C OMPOSITIO M ATHEMATICA
J.-H. E VERTSE
R. T IJDEMAN
Singular differences of powers of 2 × 2-matrices
Compositio Mathematica, tome 104, n
o2 (1996), p. 199-216
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
Singular differences of powers of 2
x2-matrices
J.-H. EVERTSE and R. TIJDEMAN
University of Leiden, Department of Mathematics and Computer Science, P. O. Box 9512, 2300 RA Leiden, The Netherlands
e-mail: evertse@wi.leidenuniv.nl. tijdeman@wi.leidenuniv.nl
Received 30 June 1995; accepted in final form 17 October 1995
Abstract. We describe the pairs of non-singular 2 x 2-matrices
(A, B)
with complex entries such that the set of pairs of integers m, n for which Am - Bn is singular is infinite.Key words: Matrices, exponential polynomial equations
@ 1996 Kluwer Academic Publishers. Printed in the Netherlands.
1. Introduction
Suppose
A and B arenon-singular
2 x 2 matrices with rationalintegral
entries. A.D.
Pollington
asked whether thefollowing
two statements are correct:(a)
Assume that for everylarge
N there are at leastN 2 / (log N) 2 pairs
ofpositive integers
m, n withmax(m, n)
N such that A’n - Bn issingular.
Then one ofthe
eigenvalues
ofA,
B is a root ofunity.
(b)
Assume that A and B have non-realeigenvalues
and that for every - > 0 and for every Nexceeding
some bound in terms of E, there are at leastNl-é pairs
of
positive integers
m, n withmax(m, n)
N such that A’n - Bn issingular.
Then there are
integers
r, s, not both zero, such that Ar = BS.Brown,
Moran andPollington [3]
needed such results for their research on aconjecture
of Schmidt[6]
onnormality
withrespect
to matrices. Some further workon this
conjecture
of Schmidt was doneby
Brown and Moran[ 1, 2] .
In the present paper we show that statements
(a)
and(b)
are correct. Moregenerally,
in statement(a)
we allowA, B
to havecomplex entries, N2/ (log N)2
may be
replaced by N f (N)
for any functionf
which is unbounded inN,
andwe show that either both A and B have an
eigenvalue equal
to a root ofunity
orone of the matrices
A,
B has twoeigenvalues equal
to a root ofunity.
In statementb),
the conclusion remains valid ifA,
B are any real 2 x 2-matrices with non-realeigenvalues
such that A"2 - B’ issingular
forinfinitely
manypairs (m, n).
Let
A, B
be two matrices inGLk(C),
i.e. the group ofnon-singular k
x k-matrices with
complex
entries. DefineDenote the transpose of a matrix C
by CT.
It is obvious that Twopairs (A, B), (A 1 , Bl)
of matrices inGLk (C)
are called similarif
Since
(1.1) implies
thatAr- Bï
=J(Am - B’)J-1
we haveSA,B
=SAI,BL
forsimilar
pairs (A, B), (A1, J3i).
We are interested in the
problem
to determine the matricesA,
B for whichSA,B
is infinite.
Clearly,
if(A, B), (Ai, B1 )
arepairs
inGLK(C)
such that(A, B)
is similar tothen
SA,B
is infinite if andonly
ifSA,,B,
is infinite. Forpairs
of matrices(A, B), (Al , B1 ) satisfying (1.2)
we say that(A, B)
is related to(Al, BI)’
In this paper we restrict our attention to 2 x 2-matrices. We describe four types of
pairs
of matrices(AI, 2?i)
inGL2(C)
for whichSA,,B,
is infinite.for certain
integers
r, s, not both zero and some nonzero issingular
forevery t
E Z.for some nonzero
is
singular
forevery t
E Z.and for some nonzero
for certain
integers
r, s withrs -#
0numbers such that
The
equality
in(1.3)
isequivalent
to deti Hence for everypair
It is easy to check that there are
pairs
of matrices(AI, Bl)
oftype I,
II or III.We do not
know,
whether there arepairs
oftype
IV. We have been able to proveonly (cf.
Section3,
Lemma8)
that(1.3) implies
Our main result is as follows:
THEOREM 1. Let
(A, B)
be apair of
matrices inGL2(C) for
whichSA,B
=( (rn , n)
E7G2:
Am - Bn issingularl
isinfinite.
Then(A, B)
is related to apair
(AI,Bt) oftypel,
II, IIIorIV.Remark. From
(1.4)
it follows that if(A, B)
is related to apair
of type IV then both A and B have adouble,
irrationaleigenvalue.
Thisimplies
that apair
ofmatrices with entries
in Q
cannot be related(over C)
to apair
of type IV.From Theorem 1 we shall derive the
positive
answer toquestion (b)
ofPollington:
COROLLARY. Let
(A, B)
be apair of nonsingular
2 x 2-matrices with real entries and with non-realeigenvalues for
which the setS A,B
isinfinite.
Then thereare
integers
r, s, not both zero, such that Ar = BS.We now consider the case that
SA,B
has’large density.’
Define the maximumnorm of h =
(hl,
...,hr)
E Crby
and for ’ put
Note that if
(A, B)
is related to apair (Ai, Bl )
of typeI,
II orIII,
thenWe now consider the
pairs (A, B)
of matrices inGL2 (C)
for which lim supv--,,1. We describe two
types
ofpairs (Al, BI)
with this property:for certain nonzero
integers
r, s. ThenAÍt - )
issingular
forevery t, u
e Z. Note that at least one of theeigenvalues
of
A1 and at least
one of theeigenvalues
ofBl
is a root ofunity.
for certain nonzero
integers
r, s and forevery t, u
E Z. Note that bothAI, BI
can bediagonalised,
that botheigenvalues
of
BI
are roots ofunity
and that theproduct
of theeigenvalues
of41
is a root ofunity.
The
following
resultimplies Pollington’s
statement(a):
THEOREM 2. Let
(A, B)
be apair of matrices in GL2(C) for
which the sequenceThen
(A, B)
is related to apair (.
Note that
SA, B
is the set of solutions of theequation det(A"2 - Bn)
= 0 in(rn, n)
EZ2.
This is aspecial type of exponential polynomial equation.
We deriveour results stated above from a theorem of Laurent
([4],
Théorème1)
on thestructure of the set of solutions of
exponential polynomial equations.
In Section 2we recall Laurent’s theorem and refine this a little bit and in Section 3 we prove
our results.
2.
Exponential polynomial equations
Let n be a
positive integer. For exponential polynomial equation
where I is a finite index set and for each i e
I, f 2 (X )
is a nonzeropolynomial
in
C[Xl,... , Xn], and Qi
=(Olil) ... ain)
is a vector with nonzerocomplex
coordinates.
For each solution of
(2.1 ),
the left-hand side of(2.1 )
can besplit
intovanishing
subsums which are minimal in the sense that each proper subsum of any of the
vanishing
subsums is non-zero. We shall deal with all solutionscorresponding
to a
given splitting
into minimalvanishing
subsums. Moreprecisely,
let P be apartition
ofI,
i.e. a collection of non-empty,pairwise disjoint
sets{PI, ... , Pt}
with
Pi
U ...U Pt =
I. For a set P we write P - P if P is a subset of one ofPl , ... , Pt,
andP j P
if P is a proper subset of one ofPl , ... , Pt .
We shall dealwith the subset of solutions of
(2.1 ),
To ’P we associate two other sets. A
pair
I such thati, j belong
to the same set of P. Define the abelian
subgroup
ofZ’,
(If
P consists ofsingletons,
i.e. sets ofcardinality 1,
thenHp
=Zn.)
Let r =rp : rank
Hp .
As is wellknown,
Zn has a basis{al, ... , an )
such that for certainpositive integers d 1, ... , dr, {dl al, ... , drar }
is a basis ofHp.
Now letThen
clearly,
every h e Zn can beexpressed uniquely
asIn what
follows,
forgiven
h EZn, h’, h"
willalways
denote the vectors definedby (2.3).
In this
section,
we write A « B or B » A ifA
c -(B + 1)
for somepositive
constant c
depending only
on thepolynomials f and
the vectors aiappearing
in(2.1).
We shall usefrequently
that for h =Ei==1 çiai with
=(Çl,’ .. , çn)
e Znwe have
with 1 - denoting
the maximum norm.LEMMA 1. Let h E
UP. Then for
the vectorh’
ESP defined by (2.3)
we haveProof.
This is astraightforward
consequence of Laurent[4],
Théorème 1.By
thistheorem,
we have h =h[
+h[’
withh[’
EH-p and Ih 1
«log 1 h (Laurent
provesthis
only
under thehypothesis
that thepartition
P does not containsingletons.
IfP does contain
singletons, f i 1 1, ... , {Ís},
say, then we can reduce to the case thatthere are no
singletons by removing i 1, ... , is
from I and{z},..., {zg}
fromP,
which makes the set
UP larger
but does not affectHP).
We havehl
=hl
+h"
with
hl
ES’P, h"
eH-P.
Thisgives
h =hl
+(h2
+h")
withh"
+h1
EHP.
Hence
h2
=h’.
As{ai,..., a}
is a basis of ZI we haveand since
h; - h[
EHp,
i.e. is a linear combination of al, ... , ar, wehave Çi
= T/i for i = r +1,...,
n.Together
with(2.4) and IÇil di, IT/il C d2
for i =1,...,
rthis
implies
thatwhich is (2.5).
0We need some more
precise
information about the setUP .
To this end we needthe
following
lemma.LEMMA 2. be a
polynomial of
totaldegree
d thatdoes not vanish
identically
onH-p.
Here r = rank
Hp
and the constantsimplied by
«ddepend only
on d andHp.
Proof.
We claim that forevery
non-zeropolynomial
For every
h’
ES-P
and each set P -P,
define thepolynomial
From
(2.2)
andfrom., H-p
and for eachpair
where 1 are the vectors with
LEMMA 3.
Proof.
If allpolynomials f
Z(i
EI)
are constant onH-p
or if r == 0 thenfor every
h’
ES-p, j = 1, ... , t
thepolynomial gl,,,pj
is constant onHP,
theconstancy
being
trivial if r = 0. HenceYh’ , p. J
is eitheridentically
zero onHp
or has no zeros in
Hp
and from(2.6)
it follows that in both casesU(l)
=0.
Suppose
that r > 0 and that not allpolynomials f i
are constant onH-p. By
For a set 9 C Zn and an abelian
subgroup
H ofZ’,
we define 9 + H ={a+h:a E Sh E H}.
LEMMA 4. There is a
finite
set S CSp
such that:with
(2.8), (2.2)
thisimplies
that h’ +h"
eU-p.
Butthen, by
Lemma 1This
implies
thatIh’l
isbounded,
i.e. that S is finite.We need the
following
result forexponential polynomial equations
in onevariable
where as
before,
I is a finite index set and for each i EI, fi (X )
EC[X]
is non-zeroand ai e C is non-zero.
LEMMA 5. Assume that
(2.9)
hasinfinitely
many solutions. Then there is apartition
Pof
I,consisting of
setsof cardinalij j
2, suchthat for
eachpair i, j
contained in one
of the
setsof P, ail aj
is a rootof unity.
Proof.
This result wasproved by Skolem-Mahler-Lech,
cf.[5].
It is astraight-
forward consequence of our Lemmas 3 and 4. Since the set of solutions of
(2.9)
is the union of
finitely
many setsUp,
there is apartition
P of I for whichUP
isinfinite. There are no
singletons
in P sincefi(h)ah
hasonly finitely
many zeros.By
Lemmas 3 and4,
rankH-p
= 0implies
that!7 = 0
andU!t)
isfinite,
hencerank 1. This
implies
Lemma 5. ~3. Proofs of the results
We first prove Theorems 1 and 2
simultaneously,
and then derive theCorollary.
A matrix N E
GL2(C)
is said to be innormal form
if either N =a 0
withA in normal
form,
(3.1 )
B =
JNJ-I
with J EGL2(C)
and N in normal form.(3.1 )
In what
follows, (A, B)
is apair
of matricessatisfying (3.1)
and we write J =a b Thus,
ad - bc = detJ =J
0. Note that(a c
cd b).
Hence, is the set of solutions of
We
distinguish
thefollowing
cases:In the
proof
of Theorems 1 and 2 we have to consider allpartitions
of the left- hand sides of(3.2a), (3.2b)
and(3.2c)
intovanishing
subsums. We can reducethe number of cases
by using
thefollowing ’symmetry considerations,’
which areconsequences of the fact that in the
proofs
of our results(A, B)
may bereplaced
by
any relatedpair satisfying (3.1 ):
Each of the above
replacements
leads to apermutation
of the terms in(3.2a, b, c), provided
that withreplacement (1),
m and n areinterchanged.
We deal with a
simple
case first:LEMMA 6. Assume that abcd = 0 in case
(a),
or ac = 0 in case(b),
or c = 0 incase
(c). If SA,B
isinfinite
then(A, B)
is related to apair of type
LIf the
sequence#5 A,B ( N) lN (N
=1, 2,..., )
isunbounded,
then(A, B)
is related to apair of type V.
Proof. By symmetry
consideration3,
it is no lossof generality
to assume bc = 0in case
(a)
and c = 0 in case(b). Further,
since in case(a), (A, B)
is related to thetransposed pair (AT, B T) = ( cg (0 a 0
,b (
bâ -a) (0 p 0) b
0, bc ) -’),
-a wemay assume that c = 0 in cases
(a), (b)
and(c). By substituting
c - 0 andusing
that ad - bc =det J 54 0, (3.2a,b,c)
become(tam - pl) (,6’ - an) = 0, (a"2 - pn) (,8m - pn )
=0, am - pn
=0, respectively.
In view of symmetry consideration4,
it suffices to prove Lemma 6 with thehypotheses SA,B infinite,
#SA,B(N)IN (N
=l, 2, ... , )
unboundedbeing replaced by
respectively,
for all three cases(a), (b), (c).
If(3.3) holds,
then takeIn the
sequel
we assume thatabcd -#
0 in case(a),
ac,-E
0 in case(b), c -#
0in case
(c).
We write h =(m, n)
and in the left-hand sides of(3.2a,b,c)
wedenote the ith term from the left
by fi(h)ah.
Forinstance,
in(3.2a)
we havef 1 (h) a ï
=(ad - bc). (a{3)m 1 n
withfl(h)
= ad -bc, gl
=(a,8, 1), f2(h)aÎ
=(ad - bc). 1 m (pa)n,
...,f6 (h) p6h
= bc,8man. Thus, (3.2a, b, c)
can be rewrittenas
EiElli(h)gr
= 0 with I =f 1, ... , 6}
in(3.2.a),
I ={1, ... , 4}
in(3.2b)
and I =
{1,2,3}
in(3.2c). By
ourassumptions
on a,b,
c, d we have thatfi(h)
isnot
identically
zero for i E I.By applying
thetheory
of Section 2 toexponential
polynomials
in n = 2 variables we infer that Theorems 1 and 2 follow from:PROPOSITION.
(i) Suppose
that in cases(a), (b)
or(c)
there is apartition
Pof I for
which rankH-p >,
1 andUP
isinfinite.
Then(A, B)
is related to apair of
typeI,II,IIIorIV.
(ii) Suppose
thatfor
somepartition
P ={PI,... , Pt 1 of
I and some a EZ 2
we have rank
HP
= 2 and every h E a +HP satisfies ¿iEPj fi (h) a4
= 0for
j
=1,
... , t. Then(A, B)
is related to apair of type
V or VI.Namely,
ifSA B
isinfinite,
then there is apartition
P of I for whichUP
is infi-nite.
By
Lemmas 3 and 4 this ispossible only
if rankH-p >,
1. Since#(S
+HP) (N)
«N ,k Hp
for any finite setS,
we haveby
Lemmas 3 and 4 thatIp (N)
«log N, p (N)
« N if rankHP
= 1 and#17(,’) (N)
« N if rankHP
= 2. Henceif #SA,B(N)/N (N
=1,2,...)
is unbounded then there must bea
partition
P of I with rankHy
= 2 andUJ;) 01 0.
Then Lemma 4implies
thatfor some a E
Z 2,
every h Ea+ Hp
satisfies¿iEPj Ji (h)a4
= 0for j = 1,
..., t. 0The
following
situation will occurfrequently:
LEMMA 7. Let P be a
partition of
I such thatfor
somepositive integer k, HP
is contained in oneof
the groups{akm
=,8km pk,1* (in
cases(a), (b»), {akm = {3km
=,knl@ 1,,km = kn = akn}, {,8km = pkn = akn} (in
case(a».
If rank Hp #
1 then(A, B)
is related to apair of
type I andif rank HP
= 2 then(A, B)
is related to apair of type V.
Proof. By
the symmetryconsiderations,
it suffices to consider the caseHp
C{akm
={3km
=pkn}.
Recall that(A, B)
is related to(Al, Bl)
withAI = J - 1 A J, B
=J B J
= N. If rankH-p >
1 then for(r, s)
EHyB{(0, 0)}
wehave
i.e.
(Ai , Bi )
is of type 1. If rankHP -
2 then there arelinearly independent (ri, SI), (r2,s2)
EHP
and froma kri
=,8kri
=pkSi
for i =1, 2
it follows that,k(rIS2-r2SI) = (3k(r1s2-r2s1) = pk(ris2-r2si) - l,
hencei.e.
(AI, ,Bi )
is of type V.* short hand
for 1 (m,
n) E Z2: akm ={3km = pkn}
Proof of
theProposition.
We first deal with case(a).
Recall that the left-hand side of(3.2a)
has six termsfi(h)gr (i
E I ={l,... , 6}), fj (h)a/ being
the ithterm from the left. If the
partition P
of I containssingletons
thenUP = 0 since
each
f i
is constant. Therefore we consideronly partitions
of I withoutsingletons.
To each such
partition P
we associate agraph
G as follows: the vertices of G areVl
={1,2}, V2
={3,4}, V3 = (5, 6)
and[%§, %J]
with i-# j
is anedge
of Gif there are k E
Vi,
1E Vj belonging
to the same set of P. Note that if[VI, V2]
is an
edge
of G thenH-p C f ce’
=p’l
orH-p g Io’ =: o,’I,
if[VI, V3]
is anedge
thenHP g {am
=an}
orH-p Ç {,8m
=pn}
and if[V2, V3]
is anedge
thenHp ç lam
=,8m}
orHp
C{pn
=an}.
Subcase
(al ).
G has at least twoedges.
Then
HP
satisfies the conditions of Lemma 7 with = 1 and theProposition
follows.
Subcase
(a2).
G has noedges.
Then P =
f f 1, 2 1, f 3, 4 1, f 5, 611. Hence Hp = {(a,8)m
=(pa)n, aman
=,8mpn, cem10nom orn 1.
For(m, rL)
eH-p
we have amanam pn
=(3mpn. {3man,
whence
a 2m a2m
andcem 10 n.,am 10 n
=,8m an . am on whence p2n =or 2n.
Thisimplies ce 4m
=(a{3)2m
=(pa)2n
=p4n .
SoHP
C{a4m
=,84m = p4n}.
Therefore,
we canagain apply
Lemma 7 and derive theProposition.
the other
possibility
can be reduced to this oneby
our symmetry considerations. With this P we haveAssuming
thatUp -# 0,
we havead/bc
=bc/ad
=(pla)n
for some n e N, hencead/ bc
= ±1. But ad - bc =detj 7 0,
so ad = -bc. Now(A, B)
is related to(AI,Bi)
withFor and
Hence
(A1, jSi)
isof type II,
with 0 =ar, K = ,8r , À
= aS.Now suppose
that rank HP
= 2.For linearly independent pairs (ml, ni), (m2, n2)
E
HP
we have(ao),in2-M2n,
=(pa)m1n2-m2n1
=l, (pla)n1
=(pla)n2
=1,
hence
a,6, p
and are rootsof unity.
Now choose(r, s)
EU-p
and letA1, Bl
be asabove. Let u be an odd
integer
with aU = ± 1.(A, B)
is related to(A2, B2)
withand we have
Hence
(A2, B2)
is apair of type
VI.the other
possibilities
can be reduced to this oneby
our symmetry considerations With this ’P we haveAssuming
thatU-p :, 0
we have(ad - bc)/ad
=ad/(ad - bc). Together
withbc #
0 thisimplies
that(ad - bc)/ad = -1,
i.e. bc = 2ad. Now(A, B)
is relatedto
(A1, BI )
withTake 1 and
hence
(AI, Bi )
is apair
of type III with 0 =,8r, K
=ar, p’
=À,
aS = 20130.Suppose
that rankHP
= 2. Chooselinearly independent (mi, ni), (m2, n2)
EHp.
Then0’i
=a-ni, (al,8)mi
=(0,/p)ni
for i =1, 2,
hence{3m1n2-m2n1
=am1n2-m2n1 =
1, (al,8)m1n2-m2n1
=(alp)m1n2-m2n1 = 1, which implies
thata,
0,
p, cr are roots ofunity. Letting k
be a nonzerointeger
withak = ,8k =10 k =
or k
=1,
we infer thatH-p ç: tcekm
=,8km
=pkn}. Together
with Lemma 7 thisimplies
that(A, B)
is related to apair
of type V.We continue with case
(b).
Recall that the left-hand side of(3.2b)
has four terms, the ith from the leftbeing
denotedby fi (h) a4. Again
we have to consider somepossibilities
for P.Subcase
(bl ).
P containssingletons.
fl(h), f2(h) areconstants, f3(h)
=0 implies that n = -(ad-bc)/acand f4(h) _
0
implies
that n =(ad - bc)/ac.
Hence ifE/p 7 0
then P does not contain{1}
or
{2}
and at most one of{3}, {4}. Thus, P = f f 1, 2, 41, {3}}
or{{1,2,3}, {4}}
and in both cases we have
HP
=a’ = p,
=pnl.
Now theProposition
followsfrom Lemma 7.
Proposition
follows from Lemma 7.as the other
possibility
can be reduced tothis one
by
our symmetry considerations. For this P we haveFor
(m, n)
eUP
we havehence n = 0.
Therefore, UP
=f am
=1, n
=O}.
IfUP
is infinite then there isa
positive integer
r withthere is no a e
Z2
such thatevery
h e a +Hp
satisfies(*),
i.e. P cannotsatisfy
the
hypothesis of part (ii)
of theProposition.
Finally,
we deal with case(c).
For eachpartition
P of{l, 2, 3} containing
asingleton, Up
isfinite; namely /i(h),/2(h)
are constants andf3(h) = -2(ad -
bc) - c2mn
hasonly finitely
many zeros(m, n)
eZ2. Therefore,
we have to dealonly
with the case P =( 1 , 2, 3 ) . (3.2c)
can be rewritten aswhere J
since we assumed thatc -#
0.Further,
The
hypothesis of part (ii)
of theProposition
cannot be satisfied.Namely,
supposethat rank HP
= 2 and that for some a =(a, b)
EZ2every (m, n)
Ea+HP
satisfies(3.5).
We haveH-p P d7G2
for somepositive integer d,
hence(a+du, b+dv)
satisfies(3.5),
i.e.for every
(u, v)
EZ’.
But this isclearly impossible.
So we have to proveonly part (i)
of theProposition.
If one of a, p, whichby
our symmetry considerations wemay assume to be a, is a root of
unity
then(A, B)
is related to apair
of type I:namely,
if ar = 1 for somepositive integer
r then andAssùme’that
cx and p are not rootsof unity,
that rankHp
=1,
and thatU-p,
i.e.the set of solutions of
(3.5)
is infinite. We recall that p-#
0. Hence(1.3)
holds.Further,
with A =
a2/(ad - bc),
{t =c2/(ad - bc).
Hence(A, B)
is of type IV. This com-pletes
theproof
of theProposition.
0We now show that
(1.3) implies (1.4):
LEMMA 8. Let a, p, M be nonzero
complex
numbers such that a and p are not rootsof unity
and such that(oz’ - pn)2
=Mmnampn
hasinfinitely
many solutions inintegers
m, n. Then J-le Q
and there areintegers
r, s such that ar =p’
=: c isa real
quadratic
unit.Proof. By
Lemmas 3 and 4(with
n =2) applied
to(3.5)
andby
the fact that aand p are not roots of
unity,
we have thatHp = {am = pn}
has rank1,
i.e.for some fixed
(rI, SI)
E7G2 with r 1 s 1 54
0. Now(m,n)
EZ2
can beexpressed uniquely
ast(ri,sl)
+(u, p)
witht, u
E7,,p
efO,...,sl - Il.
So for somep E
fO, ....
s 1 -1 }, (3.5)
hasinfinitely
many solutions(tri
+ u, tsl+ p).
In whatfollows,
we fix this p.Thus,
there areinfinitely
manypairs (u, t)
eZ2 satisfying
or,
dividing by
We first show that a, p are
algebraic.
Forgiven
u, there areonly finitely many t
sat-isfying (3.7).
Hence if(u, t)
runsthrough
all solutions of(3.7)
then u runsthrough
an infinite set. Choose solutions
(ul, tl ), (U2, t2)
of(3.7)
withUl -#
U2,aUi -# Pp
for i =
1, 2.
Put 8 :=al/SI.
Then ô= (pI/rI
for some rootof unity (.
HenceThis shows that ô is a zero of a
non-identically
zeropolynomial
withalgebraic coefficients,
i.e. ô isalgebraic.
It follows thatindeed,
a, p arealgebraic.
Let K be a finite normal extension
of Q containing
a, p.Then 4
E K. Let ube an element of the Galois group G of
K/Q.
Then every solution(u, t)
of(3.7)
satisfies
or
(3.9)
is anexponential polynomial equation
withinfinitely
many solutions u e Z.Suppose
thatu (li) 7É
y.By
Lemma5,
there is a0
e{a, a-l, a(a), a(a)-I}
such that
0/ 1
is a root ofunity.
Hence a is a root ofunity
but this isagainst
ourassumption. Therefore, a (p)
= y. This holds for every Q EG; hencey
eQ.
By inserting p e Q
in(3.8)
we infer that for every solution(u, t)
and for everyQ E G we have
From Lemma 5 and the fact that a is not a root of
unity,
it follows that eitherala(a)
oraa(a)
is a root ofunity.
So there is apositive integer
r2 such that foreach E G we have either
0,(a’2) = a’2
ora( ar2) =
a-r2. Hence G’ :=f 0,
eG :
a( ar2)
=ar2}
is asubgroup
ofindex z
2 in G and so its field of invariantsL =
Q( ar2)
iseither Q
or aquadratic
field. We infer that with r = r1 r2, 8 = S 1 r2we have ar = p s =: - e L.
We show that E is a real
quadratic
unit.Let p
be aprime
ideal of K. Theright-
hand side of
(3.7)
is a rational number with a fixed denominator. Hence there is aconstant C such that
ordp
1 for every solution(u, t)
of(3.7).
As we mentionedabove, (3.7)
has solutions witharbitrarily large
u. Henceordp (a)
= 0. Thisbeing
the case for everyprime
ideal p, it follows that a, henceE, is a unit in L.
However,
a, hence E, is not a rootof unity,
and therefore - is a realquadratic
unit. Thiscompletes
theproof
of Lemma 8. ~Proof of
theCorollary.
Assume thatA,
B arenon-singular
matrices with real entries and non-realeigenvalues
such that Am - Bn issingular
forinfinitely
many
pairs (m, n)
eZ2 .
Theeigenvalues
of A arecomplex conjugates,
a, ii, say.Similarly,
B hascomplex conjugate eigenvalues
p, p.By
Theorem1, (A, B)
isrelated to a
pair (AI, BI)
of typeI,
II,III,
or IV. Afterinterchanging A, B
ortaking transposes,
we may assume that for someWe consider all
possibilities.
Suppose
not both zero and some nonzero 0 e C. After
interchanging cx,
a or p,p ifnecessary,
an
eigenvalue of
Arand
B’ with the sameeigenvector,
a, say.By taking complex
conjugates,
we obtain a commoneigenvector
a’ of A’’ and BS witheigenvalue e.
Hence Ar and B’ have the same action on two
linearly independent
vectors, i.e.Suppose (
is not related to a
pair
of type II.Finally,
we mention that both A and B have two distincteigenvalues.
Hence(A, B)
cannot be related to apair (.41, .81 )
of type IV. This proves theCorollary.
oReferences
1. Brown, G. and Moran, W., Schmidt’s conjecture on normality for commuting matrices, Invent.
Math. 111 (1993), 4492014463.
2. Brown, G. and Moran, W., Normality with respect to matrices, Comptes Rendus, to appear.
3. Brown, G., Moran, W. and Pollington, A. D., The Schmidt conjecture on normality in two dimensions, in preparation.
4. Laurent, M.,
Équations
exponentielles-polynômes et suites récurrentes linéaires II, J. Number Theory 31 (1989), 24-53.5. Lech, C., A note on recurring series, Ark. Math. 2 (1953), 417-421.
6. Schmidt, W. M., Normalität bezüglich Matrizen, J. reine angew. Math. 214/215 (1964), 227-260.