cyclic decolnposition of e-chains -...
TRANSCRIPT
CHAPTER V
Cyclic DeCOlnposition of E-Chains
TEen induces a decomposition of V as a direct sum of certain subspace and
associates with each such decomposition of V a set of polynomial invariants in
K[:2:]. The set of polynomial invariants enable one to choo e various bases of
V relative to each of which the matrix of the given linear transformation is of
a certain type. \\'e may wish to relate the structure of the minimal E-chain
representing TEen and the properties of the minimal polynomial of T. \Y
haye already seen that there are deeper connections between the geometry of
E-chain and E-cycle and the internal structure of T. The nature of the fi ld
K will also have strong influence on the geometry of E-cycles and E-chains
which in turn will influence the structure of the biordered set En(K). Here we
give a decomposition of TEen which also depends on the nature of the field
K.
1. CYCLIC DECOMPOSITION OF E-CHAI S
In this section we follow the notations and terminologies as in [10] and [13].
The reader is supposed to be familiar with terms and theorems on 'rational
canonical forms'.
Let TEen. Then there exists a unique monic polynomial of posili\"(~
degree qT E K[x] such that qT(T) = a and CiT divides f. for every f E f\[x]
76 v Cy ':'" DEC01\IPOSITI N OF E- HAINS
such that J(T) = O. qT is known as the minimal p(lilJnominl of T . .\ ubspac
VI C V is called invariant \\'ith respect to T if VI T C "1. That is :1: E VI impli
xT E VI· T induces a X[x] module str lcture on " as follows. If f E I\[x]
and u E V then f(T) E Hom(V, V) and IlJ is defined by uf = uf(T). An
invariant subspace VI is T-invariant if and only if 'i is a K[x] submodul
of V. In particular, for v E V, the subspace V(T. u) spanned by the set
{vT i : i 2 O} is T-invariant. It is easy 10 see that {'(T. c) is precisely the cyclic
J([x] ubmodule J([x·]v g nerated by v. 1'(T,v) is called T-cycLic subspac .
Suppose that VI is a subspace of I' invariant under T. Then T indu es
a liner transformation TI on VI defined by UTI = uT for every u E iIt. If
1 = IIi C V2 :::: ... 8: Vk where each Vi is im'ariant 11 del' T and if pi(X) i the
minimal polynomial over 1\ of Ti. then Ii ear tran for ation ind uced by T on
, i. hen the minimal polynomial of T O\'f'1' 1\ is the ;easl common multipl of
Pl(X) P2(X), ... ,pdx). On the other hand suppose 1at p(x) in S[x] is th
minima:l polynomial of T O\'e1' ]{. Then \\'e can factor PlX) in K[x] in a uniqu
way as p(x) = PI(X)/lp2(.1:)/2 ... Pk(X)/k \\'h 1'e the pi(X) are distinct irreducible
polynomials in I{[x) and where II, I2, . .. . lk are positive integers. Let
Vi = {v E V : vPi(T)/i = O} for i = 1. 2.... ,k.
Then each Vi is an im'ariant subspace of T and if Ti is 1he linear transformation
induced by T on Vi, then the minimal polynomial of Ti is' Pi(X )/i. :-'Ioreover.
II = VI 9 112 6 ... ED Vk· l\ow each Vi can be decom po ed as
5,
Vi = EB \/~JJ=1
\\'h re \~)'s are cyclic invariant subspaces of lund r T. Thus \\'e have the
following.
V.I CYCLIC DECOMPOSITION OF E-CHAINS 77
THEOREM 1 Let T E 6 n . There exists monic irreducible polynomials
PI,··· ,Pk E K[x] and T-cyclic subspaces Vll , Vl2, ... , VIsll V21""1 V2s 21k Si
V31, ... , Vsks of V such that V = L L Vij and for each i there is a non-i=l j=l
increasing sequence of integers mil 2: mi2 2: ... 2: mis i 2: 0 such that p~ij
is the minimal polynomial of T I ViF Vij ~ Vij. The family of polynomials
{p~ij I 1 ::; i ::; k; 1::; j ::; Si} is uniquely determined by V and T and
p"';l1 p~21 ... p7:k1 is the minimal polynomial of T. 0
The prime power polynomials p~ij are called the elementary divisors of
T. We can find a basis of V in which the matrix of T is the direct sum of the
companion matrices of the elementary divisors p"';ll, ... , p;kSk E K[x] of T.
The matrix of T is said to be in primary rational canonical form.
DEFINITION 1 A matrix T E 6 n is said to be relatively nilpotent ifTT E En
for some integer r 2: 2.
It is easy to see that A is relatively nilpotent if and only if there exists
a nilpotent matrix B and an idempotent e such that A = B + e where eB =Be = O.
Now we show that each T E 6 n can be written as a product of two
matrices of which one is a group element and other relatively nilpotent. For
that we need the following lemma.
LEMMA 2 If T E 6 n is such that T m E He then eT = Te E He where e is
an idempotent.
PROOF By Lemma IV.13 if Tm E He then TP E He for p 2: m. Also
TmT = T m+1 E He. Hence by Greens' Lemma ([1], 2.16) PT I IIe:x ~ xT
78V CYCLIC DECOMPOSITION OF E-CHAI. -
is a bijection of Hym onto itself. Therefore. for e E HTm. cT E HTm = He.
Also TTm = T m+1E He. Hence AT I He is a bijection of HTm onto it lf.
Therefore, for e E Hym. Te E He. Now eT E He implie eTe = eT and
Te E He implies eTe = Te. Hence eT = Te E He. 0
PROPOSITION 3 Let T E 6 n . Then T
subgroup of 6 nand T2 relatively nilpotent.
PROOF Since 6 n is completely semisimple by Lemma IV.13. eyery 1 m nt
of 6 n is groupbound. That means Tm belongs to a subgroup of 6 n for some
m. Suppose that Tm E He where e is an idempotent. \\ can ee that if e is
an idempotent, 1 - e is also an idempotent.
\Ve haye T = eT + (1 - e)T. By the abO\'e lemma eT E H. \Y
will show that (1 - e)T is nilpotent. For that ((1 - e)T)m = (1 - e)mTm
(1 - e)Tm = T m - eTm = O. Also ((1 - e)T)m-1 = (1 - e)m-1Tm- 1
(1 - e)Tm- 1 = T m- I - eTm-1 :j:. O. Hence (1 - e)T is nilpot nt with index
of nilpotence m. eT E He, that means eT belongs to a subgroup of 6 n. .\lso
(1 - e)TeT = (T - eT)eT = TeT - eTeT = TeT - TeT = O. eT(l - e)T =
eT(T - eT) = eT2 - eTeT = eT2 - eT2 = O. Let eT = T{ and (1 - e)T = T~.
Then T = T{ + T~ and T{T~ = T~T{ = O.
Let 1)\(e) = VI and 1Jl(e) = V2. By definition T{ and T~ are defined on
the complementary subspaces VI and V2 of V respectivel). Define T1 and T2
as follows.
if v E VI
if v E V2
if v E h
if v E VI
V.1 CYCLIC DEco:VIP05ITION OF E-CIIAIN5
T1 E He' where e' is an idempotent defined by
79
T2 = T~ + 1Vi' which is relatively nilpotent by our D finition 1. So w g t
T = T1T2 where T1 is a group element and T2 relatively nilpotent. 0
LE:"[1IA 4 T E 6 n is nilpotent iJ and only iJ all the characteristic roots oj
T are zero.
PROOF Suppose T is nilpotent. Then T m = 0 for some m > 1 and Tm-l =!=
O. Let A be a characteristic root of T. Then for some v =!= 0 in V v(T - AI) = O.
That vT = Av. It follows that vTk = /\k v for all k 2: 1. Since Tm = O. vTm = 0
so /\m v = O. Since v =!= 0 we get A = O.
Conversely. suppose that all the characteristic roots of T are zero. If m
1S the multiplicity of 0 as a characteristic root of T then the characl ristic
equation of T is of the form xm = O. That is Tm = O. Hence T is nilpotent.
o
DEFINITION 2 T E 6 n is irreducible iJ
(1) The minimal polynomial oJT is either oj the form p(x) = x(x -l)q(x)m
or p(x) = xq(x)m where q(x) is irreducible over f{ and
(2) The invariant subspace corresponding to the irreducible factor q(x) of
p(x) is cyclic.
We now gi\Oe a decomposition of T as matrices that are irrcducibl and
one which is relatively nilpotent.
o\. yel.l DEC ~lrOSIT10, OF E- 1I.\I:'\S
Let T E 6 Tl and suppose lat p(.t:) = JlOl.I')IOp1(.t:1 11 '" pd.r { bE' th.
minimal pol~'nomial of T over X. where [).( l') l' = 0 1 I.. I' "t' .I' . , .... , t,; ale liS 111 t
irreducible factors of p(x) and li).l l .... If,; are po itive integers. Sine T i
singular one of the irreducible [actors p/x) is of the form x. \IV ma~' assum
that po(x) = .t:.
Define
lti = {v Ell: vPi\T)li = O}, i = 0,1 ... ,k.
Then each \.;. is a subspace of \' invariant under T such that the minimal
polynomial of T l\ti is Pi(X)/i. Also.
(Sa)k
\' = E9 \~i=O
Each of these \/i's can be decomposed as
(5b)5i
\; = EB \tijj=l
where ~'jS are cyclic invariant subspaces of V under T such that the minimal
polynomial of T I \tij is Pit x )mi j . The family of polynomials {p;n2J: 1 ::; i :S
k, 1::; j :S Si} form the el mentary divisors of T.
Since po(x) = x, T I Va is nilpotent with index of nilpotence 10 and so
IJt(T) = j\ ~ Va. 110reover, since .\ n Vij = {O} for i ~ 1. T I \~] is an
isomorphism of \ ij onto itself. Let Co be a complement of \' in Va· For i ~ 1
define
USc)
if v E Vij
if v E EB Vrt EB Uorii,tiJ
v E S
V.l CYCLIC DECOMPOSITION OF E-CHi\INS
"1
(5d) e = e (N; EB Vij EB uo).l<i<kl~jSSi
Also, let To E 6 n be defined by
{
vT.vTo =
v.
if v E 110k
if v E EEl Vii=l
v\ e use these notations in the following theorem.
(5e)
TIIEORE~l 5 Let T E 6 n . Then there exists To Ti) E 6 n for 1 ~ i < k
and 1 ~ j ~ Si satisfying the following
(a) T = ToTn T12 ... T1s1 T21 ... T2s2 ... Tksk .
(b) For i =f. rand j =f. t Tij commutes with Trt . .1[oreover} each Tij commutes
with To.
(c) For every i and j! Tij'He! where e is an idempotent.
(d) To is relatively nilpotent with index of nilpotence lo.
(e) There exists minimal E -cycles Oij based at e such that Tij = Toij Jor
every i and j. Also To = Too where 00 is a minimal E -chain for To
starting from e.
PROOF (a) follows from (-Sa), (5b), (5c) and (.Se). To prove· (b) suppose
that i =f. rand j =f. t. Then for v ~ Vij 8 Vrt 9 V in V VTijTrt = v = vTrtTlj.
For v E V, VTijTrt = 0 = vTrtTij· Suppose that v E Vij. Then 1 Ti) = vT E
Vij and VTijTrt = vT. Also vTrt = v and vTrtTij = vTI ) = vT. That is
VTijTrt = vTrtTi) for e\·ery v E Vij. \"ow suppose that v E Vrt · Then vTI ) = z;
:2\. l~YCLIC DEC ~IPO:'ilTIO:-l OF E- !lAIN"
a.nd I'TljTrt = I,Trt = IT..\1 a rTrt = I'T E \ 'rt and t-IrrT,) = rT. Thu
VTijTrt = vTrtTij for eyery v E "rt. From thi" we get vI,)Trt = vTrtTi] for
e\'cry E". That implies TijTrt = TrtT'i for y ry i and j.
Vve no\\" show that To commutes with Tij for every i and j. L t t/:.
Vo EB Vii in V. Then l'ToTij = v = VTijTO. For l' E Vo vTo = vT E Vo and
vToTij = vT. Also VTij = v and vTiiTo = vTo = l'T. So l'ToTij = VTijTO [or
every v E \'0' Suppose hat 'V E \'ij 1 then vTo = v and t'ToTij = vT. Also
VTij = vT E 'tij and IT,)To = vT. Thus for e\'ery v E 'i). vToTij = VTijTO'
From this we get vToTi) = VTijTO for every v E ". Hence Io commutes with
Tij. Hence (b) £ollO\\"s. from (5c) and (5d) w g (c).
T I '0 i nilpoten with index of nilpotence la. Hence (d) follows [rom
(.Se) and from Defini tion 1.
To proye (e). let ri) = ess. rank Ti) and = nullity T,) = nullity T for
1 < i :s k and 1 :s j :s Si. Let qi{S are non-negative integers alisfying
qijS < rij :s (qij + 1) . Since Ti/He. by Theorem IV.6 there exists minimal
E-cycles Oij of length 2(qij + 2) such that Tij = T ij. By our definition 2 Tij
is irreducible for 1 :s i :s k and 1 :s j :s Si·
Let ess. rank To = ro and qa be a non-negativ integer satisfying qos <
TO :s (qo + l)s. \\e ha\'e IJ1(To) = IJ1(T) = lJ1(e). Hence nullityTa = and
ToRe. To is relatively nilpaten 1 hence it follows [rom Theorems III.13 and
III.14 that there exists a minimal E-chain 00 of length 2(qa +1) starting from
e such that To = T50' o
We ha\'e thus obtain d an E-chain representation of T. which consis 5
of minimal E-cycles together wi h an £-chain. \\"e call a decomposition of
T as in Theorem .j as an irreducible decomposition. \\ e ha\'e some impor ant
observations regarding ess. rank of T, ess. rank To and ess. rank T,j for 1 S i :s
V.l Y LI DECOMPOSITIO:-J OF E- HAINS "3
A;, 1 ~ j ~ Si, where T To and Ti/S arc as in Th or m 5. \\-e O'j\' them in
the following corollaries.
COROLLARY 6 Let T = ToTn T12'" TlslT21T22'" T2s2'" Tks k be th ITT'
ducible decomposition of T E 6 n . Let ess. rank T = r J ess. rank To = 1'0 and
ess. rankTij = rij for 1 ~ i ~ k and 1 ~ j ~ Si· Then l' = 1'0 + L rij·l<i<kl~j$Si
PROOF If follows from (5c) and (5e) that ess. rankTij = ess. rankT I iij
for every i and j and ess. rank To = ess. rank T I i/o. Hence by (5a) and (5b)
we get r = 1'0 + L rij . 0l<i<kl~j$Si
COROLLARY 7 Let T = ToTuT12'" T1s1 T21'" Tks k be th irreducible d -
composition ofT E 6 n . Ifess. rankT ~ nullityT J then there xists E- quarr
Oij based at e such that Tij = TOij for 1 ~ i ~ k 1 ~ j :::; Si and To = To
where 00 is an E -chain of length 2 in the normal form.
PROOF We have from (5c) and (5e), nullity T = nullity Tij = nullity To·
By Corollary 6, ess. rank T ~ nullity T implies ess. rank Tij ~ nullity Tij and
ess. rank To ~ nullity To. Now each Tij'He, hence it follows [rom the particular
case with q = 0 of Theorem IV.6 that Tij = TOij where Oij is an E-squar
based at e for 1 ~ i ~ k, 1 ~ j ~ Si. ow To is r latively nilpotent with
IJl(To) = lJ1(e). Hence by Proposition III.9, To = Too where 00 = c(e 0,'J) is
an E-chain of length 2 in Q(En ) such that e£eoRel so that the chain product
Too is in the normal form. 0
COROLLARY 8 Let T = ToTll T12 ... Tl s1 T21 ... Tks k be the irreducible. de
composition of T E 6n
. Let T = ess. rank T and = nullity T. Suppo. c. q i.s {J
84 V CYCLIC DECO~IPOSITIONOF E-CHAINS
non negatiz'c integer satisf1jing qs < r ~ (q + 1)8. rr ess. rank Tij = rij > S for
some i and j or if ess. rank To ="0 > 5) then the length of T = 2(q + 1) 2: 4.
PROOF By Corollary 6 I' = 1'0 + L rij. Hence if rij > S th~n I' > s.l<i<kl:0~si
Or if 1'0 > s then also I' > s. Thus the non-negative integer q satisfying the
condition qs < I' ~ (q + l)s must be greater than zero. By Theorem III.14 the
length of T is 2(q + 1) 2: 4 when q > O. 0
2. SOME REMARKS ON THE TOPOLOGICAL
AND GEOMETRIC ASPECTS
In t he foregoing, \ve showed that the semigroup (5 n of singular n x n matrices
over a field J{ can be represented by E-chains. In \'iew of the topological and
geometric structures present in 6 n inherited from 0J1 n , this representation has
a number of interesting consequences. We make a few more remarks on these
(wi thout giving proofs).
In the following we shall assume that V is an n-dimensional vector space
on the field J{ where J{ = R or C. Thus 9J1n is an n 2-dimensional vector
space over J{ and 6 n is a closed subspace of 9J1n (since 6 n = 9J1n - GLn and,
it is well-known that GLn , the group of invertible linear transformations or
matrices is open in 9J1n ). After, it is easy to see that En = E(6n) is closed in
6 n and hence in 9J1n . Thus En is a complete vector space.
\Ve discuss some geometrical properties of the set En. Here, for X <;;; 6 n ,
we write E(X) = X n En· \Ve have for e E En
For if T E 6 n , (e + eT(l - e))(e + eT(l - e)) = e + eT(l - e). Hence
L + d'(l - C) E En. Also e(e + tT(l - e)) = e + eT(l - e) which implies
V.2 SOME llEMARI<S 0:-': THE TOPOI,OGICAL AND GEOMETRIC ASPECTS 5
eu/ (e+eT(l-e))and (e+eT(l-e))e=e\\'hichimplic (e+eT(L- ))w r.
Hence eR(e + eT(l - e)).
Now e6n(1 - e) is a vector subspace of 9J1n and 0 E(Re) is an affin
subspace of En with dimE(Re) = dim(e6n (1- e)). Since £- and R ar qUIV
alence relations on En we have for every e E 6 n, E(Le) and E(Re) are affine
subspaces of En and hence En is a disjoint union of its affine subspaces.
Suppose tha.t c = c(eo,el, ... ,e r ) be an E-chain in En. By definition
for each i = 1,2.... ,r, ei-l E B(LeJ or ei-l E E(R J Since E(LeJ and
E(ReJ are affine spaces, the segment
lies in E(L e;) or E(Re;). It follows that we can associate a unique polygon
in En joining eo to er · So we shall identify every E- hain with the uniq le
polygon associated with it as above. Hence every E-chain is a path in En and
so Q(En) S;;; D,(En) is a groupoid of paths in En.
Now it is well known that the set of all matrices of the same rank forms a
V-class in 6 n. We shall denote by D(k) the 'V-class of matrices of rank k and
E(k) = E(D(k)). Since 6 n is idempotent generated, e,f E En is 'V-r lated
in 6 n if and only if there is an E-chain joining e and f and hence a polygon
in E(k) S;;; En from e to f where k = rank e = rank f. Conversely it an be
shown that if there is a path joining e and f, then there is an E-chain joining
them. Thus for each 0 ::; k ::; n - 1, E( k) is a path component o[ En. lIenee
the fundamental group of En can be studied.