per(A), though no explicit bijection is known.
In contrast,little is known about d(n), despite the fact that its definition seems
at least as natural. It is clear that d(n) ≤ p(n) since every term which appears
in the determinant also appears in the permanent. However, some terms from the
permanentcouldbeabsentfromthedeterminantduetocancellation. Inthispaper
we establish the following theorem:
Theorem. If n is a prime power, d(n)=p(n).
The following table of values for d(n) and p(n) suggests that the converse may
also be true. This is still open.
n d(n) p(n) n d(n) p(n)
1 1 1 7 246 246
2 2 2 8 810 810
3 4 4 9 2704 2704
4 10 10 10 7492 9252
5 26 26 11 32066 32066
6 68 80 12 86500 112720
Background on Symmetric Functions
Theproofofthetheoremusesthetheoryofsymmetricfunctions. Inthissection,
we review the necessary definitions and results. For more detail on symmetric
functions, see Stanley [4].
Symmetricfunctions arepowerseries(inourcase,overQ)inaninfinite number
ofvariablesz ,z ,...,suchthatforany(b ,...,b )∈Nkthecoefficientofzb1...zbk
1 2 1 k i1 ik
does not depend on the choice of distinct natural numbers i ,...,i .
1 k
We write λ ⊢ q to signify that λ is a partition of q. The symmetric functions
of degree q form a vector space whose dimension is the number of partitions of q.
There are several standard bases for them. We shall need two here: {m | λ ⊢ q}
λ
and {p | λ ⊢ q}. For λ = (λ ,...,λ ) with λ ≥ λ ≥ ··· ≥ λ , m is the power
λ 1 k 1 2 k λ
series in which zλ1...zλk occurs with coefficient one for every distinct sequence
i1 ik
of natural numbers i ,...,i and no other terms occur. The symmetric function
1 k
p is defined to be zi +zi +..., and the symmetric function p is defined to be
i 1 2 λ
p ...p .
λ1 λk
As we remarked, {p | λ ⊢ q} and {m | λ ⊢ q} form bases for the symmetric
λ λ
functions of degree q. We will need to convert from the m basis to the p basis,
λ λ
which we shall do using a result of Eg˘eciog˘lu and Remmel [2], for which we must
introduce some notation.
For λ ⊢ q, let k(λ) denote the number of parts of λ. If λ = h1l12l2...qlqi, let
λ!=1!l12!l2...q!lq, and let zλ =l1!...lq!1l12l2...qlq.
The Ferrers diagram of λ consists of a left-justified column of rows of boxes of
lengths λ ,λ ,...,λ . For µ another partition of q, a filling of λ by µ is a way to
1 2 k
cover the Ferrers diagram of λ in a non-overlappingmanner by “bricks” consisting
of horizontal sequences of boxes of length µ ,µ ,.... The weight of a filling is
1 2
THE DETERMINANT OF A GENERIC CIRCULANT MATRIX 3
the product over all rows of the length of the final brick in each row. In deciding
whether two fillings are the same, bricks of the same size are considered to be
indistinguishable. Thus, the four fillings of λ = (4,2) by µ = (2,2,1,1), having
weights respectively 4, 2, 2, and 2, are:
Let w(λ,µ) be the sum over all distinct fillings of λ by µ of the weight of the
filling. Then the result we shall need from [2] is that:
(−1)k(µ)−k(λ)w(λ,µ)
m = p .
µ λ
z
λ
λ⊢q
X
Proof of Theorem
First, let us consider what terms occur in per(A). Let b = (b ,...,b ) be an
1 n
n-tuple of natural numbers summing to n. Let xb denote xb1xb2...xbn. We are
1 2 n
interested in whether xb appears in per(A) with non-zero coefficient. Suppose it
does. Then there is some permutation σ of {1,...,n} such that x = xb.
i i−σ(i)
Since (i−σ(i)) = 0, it follows that ib ≡ 0 (mod n). By the result of [3]
i i i Q
mentionedabove,thisnecessaryconditionforxbtooccurinper(A)isalsosufficient.
P P
Now we proceed to consider det(A). Let ξ be a primitive n-th root of unity. A
is diagonalizable, and its eigenvalues are c ,...,c , where c = ξijx . Thus,
1 n i j j
det(A)= c .
i
P
Again, let b = (b ,...,b ) be an n-tuple of integers summing to n. Let q =
1 n
b1 +2b2 +Q···+nbn. Let µ = h1b12b2...nbni, so µ ⊢ q. Then [xb]det(A), the
coefficient of xb in det(A), is given by
[xb]det(A)= ξ ni=1if(i).
f:{1,|.f..−,nX1}(i→)|{=1b,i...,n} P
Weobservethatthisalsoequalsm (ξ1,ξ2,...,ξn,0,0,...). (Asymmetricfunction
µ
is a power series, so it is not generally legitimate to substitute in values for the
indeterminates, but since m is homogeneous and all but finitely many of the
µ
values being substituted are zero, it is allowed in this case.)
Now we use the result of [2] mentioned above, which shows that:
(−1)k(µ)−k(λ)w(λ,µ)p (ξ1,ξ2,...,ξn,0,0,...)
(3) [xb]det(A)= λ .
z
λ
λ⊢q
X
What can we say about p (ξ1,ξ2,...,ξn,0,0,...)? Firstly, p (ξ1,ξ2,...,ξn,
λ i
0,0,...) = n if n|i and 0 otherwise. Thus, p (ξ1,ξ2,...,ξn,0,0,...) = nk(λ) if
λ
all the parts of λ are multiples of n, and 0 otherwse.
4 HUGH THOMAS
(Fromthiswecoulddeducethatif[xb]det(A)isnon-zero,q =b +2b +···+nb
1 2 n
must be a multiple of n. Of course, we already know this, by the argument given
when discussing per(A).)
To establish our result, we must now show that if q is a multiple of n and n is a
prime power, then the sum in (3) is non-zero. So assume that n = pr, p a prime.
Let v :Q× →Z denote the usual valuation with respect to p.
Forλanypartitionofq,dividethefillingsofλbyµintoequivalenceclasseswhere
two fillings areequivalent if one canbe obtainedfromthe other by rearrangingthe
bricks within each row, and by swapping the sets of bricks filling pairs of rows of
equal length.
Wewishtoshowthatthecontributionto(3)fromthepartitionhqi(allofwhose
fillings form a single equivalence class) has a smaller valuation than the sum of
weights of the fillings in any equivalence class of fillings of any other partition.
Once this is established, it follows that the sum (3) is non-zero.
Fix λ = (λ1,λ2,...) = h1l12l2...qlqi, with all the λi divisible by n, and fix an
equivalence class F of fillings of λ by µ. Write k for k(λ). Considerfirst a subclass
of fillings G, those which can be obtained from some fixed F ∈ F by rearranging
the bricks in each row, but without interchanging rows. Let the j-th row of the
filling F ofλcontainr bricks. Lete be the numberofthesebrickshavinglength
j ij
i. The number of rearrangements of this row with ending in a brick of length i is
rj−1 .
e1j,...,eij−1,...,enj
Thus, the total weight of all the rearrangements of this row is:
(cid:0) (cid:1)
n n
r −1 1 r
j j
i= ie
ij
e ,...,e −1,...,e r e ,...,e ,...,e
i=1(cid:18) 1j ij nj(cid:19) i=1 j(cid:18) 1j ij nj(cid:19)
X X
1 r
j
= λ .
j
r e ,...,e
j(cid:18) 1j nj(cid:19)
It follows that the total weight of all the fillings in G is:
k
1 r
j
λ .
j
r e ,...,e
j=1 j(cid:18) 1j nj(cid:19)
Y
Consider the l parts of λ of length i. The equivalence class F determines a
i
partition γ(F,i) of l , where the parts of γ(F,i) are the sizes of the sets of rows of
i
length i which are filled with an indistinguishable set of bricks. The sum of all the
weights over all the fillings in F is:
q k
l ! 1 r
i j
λ .
j
γ(F,i)! r e ,...,e
i=1 j=1 j(cid:18) 1j nj(cid:19)
Y Y
Writing δ(F) for the partition of k which is the sum of the γ(F,i) for all i, the
contribution to (3) from all these fillings is:
(−1)k(µ)−knk k (r −1)!
j
(4)
δ(F)! e !...e !
1j nj
j=1
Y
THE DETERMINANT OF A GENERIC CIRCULANT MATRIX 5
Now let us consider (4) in the case where λ=hqi. Here, we obtain
(−1)k(µ)−1n!
(5) .
b !...b !
1 n
We wishto showthat(4)evaluatedforanyotherequivalenceclasshasa greater
valuation with respect to p than does (5). This is equivalent to showing that, for
any λ 6= hqi, and any equivalence class of fillings F, that the following expression
has a positive valuation:
nk k (r −1)! b !...b !
j 1 n
δ(F)! e !...e ! n!
1j nj
j=1
Y
(6)  
1 n b nk−1
i
= .
δ(F)!iY=1(cid:18)ei1,...,eik(cid:19)! (n−1)...(n−k+1) r1−1n,.−..,krk−1 !
(cid:0) (cid:1)
We havewritten(6) asa productoftwoterms. We willshowthat the firstterm
is an integer, and therefore has non-negative valuation, and that the second term
has positive valuation, which will complete the proof.
For the first term, observe that if we rewrite the partition b =e +···+e as
i i1 ik
h1ci1...ncini, then ci1!...cin! divides ei1,.b.i.,eik because
1 (cid:0) b(cid:1)
i
c !...c ! e ,...,e
i1 in (cid:18) i1 ik(cid:19)
countsthenumberofwaysofdividingb objectsintosubsetsofcertainsizes,where
i
we don’t distinguish the different subsets of the same size. The first term of (6) is
now disposed of by remarking that
n
δ(F)!| c !...c !
i1 in
i=1
Y
since eachterminδ(F)!implies the existenceofatleastone equaltermamongthe
product of factorials on the right hand side.
For the second term, we need the following simple lemma:
Lemma. If m<ps,
(i) v( m )<s,
d
(ii) v( m )<(k−1)s.
(cid:0) d(cid:1)1,...,dk
Proof. (cid:0) (cid:1)
m m d (m−d) m d (m−d)
v( )= ⌊ ⌋−⌊ ⌋−⌊ ⌋ +···+ ⌊ ⌋−⌊ ⌋−⌊ ⌋
d p p p ps−1 ps−1 ps−1
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
≤1+···+1=s−1.
The second part follows immediately from repeated application of the first part,
which completes the proof of the lemma.
6 HUGH THOMAS
Now, we know that
v(nk−1)=(k−1)r,
n−k
v( )≤(k−1)(r−1),
r −1,...,r −1
(cid:18) 1 k (cid:19)
k−1
v((n−1)...(n−k+1))=v((k−1)!)=⌊k/p⌋+···< ≤k−1,
p−1
and the desired result follows.
Acknowledgements
I would like to thank Graham Denham for a useful suggestions and Jim Propp
for asking the question and for helpful comments on a previous draft.
References
1. R.BrualdiandM.Newman,Anenumeration problemforacongruence equation,J.Res.Nat.
Bur.Standards Sect.B74B(1970), 37–40.
2. O¨. Eg˘eciog˘lu and J. Remmel, Brick tabloids and the connection matrices between bases of
symmetric functions,DiscreteAppl.Math.34(1991), 107–120.
3. M. Hall, Jr., A combinatorial problem on abelian groups, Proc. Amer. Math. Soc. 3 (1952),
584–587.
4. R. Stanley, Enumerative Combinatorics, Volume 2, Cambridge University Press, Cambridge,
1999.
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