Applied Mathematics E-Notes, 14(2014), 29-36 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Some Remarks On Block Group Circulant Matrices
Pamini Thangarajahy, Petr Zizler
z
Received 29 November 2013
Abstract
Let C denote a block group circulant matrix over a …nite non-Abelian group
G. We prove results concerning the spectral properties of the matrix C. We give
an example of the spectral decomposition of a block group circulant matrix over
the symmetric group S3 .
1
Introduction
Block circulant matrices over the cyclic group Zn have been well studied, see [11]
for example. In our paper we will consider the setting where the cyclic group Zn is
replaced by a non-Abelian …nite group G. Some of the framework needed for the block
group circulant case needs to be taken from the group circulant matrix case that was
discussed in [13]. Let l2 (G) denote the …nite-dimensional Hilbert space of all complexvalued functions, with the usual inner product, for which the elements of G form the
(standard) basis. We assume that this basis (G) is ordered and make the natural
identi…cation with Cn , where jGj = n, as a linear space.
Let C[G] be the group algebra of complex-valued functions on G. Consider =
(c0 ; c1 ; : : : ; cn 1 ) 2 Cn and identify the function with its symbol = c0 1 + c1 g1 +
cn 1 gn 1 2 C[G].
b be the set of all (equivalence classes) of irreducible represenDEFINITION. Let G
b Let 2 G
b denote an
tations of the group G and let r denote the cardinality of G.
n
irreducible representation of G of degree j and let 2 C . Then the Fourier transform
of at is the j j matrix
X
b( ) =
(s) (s 1 ):
s2G
Let
and be two elements in Cn . A G-convolution of
and
following action
X
(
)( ) =
( ) ( 1 ) for 2 G:
is de…ned by the
2G
Mathematics Subject Classi…cations: 15 A57, 42C99, 43A40.
of Math/Phys/Eng, Mount Royal University, Calgary, Alberta, Canada
z Department of Math/Phys/Eng, Mount Royal University, Calgary, Alberta, Canada
y Department
29
30
Some Remarks on Block Group Circulant Matrices
We have a natural identi…cation
7!
understood with respect to the induced
group algebra multiplication. Moreover, the Fourier transform turns convolution into
(matrix) multiplication [ = b b. Thus we have a non-Abelian version of the classical
z transform. For further references on this subject we refer the reader to [1, 2, 3, 4, 5,
6, 7, 9, 10, 12].
b where
The Fourier transform gives us a natural isomorphism C[G] ) M (G)
b = Md
M (G)
1
d1 (C)
Md2
d2 (C)
Mdr
dr (C)
with d21 + d22 +
+ d2r = n. A typical element of Cn is a complex-valued function
b is the direct sum of Fourier
= (c0 ; c1 ; : : : ; cn 1 ) and the typical element of M (G)
transforms
b( ) b( )
b( ):
2
r
1
Cyclic circulant matrices are normal (hence diagonalizable) and the Fourier basis
of eigenvectors, the complex exponentials, are …xed and independent of the function .
In the Abelian setting the Fourier transform is a unitary linear transformation (proper
scaling required). In the non-Abelian setting we recapture this property if we de…ne
b Let 2 Cn and de…ne a function j by
the right inner product on the space M (G).
the following action
j (s)
=
dj
tr
jGj
j (s)
b( )
j
for s 2 G:
Pr
Note = j=1 j which constitutes the inverse Fourier transform. We are able to
decompose a function into a sum of r functions which is the number of conjugacy
classes of G.
DEFINITION. Let A = (ai;j ) be a m
by
jjAjj2F =
If we let
j
n matrix. The Frobenius norm of A is given
m X
n
X
i=1 j=1
jai;j j2 :
be given as above, then we have
h i;
and if we let
2 Cn then
ji
=
dj b
jj ( j )jj2F
jGj
ij :
r
1 X
jj jj =
dj jjb( j )jj2F :
jGj j=1
2
Thus, with proper scaling, the Fourier transform is a unitary transformation from Cn
b F .
onto M (G);
In the case of a group circulant matrix C = CG ( ) over a non-Abelian group G its
eigenvectors need not be orthogonal nor are independent in general. Moreover, the
P. Thangarajah and P. Zizler
31
matrix CG ( ) need not be diagonalizable, an example was given in [13] with G = D4 ,
the dihedral group of order 4. The group D4 is a semi-direct product of the cyclic group
Z4 and the cyclic group Z2 . Let Zn =< r > and Z2 =< s >. We have rn = s2 = 1 and
rj s = sr j for all j 2 f0; 1; : : : ; n 1g. The matrix C corresponding to the convolution
operator induced by the symbol = r + rs is not diagonalizable.
Eigenvalue analysis for group circulant matrices was studied in [8] and the eigenvector decomposition in [13]. In the Fourier domain the eigenvalue problem for a group
circulant matrix translates to AB = B where is an eigenvalue of A = b ( j ) and
the columns of B are the corresponding eigenvectors (any collection including the zero
vector).
Assume the matrix b ( j ) is diagonalizable for each j with dj eigenvalues (possibly
b ( ) = f 1;j ; : : : ; d ;j g. Consider an (unital) eigencounting multiplicities). Let
j
j
vector v s;j of b ( j ) corresponding to the eigenvalue s;j with s 2 f1; : : : ; dj g. In the
case of a multiple eigenvalue we choose linearly-independent (preferably orthogonal)
unital eigenvectors. To obtain the eigenvector decompositon of C we review some of
the developments in [13].
b
De…ne a sequence of Fourier (orthogonal) eigenvectors in M (G)
bp (
v
s;j )
= (0)d1
0
0
v
s;j
0
0
d1
(0)dr
dr
where the unital eigenvector v s;j is located in the p-th column with p 2 f1; : : : ; dj g.
The orthogonality properties are respected in the space l2 (Cn ) upon taking the
inverse Fourier Trasform which is unitary. Namely, upon taking the inverse Fourier
transform of the vectors fb
vp ( s;j )g we obtain eigenvectors
fvp (
s;j )g
for p; s 2 f1; 2; : : : ; dj g and j 2 f1; 2; : : : ; rg:
For a given s;j the eigenvectors fvp ( s;j ) j p 2 f1; 2; : : : ; dj gg are pairwise mutuallyorthogonal. Moreover vp ( s;i ) ? vq ( t;j ) for i 6= j and any choice of p; q and s; t.
The group circulant matrix CG ( ) admits pairwise mutually-orthogonal,
independent, d2j -dimensional, C-invariant subspaces
Vj
=
spanfvp (
s;j ) j p
2 f1; : : : ; dj g; s 2 f1; : : : ; dj gg
=
spanf j (k; l) j k; l 2 f1; : : : ; dj gg
where j 2 f1; 2; : : : ; rg and the action of the function j (k; l) is seen as j (k; l)(s) =
j (s)(k; l) for s 2 G. This decomposition could be su¢ cient as far as the response
of G-convolution by on functions in Cn is concerned. The functions f j (k; l)g are
the generalizations of the complex exponentials (cyclic case) as they are mutuallyorthogonal though not necessarily eigenvectors. The values fjj b ( j )jjg could act as
frequency responses.
2
Main Results: Block Group Circulant Matrices
De…ne a vector space Ck [G] consisting of elements
v = v0 1 + v1 g1 +
vn
1 gn 1
32
Some Remarks on Block Group Circulant Matrices
where vi = (v1i ; v1i ; : : : ; vki )T 2 Ck . Note that Ck [G] is not an algebra as we do not
have multiplication de…ned. We can identify the element v with
(v10 g1 +
+ v1n
1 gn 1 )
(vk0 g1 +
+ vkn
1 gn 1 )
so that v = v1
vk with vs 2 C[G]. Each vs results from collecting the same
entries in v, ranging from 1 to k. Thus Ck [G] can be identi…ed with k copies of C[G].
We refer to this as the block stacking (with respect to entry position). Undoing this
operation is referred to as block merging. To give an example we let G = Z2 with the
elements fg0 ; g1 g where g0 is the identity element and g12 = g0 . Consider
v=
1
2
g0 +
3
4
g1 = (1 ; 2 ; 3 ; 4)T :
Then v0 = g0 + 3g1 and v1 = 2g0 + 4g1 . Now de…ne a group algebra Ck k [G] over the
group G with coe¢ cients k k matrices over the complex numbers. The group algebra
Ck k [G] consists of elements
= c0 1 + c1 g1 +
cn
1 gn 1
where ci are k k matrices over the complex numbers. The element can be identi…ed
with a k k matrix [ ts ]kt;s=1 where the entry ts is an element of the group algebra
C[G]. The matrix is obtained by collecting likewise entries in the symbol similar
to the vector block stacking.
Let 2 Ck k [G] and v 2 Ck [G]. The kn kn block group circulant matrix C is
induced by the following action
w= v
where v 2 Ck [G], w 2 Ck [G] and
before. Consider
0
1
=@ 3
2 Ck
1
k
[G]. To give an example let G = Z2 as
1
0
5 6
2
4 A g0 + @ 7 8 A g1 :
Then 1;1 = g0 + 5g1 , 1;2 = 2g0 + 6g1 , 2;1 = 3g0 + 7g1 and 2;2 = 4g0 + 8g1 . Recall
f j grj=1 denote the irreducible representations of G. Let b ts (j) be the Fourier transform
(dj dj matrix) of ts evaluated at j , where j is the irreducible representation of
bi (j) is the Fourier transform (dj dj matrix) of vi evaluated
the group G. Similarly, v
at j .
The action of the block group circulant matrix C can now be lifted to the Fourier
domain and can be seen as the following action
0
1
b (j) b (j)
b (j)
0
1
11
12
1k
b1 (j)
v
B b
b (j) C
B 21 (j) b 22 (j)
C
2k
B v
C
B
B b2 (j) C
r
..
..
..
.. C
B
C
b:
B
C = rj=1 b j v
.
j=1 B
.
.
.
. C@
..
A
B
C
b (j) A
@ b k1 (j) b k2 (j)
kk
bk (j)
v
P. Thangarajah and P. Zizler
33
We will assume the matrix C is diagonalizable. This assumption is made for simplicity
reasons namely a notational one, as an extension to non-diagonalizable case can be
readily accomplished.
THEOREM 1. Let C be a diagonalizable block group circulant matrix over a …nite
non-Abelian group G. Then the eigenvalues of C are the eigenvalues f m;j g, each with
multiplicity dj , with m 2 f1; 2 : : : ; kdj g and j 2 f1; 2; : : : ; rg, of the matrices f j g.
Let m;j be given. Then we have dj corresponding (linearly-independent though not
necessarily orthogonal) eigenvectors up ( m;j ) for p 2 f1; 2; : : : ; dj g. These eigenvectors
have the following properties. Let p be given. Perform the block stacking of up ( m;j ).
Then the Fourier transform of each block usp ( m;j ), s 2 f1; 2; : : : ; kg is given by
cs p (
u
m;j )
= (0)d1
0
0
d1
us m;j
0
0
(0)dr
dr
where us m;j is the sth block (top to bottom) of the eigenvector u m;j of the matrix
s
th
column of the dj dj matrix
j with eigenvalue m;j . The vector u m;j is in the p
above.
PROOF. It is clear from the preceding discussion that the eigenvalues of the matrix
C are the eigenvalues of the matrices f j grj=1 counting multiplicities. We list these as
f m;j g with m 2 f1; 2 : : : ; kdj g and j 2 f1; 2; : : : ; rg. The eigenvectors of the matrix
C can be obtained as follows. Let j be …xed. Let u m;j be an unital eigenvector of the
matrix b j . Split the vector u m;j into k parts (top to bottom) and consider a block
us m;j , a dj 1 vector. De…ne a vector
cs p (
u
m;j )
= (0)d1
0
0
d1
us m;j
0
0
(0)dr
dr
where us m;j is located in the pth column with some …xed choice of p 2 f1; : : : ; dj g
the same for all the k blocks. Let fusp ( m;j )g be the vectors in Cn whose Fourier
cs p ( s;j )g for each s 2 f1; 2; : : : ; kg. We
transform is the given Fourier sequence fu
form the eigenvector up ( m;j ) of C for the eigenvalue m;j via block merging using the
blocks fusp ( m;j )g.
For i 6= j we have
up (
m;j )
? uq (
t;i )
for all p; q; m; t:
However, unlike the group circulant case, the eigenvector up ( m;j ) need not be
orthogonal to uq ( m;j ) for p 6= q. The group circulant matrix CG ( ) admits mutuallyorthogonal, independent, kd2j -dimensional, C-invariant subspaces
Uj
=
spanfup (
m;j ) j p
i
j (k; l) j k; l
=
spanf
2 f1; : : : ; dj g; m 2 f1; : : : ; kdj gg
2 f1; : : : ; dj g; i 2 f1; 2; : : : ; kgg
for j 2 f1; 2; : : : ; rg. The function ij (k; l) is created as follows. Consider a function
j (k; l) acting as j (k; l)(g) = j (g)(k; l) for g 2 G. Then choose a block location
34
Some Remarks on Block Group Circulant Matrices
i 2 f1; 2; : : : ; kg and merge j (k; l) from the location i with the k 1 blocks of zeros
of size n 1 to create a vector of size kn 1. Note that the functions ij (k; l) are
mutually-orthogonal though not necessarily eigenvectors.
3
Example
We now consider an example of a block group circulant matrix C over the symmetric
group S3 . The group S3 consists of elements
g0 = (1) ; g1 = (12) ; g2 = (13) ; g3 = (23) ; g4 = (123) ; g5 = (132):
We have three irreducible representations, two of which are one-dimensional, 1 is the
identity map, 2 is the map that assigns the value of 1 if the permutation is even and
the value of 1 if the permutation is odd. Finally, we have 3 de…ned by the following
assignment
0
1
g0 !
7 @ 0
1
0
0
1 A ; g1 7! @
0
1
0
0
g4 !
7 @ 1
1
0
1
1 A ; g2 7! @
1
0
1
1 A ; g5 7! @
1
0
1
1
0 A ; g3 7! @ 1
0
1
1
1
1
0
1 A
1
1
0 A:
Consider the block group circulant matrix induced by the symbol
1
1
0
0
0 1
1 0
= c0 g0 + c1 g1 with c0 = @ 0 2 A and c1 = @ 0 0 A :
The induced block circulant matrix is given by (respecting the order of elements)
0
1
c0 c1
0 0 0 0
B c1 c0
0 0 0 0 C
B
C
B 0 0 c0
0 c1
0 C
B
C
B 0 0 0 c0
0 c1 C
B
C:
B 0 0 c1
C
0
c
0
0
B
C
@ 0 0 0 c1
0 c0 A
The matrices b (j) are given by
0
1
b (1) = @ 0
1
0
1
1
2 A ; b (2) = @ 0
1
0
B
1
B
2 A ; b (3) = B
B
@
1
0
0
0
0
1
0
0
1
1 1
0 1 C
C
2 0 C
C:
0 2 A
P. Thangarajah and P. Zizler
35
The eigenvalues of b (1) are 1;1 = 1 and 2;1 = 2 with u 1;1 = (1; 0)T and u
(1; 1)T . We collect 2 corresponding eigenvectors of C (non-normalized)
u1 (
1;1 )
=
(1; 0; 1; 0; 1; 0; 1; 0; 1; 0; 1; 0)T
u1 (
2;1 )
=
(1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1)T :
2;1
=
Note that the above eigenvectors span the C-invariant subspace U1 . The eigenvalues
of b (2) are 1;2 = 1 and 2;2 = 2 with u 1;2 = (1; 0)T and u 2;2 = (1; 1)T . We collect
2 corresponding eigenvectors of C (non-normalized)
u1 (
1;2 )
=
(1; 0; 1; 0; 1; 0; 1; 0; 1; 0; 1; 0)T
u1 (
2;2 )
=
(1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1)T :
Note that the above eigenvectors span the C-invariant subspace U2 . The eigenvalues
of b (3) are 1;3 = 2;3 = 1 with multiplicity 2, and 3;3 = 4;3 = 2 with multiplicity
2 as well. We have u 1;3 = (1; 0; 0; 0)T , u 2;3 = (0; 1; 0; 0)T , u 3;3 = (1; 0; 1; 0)T
and u 4;3 = (1; 1; 0; 1)T . As a result we collect 8 eigenvectors (non-normalized) of C
corresponding to these eigenvalues
u1 (
1;3 )
=
(1; 0; 1; 0; 0; 0; 1; 0; 0; 0; 1; 0)T
u2 (
1;3 )
=
(0; 0; 0; 0; 1; 0; 1; 0; 1; 0; 1; 0)T
u1 (
2;3 )
=
(0; 0; 1; 0; 1; 0; 0; 0; 1; 0; 1; 0)T
u2 (
2;3 )
=
(1; 0; 1; 0; 0; 0; 1; 0; 1; 0; 0; 0)T
u1 (
3;3 )
=
(1; 1; 1; 1; 0; 0; 1; 1; 0; 0; 1; 1)T
u2 (
3;3 )
=
(0; 0; 0; 0; 1; 1; 1; 1; 1; 1; 1; 1)T
u1 (
4;3 )
=
(1; 0; 0; 1; 1; 1; 1; 0; 1; 1; 0; 1)T
u2 (
4;3 )
=
(1; 1; 1; 1; 1; 0; 0; 1; 0; 1; 1; 0)T :
Note that the above eigenvectors span the C-invariant subspace U3 . We will explain how we obtained u2 ( 4;3 ). Consider 4;3 = 2 and u 4;3 = (1; 1; 0; 1)T , the corc1 2 ( 4;3 ) by positioning (1; 1)T in the second
responding eigenvector of b (3). Form u
c1 2 ( 4;3 ) is
column and zero columns elsewhere. The inverse Fourier transform of u
T
c
2
given by (1; 1; 1; 0; 0; 1). Next, form u 2 ( 4;3 ) by positioning (0; 1) in the second
c2 2 ( 4;3 ) is given by
column and zeros elsewhere. The inverse Fourier transform of u
(1; 1; 0; 1; 1; 0). Now we merge and obtain
u2 (
4;3 )
= (1; 1; 1; 1; 1; 0; 0; 1; 0; 1; 1; 0)T :
Observe that for i 6= j we have up ( m;j ) ? uq ( t;i ) for all choices of p; q; m; t, but
for p 6= q up ( m;j ) need not be orthogonal to uq ( m;j ).
36
Some Remarks on Block Group Circulant Matrices
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