Counting Quasi-Idempotent Irreducible Integral Matrices

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by  Jakob Zimmermann

COUNTING QUASI-IDEMPOTENT IRREDUCIBLE
INTEGRAL MATRICES
JAKOBZIMMERMANN
7 Abstract. Westudytwoquestionconcerningirreduciblematrices. Thefirst
1 problem is a counting problem. We count the number of irreducible integral
0 matrices which areannihilated by X2−nX, for n>0. Thesecond question
2 that we study is whether the set of irreducible, integral matrices which are
annihilated bysomepolynomial p∈ [X]isfinite ornot. We show that this
n C
is true for positive integral matrices, false for non-negative integral matrices
a
andconjecturethatitmightbetrueforirreducibleintegralmatrices.
J

3
1

]
O 1. Introduction
C
. Counting problems are among the most commonly studied problems in combina-
h
torics and a lot of results in this area give rise to integer sequences which can be
t
a found on the online encyclopedia of integer sequences (OIES, [13]). In this paper
m
we are going to study the following question. Given a polynomial p∈ [X], is the
[ number positive (irreducible) integral matrices A such that p(A) = 0Cfinite? The

1 answer to this question is given by the following theorem
v
Theorem A. For any polynomial p ∈ [x], the set of positive integer matrices A
9
C
9 such that p(A)=0 is finite.
6
3 Nowthatweknowthatthissetisfiniteforanypolynomial,anaturalnextquestion
0 is, if we can count its elements. For an arbitrary polynomial, this seems to be
. difficult, however,for the polynomials f =X2−nX, we can count the number of
1 n
0 positiveintegralmatriceswhichareannihilatedbyfn. Wewillcountthesematrices
7 in two different ways. On the one hand, we will count all such matrices and, on
1 the other hand, all matrices up to an equivalence given by permutation of basis
:
v vectors. That is, here we say that two k×k-matrices A,B are equivalent if there
i exists a k×k-permutation matrix P , where σ ∈ S , such that conjugation of A
X σ k
by P yields B.
σ
r
a
The motivation to study these questions comes from higher representation theory,
more precisely, the following observation. Let A be a finite dimensional -algebra
C
of dimension n, then F := A⊗ A is an A-A-bimodule. It acts on the category
of A-A-bimodules from the left Cby taking tensor products over A, i.e. for a A-A-
bimodule M, the action of F is given by F(M) = A⊗ A⊗ M. Now, we can
A
C
observe that
F2 =F ◦F =A⊗ A⊗AA⊗ A∼=(A⊗ A)⊕n =F⊕n.
C C C
ThustheactionofF isquasi-idempotent. Therefore,ontheleveloftheGrothendieck
group, F induces a linear transformation which corresponds to a matrix [F] satis-
fying f ([F])=0. These kind of problems appear, for instance, in [11, 12].
n
Date:January16,2017.
1
2 JAKOBZIMMERMANN
Themoregeneralproblemismotivatedbyatechniquewhichisusedwhentryingto
understandcertain2-representationsoffinitary2-categories,see[10]. Themainidea
here is that a certain element whose action is given by a non-negative, irreducible
integral matrix which has to be annihilated by a certain polynomial. With this
information one then tries to find all possible such matrices. In small cases this
can be done by hand, but the question that occurs quite naturally is whether or
notthis alwaysispossible,i.e. iftherearealwaysonlyfinitely manysuchmatrices.
For more details we refer the reader to [7, 11, 12, 14].
In the next section we introduce most of the necessary notation and preliminary
resultsthatwewillneedthroughoutthepaper. InSection3wewillproveTheorem
A.Theconsecutivesectionisthenspentoncountingallirreducibleintegralmatrices
satisfying X2 = nX. Finally, in the last section we discuss generalizations of
Theorem A and use the structure theory of positively based algebras to prove a
similar result that holds for irreducible, non-negative integral matrices but for a
restricted set of polynomials.
2. Preliminaries
2.1. Basic definitions and notation. A real matrix A is called positive if all
entries of A are positive. A real matrix A is called non-negative if all entries of
A are non-negative. We say that a real square matrix A is primitive if it is non-
negative and there exists k >0 such that Ak is positive. If A is non-negative and,
for eachpair i,j,there exists some k suchthat (Ak) is positive, thenwe say that
i,j
A is irreducible. Here by (B) we denote the element in the i-th row and the j-th
i,j
column of B.
For two n×m matrices A,B we say that A≤B provided that (A) ≤(B) , for
i,j i,j
all i,j. We write A < B, if A ≤ B and A 6= B. This defines a partial order on
Mat ( )andallowsustowriteA>0toexpressthefactthatAisnon-negative,
k,k ≥0
Z
where here 0 is used as a shorthand for the k×k zero matrix.
For f ∈ [X], we define the sets
C
K>0 ={A∈Mat ( ):k >0,f(A)=0}
f k,k Z>0
K≥0 ={A∈Mat ( ):A is irreducible,k >0,f(A)=0}
f k,k Z≥0
tobe the setsofallirreduciblek×k positive(resp. non-negative)integralmatrices
whichareannihilatedbyf. NotethatwedonotrestrictthesizeofA,inparticular
A can just be a positive 1×1-matrix which we identify with the corresponding
positive integer.
Due to their appearance in, e.g. [11, 14], quasi-idempotent matrices, i.e matrices
that satisfy X2 =nX are of special interest. Set f (X):=X2−nX, K≥0 :=K≥0
n n fn
and K>0 := K>0. Then, for a positive (resp. non-negative irreducible) integral
n fn
matrix A, satisfying A2 =nA is equivalent to A∈K>0 (resp. A∈K≥0).
n n
InSection4wearegoingtocounttheelementsinK≥0intwodifferentways. Firstly,
n
we are going to simply count all of them. Secondly, we will count all matrices up
to permutation of basis vectors. By this we mean the following. Let A,B ∈
Mat ( ) and denote by S the symmetric group on k elements. Then, to each
k,k ≥0 k
Z
σ ∈ S , we assign the (permutation) matrix P which is defined by P e = e ,
k σ σ i σ(i)
on the elements of the standard basis {e } ⊆ k. Note that P is an orthogonal
i σ
matrix, i.e. P−1 = Pt. We say that A and BRare the same up to permutation of
σ σ
basis vectors, denoted A ∼ B, if there exists σ ∈ S such that P−1AP = B. We
k σ σ
COUNTING QUASI-IDEMPOTENT IRREDUCIBLE INTEGRAL MATRICES 3
will denote the set of all matrices in K≥0 up to permutation of basis vectors by
n
K≥0 :=K≥0/∼.
n n
2.2. Compositions and Partitions. Let n > 0. A composition µ of n, short
µ (cid:15) n, with k parts, is an element in k such µ +µ +···+µ = n. We call
Z>0 1 2 k
µ∈ k a partition of n, short µ⊢n, with k parts, if µ is a composition such that
Z>0
µ ≥µ ≥···≥µ . From now on, whenever we pick a composition or partition of
1 2 k
n we assume it to have k parts, unless explicitly stated otherwise, where k is some
fixed number between 1 and n.
For twoelements v,w ∈ k , we saythat v divides w andwrite v|w if v |w , for all
Z>0 i i
1≤i≤k. If v =(v ,...,v ) and σ ∈S , then
1 k k
σ(v)=σv =Pσv =(vσ−1(1),...,vσ−1(k)),
here the product of elements in k is component-wise.
Z>0
Letpbeaprimenumberandµ(cid:15)n,thenµ(p) :=(pep,1,pep,2,...,pep,k),whereeach
ep,i is defined uniquely by the properties pep,i|µi and pep,i+1 ∤ µi. We call µ(p) the
p-part of µ. Then we can write µ as follows:
(1) µ= µ(p) = (pep,1,pep,2,...,pep,k),
ppYrime ppYrime
where the product is taken component-wise.
2.3. Noetherianityof k . Fortheproofsofourresultswewillneedthefollowing
Z≥0
statement.
Theorem 2.1. The semigroup k is noetherian, for every k > 0, i.e. all of its
Z≥0
ideals are finitely generated.
Proof. We may consider α∈Zk as the exponent of a monomial Xα1Xα2···Xαk
≥0 1 2 k
in the polynomial ring [X ,...,X ]. With this identification, the result follows
1 n
immediately from [5, LeCmma 12.3, page 149]. (cid:3)
Now, we can identify the additive semigroup Mat ( ) of non-negative integral
k,k ≥0
k×k-matrices with the additive semigroup k2 andZthus we get that every ideal
Z≥0
in this semigroup is finitely generated. This will be needed in Section 3.
2.4. Perron-Frobenius Theorem. The second important theorem which we are
going to use at different points throughout this work is the Perron-Frobenius theo-
rem, more precisely the following version of it.
Theorem 2.2. Let A = (a ) ∈ Mat ( ) be a non-negative irreducible matrix.
i,j k,k
R
Then the following holds.
(1) Ahas aneigenvaluer =r>0,theso-called Perron-Frobeniuseigenvalue,
A
of algebraic multiplicity one and such that r >|λ|, for any other eigenvalue
λ of A.
(2) The Perron-Frobenius eigenvalue r satisfies
min a ≤r ≤max a .
ij ij
i i
Xj Xj
(3) If 0 ≤ A < B, then r ≤ r . Moreover, if B is irreducible, then the
A B
inequality is strict, that is r <r .
A B
4 JAKOBZIMMERMANN
Proof. A proof of these statements can be found in [2, Chapter XIII, §2]. More
precisely,(1)isTheorem2,(2)isRemark2and(3)followsfrom[1,2.1.5&2.1.10].
(cid:3)
3. K>0 is finite for general polynomials.
f
In this section we prove that the set K>0 is finite, for any f ∈ [X]. Before we
f C
can do this, we need some notation and a lemma.
Letx,y ∈ r . Wesaythatx<yprovidedthatx ≤y ,foralli,andthereexistsj
Z≥0 i i
suchthatx <y . The reasonfor this partialorderis the factthat we aregoingto
j j
study ideals I in Mat ( ) ≃ k2. By Theorem 2.1, we know that I is finitely
generated,saybysomke,kBZ,≥.0..B .ZT≥0hen,foreveryX ∈I,wehaveX = r c B .
1 r i=1 i i
So, to every X we can assign its coefficient vector cX =(ci)∈Zr≥0. NowP, we want
to compare matrices in X,Y ∈I and then we get that c <c implies X <Y, so
X Y
we can study these coefficient vectors instead.
Lemma 3.1. Let M ⊆ r be an infinite set. Then there exists and infiniteascend-
Z
ing chain in M with respect to < as defined above.
Proof. NotethatM iscountablyinfinite,sothereisanenumerationofM ={e(n)},
where e(n) =(e(n),...,e(n)). Now, since M is infinite, there exists one component
1 r
in which e(n) is unbounded. Without loss of generality assume it is the first com-
ponent. Pick a subsequence n such that the sequence in the first component of
k
e(nk) is strictly increasing, i.e. e(nk) < e(nk+1), for all k. Then there exits a sub-
1 1
sequence (n ) which is non-decreasing in the second component, i.e. such that
kl
e(nkl) ≤ e(nkl+1) for l. Similarly, by taking subsequences of subsequences we get a
2 2
subsequence (n ) of n such that e(np) < e(np+1) and e(np) ≤ e(np+1), for all p and
p k 1 1 i i
all 2≤i≤r. This yields that e(np) is an infinite ascending chain in M. (cid:3)
Now we are ready to prove the main result of this section.
Theorem 3.2. For any polynomial f ∈ [x], the set K>0 is finite.
C f
Proof. For A ∈ Mat ( ) such that f(A) = 0, we denote by µ the minimal
k,k ≥0 A
Z
polynomial of A. By definition, we have that µ divides f and thus the set of
A
eigenvalues of A is a subset of the zeros of f. Denote by x the zero of f with
0
the highest absolute value. Then, since A is a positive matrix, we can apply the
Perron-Frobenius theorem and get that the unique largest eigenvalue λ ≤ x .
A 0
Furthermore, we can apply Theorem 2.2.(2) to obtain
k ≤min a ≤λ ≤x ,
i ij A 0
Xj
since all a ≥1. Thus k is bounded.
ij
Now fix k and assume that the set Y of matrices A ∈ Mat ( ) such that
k,k ≥0
f(A) = 0 is infinite. Consider the ideal I ⊆ Mat ( ) ≃ k2 gZenerated by Y.
k,k Z≥0 Z≥0
We want to use the fact that k2 is noetherian to obtain a contradictionand thus
Z≥0
prove that Y has to be finite.
From Theorem 2.1 we get that I is finitely generated. Let B ,...,B be a set of
1 r
generators of I. Note that f does not necessarily annihilate any of the B . Then
i
COUNTING QUASI-IDEMPOTENT IRREDUCIBLE INTEGRAL MATRICES 5
we can express every A∈Y as a linear combination of the B , i.e.
i
r
A= c B .
A,i i
Xi=1
ThesetM ={c }ofcoefficientvectorsisaninfinitesubsetof r andthusLemma
A Z≥0
3.1 yields that there is an infinite ascending chain c in M. On the other hand,
Ak
we have already seen that this means that this is equivalent to having an infinite
ascending chain A of matrices in Y, with respect to <.
k
However,byTheorem2.2(3), thisyieldsthatthereisaninfinite sequenceofdiffer-
ent Perron-Frobenius eigenvalues, a contradiction, as all eigenvalues, in particular,
have to be zeros of f. (cid:3)
4. Counting quasi-idempotent matrices
A first result that we are going to use a lot is the following.
Proposition 4.1. Let M ∈ K≥0, then M has rank 1, i.e. there exist v,w ∈ k
such that M =vwt. Moreover,nM has trace n. Z>0
Proof. Let M ∈ K≥0, then M is an irreducible matrix satisfying M2 = nM, i.e.
n
the only possible eigenvalues of M are 0 and n. By Theorem 2.2, we have that n
has to be the Perron-Frobenius eigenvalue, in particular all other eigenvalues have
to be zero. The claim follows. (cid:3)
Moreover,it is easy to see that, if we have a rank 1 matrix M =vwt with trace n,
then M2−nM =0, as M2 =vwtvwt =v(wtv)wt =(wtv)M and wtv = w v =
i i
tr(M) = n. Observe that this implies, in particular, that M ∈ Kn≥0 haPs to be a
positive matrix, i.e. K≥0 =K>0.
n n
These two resultstogether showthat, ifwe wantto countmatrices in K≥0, we can
n
restrict our attention to pairs (v,w)∈ k × k such that k v w =n.
Z>0 Z>0 i=1 i i
P
4.1. Enumerating K≥0. Our strategy will be to consider all possible diagonals
n
that a matrix in K≥0 can have, in other words, all compositions of n. For each
n
µ (cid:15) n, we will consider its divisors. Note that both v and w from above are by
definition divisors of µ. Moreover, it is clear that, if we choose one divisor, say v,
k
thenthereisaunique w=(w )|µ suchthat v w =n,namely(w )=(µ /v ).
i i=1 i i i i i
Inotherwords,itseems tobe enoughto couPntalldivisorsofµ forallcompositions
of n.
Observe that, if n=4, then (2,2)⊢4 with divisors (2,2),(2,1),(1,2),(1,1). Now,
boththepair((2,2),(1,1))andthepair((1,1),(2,2))yieldthesamematrix. So,the
next goal is to find the correct equivalence relation on the set of divisors of µ such
thateachequivalenceclasscorrespondstoexactlyonematrixandvice-versa.
We say that v ,v ∈ k are equivalent, short v ∼ v , provided that there exists
a positive inte1ger2c sZuc>h0that cv = v or v =1cv1. G2 iven n > 0 and µ (cid:15) n, we
1 2 1 2
denote by
K≥0 ={A∈K≥0 :∀i∈{1,2,...,k}a =µ },
µ n ii i
D(µ)={v ∈ k :v|µ},
Z>0
D (µ)=D(µ)/ .
1 ∼1
6 JAKOBZIMMERMANN
Elements in D (µ) are equivalence classes of divisors of µ and will be denoted by
1
[v], provided that v|µ.
Lemma4.2. Letµ(cid:15)nandA∈K≥0. Thenthereexistsapair(v,w)∈D(µ)2 such
µ
that vwt =A. Moreover, for every other pair (v˜,w˜)∈D(µ) such that v˜w˜t =A, we
have v ∼ v˜ and w ∼ w˜.
1 1
Proof. Let A = (a ) ∈ K≥0. Then tr(A) = n and A has rank one. This implies
ij n
that A can be written as
−→
c · a
1
A= .. ,
.
−→
c · a
k
 
−→ −→
for some row vector a and positive integers c ,...,c . Now we can divide a by
1 k
the greatestcommondivisordofits entriesandsetv =c d. Setting v =v =(v )
i i A i
and (w ) = w = −→a/d yields A = vwt. This choice of v and w is unique up to
A
dividing the v by a common divisor e > 0 and multiplying all w by e. However,
i i
the resulting vectors are equivalent with respect to ∼ . (cid:3)
1
Intheaboveproof,weconstructedamapψˆfromK≥0 toD (µ)×D (µ)mappinga
µ 1 1
matrixAto the pair([v ],[w ])where (v ) is the greatestcommondivisorofrow
A A A i
iand(w ) the firstrowofthe matrix Adividedby (v ) . Note thatwehaveseen
A i A 1
that all rows are equal after dividing by their respective greatest common divisor.
In particular, this means that (w ) =a /(v ) .
A i ii A i
Lemma 4.3. Let µ (cid:15) n be a composition of n with k parts and v ∈ D(µ). Then
there exists a unique w ∈ D(µ) such that v(w )t ∈ K≥0. Moreover, if v ∼ v˜,
v v µ 1
then v(w )t =v˜(w )t.
v v˜
Proof. Let v = (v ,...,v ) ∈ D(µ) and set w = w = (w ) with w = µi. Then
1 k v i i vi
w ∈D(µ)and,moreover,thediagonalofthematrixvwt equalsµanditisclearthat
there exists no other w˜ ∈ k such that the diagonalof vw˜t equals µ. Furthermore,
(vwt)2 =vwtvwt =n·vwZt and thus vwt ∈K≥0.
µ
Now, let v ∼ v˜. Without loss of generality we assume that there exists c > 0
1
such that v = cv˜. Then w = µi = µi = 1w . This implies the claim, as
vwt =cv˜wt =v˜(cw )t =v˜wvt. (cid:16)vi(cid:17) (cid:16)cv˜i(cid:17) c v˜ (cid:3)
v v v v˜
Theorem 4.4. The map
ϕ:D (µ)→K≥0,
1 µ
[v]7→vwt,
v
is a bijection.
Proof. Lemma 4.3 yields that ϕ defined as above is well-defined. Now, composing
ψˆ with the projection on the first component, we get a map ψ : K≥0 → D (µ),
µ 1
mapping A to [v ].
A
Next, let A∈K≥0. Then we get that
µ
ϕ(ψ(A))=ϕ([v ])=v wt =A,
A A vA
as (w ) = µi = aii = a1i =(w ) .
vA i vi (vA)i (vA)i A i
On the other hand, if [v]∈D (µ), then
1
ψ(ϕ([v]))=ψ(vwt)=[v˜]=[v].
v
COUNTING QUASI-IDEMPOTENT IRREDUCIBLE INTEGRAL MATRICES 7
Recallthatv˜hasthegreatestcommondivisorofthei-throwofA=vwt asitsi-th
v
component. In other words, if v˜6= v, then there exists a greatest common divisor
d>1 of the entries of w . But then, by construction, we get that dv =v˜ and thus
v
v˜∼ v. (cid:3)
1
This shows that, instead of counting matrices in K≥0, we can count elements in
µ
D (µ).
1
Corollary 4.5.
|K≥0|= |K≥0|= |D (µ)|.
n µ 1
µX(cid:15)n µX(cid:15)n
4.2. Counting D (µ). Let µ (cid:15) n be a composition of n with k parts. Observe
1
thatwe mayassumethat, forevery[v]∈D (µ), the greatestcommondivisorofall
1
v is 1 due to the fact that we count up to ∼ , i.e. up to scalar multiples.
i 1
Using the p-part notation for µ, we can now prove a multiplicative property of
D (µ).
1
Lemma 4.6. Let µ(cid:15)n be a composition of n with k parts. Then
(i) |D (µ)|= |D (µ(p))|;
1 pprime 1
(ii) |D (µ(p))|Q= k (e +1)− k e .
1 i=1 i i=1 i
Q Q
Proof. Observe that every divisor of µ is divisible by some divisor of each of the
µ(p). Moreover,these divisorsofµ(p) can be chosenmaximally with respectto this
property. Then we see that, picking one divisor d , for each µ(p), and multiplying
p
all such, yields a divisor d of µ and that varying the d will lead to a result for d.
p
Thenumberofdivisorsofµ(p) equalstheproductofthenumberofdivisorsofevery
part. For every part we have e +1 divisors and thus the total number of divisors
i
k
of µ is (e +1). However, this way we count all scalar multiples which we
p i=1 i
do not wQant to count when counting D1(µ(p)). We thus have to assume that, for
every divisor v = (v ,...,v ), at least one v = 1. Equivalently, we may subtract
1 k i
all divisors v were every v ≥p. The number of those is the product of all divisors
i
differentfrom1foreachcomponent. Foreverycomponentwehavee suchdivisors
i
andthusthetotalnumberofallsuchdivisorsis k e . Thisprovestheclaim. (cid:3)
i=1 i
Q
Asanexamplewewillnowlistthenumberofmatricesforthefirstnupto13there
are. It should be remarked that up to n = 6 this has been checked both by hand
and a matlab code, the values afterwards are calculated using the same matlab
code. The corresponding integral sequence seems to be new, cf. [13].
n 1 2 3 4 5 6 7 8 9 10 11 12 13
|K≥0| 1 2 6 15 42 99 262 633 1592 3897 9697 23767 58804
n
8 JAKOBZIMMERMANN
Inthefollowingexamplewegivealistofallmatricesandthecorrespondingdivisor
of their diagonals for n = 4.
2 2 2 2
(4)=(1)(4), 1 1 1=1(1,1,1),
1 1 1 1
   
2 1 1 1
3 3 3
= (1,1), 2 1 1=1(2,1,1),
(cid:18)1 1(cid:19) (cid:18)1(cid:19)
2 1 1 1
   
1 1 1 1
3 1 1
= (3,1), 2 2 1=2(1,1,1),
(cid:18)3 1(cid:19) (cid:18)1(cid:19)
1 1 1 1
   
1 2 1 1
1 1 1
= (1,1), 1 2 1=1(1,2,1),
(cid:18)3 3(cid:19) (cid:18)3(cid:19)
1 2 1 1
   
1 1 1 1
1 3 1
= (1,3), 1 1 1=1(1,1,1),
(cid:18)1 3(cid:19) (cid:18)1(cid:19)
2 2 2 2
   
1 1 2 1
2 2 2
= (1,1), 1 1 2=1(1,1,2),
(cid:18)2 2(cid:19) (cid:18)2(cid:19)
1 1 2 1
   
1 1 1 1 1
2 4 2 1 1 1 1 1
= (1,2), = (1,1,1,1),
(cid:18)1 2(cid:19) (cid:18)1(cid:19) 1 1 1 1 1
   
1 1 1 1 1
   
2 1 1
= (2,1),
(cid:18)4 2(cid:19) (cid:18)2(cid:19)
Note that D ((2,2)) = {(2,2),(2,1),(1,2)}. So we see that, as (1,1) ∼ (2,2), we
1 1
do not include (1,1)and, in fact, (1,1)t(2,2)yields the same matrix as (2,2)t(1,1)
where the latter is counted.
4.3. Counting up to a permutation of basis vectors. In this section we will
≥0
turn our attention to elements of K ; that is positive, integral quasi-idempotent
n
matrices up to permutation of basis vectors. This means that we, instead of com-
positions of n, only need to consider partitions of n as we may assume that the
≥0
diagonal of a matrix in K is a partition of n, else we can simply permute the
n
basis to obtain a matrix which satisfies this condition.
Let µ ⊢ n be a partition of n. Then we define an equivalence relation on D(µ) as
follows: we say that v ∼ v˜, if there exists c∈ and σ ∈S such that
2 >0 k
Z
(σ(cv)=v˜ and σ(w )=cw ) or (σ(v)=cv˜ and σ(cw )=w ).
v v˜ v v˜
Example 4.7. Let µ=(4,2,2)⊢8 and consider
v =(2,2,1),w =(2,1,2),
1 1
v =(2,1,2),w =(2,2,1),
2 2
v =(1,2,2),w =(4,1,1).
3 3
Then v ,w |µ and v wt ∈K≥0. If we just study the actionof S on the v , then we
i i i i 8 3 i
see that they all lie in the same orbit. However, it is easy to see that v wt cannot
3 3
be obtainedby permutationof basis vectorsfromone of the other two matrices,as
COUNTING QUASI-IDEMPOTENT IRREDUCIBLE INTEGRAL MATRICES 9
someofitsentriesare8andsuchentriesdonotexistintheothertwomatrices. On
the other hand, we obtain v wt from v wt by swapping the second and the third
1 1 2 2
basis vector. Thus we want v ∼ v but v ,v 6∼ v . It is easy to see that this is,
1 2 2 1 2 2 3
in fact, the case.
Observe that v ∼ v˜ is equivalent to
2
(σ(cv)=v˜ or σv =cv˜) and σµ=µ,
which follows immediately as v˜v =µ, respectively vw =µ.
w˜ v
Analogously to the above, we then define
D (µ)=D(µ)/ .
2 ∼2
We can now formulate the following:
Theorem 4.8. Let µ⊢n be a partition of n. Then the map
ϕ:D (µ)→K≥0,
1 µ
[v]7→vwt,
v
≥0
induces a bijection ϕ˜:D (µ)→K .
2 µ
Proof. Let [v],[v˜] ∈ D (µ) and choose v,v˜ such that the greatest common divisor
1
of both all (w ) and all (w ) is 1. This implies that, for all c > 1, we have that
v i v˜ i
cv,cv˜∤µ.
Now v ∼ v˜ implies that there exist σ ∈ S ,c ∈ such that either σ(cv) = v˜
2 k >0
Z
and σw = cw or σv = cv˜ and σ(cw ) = w . Assume the former, the latter case
v v˜ v v˜
can be proved similarly. Note that, by our choice of v and v˜, we have that c=1.
v˜wt =σvσwt =P v(P w )t =P vwtPt =P vwtP−1,
v˜ v σ σ v σ v σ σ v σ
i.e. v˜wt ∼vwt.
v˜ v
By Theorem 4.4, we know that the inverse of ϕ is ψ. Now let A,B ∈ K≥0 and
µ
assume that A∼B. Then there exists σ ∈S such that A=P BP−1. Moreover,
k σ σ
we have
v wt =A=P BP−1 =P v wt Pt =P v (P w )t =σv σ(w )t,
A A σ σ σ B B σ σ B σ B B B
i.e. (v ,w )∼ (v ,w ). This concludes the proof. (cid:3)
A A 2 B B
≥0
Thus we get the following formula for enumeration of matrices in K .
n
Corollary 4.9.
≥0
|K |= |D (µ)|.
n 2
µX⊢n
4.4. Counting D (µ). Let µ⊢n be a partition of n with k parts and
2
µ(p)
ppYrime
its decomposition as in (1). As before, we have that µ(p) = (pe1,...,pek), where
each e is chosen maximally.
i
Now we can assign a matrix to µ with k columns where the i-th column contains
the exponent of the maximal power of the i-th prime dividing µ where we ignore
i
10 JAKOBZIMMERMANN
zero rows. This guarantees that the number of rows is bounded. For example, to
µ=(12,6,4,1)=(22,21,22,20)·(31,31,30,30) we assign the matrix
2 1 2 0
A = .
µ (cid:18)1 1 0 0(cid:19)
Note that the only permutations in S which stabilizes µ are those which swap
k
identicalcolumns. Thatmeansthat,ifwewanttocountthenumberofequivalence
classes with respect to ∼ , then we have to count the number of different classes
2
for each set of identical vectors.
First, we can observe that, for each choice that we make in one of the sets of
identicalvectors,allchoicesinthe otherclassesgiverisetonew equivalenceclasses
in D (µ). Thus we can count each set isolated and then take the product over all
2
such sets.
Now, let V be a set of r identical vectors v. In order to understand the number of
different equivalence classes,we first note that we can count the number of choices
in eachcomponent and then take the product overall components as these choices
are independent of each other. In other words, we have to count the number of
differentchoices(uptopermutation)foranon-negativesequenceoflengthr where
every element is less or equal to v . This amounts to counting all non-increasing
i
sequence of length r consisting of integers between v and 0. Counting this is the
i
sameascountingcompositionsofrwithv parts,where0isanallowedsummand
i+1
or equivalently the number of v +1-tuples of non-negative integers whose sums it
i
r. This number is given by
r+v
i
,
(cid:18) r (cid:19)
see e.g. [3, II.4.5. Lemma].
InordertowritedownaformulaforD (µ),weneedsomenotationfirst. Weassign
2
to µ the matrix A , as described above. Assume that A ∈ Mat ( ). Then, for
µ µ l,k
Z
v ∈ l, we define c as the number of times v appears as a column of A . With
v µ
Z
this, we can now formulate the following result.
Proposition 4.10. Let µ⊢n be a partition of n with k parts and A ∈Mat ( )
µ l,k
Z
the corresponding matrix of prime factors. Then we have,
l
c +v
v i
D (µ)= .
2
(cid:18) v (cid:19)
vY∈ liY=1 i
Z
Notethatc =0forallbutfinitelymanyv ∈ l andthen cv+vi =1,inparticular,
v Z vi
we use that 0 =1. (cid:0) (cid:1)
0
(cid:0) (cid:1)
Example 4.11. The following table shows how many matrices up to permutation
of basis vectors there are for n = 1,...,11. The counting has been performed by
handindependentlybytwopeople. Again,thisintegralsequencesseemstobenew,
cf [13].
n 1 2 3 4 5 6 7 8 9 10 11
|M | 1 2 4 8 16 27 51 83 140 225 369
n