GROUP MATRICES • by William T. Iwata a THESIS' SUBMITTED IN
GROUP MATRICES
• by
William T. Iwata
A THESIS' SUBMITTED IN PARTIAL FULFILMENT OF
• THE RI1QUIREMENTS FOR THE DEGREE OF
MASTER OF ARTS
in the Department of Mathematics.
We ..accept this thesis as conforming to the required standard from\andidates for the degree of MASTER OF ARTS
THE UNIVERSITY OF BRITISH COLUMBIA
September, 19^5 In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of
British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that per• mission for extensive copying of this thesis for scholarly purposes may be granted by the Head of my Department or by his representatives„ It is understood that copying or publi• cation of this thesis for financial gain shall not be allowed without my written permission.
Department of
The University of British Columbia Vancouver 8, Canada ii
ABSTRACT
A new proof is given of Newman and Taussky's result: if A is a unimodular integral n X n matrix such that A'A is a circulant, then
A = QC where Q is a generalized permutation matrix and C is a circulant.
A similar result is proved for unimodular integral skew circulants.
Certain additional new results are obtained, the most interesting of which' are: l) Given any nonsingular group matrix A there exist unique real group matrices U and H such that U is orthogonal and H is positive definite and A = UH; 2): If A is any unimodular integral k s circulant, then integers k and s exist such that A' = P A and P A is symmetric, where P is the companion matrix of the polynomial xn-l.
Finally, all the n X n positive definite integral and unimodular skew circulants are determined for values of n < 6: they are shown to be trivial for n = 1,2,3 and are explicitly described for n = k,5,6.
I hereby certify that this abstract is satisfactory. iii
TABLE OF CONTENTS
Page
1. Group Rings 1
2. Matrix Representations and Group Matrices 1
3- Units and Unimodular Group Matrices 7 k. Circulants and Skew Circulants* ' 8
5. Existence of Nontrivial Unimodular Integral
Circulants and Skew Circulants 12
6. A New Proof on Positive Definite Circulants 12
7. New Results on Group Matrices and Symmetric Circulants 20
8. Positive Definite Skew•Circulants 27
9. . Appendix 36
10. Bibliography • . 37 ACKWOm^IXJEMENTS
It is a pleasure to acknowledge my indebtedness to my supervisor Dr. R. C. Thompson for suggesting the study of skew circulants and of circulants in general and for his encouragement and advice in preparing this thesis. 1. Group Rings •
Let G be a finite group of order,, n with elements g^,...,g^ and let K be an integral domain and let'F be a field containing K as a subring. Let R(G,F). denote a vector space over F which admits the elements g^,...,g of Ef as a basis and.in which, additionally, n n n products are defined by •)' a.g. ) b.g. = ) a.b.g. . where a.,b.
L xax L J&J ' L i J&X,J x' j i=l j=l i,j=l are in F and g. . ='g.g.. It is."well known that these operations make
R(G,F): into an associative algebra. Let R„ „ denote the set of all G, A. • n elements of the form ^ aj_gi "*"n ^(^•^)" where the scalars are in K. 'i=l -
Let 1^ and 1^. be the identities of G and K respectively; and let
1=1^.' 1^ denote the identity of R^ ^ and of G and of K as well except
under anomalous situations. It is clear that R v is a subring of R(G,F):.
Since g^,...,g^ is a basis for R(G,F):, every element of R^ ^ is uniquely determined by the scalars in K. We shall refer to R„ „. as a group ring of G,JK. G over K.
2. Matrix Representations and Group Matrices.
A ^matrix representation of degree n of G is a homomorphism of G into the full linear group L^(F):, the n X n honsingular matrices over' F.
We introduce the left regular representation of G as follows. If g e G, then n
gS = 1 1 n a i I ijCs)Sj > < < (l) • j=l - 2 -
where each a. .(g) 1S 0 or 1. Let
L(g),' - (a..(g).) , (2):
the prime denoting transpose. L(g). is a permutation matrix. Moreover, h(hg) s L(h)L(g)., for h,g in G, as the following computation shows. Pre- multiply eq.. 1 by h to get n
h(gg )• = ) a (g).hg
n n
" Z C I aiJ<«>&jk^h> ) Sk k=I j=l ' = (hg)^ n ; . • - I a±^K • • k=l n
Thus aik(hg> = ^ a(g)-a k(h), hence L(hg).' = L(g)/L(h).', and so
j=l
L(hg> = L(l:)L(g). • . -
If L(g> = In, then a (g). = 0, if i 4 j, and a_(g). = 1, if
i = j; and so, ggi = ^ hence g is the identity. Thus
Lemma 1. G is isomorphic to the group of permutation matrices L(g):,
g in G, where L(g). is defined relative to the ordering g^,...}g^ of the elements of G. VI We shall call L(g)- the left regular, matrix representation of G
(relative to a particular ordering of the elements of G):. We may
extend L(g): to a representation of the group ring R '. for every G,i\.
n
k=l n
a L(s ): L(.u): = £ k k * ^ k=l
This gives us, by Lemma 1 and the rule for multiplication in R_, „, G,Ji
Lemma 2. For elements u,v in R„ _ and a and b in F . ' G,F
L(uv). = L(U)L(V>,
L(au+bv): = aL(u) + bL(v): .
For each g in G the right representation of G is given by
n
sis = X bij(g)gj' i = l,...,n. (k) 3=1
and this corresponds to the mapping
• R:g ->R(g> = (h, ,(g):):i 1 < i, j < n,
of G onto n distinct permutation matrices of degree n. Eq_. k implies
that G is isomorphic to the matrices R(g):, g in G; they form the right
regular matrix representation of G. Theorem, 1. Any linear combination of the matrices of the left regular matrix representation commutes with any linear combination of the matrices of the right regular matrix representation.
Proof. By eq.'s 1 and k- respectively we have for elements g and h in G
/ / / (gg1,...,ggn> =-L (g>(g1,...,gn> (5>
Post-multiplication of eq. 5 by h gives us
/ / / (gg1h,...-,ggnh): = L (g>(g1h,...,gnh> .
/ -> L'(g>R(hXg1,..t,gn>
where the latter result follows from eq. 6. • Premultiplying this by g and using eq. 5 produces
/ 1 1 / C&jh,..., gnh}' = L (g):R(h):(g~ g1,...,g" gn): '
1 / = L'(g)R(h)L'(g- ):(g1,...,gn) .
Comparing this with eq. 6 we get R(h): = l/(g)R(h)L'(g "'"):. Since
LCg^Cg"1): = I = L(g):L(g):% we get L(g)R(h> = R(h}L(g)., as required.
Any linear combination of the left regular matrix representation of G over K is called a group matrix of G over K. This of course - 5 -
presupposes.an ordering of the elements of G. Consider the permutation matrix L(g): in eq.. 2 in view, of eq. 1. We have a one at the (j,i) position'of L(g):''precisely when gg. = g., .hence precisely when
g = gtg- . Thus a one appears at the (±,j)' position of L(g): precisely
1 when g' - g.g. . Thus in L(u)' = / a L(.gn )., we have a appear-
1 J S k h r k ' Sk ln G • , . - ." gk
ing exactly at those'positions (i>j)- for which g, = g.g. In other words, a group matrix of G relative to g-^,...,g^ of G is of the form
IW- =' (a _i>, 1 •< 1, j < n . (7>
1 J
Theorem .2. Any matrix over' F which commutes with all matrices of the right regular matrix representation of G is a group matrix of G; that is,
it is a linear combination of the matrices of the- left .regular matrix representation of G.
Proof. Let C = (c.'.);, 1 < i, j < n, c. . in K, be such that
/ C = R(gk)CR(gk> , k = l,...,n, where R(gk> = (b (g^):): Has defined in
eq.'. k is the right regular matrix representation of G. Let u-. be the n-tuple row vector in which a one occurs in column j and O's elsewhere.
Then for fixed i,j, 1 < i, j < n, and each k, we have, - 6 -
c. . = u.Cu' = u.R(g. )CR(g, )/u.1 •
n
s,t=l
This sum may be simplified. For, by eq.. k, ^^(s^) ~ ^
-1 ' -1 * -1 where s,t are such that g. g = g = g. g . Thus g. = g g IS J£ J "G 1 SiC
1 _1 1 1 Hence bv and g = gkgt ^ so that g^" = ^B^ - > - eq. 7; C is a group matrix.
Since the matrices L(g) fc-rm a group isomorphic to GV, and since the matrices are also linearly independent over F, we have Theorem 3-
to '
Theorem 3- R(G,F): is isomorphic4the algebra over F generated by the
in J^Cg); g G. Rn „ is isomorphic to the ring generated over K by the
L(g}, g in G.
Corollary 1. The inverse and the transpose of a group matrix is a group matrix.
Proof. The inverse of any matrix is a polynomial in that matrix. Hence by. Theorem 3 the inverse of a group matrix is a group matrix.
Since L(g "^): = L(g),/, g in G, the transpose of a linear combination
of L(g )',..'. ,L(gn); over F is again a linear combination of L(g^ )•,... ,L(g^): although in a different order. This proves that the transpose of a group matrix is a group matrix. - 7 -
3. Units and Unimodular Group Matrices.
Elements u and v in R satisfying uv = 1 are called left and
right units of R& K, respectively. An element which is both a left
and right unit of R v 'is called a unit of R^ v. Any square matrix
defined over K. is said to be unimodular if its determinant is a unit the
in K.- Given the elements u,v above, Lemma 2 and*definition given, in
eq.. 3 imply;s L(U)'L(V). = (,L(UV) = L(l). = 1^. Therefore,L(u) is
unimodular. Conversely, let L(u). be unimodular over K. Then L(u):
exists and by Corollary 1 it is a group matrix with elements in F.
In fact, L(u): ^ has elements in K since, any element of L(u). ^ is of
the form S(det L(u)): in K where S is a cofactor of L(u) and
(det L(u):). is in' K. Thus an element v in R v exists such that
L(u) = L(v),,-L(uv): = L(u)L(v) = I n ; and so, by Theorem 3 uv = !• This proves "Theorem 4'A.
Theorem k. . An element is a left unit of R_, „ if and only if the ——— • • G,i>-
corresponding group matrix is unimodular.
Corollary 2. Every left.(right) unit is a unit.
Proof. L(u)L(v). « I - L(v)L(u)..
Theorem 5- The set of all units of R v under multiplication forms a — G,iv
group isomorphic"to the multiplication group of all unimodular group . .
matrices of G. over K.' - 8 -
k. Circulants and Skew Circulants.
When G is a cyclic group with an element g of order n, the group ' n-l matrix of G over K relative to the elements.l,g,...,g is called a
circulant over K. Let g. = g1 \. i = 1,. .. ,n. Then g. g . =
i-i-(j-i) _ i-j Thus, C -I = C . . and so the elements of the . ^K, g
group matrix are constant along each diagonal parallel to the main
diagonal. " n Let P be the companion matrix of the polynomial x -1. Then
Pn = I and •n '
i+l
0 . 0 i :o ..OV
0 .. 0. o .. 0 11 (9) n-i+1 1 00. . • o 0 ..0 0 0 . . . 1 0 •Ml
where 1< i < n - 1. It follows that any circulant C is a polynomial
in P. Moreover, Theorem 2 in the special case of circulants becomes
Lemma 3. The matrices of the left and right regular representations
of G-relative to the elements l,g,...,gn-1 are circulants. Any matrix
'commuting with P is a circulant..
If the first row of the circulant C is given by (c^,...,c ) we write
.C=. [c (10) n n - 9 -
for "brevity. Let the conjugate transpose of a matrix A be denoted •x- by A . .
Theorem 5. Let C = [6.,...,c ] be a circulant of order n defined 1 > n n over the complex number field. Let T = ,n"1//'2(p^i~1^^~1^), i < i, j < n where p, is a primitive nth root of unity. Then
T*CT = aiag(e1,"...,eri> (ll) where the eigenvalues, e^,..of C are given by the vector matrix equation'
1 2 / - . (e^...^)/ = n / T(c1,...,cn). . (12>
Proof. Since xn-l = (x-l)g(x) where g(x). = 1 + x + ••• + x11 \ g(pk) = 0, if n does not divide k. Thus T is unitary; that is, T T = I . For, . . -x- the (,j,i; term of T T is given by
n n n"1 I f(k-l>(j-l)p(k-l>(i-l). = n"1 I pC^Ki-j) k=l k=l
The RHS equals 1, if i = j and equals g(p'L J )• = 0, if i ^ j.
Now, since P is the companion matrix of polynomial xn-l, the eigenvalues of P are the roots of xn-l; namely, l,p, p2,...,pn \ Thus if \. equals the jth column of T we get 3 -10-
P, « n-V^pO-l p2(j-l>;...)p(n-l):(5-l). 1}
so that, the jth column of T is an eigenvector corresponding to the
i-1 * r n-l\ eigenvalue p" of P, j = l,...,n. Thus, T PT = diag(l,p,..., p ):, n
Consequently C = ^ c .P'-' implies
* (n-l)(j-l)v T CT = 1 J • > P >• 3-1
Therefore-,- if- we set
n. (i-l):(j-l). - Jo. (13)-
we get eq. 11 and 12 as.desired;
The polynomials over K in the n X n matrix
' 010 ... 0
P = ;. 'o 05 1 -1 0 .... 0
are called skew circulants of degree n over K. Skew circulants are not group matrices because in any group matrix the elements In row i are permutations of the elements in row .one, 1 < i < n. However, the powers -11-
of-P constitute a matrix representation for the cyclic group of order 2n. .
Since P_ is the companion matrix of the polynomial f (x) = x11 + 1, its eigenvalues are p,p3,...,p2n-1 where p is.the 2nth primitive root 2n of unity. If h(x) = x -1, then h(x) = (x-l)(x+l)g(x) where g(x) = 2 2n—2 k\ 1 + x + ••• + x . Therefore g(p ) = 0, if n does not divide k; otherwise g(p^) = n. Therefore, if T = n ^^(p^~^ ), 1 < i,' j < n, the (i,j) element of T$T being n • n
-1 V -(2i-l)(k-l)n(k-l)(2j-l) = ^-1 p2(k-l)(j-i) n-l ^ p(2i-l)(k-l)p(k-l)(2j-l) = n-i VY k=l • K=:.
-1 / j-i\ implies T T = 1^. Moreover, the product of P_ and A... .the jth column of T, yields
2 Pju =n-V?(p a-l,-p?(2J-l)>...,p(n-l)(2J-l)).ir
J
n n since p = -1 implies p (2j-l) _ (_]_ )^j 1 _ jn Q-^er worris} the jth column- of T is an eigenvector of P_ corresponding to its eigenvalue p2^"1, j =l;...,n. Therefore, T*PT = diag(p,p3,..:,p2n_1). Theorem 6. If A = , aj^_^ 1S a skew' circulant over K, t)hen . . j=i • . "v
TAT = diag (ei,...,en> ... ;. (l-5>
7 where (e^...,en) = n^^T'Ca^, ...,a )' .' • ' ,(l6).
5. • Existence of Nontrivial Unimodular Integral Circulants and
Skew Circulants. , • ..' v , .'.'-••
A -unimodular integral (skew), circulant- is 'called trivial, if all
elements, in any row are zero except for a single +.lj otherwise, it is
called nontrivial. We know- trivial unimodular' (skew); circulants. always
exist: see (eq_. lk) eg,. 9»'..r "It is shown in [7] that nontrivial unimodular
circulants exist if n 4 2,3,4,6. . '
•• What about nontrivial unimodular integral skew circulants? This
problem- is not settled. However, if A were .such a matrix.then, so would
be the matrix AA'. For, a diagonal element In AA' is the sum of the
squares of the elements in any row of A; and so, off diagonal elements must
occur in AA' since it is unimoduiar. Therefore, the.solution is in the
answer to another question: . For which values of n do nontrivial unimodular
skew circulants' exist when they are positive definite? This question will
.be taken up in the sequel for values of n < 7-
6. . A New Proof of a- Theorem on Positive Definite Circulants and •
, Skew Circulants." • .
In this section G is always a cyclic group of order-n and all n X n -13-
matrices are assumed to be integral and unimodular. An n. X n matrix is called a generalized permutation matrix if it has exactly one non zero element, +1 or -1, occuring in each row and column.
Theorem 7« . If G is a cyclic group of order n and A'A is a unimodular integral group matrix of G, where A is an n X n matrix of rational integers, then A = QG where Q is a generalized permutation matrix and
C is a unimodular group matrix of G.
The proof proceeds by way of Lemmas. For n > 1, let [0,1,0,...,6J-
denote the matrix in eq. lk.
Lemma k. Let P and A be n X n unimodular matrices of rational integers such that
P'A'AP = A'A. (17):
Then a generalized permutation matrix R exists such that
_1 RAPA R' = diag(Pftl,...,Pfis): (18):
where n = nn + ••• + n and for each i=l,...,s,.P isn.Xn.
1 s ' ' ' n. 11 1 and is a one rowed submatrix of the form (l). or (-l) if n^ = 1, or if n. > 1, of the form [0,1,0,... ,0] or [0,1,0,...,0]- 1 1 —1 Proof. The matrix B = APA is orthogonal since (APA~ )'APA = I : it is also an integral matrix since A is unimodular. Therefore,
B = (b. .)-, 1 < i, j < n is a generalized permutation matrix. -14-
Let T be a linear transformation of an n-dimensional space
R whose matrix is B relative to a basis ee' of R. Then, n l' ' n . '
where tr is a suitable permutation on l,...,n and b^ = + 1. Let
TT = (j(i>j(2)....j(r1>.Kj(r1+l>j(r1+2>...J(r2>).''-
•••(j(rs_1+l>j(rB_1+2)>..-j(rs_):> (20).
be a decomposition into&s disjoint cyclic products of lengths, say,
nn,...,n respectively where r, = 0, n. - r. - r. n, i = 1,...,s and 1' ' s .0 ' l l .1.-1 r m. n and where j(l)., ..., j(n): is a permutation of 1, ...,n with j(l). = 1.
This gives us another basis of R : to n
'(fl'"->fn* = K'E3(2):'"*'ej(n)): (21):
/ . - S(e1,...,en). (22): where S is some permutation matrix. Moreover by eq. 21, 19 and 20 consecutively for k = 1,...,s we get
when n = 1} and when n, > 1, k K
b r ' T(t±}- = f < 1 •< + j(i>, j(i+i) i+i- Vi k-i V -15-
with
In other words except for change in signs T. permutes
f ?-y ' +V r. 4.o>'">£ cyclically. In matrix notation this amounts k-1 rk-l rk to
(23)
where H = diag (B^. ..,Bg); is a direct sum of ^ X n^. matrices B whose typical form is the following: B = (+l). when n = 1 and .k .' k when n. > 1,
0 b.
0 V1 b 0 0
where of course the b.'s are equal to + 1. . Let Z = diag (l,bn,b,b ....,b ...b ). i ."~ ' 1' 1 s7 ' 1 n^/
Then, since b. = + 1 we get -16-
0 10
ZB,-Z's (if n^. > 1>. k
0 K...b 0 1 "k \
Thus we may construct a matrix W, a direct sum of s blocks analogous in form to Z, such that
WHW"' = ¥ aiag (B ,...,B ).W' 1 s
= diag (P ,...,P ) (2k). v n, n 1 s where P are the matrices defined in eq. 1. But n. l
/ (T(fx >,..., T(fn »' = S(T(e;L >,..., T(en»
« SB(ei,.../en):' = SBS'^,...,^).' as a result of eq.'s 22, 19, and 22 respectively. Comparing this to eq. 23 we get H = SBS'. Therefore, by eq. 2k, WSBS'W' = diag (P ,...,P > nl ns and the lemma is proved since R = FWS is a generalized permutation matrix and B =. APA-1. . v
If, in Lemma k: l);. the matrix A satisfies the hypothesis in Theorem 7
2) P is a matrix of the left regular matrix representation of G where n is' the order of P; i.e. Pn = I j' 3)- the right hand sides of eq. 18 equals
[0,1,0,...,0] , then Theorem 7 is true. For, let L(h):, h in G, be the left -ir•
regular matrix representation which define the group matrix A'A relative, say, to the ordering g^,...,g of the elements of G; let
L (h), h in G he the left regular matrie representation' relative n-l to 1, g, . ,.,g where g in G is' of order n. Then,
Lo(g) = [0,1,0,...,0]^ and L(h> = SLo(h)S', (25). h in G, where S is a permutation? matrix, such that (g^,...,g )' =
S(l,g,...,g ):'. But/conditions 2). and 3): imply RAL(g)A R" =
[0,1,0,...,0]; whence, by eq.'s 25 SRAL(g)A_1R^S^ • L(g). and so,
SRA commutes with L(h)j h in G. From Theorem .2, observing that the left and right regular matrix.representations are identical when G is abelian, we infer that SRA is group matrix C of G relative to g^,...,g .
Put Q = (SR).'. This proves Theorem 7 given assumptions l), 2) and 3) above which are justified as the next lemma.,,shows.
Lemma 5. Let G,A,n be definedras in Theorem 7- Let' A'A be a group matrix, a linear combination of the left regular representation matrices
L(h), h in G, of G. Then a generalized permutation matrix R exists such that
RAi(g)A~"*"RT = [0,1,0,...,0] (26): where g in G is of order n.
Proof. Since G is abelian Lemma 1 and 2 imply A'A and L(g): commute.
Therefore, in particular. P'A'AP = A'A, where P = L(g). This permits us to use eq. 18 in Lemma \; that is, condition l) is satisfied. Also -18-
note'that Pn - I and Pr 4 I if r .< n. • • n n
To show condition 3). holds, let D = RA in eq. 18. Then by taking
sums of powers from 1 to n and .noting that P1 = L(g1), i = 1, ...,n and n—1
L(l) + L(g) + ••• + L(g ); =• [1,1,.. .,1] we 1-get the similarity relation
_1 D[l,...,l]n D = diag (Blt...,B•)• (27) n where B. = ) P J, i = 1, ...,s. Using eq.. 18, again. Pn = I implies 1 i_i n. n >1 1
P = I , so that, for some integer m., n = n.m.. In fact, when P is n. n.. 1' 1 1 n. 11 ' ' 1 a skew circulant, 2n^ is the smallest positive integer such that
n • 2n • . 1 P 2ni = I so that, if m. = 2q., then B. = ) P J = q_. )P ^ = 0. n. n. ' 1 ^x' x L n. ^x. L n. x x . n x . ., x J=l J=l .. n n± When P is a circulant, B. = ) P .J = m. ) P ^ = m. [1,... ,1] n. ' x L n. x L n. x ' ' n x . T x . n x i
Thus rank B. is 1 or 0 according as P is a circulant or a skew circulant. x n.
x
However, the rank of the left side of eq. .27 is 1 so that on the right side one and only one non zero component exists; say, m [l,...,l] k "k arising from a circulant P . Thus n divides each element on the right V\ k side. But D is a unimodular matrix of rational integers and so m^ divides
each.element of D ^~ diag (B-^, ... ,Bg)D, hence m^. divides 1. Therefore m^. = 1, - n and eq. 27 implies RAPA-^" = P = [0,1,0,... ,0] . "k -19-
Corollary 2. If A is a unimodular integral matrix and A'A is a circulant,
then
A = QC (28). where Q is a generalized permutation matrix and C is a unimodular integral circulant.
Theorem 8. If A is a unimodular integral matrix and A'A is a skew circulant, then A = QC where Q is a generalized permutation matrix and
C is a unimodular integral skew circulant.
Proof. Let P = [0,1,0,...,0]- . Then, since A (A is by definition a linear combination of powers of P, P'A'AP = A'A. . Thus, eq. 18 in.Lemma k can be used. We shall show s = 1 and therefore P = P and this would n 1 establish Theorem 8 since a matrix which commutes with a nonderogatory matrix is a polynomial in it.
Observe that, if P is a skew circulant, then by adding the first n column of the matrix sum / -1 I-'... 1 ^
P + P + + P ni = n. n'. n. i i i 1 -1 .. -11
to every other column we get a triangular matrix with -1 as the first diagonal element and -2 for the remaining diagonal elements. . Thus
ni
P J ni ni det Y n = (-l) 2 (29>
J=l -20-
Also, ^ P^ ^ = 0, so that if m is odd
3=1 X mn-i rijL
. 1 1 . , X
Returning to eq.. 18, we see that Pn = -I implies P n = -I , n n. n. x x i = 1,..., s so that each P is a skew circulant and n equals an i odd multiple of n.. Therefore, by eq. 18, again x
n
0 det^P = det Y axag (P ,...,Pn >< J=l j=l n s „ = n det ) P 3 . , L> n. 1=1 . -> x 3=X s nl = n det Y P 3 . , L n. i=l . -, •, i
= (-1).^n_n-; s where the last two equations follow directly from eq.'s 30 and 29 respectively. But eq. 29 also implies above that the left hand side
equals (-l)n2n \ Therefore s = 1 and Theorem 8 is proved.
7. New Results on Group Matrices and Symmetric Circulants.
In what follows, the letters i, u, p, d, s stand for integral, unimodular, positive, definite, symmetric, respectively. With this notational convention Theorem 8(Theorem 7)states that an pdiu (skew) circulant of the form A'A where A is iu equals C'C where C is an iu -21-
»
(skew) circulant. In view of this it would be interesting to note for what values of n are pdiu (skew) circulants of the form C'C where C is an iu (skew); circulant.
So far, very little is known about this for circulants of degree n > 13. In an unpublished work E.C.Dadei has shown it to be true for circulants of prime order less than 100 > with one exception; in [6] it is shown to be false for n = 5 where equations 11 and 12 are used to demonstrate that the pdiu circulant [2,l,0,-l,-l,-l,0,l]g is not of the form C'C where C is an iu circulant. A result of Minkowski in [5] settles the question, in general, for n < 7; that is, if A is a pdiu n X n matrix, then A = B'B .where B is an iu n X n matrix, n < 7. A study in [7] on the uniqueness of the normal, basis for normal cyclic fields produced the result that all ui. circulants are trivial for n = 2,3,4,6. This of course is consistent with Minkowski's result.
Also for n = 5, an incomplete proof appears in [11] with corrections in
[l]. Recently in a paper presently.in press [12] R.C.Thompson solved the question for all values of n up to 13 inclusive by considering a • more general problem which we shall define in section 8.
As for skew circulants nothing has been written'on them. In fact
I am indebted to Dr. R.C.Thompson for his conjectures on skew circulants, especially for proposing Theorem 8, the parallel to Corollary 2, and the question of the existence of nontrivial pdiu' skew circulants. We shall discuss several cases in the next section.
Instead, we consider.whether every nontrivial «i circulant is of the form P^C where 1 < k < n, P = [0,1,0,... ,0] and C is a pdiu -22-
circulant; and additionally, if P^C is symmetric, then either k = 1 or n = 2k. This is only a conjecture on my part. However, in consonance with it the following facts are obtained. Let G be a group of order n.
Let (c^,...,cn) be the first row of a group matrix C of G defined over the ring of rational integers. Then, without ambiguity we may write
C = [c^,... J^-IQ*
Lemma 6. Let C = [c_....,c ] be a symmetric real nonsingular group —— 1 n G matrix with principal idempotent decomposition
C = s^+.v. + stEt (31) and let e. denote the row sum of the first row of E.. Then, for i i .
1) . E^ is a symmetric real group matrix;
2) . the diagonal element of E^ is a positive rational number equal
• to.r.n where r. is the rank of E., the number of eigenvalues
of C equal to s^; . •.
3) .! e. = e. and e.e. « 0, if i i j: xx x j 1
V) if eigenvalue sn = cn + ••• + c , then e. = 0 for j i 1 and .e, = 1. ' 1.1 n' a T 1
Note: c^+*«'+c' is always an•• eigenvalue of C.
Proof. (E'.) = E'., i = 1,...,n and E'.E'. = 0, i =}= j. Hence, since
C' = C and the principal idempotent decomposition of C is unique, eq.31 implies E'^ = E^. It is known, e.g. see [8], that for principal idempotent decompositions a matrix which commutes with C commutes with every E^.
Therefore, since by definition, C is a left regular representation, Theorem
1 implies all matrices in the right regular representations commute with the E and so, by Theorem 2, the E^ are group matrices of. G. This proves l); since the E^ are real by definition of the decomposition.
The principal idempotent decomposition requires that E^ are similar to a diagonal matrix of I's and possibly O's. By taking the trace of E^ and the corresponding diagonal matrix and taking cognizance of l), that is, the main diagonal of E^ is constant, 2): follows immediately.-. Let
X = COl (l, . . . be an n-tuple column vector all of whose elements equal 1. Then, since the E^'s are idempotents, 3): follows directly from l) and the fact that
2
E. X = E:X and E.E.x = 0. (Note: For any i,. E.x = (en , ...,en ).' = enx, i.i i. ,j > J ' i v 1' ' 1' 1 ' hence E. x = E. (e'.x).-= • e. (E.x) "'= e.e.x, whence e. = e. . These results 1 XX lv 1 . 1 1 ' 11 are a consequence:of the fact that the row sum of any group matrix is
independent of the row.).
From eq. 31, Cx = S..E x + ••• + s E x so, c.+ • • • + c = s e»+«v+s,e .
By 3); it is possible for only one of the e^'s to be non zero, say e^, whence the preceding equation reduces to
•; c, •+•••+. c =• sn en . 1 •" • , n 11.
But, since C is nonsingular the left side is non zero, so, 4 0' Since 2 •e^ is a row sum of the real matrix E^, e = >•0, and this implies = 1. -2k-
Therefore c, + * *• + c = sn. 1 • - n.. . 1 . '••
The next result is an integral circulant analogue of the polar factorization theorem.
Theorem 9. If A is an n X n nonsingular real group matrix then there are unique real matrices H and U such that A = UH where IS H is a pd group matrix and U is an orthogonal group matrix.
Proof. Let C = A'A . be the group matrix in eq. 31 and let
H = y"s1E + ••• + /s E where we note that the eigenvalues s. of C are positive since C is pd. Therefore, by l). in Lemma 6, H' is a real positive definite^ group matrix. Moreover,
A'A = K2 (33) where H is the•only positive definite matrix for which this is true by virtue of the uniqueness of equation 31• In [8] it is shown that
for nonsingular A there are unique -real matrices U and Hq such that U is
orthogonal and Hq is positive definite with A = ^-Q- But this implies
2 2 A'A = Hq = H - which by uniqueness of H in eq. 33, in turn implies,
H = HQJ and therefore U is a group matrix by Corollary 1 and multiplicative closure. Following this, the terms AyH,U in Corollaries 4,5,6 are assumed to be the group matrices in Theorem.9.
Corollary 3. If det A = + 1, then det H = 1 and det U = det A. (A,H,U are real)- 2
Proof. By eq. 33 (det H); = 1, hence det H = + 1, so, it equals + 1 since
H is positive definite. Therefore det UH = det U = det A. -25-
Corollary k. A is normal iff A = HU = UH.
Proof. Consider the commutativity property with regard to the idempotent
decomposition and the equality of ITTJ and UH2..
Corollary 5. .Tf in Theorem 9, A = [ L is an integral unimodular 1 n li
group matrix and U = [u^,...,u^\G and H = [l^,... ,h ]G, then
h = h, + • • • + h >. 1 and un + • • • + u •= a, + • • • + a = + 1. J- n 1 nl.. n —
Proof. Let u = un + ••• + u and a = a. + + a . The equation Ax = UHx 1 n 1 n • where x is an n-tuple column vector all of whose elements equal 1, implies
a = uh. Since A is unimodular and integral its row sum equals + 1; -1 -1 2 x x/ for, xAA ;.x' = naa = In = n. Since U is orthogonal, u = 1, because 2
nu = x'U'Ux = x'l^x = n. Consequently, h « + 1 which perforce equals
+1 since +lHSis positive definite. ' . .
Theorem 10.. If A is a unimodular integral circulant. then there is an
integer s such that PA is symmetric where P = [0,1,0,...,0]^.
Proof. . Let K be the n X n matrix :
K = toy WO j -1 Then KK = I and KPK = P'. Hence KA'AK = (A'A)' = A'A so that AKA' ..... n.,.
is an integral orthogonal matrix,hence a generalized permutation matrix.
In fact, if Q = AKA-1 then
KQ, = KAKA"1 = A 'A-1 so- that KQ'is a circulant, and being trivial implies there is. an integer
k' such that 1 < k < n and KQ = + PK. Thus' by eq.. 3h
K , .. A' =±P A . (35)
But since the row sum of A and A7 are equal,
• A' a PKA. (36):
Suppose n is odd. Let r be an integer such that 2r = 2n-k or
2r = n-k according as k is even or odd. Let•s be the nonnegative integer
' s = 'n-r- , (37)
This means r + k = n+ s or r> k = s according as k is even or odd.
Therefore by eq. 36
R K G P A' = Prt A = P A , '
and so by eq. 37
(PSA)/ « A'PN~S = A'PR = PRA' = PSA, '..
S which proves that' P A is symmetric. '
Mow suppose n is even. Since the trace of AKA = KPK is zero on
the left, it follows that the number of elements in the nontrivial diagonal(s):
of Pk is zero, or what is the same k is even. Hence, letting r = (n-k)/2
and. s.'== n-r, we get from eq. 36 . . .
' PV = PR+KA = PR+N"2RA = PSA,
whence
(PSA)' = A'PN"S = A.'P1" s and so, PA is symmetric. . Corollary 6. If A is a unimodular integral circulant then there is an k ' integer k such that A' = P A (where k is even if n is even).
Theorem 11. Let A he a unimodular- integral circulant. Then the eigenvalues of the symmetric matrix KA are the square roots, of the eigenvalues of the positive definite circulant A'A/.
Proof. Observe that KP1.is obviously symmetric for each i = 1,...,n.
Hence KA is symmetric. Then consider the principal idempotent decomposition of KA and A'A = (KA).'KA; and the proof follows.
8. Positive Definite Skew Circulants.
In this section B^ always denotes an n X n symmetric unimodular integral skew circulant.
• - By definition of Bn>' we may write, for k > 1,
B = [b .h. ,bn, ... ,b_ ,-h, ,-h, -b., ]~ w(38): n • cr 1' 2' ' k' k' k-1' ' In ' if n = 2k + 1, and
B B[ o i ; n vv2'"''V '"V-W "'-^; (39). if n = 2k + 2. Then, by eq. l6 if B^ = A, we get for the Ith eigenvalue of B n -
e. . " a.„(21-l)(j-1>
which, by substitutions'of the a.'s with the b's in eq. 38 or 39, yields for any n > 3, . -28-
(2l i)j (2i iMn 3) L ' --o0*i v ~ -x v ~ " (u>-
Lemma 7. If is the symmetric n X n skew circulant given by eq.'s 38 or 39; where n = 2k+l or ,2k +2 then it's eigenvalues are given by
j=i i = 1,... ,n and
' e. = e . ,, (43) vy. 1 n-i+1 ^
for i = 1,2,...,k+l.
Proof. Eq. 42, of course, follows directly from eq. 4l. • Eq. 43 follows from eq. 39 a^d the fact that the eigenvalues of a symmetric real matrix
n + are all real. For, p ( . ^ ^)' _ p-^ ^i a^ ^ substituting n-i+1 for i in eq. 40 we get,
_ Y (l-2i).( j-l). £n-i+l ~ L V j=l and so, by comparing this with eq. 40, ei= ej_ ~ en i+l ^0r = I>«".?k+1.> whether n is odd or even. This evidently implies
Lemma 8. For n > 3 2 det B = (e) e(n). * + 1 n 1 2 . k — -29-
where.
2 f e,_."k+1, , if n = 2k+2 :(n) =
ek+l - ' if n = 2k+1 V
Given a square matrix A we denote its trace by tr(A):. From eq. 15 where A-B and eq. h-3 we have n
Lemma 9. For n > 3
tr(Bn) = nbQ = 2 £ e. +6 (n> '1=1
(•2e k+l' if n = 2k + 2 where 6(n) =
ek+1,; If n •= 2k + 1
Lemma 10. If n.= 2k+l and An = [a^,..-^a^]" is a unimodular integral skew circulant with eigenvalues defined as' in eq. 15, then
n
ek+l ~ a +1. 0=2
Proof. By eq. ±6, keeping in mind that pn = -1, we get 'k+1 ~ al+ L aj p >2 •n = a1 + 1 • i=2
Therefore, ek+i is a rational integer; similarly with > k+1 eigenvalue of the inverse of A , which as with A is a unimodular n' n integral skew circulant. Therefore, since ev divides 1 r.: .x, e, . = + 1 as desired. •' ' 1 k+1 ' k+1 —
Corollary 7« if- -h - 2k+l, B^ is as in eq.. 38 then
e. . = h + 2 Y (-l)jb = + 1. k+1 O LJ ^ ' j —
Proof. Since B^ hy definition is symmetric the corollary follows directly from Lemma 10.
We now proceed to show for which values of n is B trivial or nontrivial. n Obviously it is trivial for n = 1,2.
Case 3: B^. = I .
Proof. Let = [a,b,-b]^ . By Lemma 7 and 3.3
2 e± = a + b (p-p )
e2 = a - 2b = 1 2 ' and so, since -1 + p - p = 0, = a + b, whence, by Lemma 8 -31-
2 2 det = e1 e2 = (a + b) (a - 2b > = 1.
Therefore, a = 1 + 2b implies a + b= l+3"b = + l; which holds only if b = 0 and a = + 1. Since B is pd, a =. 1.
Case k. B^ - [a,b,0,-b]^ is nontrivial for integral solutions of 2 p _ the equation a - 2b = 1 when b 4 0, a > 1. For example' [3,2,0,-2]!"V .
Proof. By Lemma 7> eq.. k-2, e^^ = a + b (p-p5) 5 9 5 .e2 = a + b (p-p)- = a + b (p-p> ^ 2 and so, since « (p-p ): = 2, by Lemma. 2, det B^. = (e^)2 = (a2 - 2b2)2 = 1
• 2 2 we have a - 2b = + 1 which equals + 1 since 3-^2 ^ ®'
Conversely if.a and-b are solutions such that a > 0, b 4 0. Then 2 2 a > 2b implies a > + /"2b so that a 4-/"2b > 0 and hence e^'e2 *> ^*
Therefore B^ is a pdsiu skew circulant and nontrivial.
Case 5' B5 = ta,b,c,-c,-b]^ is nontrivial iff a,b,c are solutions to
a2 - It-bc = 1. (kk):
(b-c):(l+b-c) = be (U5) where a > 1. For example [3,2,1,-1,-2]^.. ' -32-
Proof. By Lemma 7 and Corollary 7
e1 « a + bX - cX2
eg = a + bX2 - cX-j^ (46)-'
e,.= a - 2b + 2c = 1 3 . 4 3 2 where X-^ = p-p and Xg = p -p and p is the 10th primitive root 2 3 4 of unity. Using the fact that -1+p-p+p-p =0 a straight forward computation will show that ej_e2 ^S an in"ke6er anc^ hence from
Lemma 8 = indeed, 2 2 2 elG2 **a ~C + ab-ac-3bc = 1.
But
2 2 4£le2 + e » 5a -20bc = 5 (47)
- e^Eg = 5'(b +c - ab+ac - bc):.= 0.
The latter equation gives
(b-c)(b-c-a) = -be . which reduces to
(b-c>(l+b-c> = be by eq. 46. . Eq. 47 implies that if B<_ is nontrivial then a > 1.
Conversely if a,b,c are integral solutions to eq.'s 44 and 45 such .
that a > 1, then B,_ is a:nontrivial pd unimodular skew circulant. For,
2 2 . 4ene„ + e, = 5a - 20bc 1 2 3 which by equation 44 equals 5' "'Thus'solving for the integer e-j_e2 we
= ence get , £j.e2 • H Corollary 7,- = + 1 and so by Lemma 8 B,- is -33-
unimodular; moreover, since 5 divides the difference e-.-en e0 = e.,-1, 3 12 3'
=1; which is eq. 46. To show that all the eigenvalues of &g are-positive we note that X-^ and Xg are the-roots of the polynomial ~2 X(x): = x -x-1 which means .
So that
_ , b-e h+c „. e-i - a + o _ o /5
, h-c " h+c - — £9 a + p + o v 5 • •
We must show ' a + ~^ > + ^jp /"5 • ?y eq. 44, since a > 1, be > 0 and so by eq.'' 45 h-c > 0. Hence we only need to show a + < /5
is false when b,c > 0. . By squaring both.sides and transposing terms we get ' " ; 2 2 a + a(b-c) < (b-c) - be .
But the right hand side is negative according to eq. 45. This is a
contradiction. Hence £-^,£g > 0 and so, B,_ is pd.
Case 6. Bg = [a,b,c,0,-c,-b]g is nontrivial iff a,b,c are integral
solutions of the equations
a-2c = 1 . (48>
2 2 (^) -3h = l --(49> where a > 1. For example'[5,4,2,0,-2,-4]g Proof. We have by Lemma 7
:?5 2 H- e = a+b(p-p ). + c(p -p ):
e2 = a-2c
*5 2 ^4-
e5 = a-bCp-p?): + c(p -p );.
But by Lemma-9 and eg. '-4-3 in Lemma 7 .
2 1+ tr(B6) = 6a = 4(a+c(p -p ):> + 2(a-2c)
implies
/ 2 > c = c(p -p ) .
To show c 4'°- Let Hg = [l,0,-l,0,l,Ojg . Then
•"" .B6H6B6_1= ^B6H6>B6_1.
_1 • = (a-2c>H6B6 •• .
Since Bg is an integral matrix a"~2c divides one, and so.,
e2 = a-2c =1
2 4
since Bg is pd. Therefore, if c = 0, Bg = Ig. It follows that p -p and so, 2 2 = (a+c): - 3h . {I
By Lemma '8, e1 e . = 1. • 1 3
Conversely, suppose a,b,c satisfy equations 48 and 4-9 such that a > 1. Then a-2c = 1-implies c ^ 0 so that -35-
0 -0 +1 = 0. (51):
Therefore,, ET_E^ equals one by eq. 49 so that by Lemma 8, Bg is
unimodular. - i+/3 A solution to eq.- 51 is p = • y which is a 12th primitive root 5 i-/3 of unity. Thus p = *^ implies
= a+/3b + c
= a-/3h + c. 3
By eq. 48, ,c > 0 and hence by eq. 50 it follows that
a + c > + /3b
so that £-j_>£3 ^ 0. This proves that Bg is pd.
Case 7. For A^ = [a,b,c,d,-d,-c,-d]~ thig drily facts know.fiicar*: ^
^ = a + bTa - c?2 + d?
e2 = a + bTl^ - cTl1 + dT|2
e = a + bTl2 - c71 +• dT^
6 '52 •34''' where "0^= P~P > ^2== ^ ''^3 = ^ ' are sbuttons of the equation 3-2 x^ - -x - 2x + 1
and Tl^ - T|2 + T) = 1 when p is the 14th primitive root of unity.
solution of the cubic equation is
"i + (| /7 ) cos Q cos"1 ( ^ )) . -36-
Result: ¥e have shown that [3,2,0,-2]^, [3,2,1,-1,-2] , and
[5,4,2,0,-2,-4]^ are positive definite 'unimodular skew circulants.
The diagonal elements 3,3,5 in these matrices are minimal for this class of nontrivial matrices. Hence it is impossible for these' matrices to be of the form C'C where C is a nontrivial unimodular positive definite integral skew circulant since the diagonal elements of C'C would otherwise exceed 3,3,5-
9. Appendix.
Let C = ( m n ) where'm and n are integers. Then m + n and m - n v n m y are square integers if and only if there is a unique matrix A of rational integers of the form '' J such that CV= A'A. This comes as direct consequence of the fact: The relatively prime solutions of the equation 2 2 2 * 2 2 . 2 -2 x + y . = z . with y . even are x = r - s , y = 2rs, z = r + s , where r > s > .0, (r,s) = !..'['
The above proposition can be violated, if the conditions on m and n are relaxed. For example r.65 60 V _Y8E:J\ S4>?\ V60 65 J " V77V Vi'
The case for 2X2 skew circulants turns out to,be trivial. -37-
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