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.